*To learn about Marilyn, here's our newest video on her life (April 2024):* th-cam.com/video/F6rDygbx5Kk/w-d-xo.html Visit brilliant.org/Newsthink/ to learn math, science, computer science, and data science for FREE for 30 days
I have a question, if there are three doors, there is a 1/3 chance of winning, but if there is only two doors there is a 1/2 chance of winning. There is a 50-50 chance you will win, so what’s the point of changing. The first door you picked might be correct. Plus, isn’t it human psychology to give the person who did it correctly to pick something else so they go wrong?
I had to watch this video like ten times. I bet it could get into some mathematic gibberish. If you add them all up to 1, make a pie, and then take away 1/3, you are left with 2/3 of pie and 2 doors to give an equal slice to. You'd split them into 1/3 each, counting the total they came from, which would be half of what was left while using their logic. Help.
@@vaibhavk1342 does this make any sense? I had to watch this video like ten times. If you add them all up to 1, make a pie, and then take away 1/3, you are left with 2/3 of pie and 2 doors to give an equal slice to. You'd split them into 1/3 each, counting the total they came from, which would be half of what was left while using their logic. Help.
@@pheresy1367 That is why this question is sort of like fake news. no normal person accepts it. after the goats are shown its 50/50 of the last 2 doors. no normal person could think that in the final choice of the 2 doors THAT THEY SOMEHOW LOOK AT THE PAST.
I would have switched to door #2 as well but for a very different reason. I would have assume that the goats would need to be kept as far apart as possible so they would be less likely to incite each other into making noise and thus giving their relative positions away. Putting the car in between them would help keep them out of each other's sight. I might just have won the car because I knew more about goats than mathematics in that instant!
@@mandolinic I think statistically, door #2 was more often the right choice. Everyone playing along at home would always yell DOOR NUMBER TWO! So the trick then would be to guess door #3 and see if he shows you a goat behind door #1. If he does, you got the historical statistics and the live probability on your side.
You too?! The most remarkable thing about noshing with her is how she can pass things around the table telepathically. Oh, deary, I do hope she regaled your party with such feats. She even levitated all of us home after the dinner. Brilliant woman, that.
I am most impressed with the math professor who publicly admitted his mistake. It is so refreshing to see someone who will actually take responsibility for their errors, regardless of how embarrassing it may be. If only our politicians could show as much humility. Much respect.
Around here, when a vocal 50/50er finally figures it out, a typical response is to delete the thread so there is no evidence they were wrong. What's the opposite of humility?
"I am most impressed with the math professor who publicly admitted his mistake." Me to. How can you get a PhD in math and fail so hard at a simple probability problem?
My explanation: When you are right in the beginning , switching looses. When you are wrong in the beginning, switching always wins, as he always shows you a goat and leaves you the car. Since 2 out of 3 times your initial guess is wrong, switching wins two out of three times.
As soon as one door is totally eliminated, it changes the probably from 33% for each door to 50% for each door though. If I hand you 3 guns, say 1 is loaded and two are empty, let's play Russian roulette...then I take away 1 of them and show you it's empty, let's just use the two...you feeling more confident? 😂 Nah, because you know one is loaded and one isn't, suddenly odds shifted
@@SuperMontsta Not in this game, assuming as a rule that the host must reveal a wrong option from those that the player did not pick. That creates a disparity, because when the player's is already a losing one, the host is only left with one possible door to remove, but when the player's is the winner, he is free to reveal any of the other two, making it uncertain which he will take in that case. Using the example of the video in which you chose #1 and he opened #3, we know that the revelation of #3 was 100% mandatory in case the correct were #2, but if the correct were #1 (your choice), we are not sure if he would have opened #3 too, it was only 50% likely, as he could have opted for #2 instead. That's why it is twice as likely that the reason why he opened #3 and not #2 is because #2 is the winner, rather than because #1 is the winner (having you picked #1), and similarly occurs with all the other cases.
@@SuperMontsta The odds increase because the gamemaster always opens a door with a goat that isn't yours. 2 out of 3 times, you were wrong initially and change will always work. Think it trough!
The best thing about this video is a reminder that when people publicly stated something incorrect, they used to express accountability and humility. That never happens anymore.
Too bad idiot Vos Savant failed to acknowledge her profound error. Too bad too those MIT profs are shown to be idiots twice over. It's a new game. 1 in 2 chance, 1:1 odds. Though Hall does not say it in these words, he asked this: "There is a car behind one of two doors. There is a goat behind the other. Which do you choose?" It is irrelevant that Hall phrased it this way: "Do you wish to stay on Door 1 or switch to Door 2." Vos Savant is an idiot as are those MIT nitwits.
What? My comment has nothing to do with her achievment, its about the humble gentleman who knew how to apologize. I have no obligation what so ever to list anyones achievements, please look for an argument elsewhere.
Vos Savant is wrong still. She will be wrong even after she dies. She has applied conditional probability math skills but from the wrong premise. At the initial door opening to show one goat, the probability is 0%, odds, 0:0, chances 0 in 0. The contestant is not trying to avoid two goats, but rather only one. The probability of winning from the moment when an actual positive probability can be calculated, i.e., from the moment of two doors, is an equal probability of 50% to each door, as there are two options and no further information available. And that is the only solution, the correct answer to the Let's Make a Deal Problem (LMADP). An alike problem "The Monty Hall Problem" (MHP) is a pseudo-realistic problem derived from the Let's Make a Deal Problem that illustrates an application of conditional probability assuming a contestant can win on Round 1 but does not and gets a second chance with updated info. While the analysis of the MHP is self-referentially correct, it is inapt for the LMADP, which presents a contestant with a choice from two options, two and only two unopened doors. The MHP would be appropriate if and only if there were three doors and a constest could win right away from picking the right door of three. Yet, in the LMADP, there is no deal when there are three doors. The Rules to the LMADP are these, which are different from the MHP rules: 1. Say the name of the door. It does not matter because we're not revealing it. your odds of winning are 0:0., probability 0%, 0 in 0 chance. 2. Carol, from the doors not picked reveal a goat. 3. There could be a goat behind your door or a car. What door do you wish to name? You can name the same one as you did previously. You have a 1 in 2 chance, or 50% probability with the odds being 1:1 of getting it right. Once you pick we reveal the goat door first if you picked the car, the car door first if you picked the goat.
there's nothing to understand. She considered each door a floating variable. Each door is in absolute matter of fact, a constant. So there is no means to transfer probability from one door to another. Once the 3rd door is opened, there is no more question of what's behind door 1 or door 2. It is either a goat or a car. The feeling that you have a 33% higher chance when switching is based on the original probability you had to correctly guess between 3 doors. Once the 3rd door is eliminated you have a completely new expression. And the probability between two choices is always the same. Stay or change your mind, implies that the previous variable is still in play. It is not. You are not "switching", you are again choosing. The difference between the two choices is 50%, probability is recalculated at every choice.
@@vladimirdemirev4948 the reason why door are grouped is because they are different. One is the door you picked, the other two are the doors the host has to pick from. Your grouping doesn’t account for that. The professor admitted they were wrong because they were wrong.
@@vladimirdemirev4948Say we take a bag with ten marbles in it, one orange and nine white. Whoever gets the orange wins. You pick a random marble from the bag and I then take the remainder, open the bag and select a marble, discarding 8 white ones. Who is more likely to be holding the orange marble. Clearly the person who knows where the prize is will have a better chance of selecting it.
@@morbideddie well, I guess I am wrong. Seeing the 100 doors example changed my mind. I will go with the excuse that binge-watching TH-cam videos on different topics trains you to react first, without giving much thought ;)
What makes this problem tricky, is that people ignore the fact that they ALREADY made their first choice, when it was a 1/3 chance to get it right on the first attempt, so then when there are only 2 doors left, they treat it like it is just a simple 50-50 chance to be a car behind one of each remaining doors, but it's NOT! Door 1# is still 33% to contain the car(because you picked it before having the additional information), while switching to door 2# becomes 66%, because door 3# you know for sure that is 0%. This is why, when you run the tests with all 3 possibilities, after eliminating 1 wrong door: IF you switch => you will find a car 2 out of 3 times VS IF you stick to your first choice => you will find a car just 1 out of 3 times. Another example would be: If you could chose from a) pick 1 door from 3; or b) pick 2 doors from 3. Which option gives you more chances to find a car a) or b)? Obviously b). In our case, switching represents option b).
But the game is a 2 tier game, what people dont realise is that it's not a 1 in 3 chance because only one door will be exposed out of 3 on your first chance, so your first chance will have 2 out of 3 chance of getting it right. Which then leads onto the finally of the game. Which is 50/50 for both doors as you can keep or change your choice.
@@dominathor but if you already know they will open a goat door 100% of the time out of the 2 you didn’t pick, you aren’t making a 1/3 choice, even your first pick is 50/50 bc one of the other two doesn’t count.
@@optimizor The one that is going to be revealed is not determined. It depends on your first choice because the host is never allowed to remove your door. I mean, there are three contents: GoatA, GoatB and the Car. If you select GoatA, he will be forced to reveal GoatB, but if you select GoatB, he will be forced to reveal GoatA. So it is not like if one of the goats didn't exist, you could have chosen any of them, so more likely that in the first part you choose a goat and not the car. In fact, the average tendency is that for every 3 attempts, in 2 of them you start choosing a goat and only in one you start choosing the car. As the host always reveals a goat from the non-selected doors, in the 2 out of 3 times that your door already has a goat, the revealed goat must be the second one, so the car must have been left in the switching. Only in the 1 out of 3 times that your door has the car, the switching one will have a goat, so less often.
3:27 - "This is contingent on the host always opening a door with a goat." Yes, it is, which is why this restriction must be included within the problem as phrased, something the introduction to this video fails to do. That's actually a very common problem with this problem. It cannot be assumed that a goat had to be revealed simply because a goat was revealed unless the host's intention is incorporated into the problem. The host could have selected a door at random that simply happened to contain a goat. This legitimately changes the math to the Monty Fall/Blind Monty problem. This failure to accurately phrase the problem is frustratingly common.
well If you watch the show is oblivious, you can also assume how the show would b er if you havnt watched it. But still the author of the video has a bad taste.
Oh this makes sense now. I was under the impression that the game host always open door 3 regardless. Which is why I was confused at 3:20, when the table showed scenarios "game 3" and "game 6" having a car behind door 3 which made no sense to me at first so I excluded those scenarios. But I understand now, thanks!
Came here to say this. In video it sounds like host might have just chosen a door at random, and it happened to have a goat. It should be stated that the host will NEVER open the door with the prize when he opens his own door after the contestant chooses.
Well if the host would pick the door that was chosen by the participant and there was a goat, there would be no question of switching the choice of doors. And if it was one of the other doors and there was the car, the same thing applies. So in my opinion it's self explanatory
Simpler explanation; assume you always switch: If you initially picked a goat, you win. If you initially picked the prize you lose. What's more likely?
This video presents the presenter as being on your side, "helping you out." This is a bad assumption unless they're truly your friend, which is unlikely. Haven't you seen Slumdog Millionaire? :-) So when they select something to lure you away from your initial choice of door, as an option for you, it can be a decoy, not the real deal. After all, if you chose that door first, it would have the same amount of probability from the standpoint where the presenter may not be on your side. IQ is racist pseudo-science. Savant in name only. The reason Mongoloids in Singapore and elsewhere have the highest IQs is because their youth was sacrificed for the god of money, and they did parroting memorization including of formulae which help in many IQ tests, at least to be used to patterns. This is why a Papuan tribal won't beat them: The tribal isn't dumber; just not used to those types of tests 'cause they DON'T NEED TO BE. IQ tests use a one-size-fits-all appraoch, which is stupid, and proves IQ is stupid. If I took an IQ test, it was probably disguised as some standardized test in Florida long ago. I don't believe in that shit, so I refuse to participate. It's just like DNA testing: Different testing companies give different and contradicting results, so they're all scams. If you are truly smart, you'll know better than to let others tell you how smart you are, when those others are self-entitled narcissist establishment people trying to dictate your mind.
People learn much better when they're allowed to follow their interests. If the subject is something that bores you, you'll only retain the information for a required time (test date, usually), but if you are interested, you'll track down information and fill-out the subject more thoroughly. This is the way I educated my youngest son - he chose the subjects and the timing. He graduated top-of-class in Navy Submarine School and is now stationed on a nuclear sub based in Hawaii (his chosen profession)
ADHD compounds this problem big time. Someone with ADHD is borderline incapable of learning things that don't interest them at all, yet I believe it becomes an advantage when we really are interested in the subject because we can devote hyperfocus to it.
@@jasondashney no kidding, I'm just a teenager, and yet I can understand, and I know about, many things that adults, even in my field of interest often don't know. I was let to go my own direction, and I'm great at it!
There is a really simple explanation that doesn't require overthinking, drawing outcome tables or considering expanding or collapsing probabilities. You just have to understand that you're probably wrong to start with. 2 times out of 3, your first choice will be wrong, then the host must show the other wrong option so switching will guarantee a win. That's all you need to know to always switch because it gives you a 2/3 win chance (logically, not switching must give you a 1/3 win chance).
yes, exactly..... ........but there's still a small detail to do: now you 'only' have to make all those flatbrainers understand all this.... ....those, who come along in this comment section, either not able to understand it, or they just refuse to think..... ......good luck!🤣🤣🤣
@@KarlHeinzSpock I'm thinking as hard as I can. Can't you smell my brain cells burning? 🔥🧠 It still looks like she's left with a 50/50 chance to me and nobody's explanation has changed that yet. I want to get it. 🤔
@@websurfer5772 Think of it this way, at the start of the game the chances of winning the car by selecting a door is 1 in 3. Now why do you think it is 50/50 simply because a door has been opened? You know that they have to open a door, and it is 100% guaranteed to show a goat, so using your logic it should be 50/50 before the goat door was opened, because you will always end up with a choice of two doors.
@@asprinklingofclouds Well, I agree that you have a 1/3 chance of being right in the beginning. I had to read more about it online which says that it's because one of the doors is opened to reveal a goat _and_ the question is asked, "Would you like to change your answer?" The answer is much more likely to be right, in fact 2/3 more likely, if the contestant changes their answer to the other door.
The 1948 and 1988 encyclopedia sets of the "Book of Knowledge" were my favorite source of learning. I LOVED them! I always checked the Public library a lot. My fiction was mostly from Silver Age comics that helped me much with my art. I loved studying electronics. I didn't need school for any of that. Now days, Professionals seem to get their credentials out of a "Quaker Jack Box."
"The book of knowledge" was very pictorial and was suited for all ages. The simple facts of the topic were clearly explained and the information became gradually more complex for when you are older. It was laid out like the internet with many links in the index's. There were plenty of do it yourself projects. I build my first radio from the 1948 book at 11 and it was fantastic!
A good way to think about this problem is: You first choose one door. You are then able to change your choice to BOTH the other doors. you get a car even if one of the doors have a goat behind it. This is the exact same thing as to show the goat beforehand.
That is brilliant. And the only way to lose the switch would be because you beat the 3 to1 odds against you when you first chose. So it's always (3 to 1 against you) vs (2 to 3 for you by switching after the goat reveal). (Ooops, I think my "further clarification" only served to complicate). ;-)
That is a great way to look at it. It takes a certain personality, I think, to be encouraged about odds of 2/3. Chance is not something you can predict -- it's chance. This is the difference between stochastic (your statistical analysis) and random (what actually occurs). There is a world of words and ideals and a world of things and occurrences. Some people love to argue over words and ideals, but we each may go home with a goat. Or a car.
Unless you did pick the car on your first choice....... It's literally a 50/50 chance. You either change your answer or you don't, and you either win or you lose
exactly. or pretend there's 100 doors. now you get to pick either (a) one specific door, or (b) the combined total of any 99 doors, where if there car were in any one of the 99 doors you'd win it. would you pick choice (a) or (b) ? okay now instead of 100 total to start with, let's do 3 total to start with.
The way I approach this is by thinking that the initial choice can be one of three alternatives: Goat 1, Goat 2, or Car. This then plays out into three (and only three) possible scenarios: 1) If you chose Goat 1, and Goat 2 is revealed, then switching would have given you Car. 2) If you chose Goat 2, and Goat 1 is revealed, then switching would again have given you Car. 3) If you chose Car, and either Goat 1 or Goat 2 is revealed, then switching would have given you Goat 1 or Goat 2. Thus in 1) and 2), switching works, and in 3), it doesn't. Thus switching is right 2/3 of the time.
You are not correct, and I will explain to you why, very simply. You correctly differentiated between Goat 1 and Goat 2 in alternatives 1) and 2). In alternative 3), you do NOT differentiate between Goat 1 and Goat 2 by stating , "either Goat 1 or Goat 2" Alternative 3) should be, " If you chose car and Goat 1 is revealed, then switching would have given you Goat 1" Then alternative 4) which needs adding, "If you chose car and goat 2 is revealed, then switching would have given you Goat 2" Thus, in 1) and 2) switching works, but in 3) and 4) it doesn't. The result is 2-2.
The probability density function of any event remains unchanged...except by direct intervention in the primary predication of the probability of an event. That is to say in simple terms, if a coin were to be tossed behind a given door and it landed on heads up, it would remain on heads up irrespective of what other coins behind other doors landed up as showing to the observer...no matter how many other doors are involved.
You're still not taking into account that the host in this example already KNOWS what's behind the doors. That means he purposely avoided opening door #2 and door #1. If the Host opened door #1, that means either 1) it would reveal the prize or 2) would reveal the door as the incorrect one, which wouldn't give you an option to switch doors since the first pick was already opened. With that information, you know door #2 is definitely the correct one
First encountered this in high school. I tried to explain: "if you switch it's like picking 2 doors instead of 1", which convinced very few classmates. The teacher noted that I had good intuition and poor articulation. So true.
@@BasedGodGotenks Look at the 100 door example. The chance you chose the correct door from 100 doors is very low. The chance the winning door is amongst the other 99 doors is high. Now remove 98 of those 99 favourable doors and the remaining one has a very high probability of being the winning door. Now if you arrived at the game late and missed the above process of elimination and just had 2 doors to choose from then your chances would be 50/50 because you do not have the information that points towards the more favourable door.
it doesnt make sense still fuck, how, every door has a 1 in 3 chance of being the car, revealing one door doesnt give the other door a higher chance as it was predetermined beforehand, its still 1 in 3, it cant just change cause you revealed the other door fuck like ik it makes sense to some people but it doesnt to me. EDIT: This is how i see it, in the start you have 1 in 3 chance, after goat reveal, you are left with 2 doors, 1 has a goat one has a car, so you are back to beginning, when deciding in that moment, you have a 50 50 chance you will get it
When I first heard the "Monty Hall Experiment", I reacted the same way everyone else did. I am however a software engineer. So I decided to write a small program to prove them wrong. What I ended up doing is proving MYSELF wrong. The program played 100 random games. The first 100 games stayed on the same door whereas the second 100 games switched doors. In the end, the program that stayed on the original pick won approximately a 33% of the time whereas the program that switched won approximately 67% of the time. I couldn't believe it. You do double your chances by switching doors.
Yeah I was around ~15 when this happened and also wrote that program at the time, though I either had no idea what was correct or was pretty sure she was. Don't remember which!
@@siddharthshekhar909 the chance of your original guess being correct is 1 in 3, no matter what door you guess, the host can open an empty door, thus your original odds of 1/3 are not changed by the host opening an open door. By switching you end up with 1/2
@@siddharthshekhar909 EXACTLY!!! This is what I said. I absolutely could not believe it! This is why I wrote the program! I wanted to prove them WRONG. There is no logic to their conclusion. But as it turns out I was wrong!!! I still can't explain it. But the computer doesn't lie. I am a software developer with 30 years of professional experience. The program took an hour to write. I had to run it like 10 times before I could believe it. I still can't explain why it works. Don't feel bad. There were MIT professors who felt the same way you and I did. And she is the smartest person alive so I wouldn't sweat it if you don't understand why. I don't either.
It makes total sense when mapped out. I guess the difficulty comes in understanding why "switching" doors increases your odds at all. I got hung up on the "switching" part having any impact, instead of realizing that it's making a new selection with better odds. We used to do simple stuff like this in grade school, it's kinda crazy how a little bit of language can subvert your logical faculties.
"instead of realizing that it's making a new selection with better odds" It is not making a new selection with better odds, but swapping your lower odds for better odds.
I still don’t understand why switching will increase your odds of winning. If you take away your first selection, meaning you never made a choice, are you still going to choose the one out of the 2 doors that didn’t get eliminated?
@@jacobcutrer Switching increased odds of winning because host get to pick from two doors therefore, he gets a door with a cat twice as often than player does. When he opens his door with a goat, we know which door holds a car twice as often as the door initially picked by the player.
The key to all this is at 5:15 in the video. The improved chances only work if the guesser knows that the host is not opening one of the remaining two doors randomly. When folks pose this problem to someone who never saw Let's Make A Deal and the poser does not reveal that the host is intentionally opening a door the host knows to contain a goat... that's ruining the question and (in that case) the correct answer is "no improvement in chances by switching". The improvement ONLY comes when information is conveyed to the contestant that the host KNOWS that the door being revealed contains a goat.
@@TEK_Nemesis What he is saying is that if the host revealed a goat without knowing where everything is then there would be no advantage to switch, which is correct.
@@TEK_Nemesis "once you solve this problem, you'll realize that it is always a better idea to select the other door" Once you solve this problem you'll realize it's the knowledge of the host that is the reason switching is advantageous.
@@TEK_Nemesis If the host doesn't know where the car is, that means he has to randomly open a door. The host can just ask the player to reveal/remove a door for him. This means the player is the one who choose everything. Player: I pick door 1. Player: I want door 3 to be the last door. Player: So I choose to open door 2. What makes you think door 3 has higher winning chance than door 1?
@@TEK_Nemesis Quit pretending that you knew the reason as to why the host had to know where everything was. You should have realized my statement ......"What he is saying is that if the host revealed a goat without knowing where everything is then there would be no advantage to switch" was indeed correct. Instead you replied...."once you solve this problem, you'll realize that it is always a better idea to select the other door." You're just a fraud and couldn't admit that you were wrong.
@@TEK_Nemesis No. That's NOT what I'm saying. What I'm saying is that many people who present this problem to an unwitting 'victim' fail to explain the CRUCIAL point that the host KNOWS where the grand prize is and the host will ALWAYS reveal a door with a goat. That's precisely WHY I mention the caveat at the 5:15 point of the video. The premise upon which the logic is based relies on the fact that the host KNOWS where the grand prize is, therefore the host will ALWAYS open a door with a goat. This is what provides the extra information that tells the contestant it's better to switch. If the host does NOT know where the grand prize is, then the host will unwittingly reveal the grand prize 33% of the time, in which case there is no advantage to the contestant to switching. If the host does know where the grand prize is, but the contestant is not told that the host will ALWAYS reveal a goat, then the contestant has no justification to switch. This is the point being made at 5:15 in the video.
The Monty Hall problem and people's approach to understanding it is very interesting. Another way to think about the problem not covered explicitly in the video is the fact that only one independent choice is being made in the game. That choice is the players initial guess when there are 3 doors. The host isn't making a meaningful independent choice since they have to reveal a non-prize door only from the doors not guessed initially, and the results of the decision whether to stay or switch are entirely dependent on the initial guess(when there were 3 doors). If the player initially guessed the prize door(a 1/3 chance) and they switch they lose. If the player initially guessed a non-prize door(a 2/3 chance) and they switch they win.
"If the player initially guessed the prize door(a 1/3 chance) and they switch they lose. If the player initially guessed a non-prize door(a 2/3 chance) and they switch they win." I first heard this riddle in the 2008 movie "21", and until today I never understood why switching doors after the reveal of a goat would increase the chance of winning, but the way you phrased it made it click for me, so thank you, i finally get it :)
that is probably the only explanation i've ever heard for this that makes sense. because no other explanation, even when presenting empirical evidence, actually draws that connection.
For me the biggest mystery is why your explanation is not immediately obvious to everybody. Lots of people deny it, even after someone carefully explains it to them....
@@tonybrowneyed8277 yea I think they're missing the difference between the host revealing a random no prize door, and the actual Monty hall rules. That would give 50/50 odds and the Monty hall game looks the same on any individual round. But the host not being able to reveal the same door as the players first guess completely changes the odds and the nature of the game.
" If the player initially guessed the prize door(a 1/3 chance) and they switch they lose. If the player initially guessed a non-prize door(a 2/3 chance) and they switch they win." omg i finally get it tyy
I think that some people find hard to wrap their mind around this concept because they fail to understand the very nature of probability. It’s not about being 100% right, it’s about being more likely
@@chestnut1279 Two doors left proves that you can either switch from your car to a goat, or from a goat to the car. The host could have simply asked you that same question when you first picked a door without doing anything else.
@@chestnut1279 I'm thinking the same thing. My understanding of this brain teaser is that changing your answer after the host reveals one of the fail doors is most correct but still not a guarantee as picking the first door was 1/3 vs changing the answer now making the odds 2/3. Its less likely you got it right on your first choice but it is possible. I think the logic of "two doors means equal odds" only works if the host revealed the door you picked to be the wrong door and you still had two options. But nothing makes me feel low IQ more intensely than brain teasers. I suppose the real question is do you think its more likely you got it right on your first try when you had three options?
If the contestant was given the opportunity to swap his 1 door for the hosts 2 he would almost certainly accept the offer leaving themselves with 67% chance.Don’t forget the host knows where the car is and has to open one of his which has to be a loser leaving the remaining 1 of his 2 which still has a 67% chance
I always tell my kids four things:. 1) you panic, you die. 2) stupidity kills. 3) never do anything for which you will have to apologize for later. 4). And) (a biggie),.never take up a habit you're just going to have to break later on.
Seemed to me "they" were sort of picking on her for not using the "math stats" as they did ? Oh, of course; now let's hear it for the "Intelligence of Creative Thinking!?" It seems THERE ARE different "kinds" and "types" of IQ " Tests." Experience and Education , two possibly different types ? So we should be also asking WHICH IQ Test did Marilyn excel in, or on ? MISSING FROM the video; Does this lady write and speak in both German and Italian ?I've always felt there is an extreme indication of high intellect in regard to peoples' abilities TO express themselves in foreign languages ?Seems there is a certain "type" or "kind" of logic it seems in learning to "relate " foreign language to one's own ability to speak and write in their native language ? It appears Marilyn 's " (by assumption?) that Marilyn had TWO "Native languages" yes ? Her opinion of "public schooling" holds great merit. I remember a question required to be asked on a high school test , was "Who were the Phoenicians ?" The ABCD Answers included the answer " Venetian" . Most admitted later that they all misunderstood the word Phoenicians because they were all more "familiar" with Venetian Blinds, than historical terms of peoples and places! (Ha Welcome to American World History 101-we (all) need to repeat that one!?) Which btw lead to my last question (for you or Marilyn, ha ?) Is the inability to "spell" properly (in any language/especially ones native language ) indicate ignorance ?Duh...As a retired teacher, I submit I've become dependent on the Google Gargoyles ' offers for correction, which often just doesn't exist. The robots tell me I've misspelled something, yet/while, offering no options with which TO correct it. Good at Questions; Slow at the answers. Anyone ? "I Ain't no Middleman" Fred Gold & Lynda Faye Copyrighted 2016 by LyndaFayeSmusic@gmail.com or Yahoo, if censored for using the word " God" too often?
First heard this problem in a probability class, when posed by a professor and solved it correctly the first time. You always have a 1/3 chance if you refuse to switch. It never changes, its set from the start.
No, you have a 1/3 chance to start. When one is eliminated by the host, that means there's 2 doors with 2 possibility. 1/2. 50% for each door. Even the chart in the video at 3:25 showing all possibilities confirms. Game 3 and 6 aren't possible, since it's a goat and not a car. That leaves game 1, 2, 4, 5. 4 possibilities. 2 of them you win. Two you lose. Staying makes you win.1/2. Switching makes you win 1/2. That's 50% each and again...out of the four game possibilities, you win 2/4.
@@factormars4339 I don't know how people see that chart at 3:25, negate the ones that have door 3 as a car since we see it's a goat, and see the 4 various outcomes and see it's 50/50 My thoughts is one "smart" person got it wrong and people just jumped band wagon. It's common knowledge that probabilities are based on us not knowing something. KNOWING one door CHANGES the probabilities of the other doors.
@@SuperMontsta As I already told you in another comment: you cannot preserve Games 1 and 4 with the same probabilities as Games 2 and 5, because you don't know if the host would have opened door #3 if the car were in #1; he would have had the choice to open #2 instead. You can see it better seen in the long run. If you played a lot of times, like 900, the car would be expected to appear in each of the doors in about 1/3 of them, so in about 300 trials. Now, for simplicity assume that you always start picking #1. The games will look like: 1) In 300 games the car is in door #1 (your choice). But they are divided in two sub-cases depending on what door the host reveals then. If he takes each with the same frequency: 1.1) In 150 of those 300 games he opens #2. 1.2) In 150 of those 300 games he opens #3. 2) In 300 games the car is in door #2. In all of them he is forced to open door #3. 3) In 300 games the car is in door #3. In all of them he is forced to open door #2. If he opens #3, you could only be in case 1.2) or in case 2), that constitute a subset of 450 games. You win by staying in the 150 games of case 1.2), that are 1/3 of 450, but you win by switching in the 300 games of case 2), that are 2/3 of 450. Similarly, if he opens #2, you could only be in case 1.1) or in case 3), that are also a subset of 450 games, from which you win by staying in the 150 games of case 1.1), but you win by switching in the 300 games of case 3). Again, switching wins twice as often as staying. This adds up 300 total wins for staying and 600 for switching.
So , I've known the Monty Hall problem since 2 decades, watched 100s of youtube videos on it too. But nobody cared to tell me that not many believed Marilyn vos Savant when she gave the correct answer. (Off course Steve Selvin gave the real proofs and solution with 3 prisoners problem). *EDIT* : I had no idea that *so many dudes* would be *triggered* by this comment. I'm a dude and I'm aware that a man's ego is just *too fragile* . But imagine being triggered by a harmless comment, almost all of the replies are by butthurt young boys and men. LOL
@@anti-apathy9715 1) Nobody said it did. It didn't have to. 2) Discouraging, patronizing women in the field of science, mathematics , etc. is not a trivial matter. (Although i do believe this could happen to anyone, but being a woman made it worse for her) Point is, such incidents prove to be a hindrance for little girls and young women who are already brainwashed by the society to think that they are not as good as men.
I found the easiest way to explain the problem is: You had a 2 out of 3 chance of picking goats and only 1 out of 3 chance of picking the car. Once the host reveals a goat, there are only 3 outcomes of a switch: Scenario A) You picked the car at the start. If you switch, you lose. Scenario B) You picked Goat #1 at the start. If you switch, you win, since Goat #2 is gone. Scenario C) You picked Goat #2 at the start. If you switch, you win, since Goat #1 is gone. You can only lose if Scenario A happens, and you win if Scenario B OR C happens. Therefore, you should switch since you have 2 scenarios where you win, versus only 1 where you lose.
I find it amusing how people can be so naive. If the host reveals one of the door had a goat Then your Scenario B and Scenario C is EXACTLY THE SAME AS IT HAS 1 door with a goat and 1 door with a CAR!!!
@@shafin141 I guess you didn't fully understand the explanation. You have to think about the problem like I explained it, as in, including the odds of you picking correctly in the first place. Yes, after one door is removed, there's a 50/50 chance that the car is behind one of the remaining doors. HOWEVER. You have to take into account the fact that your INITIAL choice only had a 33% chance of being correct. Those odds have not changed. Try extrapolating it to 100 doors, and 99 goats. You have a 1% chance of picking correctly at the start. Now the host removes 98 incorrect doors, and you're left with 2 doors. It's still technically a 50/50 chance of there being a car or a goat, but you're not going to switch still? You have new information now. The host has essentially removed the majority of the wrong possibilities from the equation, so the odds of you now picking correctly if you switch have become 99/100, versus your intial 1/100 odds.
@@dchang8619 Sorry you are confusing yourself by believing that the initial 33%, 1/100 and 99/100 remain. They are gone now and dont carry over. Its now 50:50 and your original choice is just as likely as switching.
@@kenandlynboyle9300 it’s really not though. Try it yourself in practice, not in your head. Try switching 10 times, and not switching 10 times. See how many times you win by switching and how many times you win when you don’t switch. I promise you that you will win more times by switching. This has been mathematically proven, so there’s no real use trying to argue against it. Just try and understand why it’s the case.
I had never heard of her, she is so cool though!! She kind of makes me think of an old-fashioned Druid or wise person on top of a hill and people come from there and far to ask questions
This is easy...you have to keep your choice only if it was correct the first time - the odds for which is 1/3. If you have chosen wrong the first time (2/3 odds), you have to change your choice. Therefore, keeping the choice will be correct in only 33.3% of the time and changing the choice will be correct in 66.6% of the time.
Why does this seem to neglect the fact the odds of your initial choice being correct INCREASE when one of the goats is revealed? You're being given new information about what may be behind door 1 - either one of ONE goats, or a car. It used to be one of two goats, but now one of the goats is excluded. Of course the probability wouldn't stay the same.
@@epicmorphism2240 The creator probably didn't know. It was just a game. Maybe after decades of hosting the game they ended up with a "gut feeling" that is better to switch but they probably thought it was 50% too
The one-hundred door example makes it very clear. No one should assume they guessed the correct door in that scenario. The pagentry of a Game Show might make people falsely presume the host is hoping to trick the contestant into switching their pick, but unless the host is actively trying to persuade you to select one door over another there's simply no trick to be had. This line of thinking is precisely why I got this wrong upon initially hearing of it years ago. It's still shocking to see how venomous some of her detractors were.
I dont understand this, it makes no sense. if you choose a door and the other is eliminated then the probability should distribute evenly between all of the remaining doors, not all of them except for your choice. Can someone explain why it does this?
@@versenova5531 door 2 and door 3 combined add to a 2/3 probability but you know door 3 isnt it so that 2/3 probability has to be for door 2 thats how i understand it
@@versenova5531 you have 33.33...% chance to guess it right, which means you have 66.66...% to choose WRONG. Switching will automatically give you the price 66.66...% of the times. You're more likely to choose wrong, so switching is a better choice 2/3 of the times. That's why it's even more clear with 100 doors, you have 1% chance initially, but after revealing a goat behind 98 doors, it's your best option to switch 99 out of 100 times.
For people who still don't understand - Think about it with 10 trillion doors. First, you pick any random number out of 10 trillion. Then the host eliminates all the doors (numbers), except yours and one other number (remember, one of the two remaining numbers is correct). Would you switch? Obviously, because the chances you randomly picked correctly out of 10 trillion is basically zero. Some people still couldn't understand this with 100 doors, and I think it becomes even easier to understand when you increase the total number pool.
Thank you, I saw this explained on another video, I don’t understand why mathematics people are disagreeing with this simple fact of probability, it’s not that hard to understand
I still don't agree with this no matter if its correct because 1 out of 3 is not 1 out of trillion or 100 or 10 its still out of 3 so it's completely different.
@Krichnu It is the same; that's the point. It doesn't matter if you agree, because the math checks out. Find the comment on this video where the guy wrote a program and ran this simulation 10,000 times with 3 doors (or whatever the number was).
I came across this puzzle during a math course and the tutor posed it very briefly, with no indication as to whether the host had picked the door at random or whether he chose to pick one which he knew didn't contain anything of value. I asked about that at the time, saying I needed more information, but he didn't seem to understand my question. What made this problem difficult to solve was that it was not properly posed, though most people didn't realise that.
if there are 2 players and the first player chose door 1 and the second player choose door 2 and then the host revealed a goat behind door 3 . Now he asks the players do you want to switch ..now each player has a 2/3 probabilty of switching as it seems i.e both players have equal chances of switching and obviously equal chances of sticking so chances are 50 50 and we can not say one has a 1/3 chance and the other has 2/3 chance because both are under the same circumstances .Now assume the 2 players are in different rooms and they dont know anything about each other in other words each player thinks he is the only player in the game and the game is played at same time for both .Do you think now this will change the outcome .Chance is 1/2 to 1/2 for each player or for one player if the other player exists or not..This shows how people are deceived by majority oponion.
"What made this problem difficult to solve was that it was not properly posed, though most people didn't realise that." You are correct that your tutor posed it incorrectly if you couldn't know the host's selection of the door to open was deliberate. But that is not the reason most people are confused by it... the question Whittaker presented to vos Savant in Parade was worded correctly. Since we are told the host knows what is behind the doors, it is impossible for him to open a door not already knowing it was a goat.
@@alimetlak Assume the car is behind door 1. Player one pick door 1. Player two pick door 2. Host open door 3. Player 1 stay = win, switch = lose. Player 2 stay = lose, switch = win. Assume the car is behind door 2. Player one pick door 1. Player two pick door 2. Host open door 3. Player 1 stay = lose, switch = win. Player 2 stay = win, switch = lose. Assume the car is behind door 3. Player one pick door 1. Player two pick door 2. Host open door 3. Oh wait, it's the car!!! Oh yeah, you think there is some magic power making at least one player pick the door with the car. Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.
@@alimetlak Well spotted! You have put your finger right on what I feel is the weakness with how I understand this scenario. A somewhat similar principle applies in other, related puzzles. There is one classic going back a very long time which caused a lot of the same sort of disagreements as the car-goat puzzle. It involves 3 little bags, each known to contain 2 black or white counters, thus: BB, BW, WW. Someone picks up a bag at random and takes out a counter. It happens to be black. What is the probabilty that the other counter in that same bag is also black? I think this one is quite easy, but I knew someone couldn't, or maybe wouldn't see it.
@@alimetlak If there are 2 players playing the game and they can't choose the same door, then there is a 33% chance that the door the host opens has the car behind it. In the other 67% of cases, when the door the host opens has a goat behind it, the first player should switch because their probability of winning the car is 33%. The second player, however, should not switch, as their probability of winning is 67%. Explanation: First player's probability of picking the car: P1 = 1/3 Second player's probability of picking the car: P2 = 2/3 * 1/2 = 1/3 Probability that the host opens the door with the car: P3 = 2/3 * 1/2 * 1 = 1/3 2/3 of the time, when the host opens the door with the goat, the second player's probability of winning changes: P2 = 2/3 * 1 = 2/3 So, in your scenario, the first player should switch, while the second player shouldn't.
One way of wrapping your mind around this question is by thinking of it this way: Originally there are 3 doors, so your odds of picking the correct door are 1/3 and your odds of picking the wrong door are 2/3. The host is then going to reveal a goat, which seems like new information, but you already knew that at least one of the doors you didn't choose had a goat, so it's actually not telling you anything you didn't already know. Asking you to switch is really asking you if you think that your first choice was wrong, and uses the same odds as when you made that choice, so the odds that you chose wrong the first time are 2/3.
So switching comes down to thinking you made the wrong choice the first time. The odds change with the revelation of one goat. This is new information, not just "seemingly." The 1/3 or 2/3 scenario is now irrelevant.
@@diannemccarthy8685 No, the odds don't actually change, even though intuitively it may seem like they should. The way that I explained it above is the explanation that I personally find the easiest to understand, but if you don't find it convincing, just search for the "Monty Hall problem" in the search engine of your choice, and you should be able to find plenty of other explanations. When I first came across this problem a few years ago, I did a lot of reading about it myself, and I remember that there were even websites that allow you to play this game over and over, and they keep track of your win/lose record for when you stay and when you switch, and you'll see that over time the win percentage for switching averages to 66.6% while the win percentage for staying averages to 33.3%.
No, given the mind set of the Host, your chances of winning are BETTER than the problem suggests! Monty WANTS you to win! More fun, more excitement, better ratings, more sales of every item offered as a "prize" on the commercial, (ahem, "Show") The clue he gives you is that your odds got better after your first choice. Monty wants to make your chances of winning as near certain as he can, without being obvious.
@@zackreagin8384 The odds of the second door are increased after seeing the third door open, IF the game (which regulation I ignore) had the rule that BOTH a door must always been open AND the switch must always been offered. If showing an open door and/or the option of switching are not the rule of the game but may be decided by the host unpredictably, the host of the game might use the door open or the switch option as a way of deceiving the player, at times. That would not guarantee that switching to the other door carries more chances to win, in my opinion...
I can't find the comment again to give credit, but someone in the comments here helped me understand it intuitively. It gets much easier if you think about the odds of *losing* rather than the odds of winning: - If you *always* switch then you *only* lose if you picked the car to begin with. - What are the odds that you picked the car to begin with? 1/3. - So the odds that you lose if you switch are 1/3.
Yea, but, If you always stay, then you only lose if you picked the goat to begin with. What are the odds you picked a goat 2/3 (initially known to be) But after the knowledge, what are the known odds that you picked a goat from your initial pick? = 1/2! Everyone is wrong. And it's amazing.
@@nicosmavropsis459 I was mistaken. You see I missed one crucial thing = the host has an advantage, I didn't factor in. The host always keeps a good door (usually more of the time, than your door). So, you have a 1/3 chance of choosing the right door. The host has a 1/2 chance of choosing the right door. So, you always go with the host, for the best chance. I was only able to figure this out once the problem was exaggerated in my mind: Say there was 100 doors and you chose 1/100. Then the host eliminated 98/100, by choosing 1/99. You should go with the host because, you chose 1/100, but the host chose 1/2.
@@kenham6742 IN THIS EXAGGERTED PROBLEM WITH THE 100 DOORS, THE HOST ALREADY CHOSE AND ELIMINATED 98 DOORS, HE DID'T CHOOSE 1/2, RATHER THAN HE ISLEAVING THE 50/50 CHANCE TO THE PLAYER TO CHOOSE. LET'S RETURN TO THE SENARIO OF THE PROBEM PRESENTED. THE HOST WOULD ALWAYS REVEAL A DOOR WITH ONE OF THE GOATS. BECAUSE WE KNOW HE CAN'T OPEN THE DOOR WITH THE CAR, DOOR#2, SO IF HE OPENS DOOR#1, HE WOULD REVEAL THE WRONG CHOICE OF THE PLAYER AND IMPOSE THE SAME QUESTION TO THE PLAYER TO CHOOSE DOOR#2 OR DOOR#3. IT WOULD BE THE DECEPTION PLACED BY THE QUESTION NOT ONLY TO WIN, BUT THE ACTION TO STAY AND LOSE OR TO SWITCH TO WIN BY CHOOSING DOOR#2 OR DOOR#3. OF COURSE IN THIS CASE DEFINETLY SHE HAS TO SWITCH IN ORDER TO WIN. IN THE CASE THAT THE PLAYER HAD FROM THE BEGINING CHOSEN DOOR#2 WITH THE CAR, BUT HE DIDN;T KNOW THAT IT WAS THE RIGHT DOOR AND THE HOST REVEAL AS HE DID THE DOOR#3 WITH ONE OF THE GOATS, THEN IF HE PROCEEDED TO CHANGE APPLYING THE POSSIBILITIES PRESENTED IN THE VIDEO THAT HE HAD GAINED 33.3% , TOTALLING TO 2/3, HE WOULD LOOSE IF SHE WOULD FOLLOW ALWAYS THE ACTION TO SWITCH TO WIN , SO IT IS ALWAYS 50/50, BECAUSE THERE IS A DISRUPTION IN THE POSSIBILITIES AFTER THE ACTION OF THE HOST EVERYTHING CHANGES.
@@kenham6742 You are still getting it wrong. As the car can never be revealed in any of the games, then if you only pick it 1/3 of the time, the host is who is forced to leave it hidden in the other door that avoids to open in the 2/3 of the time that you start failing, not 1/2. Similarly, in the 100-doors scenario, as you only get it right in 1/100 of the attempts, his closed door will be right in 99/100. It is like if someone else looked inside all the 99 doors that you did not pick, and took which preferred from them. In that way, it is obvious that that person will find the prize in 99/100 of the cases.
It is important to note that the 1/3 to 2/3 probability is right only if the host who opens the door knows what is behind the doors. If he doesn't, then the probability is 50/50. The problem is not so buffling after all.
It is important to note that if the host reveals the car, the probability of losing goes up to 100% because the car has been removed from your choices. What this tells us, is that it's a waste of time to think about what the host knows or doesn't know, or what the host's intentions are.
@@alkinooskontopodias5919 if you know or don't know that the host knows or doesn't know. So thats like 2x2= 4 different probabilities depending on what information is known. I am sure you bring in even more play of words and rules that would increase the variables. From the original question we are limited in our knowledge. We didn't even know if he would open the right or wrong door (initially) to expose a car.
To the people who don't get it it is because the host will never open the door that has car which increases the probability of other door having the car.
In fact, no. is is 50/50 chance even it was thousand doors and only two left it is 50/50 chance. But because the host ask if you want door two is a clue. The host likes to make drama and after he will say pity you didn't change the door. it is psychology, he know most people do not want to change if he try to make them.
@@neinei5558You really missed the idea.. There is a difference between the host knowing where the goat is against showing the 2nd goat by guessing. There is no 50/50.. If you buy a lottery ticket and someone from the future says.. im going to pick 1 ticket and leave it on this table after destroying 99998 which I know wont win.. Do you want the ticket? Or do you keep the one you have before I picked one and removed the rest?
@@absolutium No matter how many tickets it originally was as long it remind two tickets one win one lose, 50/50. Some people think if you throw 3 sixs in a dice the next will most likely not be six, thats also wrong for every throw you do it is 1/6 chance to be any number no matter what the dice show before.
I've long understood the Monty Hall solution - but extrapolating the information scale to 100 doors - makes complete sense - knowing that the "one door" is hot - and that you have a 98% chance of being wrong on your first door choice.
it's 99% of being wrong and then the switch is 99% of being right. it's reversed. revealing the doors does not give any hint on your first choice, so it's still 99% and not 98%
(Car is prize). Let’s say the chooser did not pick a door at all. The chooser does not know whether there’s a car or a goat behind ANY of the three doors. The host, however, knows what’s behind each of the three doors (ALL). They both stand in front of the three doors and the chooser does not pick anything at all. Now, the host reveals that there is a goat behind door #3. That eliminates door three as an option for the prize. Bye-bye! Now there are two doors left. That means there’s a 50% chance that the car is behind either of the two remaining doors. The high IQ vos Savant is declaring the idea that the chooser having originally simply “THOUGHT” (chosen) of what the winning door might be before the host opened door 3, CAUSES the remaining 50/50 ratio to change. To me that sounds like she is either not that smart OR OR OR, she is pulling ideas from things like quantum theoretical physics such as the “double slit experiment” and such. Like, the theory that when trying to guess if there’s a cat in the box, actually there is no cat in the box until you open the box, then it either does or does not manifest, because our reality is made up of potentialities rather than actualities, until they are “looked at” or measured -to be more precise (that’s how the theory goes, which is pretty much proven at this point). So, I think she needs to clarify what level of accepted reality or physics she’s pointing to in advance of her explanation, because the description of the solution by her and this video is implying that the chooser’s THOUGHTS (manifested as a supposed “choice” or predisclosed “guess” of where the car might be, affects the outcome. It’s similar to me arguing with you that nothing actually exists because atoms don’t reeeeally touch each other and sub-atomic particles are only energy in their smallest form. In a physics class that discussion or answer to a topic might be appropriate, but in an online forum about whether I have one or two cups on my table , it isn’t necessary appropriate. We obviously have different “levels” of reality we are talking about. The one to attend to needs to be disclosed in advance because she is LITERALLY saying that the difference between whether the chooser specifically made a choice (THOUGHT), or not, before the process of door openings began, AFFECTS the outcome. That’s not really actually fair according to the standard discourse. An online basic forum is not a quantum physics class. Thanks for reading.
PS. This might be going too far, but, it also makes me think she’s purposely trying to mess with people’s heads. lol. It worked ‘cause even the PhD’s were crying but funnily enough what they missed is she subtly didn’t disclose to them all the facts of the game. If I tell you we are living in a video game, am I lying? If I walk away and leave all the PhDs to fight about it, then they cry and appologise, am I smart?? Hell yeah! Am I right? Not if the rules of said game weren’t disclosed. lol. Sounds a bit narcissistic IMO. 🤭
@@SolutionsWithin You obviously did not understand the problem and are not smarter than Savant lol. Not picking a door first is different than picking a door first, because then the host cannot pick this door and has to chose between the other 2 only. You picking a door eliminate this choice for the host. That's why it becomes a 1/3 - 2/3 and not 1/2 - 1/2
Maybe I'm not one of the smartest people around but I have a logical thinking brain. Your 1st choice has 1 in 100 chance of being correct. When 98 of the remaining doors are eliminated you end with two doors with 50/50 chance of having a car. All that math stuff don't really mean anything. The door that was not eliminated does not have greater chance of having the car than your 1st choice.
I remember reading this from her as a kid. Loved working through the logic; it's just counting. The thing people don't get is that the 'host' (originally Monty Hall) doesn't behave randomly; he always shows you a goat. Just count out the cases, which is what she did, and the answer is obvious.
If he doesn't behave randomly, the it ain't a probability question. It's more along the lines of 'form' like at the horse races . We then talk more about odds . No strict rigourous standard calculations for odds . It something settled on by experts . However the probability calculations are ½ for a choosing from two given options for which absolute randomised conditions apply.
@@philip5940 you're confused. It is a probability question, but the difference is it's about posterior versus prior probabilities. just write down all possible options; pick one at random, and consider all of the different ways that the prize might be distributed. Count what happens when you switch or stay. you have 1/3 chances of getting it right by guessing out of the box. He then shows you one of the places the prize ISN'T (that's the part that's not random). Now you have 2/3 chances of winning if you switch. Why? Because there was 2/3 of a chance at the beginning that it was in one of the spots you DIDN'T pick, and he showed you one of the places where it isn't, so those odds still apply; you have 2/3 chances of getting the prize if you switch. It's a classic problem that is now commonly taught in probability, known as the Monte Hall problem (he was the host of the game show). If you don't believe me, write down 3 doors, put the prize behind one of the (doesn't matter which), let's say it's behind door 1. Now say you pick door #1, and it's behind door #1. He shows you either 2 or 3 (where there's a goat), if you switch you lose. Say you originally pick door 2. He has to show you door 3, where there's a goat, if you switch, (to door 1) you win. Let's say you pick door #3; he has to show you the goat behind door 2, you switch, you win. Count it up; 2/3 chance of winning if you switch. It's also ironic you're arguing about his given that the entire point of this video is to explain to you how your answer is wrong.
@@bartonanderson1106 seems that the entire point of the video is to show the power of spin and to highlight mass gullibility . ⅓ and ⅔ probability when given three choices are now a phantom that have ceased to exist The new state is two choices for which the probability is 50/50 .
@@philip5940 I just explained the entire logic to you; you're obviously not capable of understanding this, but here's another chance: en.wikipedia.org/wiki/Monty_Hall_problem
@@philip5940 Since the host always removes a goat that is not which the player picked and neither which contains the car, then that means that the other he leaves closed is the winner one as long as the player starts failing, and that occurs 2 out of 3 times on average, not 1 out of 2. This is better seen in the long run: imagine you played 900 times. In the first selection you are equally likely to select the option that has each content, so in about 300 times you would get which has goat1, in 300 which has goat2, and in 300 which has the car. In total, 600 times a goat and 300 times the car. As the host will reveal a goat from the two doors that you did not pick in all the 900 games, in the 600 that you already have a goat, the revealed goat must be the second one, and so the car is in the switching door. Only in the 300 games that you had the car behind yours, the switching door will have a goat. So, always two doors left, but yours results to be correct 300 times, and the other that the host leaves, 600 times.
When you initially explained that the other door would have a 2/3rd probability of a car being behind it, i couldn't understand it one bit. But i loved the explaination including a 100 doors where 98 were removed. That explaination immediately clicked to me and now I get it! What an interesting question. I always love these kinds of probability questions cuz they make me use my brain in ways I don't get to use while studying 😅
Im really curious about why/how the second example changed your mind since its exactly the same explanation, just different numbers. Was it the fact that it sounds more absurd that your 1% turned into a 50% instead of the 33%?
My initial issue with the concept was my assumption that the host could reveal your initial choice and it’d be allowed to be wrong lol I didn’t even understand the game apparently
@@BillyViBritannia Why wouldn't extreme exaggeration of an effect help people prove to their own intuitions that the effect exists? Guessing right/wrong in the first question will "feel" like it could be a result of natural RNG because the discrepancy of a single outcome between 33% and 50% can be masked by luck. That won't intuitively update your brain's model of prior probabilities. The rational explanation of Monty Hall will be easier to wrap your head around when it is ALSO reinforced by your intuition that you would almost definitely lose by staying.
@@jacobkdunn most people who don't understand the problem - as I've found out from the hundreds of replies under my attempt at explaining this - believe that anything that happens before you are left with 2 doors is irrelevant. And if you believe this then it shouldn't matter whether you open 1 or a million doors, final choice is always between two, and thus 50-50. That's why I was genuinely curious at this comment, I was not trying to bash them like the next person did. And it's also a huge pet peeve of mine when someone tries to explain something in a different way and then proceeds to give the exact same example with different numbers, so I guess I was kind of annoyed that it actually worked.
There are two important things one has to realize, in order to understand this problem (this is not a puzzle at all): - player picks only 1 door, leaving 2 doors to a host - and that's exactly the reason why the door offered by the host will hold a car twice as likely as the door player picked - host knows where the car is, and never opens it - which is needed for a car to always be behind the door offered for a switch, if player haven't already picked it
Here's a clear way to think about this There are 3 cases C-car G-goat C - G - G G - C - G G - G - C There's 1/3rd chance you'll pick the C door in first trial, and 2/3rd chance of G door Because the opened door is not random and the host always opens a G door, you have essentially "flipped" the odds if you picked door 1 (or 2 or 3) in all these cases, in 2 of them you will pick the G door, host opens another G door, so the remaining door has C, switching gives you C in 2 of 3 cases you lose by switching only in the one case where you picked C door right away, which happens 1 in 3 times
Put it this way: Everyone would switch in the 3-door case when the host does NOT open one of the doors but offers you to take the two doors instead of the one you took.
Wow! That is the clearest way of describing the situation I've seen. "Stay with YOUR original 1:3 chance or take OUR 2:3 chance". The revealing of the goat was just a trick to make it seem as if it's become 50/50 to switch or stay.
What a great explanation, well said! It's obvious explained this way. I also like how Savant's analysis of all outcomes is so simple and really proves it, and yet the top mathematicians who criticized her didn't think to do such a simple analysis.
The host knows what is behind every door. The offer to change my choice is only given after I picked a door already. Obviously the host is trying to fool me. If I was given 2 choices from the start: pick a door, or pick 2 doors, the answer is obvious.
@@Tombecho If you think the host is benevolent: Switch. If you think he doesn't know where the car is but just got told where one goat is: Switch. If you think the host is mean: Stick with your choice.
Just reverse engineer it. If you always switch, you will only lose if you had first selected the car. There is a 33% chance to lose when you do select the car on the first try. Leaving you with a 66% chance to win if you always switch.
But having a chance and actually proving you are correct is two different things. The car could have been behind door #1 and then everyone using their math ability to choose #2 would lose. Sometimes you need luck too. That is how nature works. You cannot predict anything definately. You can only guess.
@@edwardspencer9397 that's why it's called probability. Out of three doors you will win one in three times if you stay with your original door. Out of 100 Doors you will win one in 100 times if you stick with your original door. It would be ridiculous to not switch unless you didn't want to win
@@edwardspencer9397 yes you need 'luck', but by always switching you will be 'lucky' 2 out of 3 times. Whereas, if you do not switch, you will be 'lucky' 1/3 times. Therefore, by switching you increase your 'luckiness'. Considering this, you may be less fortunate if you merely guessed at what to do.
The explaination is quite simple : Only one car and TWO goats. So : 1/3 : you pick the car, then you must stay to win. 2/3 : you pick goat 1 or goat 2, then you must switch to win. Saying that the final choice is a 50/50 chance means that your initial choice is 1/2 to pick the car and 1/2 to pick a goat, which is mathematically wrong. Then : ALWAYS SWITCH. CQFD as we say in France :)
This is the problem i have with it. In math if you take something away it is removed from the equation. Yet here it stays with the equation. Not very math like.
@@kevinriffey9970 You are confused because you don't understand where does the 1/3 vs 2/3 comes from. You think that it is because they are still keeping the original 1/3 of the door that was removed. That's not true; it does not come from that. That 1/3 was in fact removed. The thing is that also half of the 1/3 of your original door was removed, and that's why it is now half as likely as the other closed door. Remember that the host must reveal a losing option from the two that you did not pick. That means that if yours is already a losing one, he is only left one possible losing option to remove, but if yours is the winner, he can reveal any of the other two, making it uncertain which he will take in that case. So, the original 1/3 chances of your door, that is #1 in the example of the video, are actually divided on two halves of 1/6 each depending on which of the others is revealed then, #2 or #3, so once #3 was revealed you must discard the 1/6 in which the revealed one would have been #2. Therefore, your door #1 is only left with 1/6 chance, while #2 remains with its entire 1/3, as we are sure that if the correct were #2 the host would have opened #3. Now, since the total probabilities must add up 1 = 100%, you must scale those chances 1/6 and 1/3. Applying rule of three, you get that the original 1/6 represents 1/3 now, and the original 1/3 represents 2/3 now.
@@kevinriffey9970 Continuing with my previous example, the confusion comes because the final proportions end up being the same as in the beginning, which gives the impression that the information was not changed, but that is not the case. It is just that when you make the new calculation, the proportions happen to be the same. If you think this is strange, think about a soccer match. We know that each team starts with 11 players, so each has 50% of the total. Now suppose that during the match, a player from each team is ejected from the game. Each team is left with 10 players, so if you calculate the current proportion, you get that each team still has 50% of the total. That does not mean that they are still counting the players that were ejected, it is just that the proportions with the remaining ones happen to be the same.
@@RonaldABG No I'm saying she is using a hypothetical of a hypothetical. For example, try using her logic if you fix the car to 1 door and make the game show host open only 1 door as the question is worded. Most people see the question as a real life scenario in which there is only 1 attempt and a set door.
@@RonaldABG If you look at the possibilities listed its a 50/50 at 3:21 She is just picking a subset of possibilities to validate the switch. In reality its a 50/50 in a statistics world that does not exist its different. In reality there is only 1 possibility which is a 50/50. Fix the car to 1 door like it would be in real life and its only a 50/50 which is what most logical people assume the question is asking.
Nah! They were more worried about how they would be received if they did not. The humble would not rebuke someone in public, as in a letter to the editor!
@@Elle-ht3km There was an extreme amount of sexism back then, and there still is a shocking amount now, but I don't think all of them were because she was a woman. At least a couple acknowledged that she still had higher IQ than them, and thought she simply made an obvious mistake. Even the cleverest people can do incredibly dumb things sometimes. I'm sure even Marilyn oversaw something really dumb in her private life at some point; and they probably thought it was one of those, but in public. They did grew heated to make sure she "corrects" the mistake and thought it was so obvious that it would severely undermine the population's understanding of probability. It was the exact opposite, but I can definitely see why they'd think it was a mistake. At least one of those letters was definitely sexist though, and knowing those times, probably a lot more. There's probably a lot of enchancement on it too, after having been told she's more intelligent than you are and then given actual proof of that statement. Thinking someone make a dumb mistake when you're both sexist and jealous would probably compound and get you extremely heated up. And then after she proves you wrong, and you get personal proof that you're dumber, you'd either get more embarassed and angry or you'd be embarassed and start respecting her.
The easiest way to understand this: The probability of you choosing the right door on the first try is only 1/3, but choosing a goat is 2/3. There is a higher chance that the door you first chose has a goat behind it than a car. And since door C has already been revealed to contain a goat, door B receives the rest of the probability of getting a car, which is a 2/3 chance (because there was only a 1/3 chance it was in door A).
I can understand the confusion. There are three possible interpretations: 1. You randomly choose one out of three doors and stay - 1/3 chance of winning every time 2. You randomly choose to switch or stay - 1/2 chance of winning every time 3. You always switch - 2/3 chance of winning every time And the funny thing is, they are all correct, which is what makes this problem interesting and controversial.
I still can't get the "correct" answer to sink in... When the goat is revealed, you're starting all over, but instead of 3 doors you only have with a 50/50 chance. Why must the calculation continued to have anything to do with 3 doors when it gets reduced to 2 doors and 2 choices? Is there a reason why a person is forced to choose from the original odds after they change to different odds?
Exactly. The thing people don't seem to understand is that the host must open one of the two wrong doors. He cannot open the door with the prize in it (this defeats the point of the game) or your door (which would render the decision obvious). If the host were randomly opening doors (and could open your door or the one with the prize in it), then this wouldn't work.
The reason it is not 50/50 is because the host ALWAYS opens a LOSING door. He does not pick the door at random - it's always a losing door he exposes - this changes everything. Therefore you will be correct 2/3rd of the time if you change your pick.
"The reason it is not 50/50 is because the host ALWAYS opens a LOSING door." So what was the contestant's probability in the host's FIRST game that he did?
@@klaus7443 If you keep your original pick - you have a 1/3 chance of being right. If you switch after host purposely opens a losing door - it's 2/3rd chance of being right.
The best way to understand this is to imagine doing it many times. If you employ the strategy of always staying with your original decision, you'll only win a third of the time.
It's not hard to understand the pyramids !! By vibrating within the earth's "B" field they could communicate as well as travel intergalactically. My goal is to educate others. Just for clarification. The center of a magnet is by far the most powerful. AC current Radio frequency signals could not travel without the "B" field. The oscilloscope never shows the "B" field but it is there. otherwise it would show a short circuit as the two polarities collide. The "B" field pushes everything forward. Things you can do to see the "B" field:1> vibrate the "B" field of a magnet near a trickle of water and watch the water repel the same.2> Vibrate the "B" field on plastic or glass and watch the item lose weight!! You must understand, don't try this on iron or metal; it will distort the magnetic field . 3> Vibrate on Granite rock and it will become weak. and You can cut and shape ONLY with copper tools. AGAIN pay attention to the magnetic field you can't use iron tools for this action. . 4> I have not proven this part yet, but you should be able at the correct Two frequencies to levitate non metallic objects. FYI I think that's why most UFO's don't show on radar they are not metallic. I have come to believe we are all operating on the magnetic "B" field. The entire universe is 100% magnetic, we are an algorithm of the same. Think about this : The pyramids were communication and transportation device's of yesteryear: When you install such a huge mass within the "B" field of earth they intercommunicate via the "B" field. FYI I have teaching credentials and much much more. Want some Proof, Check out Coral Castle in Florida on TH-cam, all his generators were set up in the repel mode to maximize the "B" field into the earth FACT!! The stones at Stonehenge and KT were part of the "B" field system, The stones would vibrate at frequencies that would stimulate the crops and many other things. Those were the true power plants of yesteryear. WB6HUN/1958
I remember this problem from quite awhile ago. I was able to get it by the method used in the middle which involved the hundred doors, in which case one’s intuition works properly. It’s interesting that if you think of the given case with only two remaining doors, it’s easy for your intuition to be wrong, but in the hundred door case, anyone would switch. In probability problems our intuition is often wrong, so if you generalize the problem by considering what happens as you increase the numbers and look for a pattern, that’s often helpful.
What if you didn’t pick any door and I eliminated all but two doors from the 100 and told you to pick. Would switching between them change your odds? I don’t think so.
Doh … with the 100 door example boths doors have a 1/100 chance of having the car, the same odds. After 98 are eliminated two remaining doors each have a 1/2 chance of having the car…the SAME odds relative odds. The absolute addos of both doors changes but the relative odds of those two doors remain unchanged. The premis is simply wrong and the big joke is that it is just a con trick like the hare and the tortoise, which clearly never catches up. Also some people might actually prefer the goat, for a lifetime supply of goat milk. The woman was taking the piss and most people are too dumb to spot the trick, including the narrator.
The problem is that people confuse what the question is based on. It is based on knowns and unknowns. as soon as a door is known it is removed from the equation AS AN UKNOWN because it is no longer an unknown door. to keep the door in the equation you have to PRETEND you don't know what is there which throws people off. so you have 3 doors and open one so now you have 2 doors and a goat standing there NOT 3 doors as is the mistake here. from 100 unknown doors if you open 98 you then have 2 doors of unknowns and 98 'objects or nothing' the 98 'things' are NO LONGER DOORS. Why the people are insisting on keeping the *known* doors in the math as though they are *unknown* doors is the real stupidness here .. lol as soon as an unknown door becomes known it isn't unknown anymore and should be removed entirely or if you pretend it is still unknown then you should also pretend it was never opened. lmfao i think people miss the fact that the room with the door only exists because it's unknown and should disappear from being unknown when it IS known. how can you possibly include the third door as an unknown in calculating odds WHEN IT IS STATED THAT IT WAS TRANSFORMED INTO A KNOWN.
@@spivvo it seems few people see the root of the issue here. after the door is discovered people are still including it in the math as an 'unknown'. If the door has been opened it is no longer a door so how can you possibly include it. you have to PRETEND it's still a door even though it's been destroyed by becoming a 'known' i would like to talk to the people who said she was wrong then said she was right.. lol
@@jaybefaulky4902 You are not using the information given by the host. This is a conditional probability problem and as such is mathematically and logically solved with the doors closed. There are three ways it could have been worded, two of them with the doors closed. The author of this puzzle chose the wording to include an open door only for the purpose of fooling the most readers. If you read the rules carefully you would know where a goat is even if it wasn't revealed. Having said that the door that must have a goat is useless to both the contestant and the host. To show it's of no value the host could have given that whole door to the contestant and his chances of winning by staying is still 1/3+0=1/3.
It’s one of those problems, which one can understand better if one writes down the possibilities of each choice and calculate the probability of switching. The constant in each game is the host opening the door with a goat in the second step. Game 1 and variations Door 1 chosen by the player 1 Car 2 Goat 3 Goat (2nd or 3rd door reveals the goat) - Switching makes the player lose the game 1 Goat 2 Car 3 Goat (3rd door reveals the goat) - Switching makes the player win the game 1 Goat 2 Goat 3 Car (2nd door reveals the goat) - Switching makes the player win the game Switching wins 2 out of 3 games. In other words, there is a 66.67% chance of winning when you switch the doors. Game 2 and variations Door 2 chosen by the player 1 Car 2 Goat 3 Goat (3rd door reveals the goat) - Switching makes the player win the game 1 Goat 2 Car 3 Goat (1st or 3rd door reveals the goat) - Switching makes the player lose the game 1 Goat 2 Goat 3 Car (1st door reveals the goat) - Switching makes the player win the game Switching wins 2 out of 3 games. In other words, there is a 66.67% chance of winning when you switch the doors. Game 3 and variations Door 3 chosen by the player 1 Car 2 Goat 3 Goat (2nd door reveals the goat) - Switching makes the player win the game 1 Goat 2 Car 3 Goat (1st door reveals the goat) - Switching makes the player win the game 1 Goat 2 Goat 3 Car (1st or 2nd door reveals the goat) - Switching makes the player lose the game Switching wins 2 out of 3 games. In other words, there is a 66.67% chance of winning when you switch the doors. There are nine total games possible (technically 12, but that’s irrelevant to the problem). Switching will win you 6, and not switching will win you 3. Switching 2/3 (66.7%) Not switching 1/3 (33.33%) What we’re really calculating here is not the probability of what is behind the door, but whether switching makes a win or not.
The or is what makes this statistical assumption wrong. There are 12 distinct possible outcomes. you can't put an or on the ones that make the player lose and treat the ones that make the player win as separate. for each variation, there are 2 that will win and 2 that will loose. 50/50. They even show it at the 3:23 in the video that its equal while saying that it's not. hilarious.
There seems to be an error, because in the example in the video door 1 was chosen (your GAME 1) and exactly door 3 was shown. That only leaves the first to lines/options as remaining possibilities, otherwise door 2 had been opened to reveal the goat there and still hiding the car. Do you anderstand, that you cannot "switch" here and count that, because "2nd door reveals the goat" did not happen. There are only options left that could reveal the goat behind door 3, and this is not goat, goat, car (with the information of a goat behind door 3, this is eliminated. Likewise, showing a goat behind door 2, completely eliminates goat, car, goat - and so on).
God how we hated Statistics 101, just to get our degrees in Sociology. YOU PROVING MY "theory" MATH MATH MATH... Ah, maybe YOU can answer my question then ? I'd heard of her before, but "as a mathematical expert ?!" I'd like to see the BRAINY DOSE VIDEO MAKERS get rid of their mechanical narrators and DO AN INTERVIEW WITH Marilyn..wouldn' tyou ? Smart? Intuitive? Knowledgeable ? Verbally expressive of her thoughts ? The list could go on, yes ? Seemed to me "they" were sort of picking on her for not using the "math stats" as they did ? Oh, of course; now let's hear it for the "Intelligence of Creative Thinking!?" It seems THERE ARE different "kinds" and "types" of IQ " Tests." Experience and Education , two possibly different types ? So we should be also asking WHICH IQ Test did Marilyn excel in, or on ? MISSING FROM the video; Does this lady write and speak in both German and Italian ?I've always felt there is an extreme indication of high intellect in regard to peoples' abilities TO express themselves in foreign languages ?Seems there is a certain "type" or "kind" of logic it seems in learning to "relate " foreign language to one's own ability to speak and write in their native language ? It appears Marilyn 's " (by assumption?) that Marilyn had TWO "Native languages" yes ? Her opinion of "public schooling" holds great merit. I remember a question required to be asked on a high school test , was "Who were the Phoenicians ?" The ABCD Answers included the answer " Venetian" . Most admitted later that they all misunderstood the word Phoenicians because they were all more "familiar" with Venetian Blinds, than historical terms of peoples and places! (Ha Welcome to American World History 101-we (all) need to repeat that one!?) Which btw lead to my last question (for you or Marilyn, ha ?) Is the inability to "spell" properly (in any language/especially ones native language ) indicate ignorance ?Duh...As a retired teacher, I submit I've become dependent on the Google Gargoyles ' offers for correction, which often just doesn't exist. The robots tell me I've misspelled something, yet/while, offering no options with which TO correct it. Good at Questions; Slow at the answers. Anyone ? "I Ain't no Middleman" Fred Gold & Lynda Faye Copyrighted 2016 by LyndaFayeSmusic@gmail.com or Yahoo, if censored for using the word " God" too often?
@@jameslinmd 1/3x1/2=1/6 Pick Car, host shows Goat A 1/3x1/2=1/6 Pick Car, host shows Goat B 1/3 Pick Goat A, host shows Goat B 1/3 Pick Goat B, host shows Goat A Probability of winning by staying 1/3 Probability of winning by switching 2/3 This is the second time your faulty theory has been debunked.
I am amazed at this! I consider myself a smart person, but I'm not a genius. This made perfect sense after listening to the explanation. I never would've come to that conclusion. The one with 100 doors was clear as day. It's HIGHLY unlikely that you would choose the correct door out of 100 possible choices. After they eliminate 98 doors, it pretty much tells you which door has the car. You would've had to be extremely lucky (or unlucky, as it turns out) to select the correct door out of 100. Brilliant!
Great. Now note that if you select door #1, switching offers no advantage at all unless the host opens door #2 and there's a goat behind it. Then, if you don't switch, you lose. Do you understand why this is?
@@klocke5247 In the case of 3 doors, it gives you a very slight advantage if you switch. If you look at the scenarios in this image (3:20) The top three show that you would win 2 out of 3 times if you switch. The bottom three show that you would lose 2 out of 3 times if you stay. In the case of 100 doors, your odds are 99/100 if you switch. For 3 doors, it's 66.6/100 if you switch. It seems like it shouldn't be that high of a probability with 3 doors. I guess 66.6/33.3 is a lot closer to 50/50. I wouldn't be as confident switching with just 3 doors.
We Russians knew of several men and women with IQs higher than 235 to 239 and you boast about her? If I were you I keep them as a group of natural security assets. But one of them, a woman said this, IQ IS NOT EVERYTHING! You may have a higher IQ but still remain a complete utter boop if you did not receive the required proper education to make use of it. In fact, she says, proper educational environments can practically increase the IQ of EVERYBODY, man, womand, and child if we are willing to have the political will to do it!
The confusion people have, is that the door being opened by the host is NOT RANDOM. The host opening the door KNOWS it's NOT the winning door. If the host opened the door randomly then the odds wouldn't change.
Not really. Because if the car was behind door 1 and you chose door 1, then the host could have opened either of the other doors at random. In that case, switching would have caused you to lose. However, that doesn't change the answer to the question. The question simply stipulated if your chances would be better for you to change doors, not if changing doors would guarantee you the win. And the answer is still yes, that technically, logically, and mathematically, switching, even in that specific circumstance, would have increased your chances of winning. For example, if you look at the example the narrator gives with 100 doors. If you just so happen to pick the door with the car as your first choice, and they eliminated 98 of the remaining doors, that other door still would have a higher chance of being the correct one, except for the fact that you had the dumb luck to happen to pick the one door that wouldn't have helped you.
@@TalkingHands308 The host doesn't randomly reveal the car if you picked a goat. That's what they meant. This makes your odds after switching 2/3 instead of 1/2 or 1/3. You will still lose if you switch after originally picking the car.
3:33 Yes, because the host would not open a winning door as in game #3 with door #3 and would have therefore opened door #2, since she had already chosen door #1. And switching to #3 in game 3 is a winner.
I am so happy about it too. I knew that you could easily write it out but the idea to do it with 100 doors blew my mind. Then without writing it out, it will make more sense to many people.
Probability is a coin does it fall heads or tails?? and may be one in two thousand flips it lands on its edge ?? play roulette on red or green?? sit and wait for a run of 7 red?? then bet on green? logic says it as a good chance of falling on a red??? Yet really every time is a new happening and as a chance of red or green or may be the bankers zero's You just got to feel lucky
The speed of Light is a very insignificant item. Has little to do with the universe we live in. The pyramids were used for not only instantaneous communications but also for instantaneous physical transportation to other galaxy's. E=MC2 has absolutely no relevance to anything!! Wake up please!! We live in a 100%magnetic universe! The center of a magnet will show you anti gravity, It also will show you how to repel water, we will no longer have a problem going to extreme depths. Do some basic experimentation using the center of a magnet it will blow your mind. However, Keep in mind the magnetic field you cannot use any iron anywhere near, it would disrupt the center field of the magnet. As you know I refer to the center of a magnetic force the "B" field. You can tap glass plastic anything other then iron and it will loose weight. That is exactly why they used copper tools to build the Pyramid's Please review "CORAL CASTLE Florida on TH-cam> What he did was transmit the "B"field into the ground via his generator that was set up with 25 "V" magnets five deep, all in the repel mode to maximize the "B"field energy about his property and he also had a horizontal stone similar to Stone Henge on his property. Don't waste your time on E=mc2 it has no relevance to anything!! If you take a granite rock just like those used on the pyramids and "VIBRATE the "B" field into same at the correct frequency it will become easy to cut with a copper tool. FACT!!!! Please get the tools you need to learn the facts.!! Thanks
The Monty Hall problem is flawed. The question poses two completely different scenarios 1 being "If you choose a door #1 and the Host opens 'another door'" ...then it proceeds to add information thus changing the question..."If you choose door #` and then opens the third door" So there's nothing mystical or profound about this quesiton, it's simply flawed. Yes the player might have an advantage with the first scenario as seen by Marylin's Graph, but the Second question, once that third door is opened the problem negates the graph because she's showing the third door not beign opened as a variable xD
If I remember correctly, the way the problem was originally worded in the newspaper didn't have one key condition - the host being required to open a non-winning door. It only said the host opened one of the remaining doors. So her answer was actually wrong for the specific problem, and it really was 50/50. She gave the right answer to the wrong problem.
@@weevil601 "It doesn't matter if he simply guessed or if he knew where the goat was." It does matter. If he didn't know where the car was and opened another door that has a goat then it's 50/50.
@@weevil601 If the host does not where the car is and randomly opens another door with a goat then here are all the cases, each of them with the same 1/6 probability. Pick Car, host shows Goat A Pick Car, host shows Goat B Pick Goat A, host shows Goat B Pick Goat B, host shows Goat A Probability of winning by staying and switching is equal, 50/50.
@@weevil601 Even if you perform 99 trials the car would still be randomly placed behind the host's door 33 times, behind the contestant's door 33 times, and behind the door no one picked 33 times.
@1978ajax No she achieved the Mensa 228 IQ test in a supervised manner (they are always monitored tests). However, she spent a lot of time actually studying IQ tests and sitting them privately which makes you much more versed in the problems and questioning style and can help achieve much higher scores, than just being thrown an IQ test out of the blue as most people are when tested at school etc. You must still read and comprehend the questions however, as they may slip in updates / variance or even completely new problems. She almost certainly has a photographic / Eidetic memory and can memorize anything. She writes and publishes books. So for all the massive IQ she has, she hasn't really invented a space warp drive or anything, which a lot of people perhaps expect. That's not how people's intelligence and lives work really.
yes same as people knew that earth is center of universe. :D ..... To avoid being burned at the stake, the 69-year-old was forced to renounce his belief in a heliocentric model of the universe. And so I see you, Trying to burn those with different oppinion :)
Marilyn has been one of my idols since I was 17... that was 50 years ago. I used to read her column in the paper every Sunday and loved it! She's a national treasure I rank right up there with Dr. Thomas Sowell and John Stoffel.
By reading the comments I thought of my own way of understanding the problem. You have higher chance to pick the wrong door in first question. So when the second question arrives, you have higher chance of winning by changing doors because it is dependent on the chance of you picking the wrong door, and since you have higher chance of picking the wrong door, you have higher chance of winning if you switch doors.
You understand it right. Since the winning door is 2/3 likely to be among the 2 doors you didn't pick, and the host eliminates one of them, you should switch and double your chances (1/3→2/3)
The professional mathematicians who pointed out her error are correct and vos savant is wrong . She may perhaps be confusing the odds that relate to coin tosses .
Ah, I get it now. You pick a door: Outcome 1: The prize is in the door you picked. So switching would lose the game. Outcome 2: The prize is in the 2nd door you did not pick. And since monty hall only removes the goat door, switching would win the game. Outcome 3: The prize is in the 3rd door you did not pick. And since monty hall only removes the goat door, switching would win the game. Switching has 1/3 chance of losing and 2/3 chance of winning!
@@tactic2569 well if the total num of doors is 3 + x, with x being more doors added. If monty hall removes 1+x amount of goat doors, Then the chance of winning is 2+x/3+x and losing is 1/3+x. Meaning it would be even more logical to switch if more doors are added and monty removes more goat doors.
I got it a different way-- in the 100 door example, you pick a door randomly, then 98 wrong doors are removed. Your door can't be touched, but the right door also can't be touched. Your door was originally chosen at random, so switching doors would be better, since you have more information to make the choice with. Although this method probably wouldn't work for any other problem and I just didn't know any other way to rationalize it.. Here's an extra thing: I also thought, "hey, the probabilities just merged into one door! That's cool." Without really understanding why it happened.. :/
I remember when she presented this puzzle in the newspaper. I was one of those who was skeptical, so I set up a quicky computer program to simulate it over many trials to see what would happen. I was surprised to find that she was absolutely right! lol
What about intention in this equation. The people at Let's Make A Deal know where the winning door is, so their intention could be to help or to harm. Is that not a variable.
ok i never got this. i thought it fell under the gamblers fallacy, that the first choice is irrelevant to the second choice. but after thinking about the hundred doors, i get it now. thanks for the video
It's a very easy problem actually. There is a (1/3) chance you pick the right door. In that case you are wrong if you switch. However, there is a (2/3) chance you pick the wrong door and the presenter opens up another wrong door for you. In that case there is a 100 percent chance you pick the right door when you switch. 0+(2/3)*1 is 2/3. I think half the schooled population could figure this out as well.
If the presenter opened up another wrong door you are implying a 3rd chance for the 100% pick. Or more than half of course could figure this out mathematically. There's always someone that says "it's a very easy problem actually". There's only two chances offered and there is no 100% chance.
But when you stay with the first door, you are also playing the new odds. So this debate is irrelevant. I'd expect more from this smartest person in the world not to realize this.
@@FathomlessJoy no the odds don’t change at all if you stick with the first door you choose, that’s like asking someone to pick a number between 1 and a million trying to guess the number you’re thinking of and then revealing the 999998 wrong numbers leaving two numbers left and he says ”It’s a 50% chance i choose the right number initially!” when in fact it’s still only a one in a million chance like before.
Another way to try to grap it with intuition is to think about it the other way around. Initially the Chance to pick a goat is 2 out of 3 and to pick the car is 1 out of 3 So the chance to pick the wrong door is 2:3. At this time the host has only ONE door left to open, as YOU already picked the one with a goat. Therefore by switching the door you CONTAIN this 2:3 probability. In 2 out of 3 cases you picked the wrong door and the only one left is the one with the car as the host ALWAYS opens one with a goat.
Unfortunately it's not clear how it's not 50%. Even if I picked the right door the first time (say door 2), the host would have picked the wrong door (say door 3) and offered me door 1 or 2. The same applies if I'd picked door 1. So to me I don't see how the hosts actions increases the probability of the unpicked door having the car.
@@nitinullas it's not 50% because there is a much deeper intelligence at work, you have the social norm mathematical equation that would give door one the better chance. 😉 and then you have the social norm that only sees what's right in front of them. Think deeper or overthinking makes for a much wider probability of intelligence. Hint: shallow thinkers of black and white only see 50% chance.
@@nitinullas It is quite clear. 1/3 times you pick the car. This means your odds of winning by staying are 1/3. By staying you only win when you initially chose the car. 2/3 times you pick the goat. This means your odds of winning by switching are 2/3. By switching you win by initially choosing a goat and switching to a car.
@@nitinullas Well, the probability does not change. That is part of the reason. And the host itself is not free to do what he wants - he is restricted in his actions as he is only allowed to open a door with a goat. If he would open doors freely and the doors would contain like envelopes with a key or a "you lost" note the probability would not change. Think about it this way. Your chance to pick the correct door is one out of three. In this case the chance that the car is behind one of the other two doors is two out of three. This does not change when the host opens a door. The chance you chose the correct door is still one out of three and the chance that the car is behind one of the other two doors is still two out of three BUT the host opened a door with a goat as he is not allowed to open the door with the car. Therefore the chance that the door you did not chose contains the car is two out of three. The point is. You chose door A. Car can be behind A or B or C. If it is behind A the host can either open door B or C. If you switch, you lose. If the car is behind door B the host can ONLY open door C. If you switch to B you win. If the car is behind door C the host can ONLY open door B. If you switch to C you win. Even in this small example you can already see that you win in two of the three cases if you switch.
The trick, for me, to intuitively understand that she was right is the realization that, when you first picked a door, the odds were against you. You only had a 1 in 3 chance of picking the correct door - probabilistically speaking, you were more likely to choose the wrong door than you were to choose the correct door and so, when more information is received, narrowing your ALTERNATE choices down to only one, you should probably switch (because you were more likely wrong than right the first time). That's it, that's the whole thing (the probability of your first door pick being incorrect). ... Not convinced yet? Here's the real kicker, the thing that "bends the odds" in your simple choice of two doors at the end. Because you have three doors to choose from up front, in only 1 out of 3 play outs of this game (probablistically speaking) are you going to be in the scenario where you picked the door with the prize and it doesn't matter at all what door the game host opens because they both are going to have goats behind them. The other 2 out of 3 times, you're going to be in the scenario where you didn't pick the correct door at first, and so it ABSOLUTELY MATTERS what door the game hosts leaves closed. 2 out 3 times, what the game host does provides new and relevant information, 2 out of three times the door they leave closed will be the door with the prize. This is the real world world explanation for why Savant's math is probablistically right. It took me longer than I care to admit to understand all of this. Intuitively, the problem as first stated, is hard because I think there is something inherently "real" to the belief that you have a 50% chance of being right or wrong at the final choice. What's "real" about that instinct is YOUR LEVEL OF "REAL" KNOWLEDGE when the choice has been narrowed down to only two doors. At that moment, you do not know what actually happened when you picked your first door. You don't know if you beat the odds or if you didn't. Also, the fact that you can now choose a new door kind of changes everything (you can argue it sets up a "new coin flip" and a new probability). As always, probability is best observed at scale. When you only get one shot at this game of three doors and a single prize (not 100 chances for 100 prizes in 300 doors, or as the video suggests 1 prize in 100 doors), you're, seemingly, in the same position as someone who has just flipped heads on a coin 9 times in a row. Regardless of those previous outcomes, what happens next is a 50/50 chance. All that said, she is still absolutely right and the math is sound. Your first choice of door was made at an odds of 1/3 (not 1/2). It, probablistically speaking, is slightly less likely to have the prize than the door that survived elimination by the game host.
"Your first choice of door was made at an odds of 1/3 (not 1/2). It, probablistically speaking, is slightly less likely to have the prize than the door that survived elimination by the game host." I figured this was a very easy thing to work out (?). First choice most likely wrong, one correct choice left, better to switch. Logical no?
@@SicketMog No, not logical to me at all. You pick the first door with a one in three chance...the second door is also sitting there with a one in three chance. The third door is opened and there is a goat behind it so, with no information to go on with the other two doors, they are equally apt to have a car. I don't get why it is door two that gets the extra kick, you have no information on either door, you can just as easily say door number one gets the kick.
@@mercurysmith563 When you first picked with no information you had only a 33% chance of being right. You are most likely wrong on your first guess; THAT's the information you have to go on. Yes you COULD have picked the correct door from the beginning, could happen (if so you were ironically unlucky), but there's a smaller chance of that having happened (33% to 66% chance being wrong). It's about improving the odds not wether or not you ACTUALLY win. To me this is super logical, it was when I first heard of the problem as a dumb kid too (without decent math), but maybe it isn't? 🤷♂
@mercurysmith563 Maybe it would help if you actually "played the game" at home with 3 doors in your residence. Under this scenario always make Door 1 your first pick. Run through the 6 scenarios and compare how often you win if you switch to how often you win if you don't. You will find that you will win twice if you switch and only once if you don't. If those aren't your results - you have a flaw in the way you've created the scenario.
My other posts were wrong . I finally found an explanation that made me see that switching is better. When you first choose you have a one in three chance of picking the car, 2 in 3 chance of picking a goat. If you picked a goat door the host opens the other goat door meaning that no matter which goat door you picked you win by switching. Picking the car door first and switching is the only way you lose by switching. This works in all three possible scenarios. Thanks to Bai Su Zhen for the helpful post. It's really just logic but you have to find the right way to look at it.
i finally manage to understood this problem thanks to you. is this how you interpreted it? "i think most people that are like me, got confused by this, is because nobody state the underlying rule. - that the host would always open the goat door. through this rule, i manage to understood that the only way to loose is by picking the car first, then switching to the remaining goat door after the reveal, which is 33% compared to picking the goat door first which is 66%, have the host eliminate the other goat door and then switching to the car door. now if you didn't switch at all, then it would remain 33% each (as if making the choice without taking note of the underlying rule). which is why the video tries to explain that information affects the probability of your choices."
So if it was "if you can switch to a door that hide a different thing from your door whatever it is", I think everyone would get it right. It is mesmerising to think about
tell me what formula you used to pick the first door you do not know what is behind any door. The host knows were all the goats are you do not . If you pick 1 door out of 3 doors you have 33.inf % chance to be right. If the host shows you were one goat is he has removed that door from the equation. You now have 2 doors to pick not the same equation you had 1 choice with 3 doors n ow you have 1 choice with 2 doors 50% chance.
Isn’t that implying the host will only reveal a goat door if you chose a goat door initially? The host is to open a goat door regardless if you chose a car door or a goat door. So how does one have a higher percentage considering they are unaware if they chose the goat and the host will open a goat door regardless.
@@emptyempty4238 Because you had a 2/3 chance of picking a goat originally, and if you did, switching guarantees a car. (and if you didn't, switching guarantees a goat)
The way I always thought about the problem was this: choosing the right door out of three, you’ll be wrong two times out of three so, two times out of three, you’d better better off switching to one of the other doors. If neither of the other doors is opened, you have no idea which one to switch to. The host takes care of that problem and shows you which other door it _isn’t_ - if you’ve chosen Door 1, he’ll open Door 2 if the right door is Door 3 and Door 3 if the right door is Door 2. So in those two-thirds of the time when you’ve chosen the wrong door, you’re taken care of and you’re better off switching. (And, of course, if Door 1 is the right one, he’ll open _either_ Door 2 or Door 3 and you’re better off _not_ switching but you have no way of knowing that.)
Exactly (you have no way of knowing) which renders this whole exercise pointless. When you pick your initial door and the host removes one with a goat, your chances are now 1/2. Since no new information was presented. You can consider the the one the host removed was never there in the 1st place.
@@DMZRPG “When you pick your initial door and the host removes one with a goat, your chances are now 1/2. Since no new information was presented.” That’s false. Again, if you choose _any_ door, that initial choice will be incorrect two times out of three on average. If no other door were revealed, you would not know _which_ of the _other_ two doors was the right one. But, by the terms of the problem, the host will _always_ open a door that _isn’t_ the right one so, in those two out of three times, the right door will be the one he _doesn’t_ open and you’re better off switching. So, if you choose, say, Door A and the right door is _either_ Door B or Door C (again, two out of three times), the host will _always_ open the _other_ door (Door C if the right door is B, and Door B if the right door is C). That’s new information. Switching leads to picking the right door in those two out of three times. If you happen to pick the right door initially (one out of three times), the host can (and will) open either of the other doors and switching will result in you choosing the wrong door. But that occurs in, again, only one out of three times on average.
@@DMZRPG thats what Im thinking. The odds reset when one door is eliminated. There are no longer 3 doors, there are only 2 doors. Why do the odds carry on ? In my view it becomes a new probability.
@@ScreamingEagleFTW “The odds reset when one door is eliminated.” They don’t. Again, when you have three doors, whether you’re dealing with the Monty Hall problem or not, your initial guess is, on average, wrong two out of three times. In those two out three times, if no other doors are opened, you don’t know _which_ of the two doors you didn’t pick is the correct one. In the Monty Hall problem, in those two out of three times, the host, in effect, _tells_ you by opening the _other_ incorrect door-you’ve chosen (incorrect) Door 1; with _either_ Door 2 _or_ Door 3 being correct, the host will open the other one. (By the terms of the problem, he _has to_ because he _has_ to reveal one incorrect door-and, again, two out of three times, you’ll have already chosen the other one.) So, again, if you’ve initially chosen the wrong door, which happens, on average, two out of three times, switching to the _other_ unopened door gives you the right one. You _don’t know,_ of course, _if_ you’re in the situation where you’ve initially chosen the right door (on average, one out of three times) or in the situation where you’ve initially chosen the wrong one (on average, two out of three times) but the odds remain the same, regardless. (If you knew that-there _could_ be a “foolproof” version of the game where the host opens one of the doors you didn’t choose and then says “And, BTW, your choice was wrong (or right)”-you could win 100% of the time: switch to the _other_ unopened door if you initially chose the wrong one, which would happen, on average, two-thirds of the time, and stick with the one you initially chose if that was the right one, which would happen, on average, one-third of the time.)
I actually read the Parade article. I had 5 classes of math in high school and one chapter was on probability. One thing that stuck in my mind was 'All RANDOM choices must add to 1.' So if I had a 1 in 3 chance the first round, and Monty's choices are not random, then my second choice is actually a 2 in 3 chance.
@@MrSanford65 No, you already chose that door and it had a 1 in 3 chance, not choosing the other door means you still only have a 1 in 3 chance. It only changes to 2 in 3 if you switch doors.
The table is flawed logic because of the labeling of the doors. And that's why it becomes so confusing. Only the first two entries of the table are valid. The rest of the entries are actually just permutations of the first two. Imagine the doors had no numbers or arrangements and you didn't even know which one you picked, only that you picked one (of three). Now you were told, you thought you picked one of three, but, it's been reduced to one of two. Would you like to change doors? There is literally no sense in changing. I'm saying that I look at it this way: Instead of picking door 1 of three... I pick one door out of three. This is where the table becomes redundant. The first two entries are the only useful ones.
After the first door is opened, the problem is changed. The remaining doors have a 50% chance of having the car. But when you chose the first door, it had a 1/3 chance of having the car. So you get to 50% when you choose the other door.
Let's say you buy a lottery (1,000,000 numbers with 1 winning number). After the result is out (you don't know the result yet), your friend (who know the winning number) say your number and another number. One of them is the winning number. What is your winning chance? Matt: I pick the lottery number 000000. My friend revealed that all the other numbers but 415896 are loser numbers. Matt: There are only 2 options left: My number is the winner number or the other number is the winner number. Matt: Those 999,998 choices are gone. At this point, my number has a 50% chance to be a winner number. Matt: When my friend remove 999,998 loser numbers, he makes my winning chance becomes 50%. Matt: This means my winning chance will be 50% no matter which ticket/number I bought. Matt: I just found an easy way to win lottery. I am a genius. Hoooraaay!!! Matt's parents: Hoooraaay!!!
5:11 Yes, information matters. I believe the misconception comes from the idea that the host is working against your interest. Subconsciously you may think he will only open a door in order to confuse you and is only offering you to switch, because he knows you have chosen the car. The way the problem is communicated, you are offered a switch, you are not entitled to one. So game theory kicks in and your brain pushes you to refuse the switch, overriding your base math logic in order to protect you from fraud - a mechanism we have evolved to trust :)
Is more complex than that , psychological game he already wants you to keep number one making you think he's asking you and wants you to switch is obviously asking to switch and pushing you to think you have a winner on one , the obvious game here is to switch
@@makalribera6742 It all depends on how often he is working against you and how often he is working in your favor. You can add that to the equation and figure out how the numbers add up :)
And another level to this that no one has said yet is the unspoken rule of the host not to open the first door if it had a goat. If the host could do that, then the game would reset immediately and it would be a 50-50%. And that's unspoken, just a structure of the show
@@shaft9000 None of that matters. This is a straight mathematical problem, which does have a strange answer thanks to the weird world of probabilities. Mind you, picking door 2 gives you a higher probability of being right, but not a guarantee. There is still a 33% probability it could be behind door 1. But the odds are in your favour with door 2, as she said.
Be not misled by megalomanic schizophrenics who in their grandiosity associate themselves with these levels of intelligence based on delusional claims regarding ability, achievement, or test performance. This helps them write books, become famous and make money. Be assured that no one has ever come close to I.Q. 230, that the highest I.Q. to be realistically expected on Earth is about 185, and that only through tentative challenging norms in unexplored score ranges someone will one day qualify.
The real heroes are not those that are born lucky beiy insanely smart. The real heroes are those that use their intelligence (no matter how smart they are) and make the world a better place.
...as you are doing right now. Intelligence without wisdom is like a boat without water. And intelligence without wisdom enables one to outsmart one's common sense. In this case, with MATH. If the car is behind door number one or number two has absolutely nothing to do with a goat being behind door number three, no matter how many math wizards look at the issue. And the 100 door example is a completely different scenario. It doesn't take a high IQ or math wizard to know that. But it does take common sense. GOD's prophet has spoken.
Basically you switch because the door you picked could’ve also been eliminated by the host if you didn’t pick it but the door that remained closed would never, so the unopened door is the best choice
The better answer is simply that since every door has a 33% chance, thus those 2 other doors have a 67% chance of the car being behind one of them. And if you switch, you will get the prize in every game where the car happens to be under one of those 2 other doors. So switching gives a 67% chance The real intriguing question is one of why this very simple problem stumps over 90% of people at first glance. But it certainly does
Here’s how ChatGPT explained the Monty Hall Problem: The Marilyn vos Savant question refers to a famous logic problem she addressed in her "Ask Marilyn" column, specifically the Monty Hall Problem, which generated significant public debate. The Problem: You're on a game show with three doors: 1. Behind one door is a car (the prize). 2. Behind the other two doors are goats (not a prize). You choose a door. The host (Monty Hall), who knows what is behind each door: 1. Opens one of the other two doors, revealing a goat. 2. Asks if you'd like to switch your choice to the remaining unopened door or stick with your original choice. The question is: Should you switch, stick, or does it not matter? --- Marilyn's Answer: She advised always switching, because it increases the probability of winning the car from 1/3 (if you stick) to 2/3 (if you switch). --- Explanation: 1. When you pick a door initially, the probability of choosing the car is 1/3, and the probability of choosing a goat is 2/3. 2. If you stick with your original choice, you retain the 1/3 chance of winning the car. 3. If you switch, you win the car in the cases where your original choice was a goat (2/3 probability). Monty's action of revealing a goat doesn't change the initial probabilities but shifts the odds in favor of switching. --- Controversy: Many people, including mathematicians, initially disagreed, believing the odds were 50/50 after Monty reveals a goat. However, mathematical simulations and proofs consistently supported Marilyn's explanation.
I was skeptical until I wrote this: There are 3 options: I select a car, then the host reveals a goat and the remaining door must contain a goat. I select a goat, then the host reveals another goat and the remaining door must contain a car. I select a goat, then the host reveals another goat and the remaining door must contain a car. In 2 of 3 cases the remaining door contains a car... so I should switch to it.
@@kelseysmart9472 the point of the game is that the host is eliminating one of the doors from the equation. From there you are choosing to switch your answer to the remaining door or not. Wether that is door 2 or 3 I do not believe matters.
I think a more intuitive way of understanding this problem is recognising how the question has changed from the first choice when offered the second choice. The chance of being correct (selecting the car) on the second choice is dependent on being incorrect on the first choice (having selected a either goat). Therefore, the chance of selecting the car in the second choice is the chance of NOT selecting the car in the first choice, which is the chance of selecting a goat on the first choice. This is 2/3.
@@R391s Let’s run through the different scenarios/outcomes (allowing for where the car is and whether you choose to stay or change your answer): Car behind door 1 and you keep your 1st answer - Choose Door 1. Shows Door 2 or 3. You stay at 1. Therefore: WIN - Choose Door 2. Shows Door 3. You stay at 2. Therefore: LOSE - Choose Door 3. Shows Door 2. You stay at 3. Therefore: LOSE Car behind door 1 and you change your 1st answer - Choose Door 1. Shows Door 2 or 3. You change to 3 or 2. Therefore: LOSE - Choose Door 2. Shows Door 3. You move to 1. Therefore: WIN - Choose Door 3. Shows Door 2. You move to 1. Therefore: WIN. The same outcomes will occur if the Car was behind door 2 or door 3. You may notice that if you stay with you answer you LOSE twice and WIN once. Therefore, your probability remains at 33,33% (as before the door was revealed). However, if you change your answer you WIN twice and LOSE once. Therefore, your probability of choosing the car goes up to 66,67% when you use the strategy of changing your answer. Showing the above by force is not ideal, but it illustrates it visually very well (imo). However, I would implore you to consider looking at the Bayesian Statistics proof or running a simulation (with Python for example) if you don’t trust the above at face value.
It's easy af, by switching you pick 2 doors, the wrong one and the unknown one, that's 2/3 or 66%. If you don't switch you pick only 1 unknown door, that's 1/3 or 33%. Some people believe it's always 50% but you have to start counting your propability from the very beginning when there were 3 doors, not after one was openend. Because the first choice also matters.
disagree. You flip a coin, neglecting the edge probability, you have either tails or heads. When you flip you always have 50% chance probability of landing either heads or tails. However, flipping 50 heads in a row is extremely unlikely and thus getting a 51st head in a row is, too, extremely unlikely, despite having only 2 outcomes. Lavant's high IQ is garbage and false or she cheated on her test. IQ test measures several things: verbal, reasoning, memory, processing speed. Yes, those 4 combine into 1 IQ score, still, I would love to see her true published IQ test. Until then I call bullshit.
@@vkaa3k190 Chance for any of the doors to have a car is 1/3. Then the host is opening one door. But host is not allowed to open your door -> your door is a separate group of propabilities -> the door that is revealed transfer its probability to the other door in its separate group... Or You make decision to stick with your initail choice(when all 3 doors were unknown) = you are ignoring that extra information provided to you by the host so your chances do not change...
@@fleurdemaiko lots of mathematicians have called out her bs, so you are the one making a n**ger out of yourself. And you don't have to have a PhD in math to figure out the probabilities.
@@vkaa3k190 The fact that it is astronomically unlikely that you might flip 50 heads on a row is irrelevant to the MHP. But if you did that , the 51st flip has a 1 in 2 chance, The past, in this instance has exactly nothing to do with the future If you stick to the first choice, the chances are 33%, opening other doors cannot change that fact. If you switch, you get the car if it is behind either of the 2 other doors, which happens 67% of the games.
Huh, interesting. I didn't know about her. She married Jarvik, of artificial heart fame, and once publicly questioned Wiles for relying on hyperbolic geometry for Fermat's last theorem.
The key here is that we know that the host will always open a door with a goat. Because if the host picks one of the doors at random that happens to be a goat, the odds will be 50/50. But as stated, the host will always open a door with a goat and then it’s is better to switch. The easiest ways to see this is that if we pick door #1 and then switch, we will win if the prize was behind door #2 or #3, while by not switching we only win if the prize is behind door #1.
"Another way to look at it is from Monty's perspective: Monty can only open a door that you didn't pick, and he can only open a door that has a Zonk behind it. There's a 1 in 3 chance that you picked the car. So there's a 2 in 3 chance that you picked a Zonk. So 2 out of 3 times, Monty has only one door that he's allowed to open. Because one is your Zonk and the other is the car." - @motorsheep
Wrong. Nearly everyone gets this wrong. Let me explain the true flaw in the Monty Hall problem. There's an implicit assumption that few seem to notice, and that's that the host is required to offer a door switch. What if the offer is made only when the host knows that the contestant has chosen the prize? If this is so, then it's best not to switch. Because we don't know the rule the host is using, or the odds of the host adhering to any rule, we cannot say whether it's best to switch or not. Sorry Marilyn, but you blew this one, but you're hardly alone.
@@mrwilkdfly are you familiar with the saying in physics, “let’s assume the cow is a square”? The point is that you are always working within a model that is not reality. Here it is the same, you can bring up any number of valid points that show the model is not a perfect representation of reality, but that is not the point. In the original problem it was the assumption that always a goat door is opened by the host, that creates an interesting problem that many people got wrong as they thought it was 50/50 to switch.
@@mrwilkdfly Your math skills are really poor. If it's not in the question then it's not in the equation either. Math problems are solved with the information that is there, and not from the imagination coming out of your arse.
An example showing that changing doors increase your chances: Door 1: goat Door 2: goat Door 3: car You chose 1, it’s wrong, if you changed you would be on door 3, win. You chose 2, it’s wrong, if you changed you would be on door 3, win. You chose 3, it’s correct, if you changed you would be on door 1 or 2, loss. Here you can see that by keeping the first choice you have 1 in 3 chance, and by changing you have 2 in 3 chance.
What if you pick 3 in the first place? Its a bit confusing. Maybe because I just got introduced to the problem. Maybe I will get some clarity in a few days
I look at it this way. There are only three possibilities: 1- you pick the car, so the host shows one of the 2 goats - then you should not switch door 2- you pick goat 1, so the host shows goat 2 - you must switch door 3 - you pick goat 2, so the host shows goat 1 - you must switch door Therefore there is 2 out of 3 chance that switching your door choice will get you the car
@@user-ej9nl1ng9dMost of the people who argue about it can't read a table. Many can't even follow John's simple logic. The biggest problem is getting them to realize 50/50 is wrong or they don't even listen to explanations of the right answer.
thank you - that is a good explanation the key is that *the host always eliminates a losing choice* the odds that the original choice was correct don't improve but the odds that the remaining non-eliminated door is correct do increase it is easier to grasp for most people if the door count is say 10 or 20 or some other larger number if all the *wrong* choices are eliminated except the initial choice (which may or may not be wrong) - plus one other last remaining choice - there is a very high probability that the correct door is the one remaining door (the one not eliminated)
The easiest explanation is: your 1st choice has a probability of 1/3. The revealed door a probability of 0/3. Therefore the final door a probability of 2/3. Is that easy or what?
It goes back to probability. Based on your information, what is the probability that door number one will be the correct one. Probably zero. It also highly depends on your approach too, whether it's thinking of the goats being animals not wanting to make too much noise Near each other, or maybe it's the car, keeping it further away from the goats because goats can be destructive. So keeping the car safe is most important, giving you the answer straight away.
As a french native. "Savant" means someone who is educated and smart, even someone like a professor or a researcher. Or something that is highly intellectual and sophisticated. The equivalence of "learn" would be "apprendre". Savoir, savant are more an equivalence of "knowing"
There are two ways to say the word learned which have different meanings. One is past tense for learn. The other means knowledgeable. Savant does mean learned but the narrator said the word incorrectly.
@@Demigodess42 This is twice wrong. First, "Savoir" is to know, not to learn (which is apprendre). Second, there is no "savant" conjugaison of "savoir "(the closest would be its present participle, "sachant"). Savant is always a substantive, which I would translate loosely as scholar. I am also a French native.
@@matthieutricottet You haven't understood the difference between the two pronunciations of 'learned'. If pronounced 'lernd' it is the past tense of learn, but if pronounced 'lern-id' it means educated or knowledgeable.
@@jdb47games OK, now I read again your comment and I understand what you meant. Having heard "learnd" in the video, I did not make the connection with the substantive learned (which actually I knew), and totally misread your remark, sorry about that.
I believe one of the circles of hell in Dante's inferno is trying to explain the Monty Hall problem to people who can't understand it and who smugly insult you for saying the correct answer
Have the person trying to understand the problem play the role of Monty Hall using playing cards or something with someone who will always switch. They should soon realize that Monty Hall will have the car 2/3 of the time.
Host is not allowed to interact with door that was initialy selected that is why its chance for having a car do not change. So you ending with tree doors one with 1/3 chance(the one selected when no information about one of the doors was known), one with zero chance as the host is allowed to open only the door whithout car and the third door takes the whole propability of the group of doors that host is allowed to interact with => 2/3.
I think there's a simpler way of explaining this puzzle. If you choose the correct door and switch you will be wrong. If you choose the incorrect door and switch you will be right ( because the other wrong door gets eliminated), and you choose the wrong door 2/3 times
@@nnnnnn3647 It's pretty easy to test it for yourself. Get someone to put a matchstick in one of three boxes. Then choose. Do it about a hundred times. Then see how many times you chose the wrong box. It should be about 67 times. So on each of those times if you'd then gone on to play the game, your odds of winning if you stick are only about 33%, but if you switch, your odds will be about 67%. Not 50%, because it's more likely that you will already have made the wrong choice. So 66/33, not 50/50. Your previous decision has already shifted the odds away from an even chance.
@@owenorders5202 the problem is that you only play once and you have 1/3 to choose from. If the game consisted of 100 draws, the changing door strategy would be better.
Think of the door question this way. When you pick a door, there is a 1 in 3 chance that you are right. That means that there is a 66.66% chance that you are wrong. In other words, you're PROBABLY wrong. When another door is opened to reveal a goat, you now now that it's in your best interest to switch, since you were probably wrong from the beginning. Since your first choice was probably wrong, and you've been shown another door with a goat behind it, that means that switching doors would make you, probably right.
My explanation for the Monty Hall problem: Suppose Monty does NOT open one of the doors you didn't pick, but instead offers that you can swap your door for both the others. Of course you swap your 1/3 chance for a 2/3 chance. Now one of them must have a goat - so what?. Monty just showed you that one, and you take the other.
Well if the door was not opened essentially yes, but door 3 is opened, no one wants the goat, so 2/3 of a chance, but why would it be a better chance for 2 to be the door? Between 1 and 2 is a 50/50 shot. Both being 2/3 chance
@@anthonyharper4877 Swapping for both of the other doors, will always give you a better chance than staying with just the one - that's beyond question. The only differences between how the game is normally played and that are:- 1/ You are shown a goat (which must be behind at least one of your two doors) 2/ You don't have to take that goat home, just the other door's content (hopefully the car)
@bertoid as it is stated, the goat is behind door number three, I'm implementing a conclusion that 1 and 2 has equal chances, switching or not does Increase the chances since 3 is not what we would chosen for the car, sure switching and there is the said car, but also the chances of switching and there's not to be.
Clearly the biggest thing people can't wrap their heads around is why the fact that Monty knows which door is right, and will never obviously open that one makes a difference. People are seeing Monty's choice as random (which would make it not any better to switch) when it isn't.
I think I just figured out a unique concise way of explaining this that will be more intuitive for people. If you start out with the wrong door (which is more likely than not), then there is no other wrong door to switch to because the host has opened it.
@@dan20020 You don't know if you picked the wrong door but it is pretty easy to see that, with two goats and one car, you are twice as likely to pick a goat door as you are the car door with your first pick.
@@dan20020 It is all about the way you approach the problem... stop thinking about the second choice as a new 2-door game, that is not correct and it will be obvious once you understand it. B and C are grouped together because they are both the host's doors. This game is all about the contestant's first pick splitting the doors into two groups, the contestant's door with 1/3 probability and the host's two doors with 2/3 probability. Would you switch your one door for the host's two doors with both of them still closed?
@@dan20020 look at the 100 door example. Is the prize more likely to be behind your door? Or the other 99. With three, is the prize more likely to be behind your door, or the other two?
It comes down to this: One of these three doors has a prize behind it. Do you want to pick one door or two doors? And if you want to pick two doors but are only given the opportunity to choose one, switch when the host gives you the option after showing the empty door (which is just the empty side of the 2/3 chance).
2011CaRe that is a very good way to describe the problem. That should be a very effective way to get stubborn people to change their minds about insisting there’s still a 50-50 chance. However, I think you would still have trouble convincing them that your description is accurate. Many people will bend over backwards to rationalize that what you’re saying isn’t true. So I would recommend starting with the statement “the host has no choice which door he is going to open.” Then say the host must open a door that has a goat. And then say, if both of the doors that the host may open have a goat, it doesn’t matter which one the host chooses. If the person you were trying to convince doesn’t agree with all of this, then you cannot reason with them. If you both have the same understanding, then you can ask them your rhetorical question. If I were a college professor, I would give a lecture like this : there is one prize behind one of three doors. The other doors have a ‘goat’. I’m going to let one student ‘own’ one door and I’m going to give both of the other doors to another student. I’m sure you all agree that the student with two doors is twice as likely to get the prize, but won’t necessarily get it. OK, now I’m going to open one of the two doors that the second student controls, revealing that it does not have the prize. Please carefully take note that it doesn’t matter which door the prize is actually behind . There is always at least one goat available for me to choose. Mr. first student, would you like to change places with the second student now?
If you have 100 doors the probability of picking the car is 1/100. The host opens 98 doors all goats leaving one closed door asking you if you want to switch. Obviously you want because now you have 99/100 of winning. It reverses the probability. 3 is the least amount of doors you need to for this to work.
I never got the Monty Hall problem, because it seems like your odds go from 1-in-3 to 1-in-2... why switch? It's 50/50, right? So I got some paper and dice out, and using a 6-sided-die, I assigned a roll of a 1 or 2 to door number one, and so on. - The point of the dice was to eliminate my own knowledge, because since I was playing both the 'contestant' as well as the 'game-show-host', I always knew where the prize was. Anyway, a little ways into this, it was true that switching improved the odds of winning.... but I still couldn't figure out why. Then I figured it out: it was because the 'game-show-host' *never eliminates the grand-prize* -- so it is a *biased* decision, and *not random.* - Anyway, hopefully you guys reading this are able to finally grasp *why* this problem works the way that it does, because it took me a long time and some actual sit-down-and-experiment both to verify it, as well as see why it worked.
It's actually just nunber cranking. The true response would depend on the behaviour of the host. If he always does that, then it makes no difference, if not, and he has a bias towards doing that when you choose right, then you should switch. You can crank numbers on hipothetical number world, but that doesn't correlate to the real world.
@@klaus7443 That's not true. In the real world. . . The car is in the same door. Assuming switching increases your odds when the car is in the same place hold no sense in the physical world. This is a fun abstract logic exercise, but it holds no real world logic. Just like you can prove different numbers are the same if you crank the math up. But It won't make you Jesus in the real world.
@@arthurrosa9403 Your mathematics is terrible, If the contestant has a 2/3 of picking a door with a goat and the host always picks a door with a goat then there is a 2/3x1=2/3 chance that they both did.
Edit: Adding shorter explanation to the top: In the mentioned scenario, when you initially pick one from 10 doors, you have a 1/10 or 10% chance of success. None of the variables that are changed are ones that would affect your chances of picking the right door the first time, and unlike the rest of the doors which could be removed, the door you picked had no chance of being removed even if it was wrong. Since the correct door hasn't been randomly shuffled again between the two remaining doors, in order for you to be right the second time by picking the same door you did the first time, you would have had to have been already been right the first time before any doors were removed, and that only had a 10% chance of success. For those odds to be able to change, it would have to be possible for a different door to become correct than at first, or your door would need equal chance of being removed compared to the rest of the doors, instead of being guaranteed to still be there even when it's wrong; or you'd have to have extra information available to influence your first choice. If only the correct door and one random door were guaranteed to be there, and your first choice was not guaranteed to stay, then your door would be gone 80% of the time just like the other doors, it would be right 10% of the time, and it would be wrong 10% of the time; then in that scenario, if your door was still there after you picked it the first time, then it would have a 1/2 chance of success, but even that would be 10% of the total, because 80% of the time the door you picked was gone. However, since the door you picked can't be removed, it's still there as an option all of the times that it was the wrong door. Original comment: I spent some time pondering this, and at first I still didn't believe it, but then I thought about it. If there were 3 doors, you only have a 1/3 chance of picking the desired door, and even after one door is eliminated, that still doesn't change the fact that you had a 1/3 chance of picking the right door. And you are right that using more doors in an example makes it more obvious. Your door is no less likely to be unsuccessful than it was before, because your door was ignored only because it is the door you chose, whether it is successful or not. However the other door was only left available either because it contains success, or, on the 1/3 chance that you already picked the right door, then the other door was randomly left available merely to have an extra door available. For people who still aren't convinced, think about it from these two different realities of the same scenario: Example: Without you seeing, a person puts a coin under one cup out of 10. You choose a cup, but instead of revealing it, the person lifts 8 cups that as the dealer they knew were empty, leaving two remaining cups, including the one you had chosen, and they do not move the cups around (because doors don't move) so you still know which one was your first choice, and which one wasn't. Reality A: The cup you chose has a 1/10 chance of hiding a coin, and the other remaining cup has a 9/10 chance of hiding a coin. This sounds weird to you. Reality B: Both of the remaining cups have a 1/2 chance of hiding a coin. If this were true, that would mean that the cup you originally chose when there were 10 to choose from, would have the coin, 50% of the time. But revealing the other cups AFTER you made your decision does not increase the odds of your original decision being correct. If you repeat this experiment 100 times, it is highly unlikely that your first pick would be the cup with the coin at least 50 times out of a hundred. See how Reality A makes more sense now?
@@amyflana With 3 doors, there is only a 1/3 chance that you originally picked the right door. With 10 doors, there is only a 1/10 chance that you originally picked the right door. With 100 doors, there is only a 1/100 chance that you originally picked the right door. So it is more likely that the right door is the other remaining door. If the dealer was sabotaging people by only revealing doors when people pick the right door to scare them, then before long people would catch onto that, and it wouldn't work. Any decision-making on their part would only turn into clues for people in the future. In order to not make it obvious, they have to be consistent and reveal doors regardless of whether or not the person picked the right door to begin with. Sure, there's a chance you picked the right door and then switch to the wrong one. It's just much more likely that you picked the wrong one.
I still don't get it. The probability changes as the number of options changes. As the number of choices decreases, so too does the number of possible outcomes.
@@SeanPAllen Yes, but you made your choice before the amount of options changed. With ten doors, and zero information to go on to help you determine which door hides what you want, you have a 1/10 chance of choosing the correct door. What happens in the future doesn't change what the odds were when you made your decision. Think about an example with 100 doors. You pick one door, and 98 other doors are revealed, but the door you picked, and one other door, still remain unrevealed. Imagine repeating this experiment 100 times. If you had a 1/2 chance of picking the correct door the first time, that would mean that on average you would have chosen the correct door 50 out of 100 times, despite your choice being blind. Unless you can see the future, the doors being revealed afterwards doesn't change the odds of you originally picking the right door beforehand. Do an experiment with an object under cups, using as many cups as you have available; they don't even need to be identical. Have another person hide something under one of the cups while you cannot see. Wear noise cancelling headphones or ear plugs if you have to, to eliminate auditory clues, or leave the room. Tell them which cup you chose, then have them reveal all but two cups, being sure to leave out the cup you chose, and the cup with the object (and if these are one in the same, have them still keep one other cup down, for a total of two remaining). Then have them reveal your original choice. Repeat this experiment at least 10 times and see how often you get it right. It will probably end up a fraction close to 1 out of the amount of cups used. If someone else were to enter the room knowing nothing, not knowing your original choice, and pick from the two remaining cups out of ten, they would have a 50/50 chance of picking the right cup. But that still wouldn't change the fact that the cup you originally chose has a 1/10 chance of being correct, and the other remaining cup has a 9/10 chance of being correct; but the person not knowing this has a 1/2 chance of picking your cup, and a 1/2 chance of picking the other cup. The thought process leading you to believe it is still a 50/50 chance for you, the person who made the original choice and saw the reveals, is that you think you still have zero information and that new odds are calculated based on the remaining choices. If you truly made a random choice with the two remaining cups, you really would have a 50/50 chance, however you do not actually have zero information and the choice is not random anymore.
I have always enjoyed reading her column and agree with what she said about education vs intelligence. I know plenty of people who have college degrees and are complete morons outside of their very small realm of study and their profession. I work with engineers who couldn't engineer themselves out of a cardboard box. Studying and passing test doesn't make you a genius or guarantee that you even have an ounce of common sense. You go through their well oiled machine and don't make waves by questioning and you pass. I love that Marilyn was found to be correct. Ha ha ha. You rock girl! and I'm a guy.
@@charlesdickens6706 but the professional mathematicians agree with Savant. The mathematicians that disagreed were proven wrong and now agree with her, because she was correct.
@@Hank254 .. and if you find someone that chooses to use his energy to understand the problem instead of doubling down, then actually understands what is happening (ie. each door has 1/3 chance because: 3 door, 1 car.. quite simple really) more 50/50 folks come out of the weeds, once one understands and even says thank you, wow, this was great.. 20 more will come out and "debunk" whole conversation of 300 replies above resetting the conversation back to social media standards... sigh
For example my problem is that I actually enjoy explaining things and when someone understands or learns something new it's great.. but.. the manners of stupid people are such that they insult you, call you stupid, ask if you ever graduated from elementary school, this sort of petty things.. I came to conclusion that some people are just beyond salvage.. these videos where they ask geography and other questions from college students.. depressing.. are we really this stupid as a whole? Depressing.. that said, I *never* call someone stupid even if it is plain obvious that they are, it's not because I am polite.. far from it.. the danger is that the discussion is de-railed into direction I am not interested in, so, when they insult me I don't react just keep explaining as clearly as possible and once in a while someone takes their time to read with thought and Bob's your uncle, something nice happens.. if not.. then I will keep explaining until number of comments run out (500) or they don't reply anymore.. just remain calm and rational, keep yappin' guys..
@@n00blamer The number of people that are willing to listen seems to be decreasing, on MHP videos at least, but I suspect it isn't only here. Strange as it might seem, some of those that seem beyond salvage actually snap back when they are ridiculed. I've seen it happen plenty of times in here, sometimes it is the only way to get them to stop trying to explain the 'new two door problem' and actually think. Negative reinforcement is not the best tool, certainly not the first choice, but it can work. It's totally useless for the ones only trolling but I think the reason most of us are here is because we enjoy explaining it to people who are interested in learning it.
She literally did a move whose only logic is used to the fullest in Minesweeper. And that is the true genius behind it. We make these mathematically rational decisions when playing minesweeper and don't even realize we are doing it. She points out how that basic algorithm can apply to a gameshow with only a 1 x 3 row of potentially mined tiles, and everyone looses their minds. The first tile clicked in minesweeper is the 2nd safest tile on the board, and is never mined. Making the chosen door a door that reveals other doors' contents.
I was amazing at Minesweeper and yes, I remember using similar logic to determine the best odds of not choosing a mine. I was wrong some of the time using what I thought was 'stellar logic', but most of the time I was correct.
The kicker seems to be the game operator's knowledge. After I choose, the operator opened one that he knew didn't have the prize. That would mean that he also knew NOT to open the other one that the contestant didn't choose, perhaps because it was the winner.
Precisely. The host's knowledge and how they operate is the _key_ to the Monty Hall problem. If the host does not have knowledge of what each door contains, and randomly opens a door other than that which the contest picked, and it just so happens to be a goat, then there's a 50/50 chance that switching wins.
That's what I realised, finally. If I'm on the correct door, then the host has 2 options to open. If I'm on the wrong door, the host only has one option for opening. That's not the whole explanation, but it's part of it.
That is cheating. Serious answer is Math only. There might not be any car at all, only 3 goats. He does not NEED to open all 3, once 2 goats appear he says "you had 2 chances, you lost, contest/game ends here"
@@MuffinsAPlenty what??? No! Even if the moderator doesn’t know and by pure chance picked a door with a goat (that is not your door), you are still better off switching! What are you saying here?
Public scrutiny of such a thing was less back then. Yes you would be scrutinised by those in your field and if it was particularly egregious someone in the media might have done it but we have social media these days and people in these fields also tend to become more famous now because there are more people than back in 1980s and thee are more ways for people to acces information through the internet so you also have more armchair experts who have no knowledge, experience, logic, smarts, training think they do know better.
Today all the US has is what passes for mainstream media ravaged by gems the truth in broadcasting act of 2005 and smith-mundt modernization act of 2012. Then there’s that ‘87 reject Biden who has had mental illness issues for decades.. Even the DOJ doubts his fitness to speak on basic topics. It’s awful that during the largest world crisis, Covid-19 neither she nor any mensans or other supposed “high iq” people stepped up to guide or lead during a significant time of need. Lost respect for the lot of them for not speaking out.
*To learn about Marilyn, here's our newest video on her life (April 2024):* th-cam.com/video/F6rDygbx5Kk/w-d-xo.html
Visit brilliant.org/Newsthink/ to learn math, science, computer science, and data science for FREE for 30 days
I have a question, if there are three doors, there is a 1/3 chance of winning, but if there is only two doors there is a 1/2 chance of winning. There is a 50-50 chance you will win, so what’s the point of changing. The first door you picked might be correct. Plus, isn’t it human psychology to give the person who did it correctly to pick something else so they go wrong?
The chart at 3:40 is wrong. Only getting a car is considered winning
I had to watch this video like ten times. I bet it could get into some mathematic gibberish. If you add them all up to 1, make a pie, and then take away 1/3, you are left with 2/3 of pie and 2 doors to give an equal slice to. You'd split them into 1/3 each, counting the total they came from, which would be half of what was left while using their logic. Help.
@@vaibhavk1342 does this make any sense? I had to watch this video like ten times. If you add them all up to 1, make a pie, and then take away 1/3, you are left with 2/3 of pie and 2 doors to give an equal slice to. You'd split them into 1/3 each, counting the total they came from, which would be half of what was left while using their logic. Help.
@@pheresy1367 That is why this question is sort of like fake news. no normal person accepts it. after the goats are shown its 50/50 of the last 2 doors. no normal person could think that in the final choice of the 2 doors THAT THEY SOMEHOW LOOK AT THE PAST.
I would have switched to door #2 as well but for a very different reason. I would have assume that the goats would need to be kept as far apart as possible so they would be less likely to incite each other into making noise and thus giving their relative positions away. Putting the car in between them would help keep them out of each other's sight. I might just have won the car because I knew more about goats than mathematics in that instant!
I had to give you a thumbs up, not because it was correct, but because it was damn good.
However, on a show that happens every week, the viewers would soon get wise if the car was always behind door 2.
lol!
@@mandolinic I think statistically, door #2 was more often the right choice. Everyone playing along at home would always yell DOOR NUMBER TWO!
So the trick then would be to guess door #3 and see if he shows you a goat behind door #1. If he does, you got the historical statistics and the live probability on your side.
They were schroedinger's goats
I had the honor of having dinner with this lady while I was in college. Smart as hell, but very down-to-earth.
Wow, that's amazing.
She probably picked up the bill knowing there was a 100 % chance you would take it from her 😆
If youre the smartest person in the room? You're in the wrong room.
Lucky you. I'd love'd to speak to her for just few minutes.
You too?! The most remarkable thing about noshing with her is how she can pass things around the table telepathically. Oh, deary, I do hope she regaled your party with such feats. She even levitated all of us home after the dinner. Brilliant woman, that.
@@roberttyrrell2250 So what you're saying is, that she's always in wrong rooms?
I am most impressed with the math professor who publicly admitted his mistake. It is so refreshing to see someone who will actually take responsibility for their errors, regardless of how embarrassing it may be. If only our politicians could show as much humility. Much respect.
True
Around here, when a vocal 50/50er finally figures it out, a typical response is to delete the thread so there is no evidence they were wrong. What's the opposite of humility?
@@Hank254hubris?
Or don't respond like a prick in the first place.
"I am most impressed with the math professor who publicly admitted his mistake."
Me to. How can you get a PhD in math and fail so hard at a simple probability problem?
My explanation: When you are right in the beginning , switching looses. When you are wrong in the beginning, switching always wins, as he always shows you a goat and leaves you the car. Since 2 out of 3 times your initial guess is wrong, switching wins two out of three times.
As soon as one door is totally eliminated, it changes the probably from 33% for each door to 50% for each door though.
If I hand you 3 guns, say 1 is loaded and two are empty, let's play Russian roulette...then I take away 1 of them and show you it's empty, let's just use the two...you feeling more confident? 😂 Nah, because you know one is loaded and one isn't, suddenly odds shifted
@@SuperMontsta Not in this game, assuming as a rule that the host must reveal a wrong option from those that the player did not pick. That creates a disparity, because when the player's is already a losing one, the host is only left with one possible door to remove, but when the player's is the winner, he is free to reveal any of the other two, making it uncertain which he will take in that case.
Using the example of the video in which you chose #1 and he opened #3, we know that the revelation of #3 was 100% mandatory in case the correct were #2, but if the correct were #1 (your choice), we are not sure if he would have opened #3 too, it was only 50% likely, as he could have opted for #2 instead.
That's why it is twice as likely that the reason why he opened #3 and not #2 is because #2 is the winner, rather than because #1 is the winner (having you picked #1), and similarly occurs with all the other cases.
@@SuperMontsta The odds increase because the gamemaster always opens a door with a goat that isn't yours. 2 out of 3 times, you were wrong initially and change will always work. Think it trough!
@@leonda4817 Pause it at 3:22 and type out the "results" from Game 1, 2, 4 and 5. Give the ratio on those results.
@@RonaldABG Pause it at 3:22 and type out the "results" from Game 1, 2, 4 and 5. Give the ratio on those results.
The best thing about this video is a reminder that when people publicly stated something incorrect, they used to express accountability and humility. That never happens anymore.
I know, and I hate it, we need better people in this world!
It doesn't matter what people think. it is what it is
in your imaginary world? sure
@@reformed_attempt_1 wdym?
@@reformed_attempt_1 without more context, your comment means nothing.
People humbly and publicly admitting to be wrong, if only that existed today.
Too bad idiot Vos Savant failed to acknowledge her profound error. Too bad too those MIT profs are shown to be idiots twice over.
It's a new game. 1 in 2 chance, 1:1 odds.
Though Hall does not say it in these words, he asked this: "There is a car behind one of two doors. There is a goat behind the other. Which do you choose?"
It is irrelevant that Hall phrased it this way: "Do you wish to stay on Door 1 or switch to Door 2."
Vos Savant is an idiot as are those MIT nitwits.
@@DonLicuala It is almost a psy op her exist right down to her name "Savant".
What? My comment has nothing to do with her achievment, its about the humble gentleman who knew how to apologize.
I have no obligation what so ever to list anyones achievements, please look for an argument elsewhere.
Vos Savant is wrong still. She will be wrong even after she dies.
She has applied conditional probability math skills but from the wrong premise.
At the initial door opening to show one goat, the probability is 0%, odds, 0:0, chances 0 in 0.
The contestant is not trying to avoid two goats, but rather only one.
The probability of winning from the moment when an actual positive probability can be calculated, i.e., from the moment of two doors, is an equal probability of 50% to each door, as there are two options and no further information available.
And that is the only solution, the correct answer to the Let's Make a Deal Problem (LMADP). An alike problem "The Monty Hall Problem" (MHP) is a pseudo-realistic problem derived from the Let's Make a Deal Problem that illustrates an application of conditional probability assuming a contestant can win on Round 1 but does not and gets a second chance with updated info.
While the analysis of the MHP is self-referentially correct, it is inapt for the LMADP, which presents a contestant with a choice from two options, two and only two unopened doors.
The MHP would be appropriate if and only if there were three doors and a constest could win right away from picking the right door of three. Yet, in the LMADP, there is no deal when there are three doors.
The Rules to the LMADP are these, which are different from the MHP rules:
1. Say the name of the door. It does not matter because we're not revealing it. your odds of winning are 0:0., probability 0%, 0 in 0 chance.
2. Carol, from the doors not picked reveal a goat.
3. There could be a goat behind your door or a car. What door do you wish to name? You can name the same one as you did previously. You have a 1 in 2 chance, or 50% probability with the odds being 1:1 of getting it right. Once you pick we reveal the goat door first if you picked the car, the car door first if you picked the goat.
The question which wasn't answered here is that "why would the host open a door if they had the wrong option"?
This hurts my brain. But even high level mathematicians didn't understand it at first so I can't feel too bad for not getting it.
there's nothing to understand. She considered each door a floating variable. Each door is in absolute matter of fact, a constant. So there is no means to transfer probability from one door to another. Once the 3rd door is opened, there is no more question of what's behind door 1 or door 2. It is either a goat or a car. The feeling that you have a 33% higher chance when switching is based on the original probability you had to correctly guess between 3 doors. Once the 3rd door is eliminated you have a completely new expression. And the probability between two choices is always the same. Stay or change your mind, implies that the previous variable is still in play. It is not. You are not "switching", you are again choosing.
The difference between the two choices is 50%, probability is recalculated at every choice.
@@ZennExile incorrect.
@@vladimirdemirev4948 the reason why door are grouped is because they are different. One is the door you picked, the other two are the doors the host has to pick from. Your grouping doesn’t account for that.
The professor admitted they were wrong because they were wrong.
@@vladimirdemirev4948Say we take a bag with ten marbles in it, one orange and nine white. Whoever gets the orange wins. You pick a random marble from the bag and I then take the remainder, open the bag and select a marble, discarding 8 white ones. Who is more likely to be holding the orange marble.
Clearly the person who knows where the prize is will have a better chance of selecting it.
@@morbideddie well, I guess I am wrong. Seeing the 100 doors example changed my mind.
I will go with the excuse that binge-watching TH-cam videos on different topics trains you to react first, without giving much thought ;)
What makes this problem tricky, is that people ignore the fact that they ALREADY made their first choice, when it was a 1/3 chance to get it right on the first attempt, so then when there are only 2 doors left, they treat it like it is just a simple 50-50 chance to be a car behind one of each remaining doors, but it's NOT! Door 1# is still 33% to contain the car(because you picked it before having the additional information), while switching to door 2# becomes 66%, because door 3# you know for sure that is 0%. This is why, when you run the tests with all 3 possibilities, after eliminating 1 wrong door: IF you switch => you will find a car 2 out of 3 times VS IF you stick to your first choice => you will find a car just 1 out of 3 times. Another example would be: If you could chose from a) pick 1 door from 3; or b) pick 2 doors from 3. Which option gives you more chances to find a car a) or b)? Obviously b). In our case, switching represents option b).
But the game is a 2 tier game, what people dont realise is that it's not a 1 in 3 chance because only one door will be exposed out of 3 on your first chance, so your first chance will have 2 out of 3 chance of getting it right. Which then leads onto the finally of the game. Which is 50/50 for both doors as you can keep or change your choice.
@@dominathor but if you already know they will open a goat door 100% of the time out of the 2 you didn’t pick, you aren’t making a 1/3 choice, even your first pick is 50/50 bc one of the other two doesn’t count.
@@optimizor The one that is going to be revealed is not determined. It depends on your first choice because the host is never allowed to remove your door. I mean, there are three contents: GoatA, GoatB and the Car. If you select GoatA, he will be forced to reveal GoatB, but if you select GoatB, he will be forced to reveal GoatA.
So it is not like if one of the goats didn't exist, you could have chosen any of them, so more likely that in the first part you choose a goat and not the car. In fact, the average tendency is that for every 3 attempts, in 2 of them you start choosing a goat and only in one you start choosing the car.
As the host always reveals a goat from the non-selected doors, in the 2 out of 3 times that your door already has a goat, the revealed goat must be the second one, so the car must have been left in the switching. Only in the 1 out of 3 times that your door has the car, the switching one will have a goat, so less often.
@@RonaldABG exactly! Thank you
3:27 - "This is contingent on the host always opening a door with a goat."
Yes, it is, which is why this restriction must be included within the problem as phrased, something the introduction to this video fails to do. That's actually a very common problem with this problem. It cannot be assumed that a goat had to be revealed simply because a goat was revealed unless the host's intention is incorporated into the problem. The host could have selected a door at random that simply happened to contain a goat. This legitimately changes the math to the Monty Fall/Blind Monty problem.
This failure to accurately phrase the problem is frustratingly common.
well If you watch the show is oblivious, you can also assume how the show would b er if you havnt watched it. But still the author of the video has a bad taste.
ohhhhhhhh ok this makes sense now
Oh this makes sense now. I was under the impression that the game host always open door 3 regardless. Which is why I was confused at 3:20, when the table showed scenarios "game 3" and "game 6" having a car behind door 3 which made no sense to me at first so I excluded those scenarios. But I understand now, thanks!
Came here to say this. In video it sounds like host might have just chosen a door at random, and it happened to have a goat. It should be stated that the host will NEVER open the door with the prize when he opens his own door after the contestant chooses.
Well if the host would pick the door that was chosen by the participant and there was a goat, there would be no question of switching the choice of doors.
And if it was one of the other doors and there was the car, the same thing applies.
So in my opinion it's self explanatory
Simpler explanation; assume you always switch:
If you initially picked a goat, you win. If you initially picked the prize you lose.
What's more likely?
This is the only thing I understood
@@thecoweggs i know right
I love goats. I can no longer afford to run a car... :-)
This video presents the presenter as being on your side, "helping you out." This is a bad assumption unless they're truly your friend, which is unlikely. Haven't you seen Slumdog Millionaire? :-)
So when they select something to lure you away from your initial choice of door, as an option for you, it can be a decoy, not the real deal. After all, if you chose that door first, it would have the same amount of probability from the standpoint where the presenter may not be on your side.
IQ is racist pseudo-science. Savant in name only.
The reason Mongoloids in Singapore and elsewhere have the highest IQs is because their youth was sacrificed for the god of money, and they did parroting memorization including of formulae which help in many IQ tests, at least to be used to patterns. This is why a Papuan tribal won't beat them: The tribal isn't dumber; just not used to those types of tests 'cause they DON'T NEED TO BE.
IQ tests use a one-size-fits-all appraoch, which is stupid, and proves IQ is stupid.
If I took an IQ test, it was probably disguised as some standardized test in Florida long ago.
I don't believe in that shit, so I refuse to participate. It's just like DNA testing: Different testing companies give different and contradicting results, so they're all scams.
If you are truly smart, you'll know better than to let others tell you how smart you are, when those others are self-entitled narcissist establishment people trying to dictate your mind.
@@scintillam_dei very very wrong
People learn much better when they're allowed to follow their interests. If the subject is something that bores you, you'll only retain the information for a required time (test date, usually), but if you are interested, you'll track down information and fill-out the subject more thoroughly. This is the way I educated my youngest son - he chose the subjects and the timing. He graduated top-of-class in Navy Submarine School and is now stationed on a nuclear sub based in Hawaii (his chosen profession)
ADHD compounds this problem big time. Someone with ADHD is borderline incapable of learning things that don't interest them at all, yet I believe it becomes an advantage when we really are interested in the subject because we can devote hyperfocus to it.
@@jasondashney no kidding, I'm just a teenager, and yet I can understand, and I know about, many things that adults, even in my field of interest often don't know.
I was let to go my own direction, and I'm great at it!
I believe it. That's great you understand that about yourself. Keep that in mind when you decide what to do with your life.@@Metal_Master_YT
@@jasondashney thank you! :D
I also have ADHD, but I feel like I can "tame" it and use it to my advantage, kind of like you said.
This is interesting to hear. The school-system bothers me alot. For me, voluntariness is essential for sustainable learning.
There is a really simple explanation that doesn't require overthinking, drawing outcome tables or considering expanding or collapsing probabilities. You just have to understand that you're probably wrong to start with.
2 times out of 3, your first choice will be wrong, then the host must show the other wrong option so switching will guarantee a win. That's all you need to know to always switch because it gives you a 2/3 win chance (logically, not switching must give you a 1/3 win chance).
yes, exactly.....
........but there's still a small detail to do:
now you 'only' have to make all those flatbrainers understand all this....
....those, who come along in this comment section, either not able to understand it, or they just refuse to think.....
......good luck!🤣🤣🤣
@@KarlHeinzSpock I'm thinking as hard as I can. Can't you smell my brain cells burning? 🔥🧠
It still looks like she's left with a 50/50 chance to me and nobody's explanation has changed that yet. I want to get it. 🤔
@@websurfer5772 Think of it this way, at the start of the game the chances of winning the car by selecting a door is 1 in 3. Now why do you think it is 50/50 simply because a door has been opened? You know that they have to open a door, and it is 100% guaranteed to show a goat, so using your logic it should be 50/50 before the goat door was opened, because you will always end up with a choice of two doors.
@@asprinklingofclouds Well, I agree that you have a 1/3 chance of being right in the beginning. I had to read more about it online which says that it's because one of the doors is opened to reveal a goat _and_ the question is asked, "Would you like to change your answer?" The answer is much more likely to be right, in fact 2/3 more likely, if the contestant changes their answer to the other door.
Ok. So which door do you give the1/3 chance to and which do you give the 2/3 chance to??
Almost every smart or educated person says that school is not the best way to learn and still nobody tries to change it.
It's the only affordable way for most.
The 1948 and 1988 encyclopedia sets of the "Book of Knowledge" were my favorite source of learning. I LOVED them! I always checked the Public library a lot. My fiction was mostly from Silver Age comics that helped me much with my art. I loved studying electronics. I didn't need school for any of that.
Now days, Professionals seem to get their credentials out of a "Quaker Jack Box."
I prefer self-teaching.
@@karangupta1825 Yes. Expand your passions and have creative hobbies.
"The book of knowledge" was very pictorial and was suited for all ages. The simple facts of the topic were clearly explained and the information became gradually more complex for when you are older. It was laid out like the internet with many links in the index's. There were plenty of do it yourself projects. I build my first radio from the 1948 book at 11 and it was fantastic!
A good way to think about this problem is: You first choose one door. You are then able to change your choice to BOTH the other doors. you get a car even if one of the doors have a goat behind it. This is the exact same thing as to show the goat beforehand.
This is an excellent way of looking at it
That is brilliant.
And the only way to lose the switch would be because you beat the 3 to1 odds against you when you first chose.
So it's always (3 to 1 against you) vs (2 to 3 for you by switching after the goat reveal).
(Ooops, I think my "further clarification" only served to complicate).
;-)
That is a great way to look at it. It takes a certain personality, I think, to be encouraged about odds of 2/3. Chance is not something you can predict -- it's chance. This is the difference between stochastic (your statistical analysis) and random (what actually occurs). There is a world of words and ideals and a world of things and occurrences. Some people love to argue over words and ideals, but we each may go home with a goat. Or a car.
Unless you did pick the car on your first choice.......
It's literally a 50/50 chance.
You either change your answer or you don't, and you either win or you lose
exactly. or pretend there's 100 doors. now you get to pick either (a) one specific door, or (b) the combined total of any 99 doors, where if there car were in any one of the 99 doors you'd win it. would you pick choice (a) or (b) ? okay now instead of 100 total to start with, let's do 3 total to start with.
The way I approach this is by thinking that the initial choice can be one of three alternatives: Goat 1, Goat 2, or Car. This then plays out into three (and only three) possible scenarios:
1) If you chose Goat 1, and Goat 2 is revealed, then switching would have given you Car.
2) If you chose Goat 2, and Goat 1 is revealed, then switching would again have given you Car.
3) If you chose Car, and either Goat 1 or Goat 2 is revealed, then switching would have given you Goat 1 or Goat 2.
Thus in 1) and 2), switching works, and in 3), it doesn't. Thus switching is right 2/3 of the time.
Yup, it really is that simple but people still get locked in to 50/50 for some reason and hold on to it like a pit bull.
You are not correct, and I will explain to you why, very simply.
You correctly differentiated between Goat 1 and Goat 2 in alternatives 1) and 2).
In alternative 3), you do NOT differentiate between Goat 1 and Goat 2 by stating , "either Goat 1 or Goat 2"
Alternative 3) should be, " If you chose car and Goat 1 is revealed, then switching would have given you Goat 1"
Then alternative 4) which needs adding, "If you chose car and goat 2 is revealed, then switching would have given you Goat 2"
Thus, in 1) and 2) switching works, but in 3) and 4) it doesn't.
The result is 2-2.
Speaking of a pit bull...
@@Hank254 LOL!!!
@@georgebliss964 How can you NOT make a probability tree for this problem? Contestant pick car, host leaves goat. It has only ONE branch!!!
The probability density function of any event remains unchanged...except by direct intervention in the primary predication of the probability of an event. That is to say in simple terms, if a coin were to be tossed behind a given door and it landed on heads up, it would remain on heads up irrespective of what other coins behind other doors landed up as showing to the observer...no matter how many other doors are involved.
😂 throwing coins behind a door:
nice opportunity for all those 'experts' coming along to develop some new weird theories.....
You're still not taking into account that the host in this example already KNOWS what's behind the doors. That means he purposely avoided opening door #2 and door #1. If the Host opened door #1, that means either 1) it would reveal the prize or 2) would reveal the door as the incorrect one, which wouldn't give you an option to switch doors since the first pick was already opened. With that information, you know door #2 is definitely the correct one
First encountered this in high school. I tried to explain: "if you switch it's like picking 2 doors instead of 1", which convinced very few classmates. The teacher noted that I had good intuition and poor articulation. So true.
You explained it in simple to understand language. In fact, the best articulation!
But you are wrong.
I don’t understand how the extra odds don’t also add to the door you chose. They’re both still closed and options.
@@BasedGodGotenks Look at the 100 door example. The chance you chose the correct door from 100 doors is very low. The chance the winning door is amongst the other 99 doors is high. Now remove 98 of those 99 favourable doors and the remaining one has a very high probability of being the winning door.
Now if you arrived at the game late and missed the above process of elimination and just had 2 doors to choose from then your chances would be 50/50 because you do not have the information that points towards the more favourable door.
it doesnt make sense still fuck, how, every door has a 1 in 3 chance of being the car, revealing one door doesnt give the other door a higher chance as it was predetermined beforehand, its still 1 in 3, it cant just change cause you revealed the other door fuck like ik it makes sense to some people but it doesnt to me. EDIT: This is how i see it, in the start you have 1 in 3 chance, after goat reveal, you are left with 2 doors, 1 has a goat one has a car, so you are back to beginning, when deciding in that moment, you have a 50 50 chance you will get it
When I first heard the "Monty Hall Experiment", I reacted the same way everyone else did. I am however a software engineer. So I decided to write a small program to prove them wrong. What I ended up doing is proving MYSELF wrong.
The program played 100 random games. The first 100 games stayed on the same door whereas the second 100 games switched doors. In the end, the program that stayed on the original pick won approximately a 33% of the time whereas the program that switched won approximately 67% of the time. I couldn't believe it. You do double your chances by switching doors.
Omg, I was just about to write the code after seeing this random puzzle. Unbelievable conclusion.
Yeah I was around ~15 when this happened and also wrote that program at the time, though I either had no idea what was correct or was pretty sure she was. Don't remember which!
I still don't understand how. You don't know what is behind the two closed doors . So the probability for you ( any subject) is 1/2 .
@@siddharthshekhar909 the chance of your original guess being correct is 1 in 3, no matter what door you guess, the host can open an empty door, thus your original odds of 1/3 are not changed by the host opening an open door. By switching you end up with 1/2
@@siddharthshekhar909 EXACTLY!!! This is what I said. I absolutely could not believe it! This is why I wrote the program! I wanted to prove them WRONG. There is no logic to their conclusion. But as it turns out I was wrong!!! I still can't explain it. But the computer doesn't lie. I am a software developer with 30 years of professional experience. The program took an hour to write. I had to run it like 10 times before I could believe it. I still can't explain why it works. Don't feel bad. There were MIT professors who felt the same way you and I did. And she is the smartest person alive so I wouldn't sweat it if you don't understand why. I don't either.
It makes total sense when mapped out. I guess the difficulty comes in understanding why "switching" doors increases your odds at all. I got hung up on the "switching" part having any impact, instead of realizing that it's making a new selection with better odds. We used to do simple stuff like this in grade school, it's kinda crazy how a little bit of language can subvert your logical faculties.
"instead of realizing that it's making a new selection with better odds"
It is not making a new selection with better odds, but swapping your lower odds for better odds.
@@max5250 Now I get it, thanks max ; )
@@anthonydenn4345 Welcome back dude.
I still don’t understand why switching will increase your odds of winning. If you take away your first selection, meaning you never made a choice, are you still going to choose the one out of the 2 doors that didn’t get eliminated?
@@jacobcutrer
Switching increased odds of winning because host get to pick from two doors therefore, he gets a door with a cat twice as often than player does.
When he opens his door with a goat, we know which door holds a car twice as often as the door initially picked by the player.
The key to all this is at 5:15 in the video. The improved chances only work if the guesser knows that the host is not opening one of the remaining two doors randomly. When folks pose this problem to someone who never saw Let's Make A Deal and the poser does not reveal that the host is intentionally opening a door the host knows to contain a goat... that's ruining the question and (in that case) the correct answer is "no improvement in chances by switching". The improvement ONLY comes when information is conveyed to the contestant that the host KNOWS that the door being revealed contains a goat.
@@TEK_Nemesis
What he is saying is that if the host revealed a goat without knowing where everything is then there would be no advantage to switch, which is correct.
@@TEK_Nemesis
"once you solve this problem, you'll realize that it is always a better idea to select the other door"
Once you solve this problem you'll realize it's the knowledge of the host that is the reason switching is advantageous.
@@TEK_Nemesis
If the host doesn't know where the car is, that means he has to randomly open a door.
The host can just ask the player to reveal/remove a door for him.
This means the player is the one who choose everything.
Player: I pick door 1.
Player: I want door 3 to be the last door.
Player: So I choose to open door 2.
What makes you think door 3 has higher winning chance than door 1?
@@TEK_Nemesis
Quit pretending that you knew the reason as to why the host had to know where everything was. You should have realized my statement ......"What he is saying is that if the host revealed a goat without knowing where everything is then there would be no advantage to switch" was indeed correct. Instead you replied...."once you solve this problem, you'll realize that it is always a better idea to select the other door."
You're just a fraud and couldn't admit that you were wrong.
@@TEK_Nemesis No. That's NOT what I'm saying.
What I'm saying is that many people who present this problem to an unwitting 'victim' fail to explain the CRUCIAL point that the host KNOWS where the grand prize is and the host will ALWAYS reveal a door with a goat. That's precisely WHY I mention the caveat at the 5:15 point of the video.
The premise upon which the logic is based relies on the fact that the host KNOWS where the grand prize is, therefore the host will ALWAYS open a door with a goat. This is what provides the extra information that tells the contestant it's better to switch.
If the host does NOT know where the grand prize is, then the host will unwittingly reveal the grand prize 33% of the time, in which case there is no advantage to the contestant to switching. If the host does know where the grand prize is, but the contestant is not told that the host will ALWAYS reveal a goat, then the contestant has no justification to switch. This is the point being made at 5:15 in the video.
The Monty Hall problem and people's approach to understanding it is very interesting. Another way to think about the problem not covered explicitly in the video is the fact that only one independent choice is being made in the game. That choice is the players initial guess when there are 3 doors. The host isn't making a meaningful independent choice since they have to reveal a non-prize door only from the doors not guessed initially, and the results of the decision whether to stay or switch are entirely dependent on the initial guess(when there were 3 doors). If the player initially guessed the prize door(a 1/3 chance) and they switch they lose. If the player initially guessed a non-prize door(a 2/3 chance) and they switch they win.
"If the player initially guessed the prize door(a 1/3 chance) and they switch they lose. If the player initially guessed a non-prize door(a 2/3 chance) and they switch they win."
I first heard this riddle in the 2008 movie "21", and until today I never understood why switching doors after the reveal of a goat would increase the chance of winning, but the way you phrased it made it click for me, so thank you, i finally get it :)
that is probably the only explanation i've ever heard for this that makes sense. because no other explanation, even when presenting empirical evidence, actually draws that connection.
For me the biggest mystery is why your explanation is not immediately obvious to everybody. Lots of people deny it, even after someone carefully explains it to them....
@@tonybrowneyed8277 yea I think they're missing the difference between the host revealing a random no prize door, and the actual Monty hall rules. That would give 50/50 odds and the Monty hall game looks the same on any individual round. But the host not being able to reveal the same door as the players first guess completely changes the odds and the nature of the game.
" If the player initially guessed the prize door(a 1/3 chance) and they switch they lose. If the player initially guessed a non-prize door(a 2/3 chance) and they switch they win." omg i finally get it tyy
I think that some people find hard to wrap their mind around this concept because they fail to understand the very nature of probability. It’s not about being 100% right, it’s about being more likely
i just don't get it. if there are still only two doors wouldn't the odds be the same.
@@chestnut1279 Two doors left proves that you can either switch from your car to a goat, or from a goat to the car. The host could have simply asked you that same question when you first picked a door without doing anything else.
@@chestnut1279 I'm thinking the same thing. My understanding of this brain teaser is that changing your answer after the host reveals one of the fail doors is most correct but still not a guarantee as picking the first door was 1/3 vs changing the answer now making the odds 2/3. Its less likely you got it right on your first choice but it is possible. I think the logic of "two doors means equal odds" only works if the host revealed the door you picked to be the wrong door and you still had two options. But nothing makes me feel low IQ more intensely than brain teasers. I suppose the real question is do you think its more likely you got it right on your first try when you had three options?
@@chestnut1279 Yes. Sheep being lead.
If the contestant was given the opportunity to swap his 1 door for the hosts 2 he would almost certainly accept the offer leaving themselves with 67% chance.Don’t forget the host knows where the car is and has to open one of his which has to be a loser leaving the remaining 1 of his 2 which still has a 67% chance
Really impressed with the individuals that took accountability for ridiculing her and publicly apologized. That sort of character is in short supply
I always tell my kids four things:. 1) you panic, you die. 2) stupidity kills. 3) never do anything for which you will have to apologize for later. 4). And) (a biggie),.never take up a habit you're just going to have to break later on.
@@sheilalopez3983 Good word
Seemed to me "they" were sort of picking on her for not using the "math stats" as they did ?
Oh, of course; now let's hear it for the "Intelligence of Creative Thinking!?"
It seems THERE ARE different "kinds" and "types" of IQ " Tests." Experience and Education , two possibly different types ?
So we should be also asking WHICH IQ Test did Marilyn excel in, or on ?
MISSING FROM the video; Does this lady write and speak in both German and Italian ?I've always felt there is an extreme indication of high intellect in regard to peoples' abilities TO express themselves in foreign languages ?Seems there is a certain "type" or "kind" of logic it seems in learning to "relate " foreign language to one's own ability to speak and write in their native language ? It appears Marilyn 's " (by assumption?) that Marilyn had TWO "Native languages" yes ?
Her opinion of "public schooling" holds great merit. I remember a question required to be asked on a high school test , was "Who were the Phoenicians ?" The ABCD Answers included the answer " Venetian" . Most admitted later that they all misunderstood the word Phoenicians because they were all more "familiar" with Venetian Blinds, than historical terms of peoples and places! (Ha Welcome to American World History 101-we (all) need to repeat that one!?) Which btw lead to my last question (for you or Marilyn, ha ?) Is the inability to "spell" properly (in any language/especially ones native language ) indicate ignorance ?Duh...As a retired teacher, I submit I've become dependent on the Google Gargoyles ' offers for correction, which often just doesn't exist.
The robots tell me I've misspelled something, yet/while, offering no options with which TO correct it.
Good at Questions; Slow at the answers. Anyone ?
"I Ain't no Middleman"
Fred Gold & Lynda Faye
Copyrighted 2016 by LyndaFayeSmusic@gmail.com or Yahoo, if censored for using the word " God" too often?
I don't because they made her same mistake she did when they followed her idea.
@@gerardcote8391 ? Please explain! Thanks!
First heard this problem in a probability class, when posed by a professor and solved it correctly the first time. You always have a 1/3 chance if you refuse to switch. It never changes, its set from the start.
missing in the video's explanation!
No, you have a 1/3 chance to start. When one is eliminated by the host, that means there's 2 doors with 2 possibility. 1/2. 50% for each door. Even the chart in the video at 3:25 showing all possibilities confirms. Game 3 and 6 aren't possible, since it's a goat and not a car. That leaves game 1, 2, 4, 5.
4 possibilities. 2 of them you win. Two you lose. Staying makes you win.1/2. Switching makes you win 1/2. That's 50% each and again...out of the four game possibilities, you win 2/4.
@@SuperMontstafinally,ohhh thk you.
@@factormars4339 I don't know how people see that chart at 3:25, negate the ones that have door 3 as a car since we see it's a goat, and see the 4 various outcomes and see it's 50/50
My thoughts is one "smart" person got it wrong and people just jumped band wagon. It's common knowledge that probabilities are based on us not knowing something. KNOWING one door CHANGES the probabilities of the other doors.
@@SuperMontsta As I already told you in another comment: you cannot preserve Games 1 and 4 with the same probabilities as Games 2 and 5, because you don't know if the host would have opened door #3 if the car were in #1; he would have had the choice to open #2 instead.
You can see it better seen in the long run. If you played a lot of times, like 900, the car would be expected to appear in each of the doors in about 1/3 of them, so in about 300 trials. Now, for simplicity assume that you always start picking #1. The games will look like:
1) In 300 games the car is in door #1 (your choice). But they are divided in two sub-cases depending on what door the host reveals then. If he takes each with the same frequency:
1.1) In 150 of those 300 games he opens #2.
1.2) In 150 of those 300 games he opens #3.
2) In 300 games the car is in door #2. In all of them he is forced to open door #3.
3) In 300 games the car is in door #3. In all of them he is forced to open door #2.
If he opens #3, you could only be in case 1.2) or in case 2), that constitute a subset of 450 games. You win by staying in the 150 games of case 1.2), that are 1/3 of 450, but you win by switching in the 300 games of case 2), that are 2/3 of 450.
Similarly, if he opens #2, you could only be in case 1.1) or in case 3), that are also a subset of 450 games, from which you win by staying in the 150 games of case 1.1), but you win by switching in the 300 games of case 3). Again, switching wins twice as often as staying.
This adds up 300 total wins for staying and 600 for switching.
So , I've known the Monty Hall problem since 2 decades, watched 100s of youtube videos on it too. But nobody cared to tell me that not many believed Marilyn vos Savant when she gave the correct answer. (Off course Steve Selvin gave the real proofs and solution with 3 prisoners problem).
*EDIT* : I had no idea that *so many dudes* would be *triggered* by this comment. I'm a dude and I'm aware that a man's ego is just *too fragile* . But imagine being triggered by a harmless comment, almost all of the replies are by butthurt young boys and men. LOL
we already knew the ans if we watched other videos about that
I picked the 2nd door also. They said the car was behind #1 at the beginning. I listen very well
And this helped move mankind forward...how?
@@anti-apathy9715 1) Nobody said it did. It didn't have to.
2) Discouraging, patronizing women in the field of science, mathematics , etc. is not a trivial matter.
(Although i do believe this could happen to anyone, but being a woman made it worse for her)
Point is, such incidents prove to be a hindrance for little girls and young women who are already brainwashed by the society to think that they are not as good as men.
@@anti-apathy9715 ego trip
I found the easiest way to explain the problem is:
You had a 2 out of 3 chance of picking goats and only 1 out of 3 chance of picking the car.
Once the host reveals a goat, there are only 3 outcomes of a switch:
Scenario A) You picked the car at the start. If you switch, you lose.
Scenario B) You picked Goat #1 at the start. If you switch, you win, since Goat #2 is gone.
Scenario C) You picked Goat #2 at the start. If you switch, you win, since Goat #1 is gone.
You can only lose if Scenario A happens, and you win if Scenario B OR C happens. Therefore, you should switch since you have 2 scenarios where you win, versus only 1 where you lose.
Thank you so much,your explanation made it make sense to me!
I find it amusing how people can be so naive. If the host reveals one of the door had a goat
Then your Scenario B and Scenario C is EXACTLY THE SAME AS IT HAS 1 door with a goat and 1 door with a CAR!!!
@@shafin141 I guess you didn't fully understand the explanation. You have to think about the problem like I explained it, as in, including the odds of you picking correctly in the first place. Yes, after one door is removed, there's a 50/50 chance that the car is behind one of the remaining doors. HOWEVER. You have to take into account the fact that your INITIAL choice only had a 33% chance of being correct. Those odds have not changed.
Try extrapolating it to 100 doors, and 99 goats. You have a 1% chance of picking correctly at the start. Now the host removes 98 incorrect doors, and you're left with 2 doors. It's still technically a 50/50 chance of there being a car or a goat, but you're not going to switch still? You have new information now. The host has essentially removed the majority of the wrong possibilities from the equation, so the odds of you now picking correctly if you switch have become 99/100, versus your intial 1/100 odds.
@@dchang8619 Sorry you are confusing yourself by believing that the initial 33%, 1/100 and 99/100 remain. They are gone now and dont carry over. Its now 50:50 and your original choice is just as likely as switching.
@@kenandlynboyle9300 it’s really not though. Try it yourself in practice, not in your head. Try switching 10 times, and not switching 10 times. See how many times you win by switching and how many times you win when you don’t switch. I promise you that you will win more times by switching. This has been mathematically proven, so there’s no real use trying to argue against it. Just try and understand why it’s the case.
I used to read her section in the PARADE magazine every Sunday morning. It was one of the few things that made me look forward to sundays.
I had never heard of her, she is so cool though!! She kind of makes me think of an old-fashioned Druid or wise person on top of a hill and people come from there and far to ask questions
This is easy...you have to keep your choice only if it was correct the first time - the odds for which is 1/3. If you have chosen wrong the first time (2/3 odds), you have to change your choice. Therefore, keeping the choice will be correct in only 33.3% of the time and changing the choice will be correct in 66.6% of the time.
Why does this seem to neglect the fact the odds of your initial choice being correct INCREASE when one of the goats is revealed? You're being given new information about what may be behind door 1 - either one of ONE goats, or a car. It used to be one of two goats, but now one of the goats is excluded. Of course the probability wouldn't stay the same.
@@lentilswoo I don't think that is correct, As the host Will never open the door with the car ...
Imagine how intelligent the person who created this problem would be
probably in the 125 - 150 range.
@@arandomguy46 wtf
@@epicmorphism2240 The creator probably didn't know. It was just a game. Maybe after decades of hosting the game they ended up with a "gut feeling" that is better to switch but they probably thought it was 50% too
@@andressoto739 i was commenting ln CRB‘s ridiculous comment
noobs create problem pros solve it
The one-hundred door example makes it very clear. No one should assume they guessed the correct door in that scenario. The pagentry of a Game Show might make people falsely presume the host is hoping to trick the contestant into switching their pick, but unless the host is actively trying to persuade you to select one door over another there's simply no trick to be had. This line of thinking is precisely why I got this wrong upon initially hearing of it years ago. It's still shocking to see how venomous some of her detractors were.
I thought the host was wanting the contestant to lose the car, thus influencing my answer.
I dont understand this, it makes no sense. if you choose a door and the other is eliminated then the probability should distribute evenly between all of the remaining doors, not all of them except for your choice. Can someone explain why it does this?
@@versenova5531 door 2 and door 3 combined add to a 2/3 probability but you know door 3 isnt it so that 2/3 probability has to be for door 2 thats how i understand it
Exactly, saying that the host will open a door with a goat, before you make your pick, is must for it not to influence your answer in this scenario.
@@versenova5531 you have 33.33...% chance to guess it right, which means you have 66.66...% to choose WRONG. Switching will automatically give you the price 66.66...% of the times. You're more likely to choose wrong, so switching is a better choice 2/3 of the times.
That's why it's even more clear with 100 doors, you have 1% chance initially, but after revealing a goat behind 98 doors, it's your best option to switch 99 out of 100 times.
For people who still don't understand - Think about it with 10 trillion doors. First, you pick any random number out of 10 trillion. Then the host eliminates all the doors (numbers), except yours and one other number (remember, one of the two remaining numbers is correct). Would you switch? Obviously, because the chances you randomly picked correctly out of 10 trillion is basically zero.
Some people still couldn't understand this with 100 doors, and I think it becomes even easier to understand when you increase the total number pool.
Oh that makes sense
Thank you, I saw this explained on another video, I don’t understand why mathematics people are disagreeing with this simple fact of probability, it’s not that hard to understand
I still don't agree with this no matter if its correct because 1 out of 3 is not 1 out of trillion or 100 or 10 its still out of 3 so it's completely different.
@Krichnu It is the same; that's the point. It doesn't matter if you agree, because the math checks out.
Find the comment on this video where the guy wrote a program and ran this simulation 10,000 times with 3 doors (or whatever the number was).
Can you imagine how you would feel if you switched and were wrong? @@MrMcSnuffyFluffy
I came across this puzzle during a math course and the tutor posed it very briefly, with no indication as to whether the host had picked the door at random or whether he chose to pick one which he knew didn't contain anything of value. I asked about that at the time, saying I needed more information, but he didn't seem to understand my question. What made this problem difficult to solve was that it was not properly posed, though most people didn't realise that.
if there are 2 players and the first player chose door 1 and the second player choose door 2 and then the host revealed a goat behind door 3 . Now he asks the players do you want to switch ..now each player has a 2/3 probabilty of switching as it seems i.e both players have equal chances of switching and obviously equal chances of sticking so chances are 50 50 and we can not say one has a 1/3 chance and the other has 2/3 chance because both are under the same circumstances .Now assume the 2 players are in different rooms and they dont know anything about each other in other words each player thinks he is the only player in the game and the game is played at same time for both .Do you think now this will change the outcome .Chance is 1/2 to 1/2 for each player or for one player if the other player exists or not..This shows how people are deceived by majority oponion.
"What made this problem difficult to solve was that it was not properly posed, though most people didn't realise that."
You are correct that your tutor posed it incorrectly if you couldn't know the host's selection of the door to open was deliberate. But that is not the reason most people are confused by it... the question Whittaker presented to vos Savant in Parade was worded correctly. Since we are told the host knows what is behind the doors, it is impossible for him to open a door not already knowing it was a goat.
@@alimetlak
Assume the car is behind door 1.
Player one pick door 1.
Player two pick door 2.
Host open door 3.
Player 1 stay = win, switch = lose.
Player 2 stay = lose, switch = win.
Assume the car is behind door 2.
Player one pick door 1.
Player two pick door 2.
Host open door 3.
Player 1 stay = lose, switch = win.
Player 2 stay = win, switch = lose.
Assume the car is behind door 3.
Player one pick door 1.
Player two pick door 2.
Host open door 3. Oh wait, it's the car!!!
Oh yeah, you think there is some magic power making at least one player pick the door with the car.
Assume you stay with your first pick.
If your first pick is Goat A, you get Goat A.
If your first pick is Goat B, you get Goat B.
If your first pick is the car, you get the car.
You only win 1 out of 3 games if you stay with your first pick.
Switching means the opposite.
It's just basic math/logic kids understand.
Sadly, it's far too hard for idiots.
@@alimetlak Well spotted! You have put your finger right on what I feel is the weakness with how I understand this scenario.
A somewhat similar principle applies in other, related puzzles. There is one classic going back a very long time which caused a lot of the same sort of disagreements as the car-goat puzzle. It involves 3 little bags, each known to contain 2 black or white counters, thus: BB, BW, WW. Someone picks up a bag at random and takes out a counter. It happens to be black. What is the probabilty that the other counter in that same bag is also black?
I think this one is quite easy, but I knew someone couldn't, or maybe wouldn't see it.
@@alimetlak If there are 2 players playing the game and they can't choose the same door, then there is a 33% chance that the door the host opens has the car behind it. In the other 67% of cases, when the door the host opens has a goat behind it, the first player should switch because their probability of winning the car is 33%. The second player, however, should not switch, as their probability of winning is 67%.
Explanation:
First player's probability of picking the car: P1 = 1/3
Second player's probability of picking the car: P2 = 2/3 * 1/2 = 1/3
Probability that the host opens the door with the car: P3 = 2/3 * 1/2 * 1 = 1/3
2/3 of the time, when the host opens the door with the goat, the second player's probability of winning changes: P2 = 2/3 * 1 = 2/3
So, in your scenario, the first player should switch, while the second player shouldn't.
One way of wrapping your mind around this question is by thinking of it this way: Originally there are 3 doors, so your odds of picking the correct door are 1/3 and your odds of picking the wrong door are 2/3. The host is then going to reveal a goat, which seems like new information, but you already knew that at least one of the doors you didn't choose had a goat, so it's actually not telling you anything you didn't already know. Asking you to switch is really asking you if you think that your first choice was wrong, and uses the same odds as when you made that choice, so the odds that you chose wrong the first time are 2/3.
So switching comes down to thinking you made the wrong choice the first time. The odds change with the revelation of one goat. This is new information, not just "seemingly." The 1/3 or 2/3 scenario is now irrelevant.
@@diannemccarthy8685 No, the odds don't actually change, even though intuitively it may seem like they should. The way that I explained it above is the explanation that I personally find the easiest to understand, but if you don't find it convincing, just search for the "Monty Hall problem" in the search engine of your choice, and you should be able to find plenty of other explanations. When I first came across this problem a few years ago, I did a lot of reading about it myself, and I remember that there were even websites that allow you to play this game over and over, and they keep track of your win/lose record for when you stay and when you switch, and you'll see that over time the win percentage for switching averages to 66.6% while the win percentage for staying averages to 33.3%.
No, given the mind set of the Host, your chances of winning are BETTER than the problem suggests! Monty WANTS you to win! More fun, more excitement, better ratings, more sales of every item offered as a "prize" on the commercial, (ahem, "Show") The clue he gives you is that your odds got better after your first choice. Monty wants to make your chances of winning as near certain as he can, without being obvious.
@@lindsaymitchell9300 The show producers are always telling Monty what his next move should be. Anything that makes mo money.
@@zackreagin8384 The odds of the second door are increased after seeing the third door open, IF the game (which regulation I ignore) had the rule that BOTH a door must always been open AND the switch must always been offered. If showing an open door and/or the option of switching are not the rule of the game but may be decided by the host unpredictably, the host of the game might use the door open or the switch option as a way of deceiving the player, at times. That would not guarantee that switching to the other door carries more chances to win, in my opinion...
I can't find the comment again to give credit, but someone in the comments here helped me understand it intuitively. It gets much easier if you think about the odds of *losing* rather than the odds of winning:
- If you *always* switch then you *only* lose if you picked the car to begin with.
- What are the odds that you picked the car to begin with? 1/3.
- So the odds that you lose if you switch are 1/3.
Yea, but, If you always stay, then you only lose if you picked the goat to begin with.
What are the odds you picked a goat 2/3 (initially known to be)
But after the knowledge, what are the known odds that you picked a goat from your initial pick? = 1/2! Everyone is wrong. And it's amazing.
@@kenham6742 YOU ARE RIGHT 50/50
@@nicosmavropsis459 I was mistaken. You see I missed one crucial thing = the host has an advantage, I didn't factor in. The host always keeps a good door (usually more of the time, than your door).
So, you have a 1/3 chance of choosing the right door.
The host has a 1/2 chance of choosing the right door.
So, you always go with the host, for the best chance.
I was only able to figure this out once the problem was exaggerated in my mind:
Say there was 100 doors and you chose 1/100.
Then the host eliminated 98/100, by choosing 1/99.
You should go with the host because, you chose 1/100, but the host chose 1/2.
@@kenham6742 IN THIS EXAGGERTED PROBLEM WITH THE 100 DOORS, THE HOST ALREADY CHOSE AND ELIMINATED 98 DOORS, HE DID'T CHOOSE 1/2, RATHER THAN HE ISLEAVING THE 50/50 CHANCE TO THE PLAYER TO CHOOSE.
LET'S RETURN TO THE SENARIO OF THE PROBEM PRESENTED. THE HOST WOULD ALWAYS REVEAL A DOOR WITH ONE OF THE GOATS. BECAUSE WE KNOW HE CAN'T OPEN THE DOOR WITH THE CAR, DOOR#2, SO IF HE OPENS DOOR#1, HE WOULD REVEAL THE WRONG CHOICE OF THE PLAYER AND IMPOSE THE SAME QUESTION TO THE PLAYER TO CHOOSE DOOR#2 OR DOOR#3. IT WOULD BE THE DECEPTION PLACED BY THE QUESTION NOT ONLY TO WIN, BUT THE ACTION TO STAY AND LOSE OR TO SWITCH TO WIN BY CHOOSING DOOR#2 OR DOOR#3. OF COURSE IN THIS CASE DEFINETLY SHE HAS TO SWITCH IN ORDER TO WIN.
IN THE CASE THAT THE PLAYER HAD FROM THE BEGINING CHOSEN DOOR#2 WITH THE CAR, BUT HE DIDN;T KNOW THAT IT WAS THE RIGHT DOOR AND THE HOST REVEAL AS HE DID THE DOOR#3 WITH ONE OF THE GOATS, THEN IF HE PROCEEDED TO CHANGE APPLYING THE POSSIBILITIES PRESENTED IN THE VIDEO THAT HE HAD GAINED 33.3% , TOTALLING TO 2/3, HE WOULD LOOSE IF SHE WOULD FOLLOW ALWAYS THE ACTION TO SWITCH TO WIN , SO IT IS ALWAYS 50/50, BECAUSE THERE IS A DISRUPTION IN THE POSSIBILITIES AFTER THE ACTION OF THE HOST EVERYTHING CHANGES.
@@kenham6742 You are still getting it wrong. As the car can never be revealed in any of the games, then if you only pick it 1/3 of the time, the host is who is forced to leave it hidden in the other door that avoids to open in the 2/3 of the time that you start failing, not 1/2.
Similarly, in the 100-doors scenario, as you only get it right in 1/100 of the attempts, his closed door will be right in 99/100.
It is like if someone else looked inside all the 99 doors that you did not pick, and took which preferred from them. In that way, it is obvious that that person will find the prize in 99/100 of the cases.
It is important to note that the 1/3 to 2/3 probability is right only if the host who opens the door knows what is behind the doors. If he doesn't, then the probability is 50/50.
The problem is not so buffling after all.
Its also important to know if you also know this information prior to picking the door.. Because that also changes the probability.
@@bodybuilder6350 i do not think so. Could you explain?
Now i understand what you mean. You are right.
It is important to note that if the host reveals the car, the probability of losing goes up to 100% because the car has been removed from your choices.
What this tells us, is that it's a waste of time to think about what the host knows or doesn't know, or what the host's intentions are.
@@alkinooskontopodias5919 if you know or don't know that the host knows or doesn't know. So thats like 2x2= 4 different probabilities depending on what information is known. I am sure you bring in even more play of words and rules that would increase the variables. From the original question we are limited in our knowledge. We didn't even know if he would open the right or wrong door (initially) to expose a car.
To the people who don't get it it is because the host will never open the door that has car which increases the probability of other door having the car.
In fact, no. is is 50/50 chance even it was thousand doors and only two left it is 50/50 chance.
But because the host ask if you want door two is a clue. The host likes to make drama and after he will say pity you didn't change the door. it is psychology, he know most people do not want to change if he try to make them.
@@neinei5558You really missed the idea..
There is a difference between the host knowing where the goat is against showing the 2nd goat by guessing.
There is no 50/50..
If you buy a lottery ticket and someone from the future says.. im going to pick 1 ticket and leave it on this table after destroying 99998 which I know wont win..
Do you want the ticket?
Or do you keep the one you have before I picked one and removed the rest?
No, that's not her reasoning.
@@absolutium No matter how many tickets it originally was as long it remind two tickets one win one lose, 50/50.
Some people think if you throw 3 sixs in a dice the next will most likely not be six, thats also wrong for every throw you do it is 1/6 chance to be any number no matter what the dice show before.
@@neinei5558 hahahahahaha
I've long understood the Monty Hall solution - but extrapolating the information scale to 100 doors - makes complete sense - knowing that the "one door" is hot - and that you have a 98% chance of being wrong on your first door choice.
it's 99% of being wrong and then the switch is 99% of being right. it's reversed.
revealing the doors does not give any hint on your first choice, so it's still 99% and not 98%
(Car is prize). Let’s say the chooser did not pick a door at all. The chooser does not know whether there’s a car or a goat behind ANY of the three doors. The host, however, knows what’s behind each of the three doors (ALL). They both stand in front of the three doors and the chooser does not pick anything at all. Now, the host reveals that there is a goat behind door #3. That eliminates door three as an option for the prize. Bye-bye! Now there are two doors left. That means there’s a 50% chance that the car is behind either of the two remaining doors. The high IQ vos Savant is declaring the idea that the chooser having originally simply “THOUGHT” (chosen) of what the winning door might be before the host opened door 3, CAUSES the remaining 50/50 ratio to change. To me that sounds like she is either not that smart OR OR OR, she is pulling ideas from things like quantum theoretical physics such as the “double slit experiment” and such. Like, the theory that when trying to guess if there’s a cat in the box, actually there is no cat in the box until you open the box, then it either does or does not manifest, because our reality is made up of potentialities rather than actualities, until they are “looked at” or measured -to be more precise (that’s how the theory goes, which is pretty much proven at this point). So, I think she needs to clarify what level of accepted reality or physics she’s pointing to in advance of her explanation, because the description of the solution by her and this video is implying that the chooser’s THOUGHTS (manifested as a supposed “choice” or predisclosed “guess” of where the car might be, affects the outcome. It’s similar to me arguing with you that nothing actually exists because atoms don’t reeeeally touch each other and sub-atomic particles are only energy in their smallest form. In a physics class that discussion or answer to a topic might be appropriate, but in an online forum about whether I have one or two cups on my table , it isn’t necessary appropriate. We obviously have different “levels” of reality we are talking about. The one to attend to needs to be disclosed in advance because she is LITERALLY saying that the difference between whether the chooser specifically made a choice (THOUGHT), or not, before the process of door openings began, AFFECTS the outcome. That’s not really actually fair according to the standard discourse. An online basic forum is not a quantum physics class. Thanks for reading.
PS. This might be going too far, but, it also makes me think she’s purposely trying to mess with people’s heads. lol. It worked ‘cause even the PhD’s were crying but funnily enough what they missed is she subtly didn’t disclose to them all the facts of the game. If I tell you we are living in a video game, am I lying? If I walk away and leave all the PhDs to fight about it, then they cry and appologise, am I smart?? Hell yeah! Am I right? Not if the rules of said game weren’t disclosed. lol. Sounds a bit narcissistic IMO. 🤭
@@SolutionsWithin You obviously did not understand the problem and are not smarter than Savant lol.
Not picking a door first is different than picking a door first, because then the host cannot pick this door and has to chose between the other 2 only. You picking a door eliminate this choice for the host.
That's why it becomes a 1/3 - 2/3 and not 1/2 - 1/2
Maybe I'm not one of the smartest people around but I have a logical thinking brain. Your 1st choice has 1 in 100 chance of being correct. When 98 of the remaining doors are eliminated you end with two doors with 50/50 chance of having a car. All that math stuff don't really mean anything. The door that was not eliminated does not have greater chance of having the car than your 1st choice.
I remember reading this from her as a kid. Loved working through the logic; it's just counting. The thing people don't get is that the 'host' (originally Monty Hall) doesn't behave randomly; he always shows you a goat. Just count out the cases, which is what she did, and the answer is obvious.
If he doesn't behave randomly, the it ain't a probability question. It's more along the lines of 'form' like at the horse races . We then talk more about odds . No strict rigourous standard calculations for odds . It something settled on by experts . However the probability calculations are ½ for a choosing from two given options for which absolute randomised conditions apply.
@@philip5940 you're confused. It is a probability question, but the difference is it's about posterior versus prior probabilities. just write down all possible options; pick one at random, and consider all of the different ways that the prize might be distributed. Count what happens when you switch or stay. you have 1/3 chances of getting it right by guessing out of the box. He then shows you one of the places the prize ISN'T (that's the part that's not random). Now you have 2/3 chances of winning if you switch. Why? Because there was 2/3 of a chance at the beginning that it was in one of the spots you DIDN'T pick, and he showed you one of the places where it isn't, so those odds still apply; you have 2/3 chances of getting the prize if you switch. It's a classic problem that is now commonly taught in probability, known as the Monte Hall problem (he was the host of the game show). If you don't believe me, write down 3 doors, put the prize behind one of the (doesn't matter which), let's say it's behind door 1. Now say you pick door #1, and it's behind door #1. He shows you either 2 or 3 (where there's a goat), if you switch you lose. Say you originally pick door 2. He has to show you door 3, where there's a goat, if you switch, (to door 1) you win. Let's say you pick door #3; he has to show you the goat behind door 2, you switch, you win. Count it up; 2/3 chance of winning if you switch.
It's also ironic you're arguing about his given that the entire point of this video is to explain to you how your answer is wrong.
@@bartonanderson1106 seems that the entire point of the video is to show the power of spin and to highlight mass gullibility . ⅓ and ⅔ probability when given three choices are now a phantom that have ceased to exist The new state is two choices for which the probability is 50/50 .
@@philip5940 I just explained the entire logic to you; you're obviously not capable of understanding this, but here's another chance: en.wikipedia.org/wiki/Monty_Hall_problem
@@philip5940 Since the host always removes a goat that is not which the player picked and neither which contains the car, then that means that the other he leaves closed is the winner one as long as the player starts failing, and that occurs 2 out of 3 times on average, not 1 out of 2.
This is better seen in the long run: imagine you played 900 times. In the first selection you are equally likely to select the option that has each content, so in about 300 times you would get which has goat1, in 300 which has goat2, and in 300 which has the car. In total, 600 times a goat and 300 times the car.
As the host will reveal a goat from the two doors that you did not pick in all the 900 games, in the 600 that you already have a goat, the revealed goat must be the second one, and so the car is in the switching door. Only in the 300 games that you had the car behind yours, the switching door will have a goat.
So, always two doors left, but yours results to be correct 300 times, and the other that the host leaves, 600 times.
When you initially explained that the other door would have a 2/3rd probability of a car being behind it, i couldn't understand it one bit. But i loved the explaination including a 100 doors where 98 were removed. That explaination immediately clicked to me and now I get it! What an interesting question. I always love these kinds of probability questions cuz they make me use my brain in ways I don't get to use while studying 😅
Im really curious about why/how the second example changed your mind since its exactly the same explanation, just different numbers.
Was it the fact that it sounds more absurd that your 1% turned into a 50% instead of the 33%?
@@BillyViBritanniaprobably because he/she is less intellegent than you
My initial issue with the concept was my assumption that the host could reveal your initial choice and it’d be allowed to be wrong lol I didn’t even understand the game apparently
@@BillyViBritannia Why wouldn't extreme exaggeration of an effect help people prove to their own intuitions that the effect exists? Guessing right/wrong in the first question will "feel" like it could be a result of natural RNG because the discrepancy of a single outcome between 33% and 50% can be masked by luck. That won't intuitively update your brain's model of prior probabilities.
The rational explanation of Monty Hall will be easier to wrap your head around when it is ALSO reinforced by your intuition that you would almost definitely lose by staying.
@@jacobkdunn most people who don't understand the problem - as I've found out from the hundreds of replies under my attempt at explaining this - believe that anything that happens before you are left with 2 doors is irrelevant.
And if you believe this then it shouldn't matter whether you open 1 or a million doors, final choice is always between two, and thus 50-50.
That's why I was genuinely curious at this comment, I was not trying to bash them like the next person did.
And it's also a huge pet peeve of mine when someone tries to explain something in a different way and then proceeds to give the exact same example with different numbers, so I guess I was kind of annoyed that it actually worked.
The key to this puzzle is that the host knows which door has the car and never picks it allowing the contestant to improve their odds.
There are two important things one has to realize, in order to understand this problem (this is not a puzzle at all):
- player picks only 1 door, leaving 2 doors to a host - and that's exactly the reason why the door offered by the host will hold a car twice as likely as the door player picked
- host knows where the car is, and never opens it - which is needed for a car to always be behind the door offered for a switch, if player haven't already picked it
Here's a clear way to think about this
There are 3 cases
C-car
G-goat
C - G - G
G - C - G
G - G - C
There's 1/3rd chance you'll pick the C door in first trial, and 2/3rd chance of G door
Because the opened door is not random and the host always opens a G door, you have essentially "flipped" the odds
if you picked door 1 (or 2 or 3) in all these cases, in 2 of them you will pick the G door, host opens another G door, so the remaining door has C, switching gives you C in 2 of 3 cases
you lose by switching only in the one case where you picked C door right away, which happens 1 in 3 times
What if you do not picked one?
@@1984Kojot picking one is equivalent to not switching after the host opens a door
Put it this way: Everyone would switch in the 3-door case when the host does NOT open one of the doors but offers you to take the two doors instead of the one you took.
What I didn't realize was that the host obviously only opens a door different from the one I picked. So the odds remain 2/3 for goat.
Wow! That is the clearest way of describing the situation I've seen.
"Stay with YOUR original 1:3 chance or take OUR 2:3 chance". The revealing of the goat was just a trick to make it seem as if it's become 50/50 to switch or stay.
What a great explanation, well said! It's obvious explained this way. I also like how Savant's analysis of all outcomes is so simple and really proves it, and yet the top mathematicians who criticized her didn't think to do such a simple analysis.
The host knows what is behind every door. The offer to change my choice is only given after I picked a door already. Obviously the host is trying to fool me.
If I was given 2 choices from the start: pick a door, or pick 2 doors, the answer is obvious.
@@Tombecho If you think the host is benevolent: Switch. If you think he doesn't know where the car is but just got told where one goat is: Switch. If you think the host is mean: Stick with your choice.
Just reverse engineer it.
If you always switch, you will only lose if you had first selected the car.
There is a 33% chance to lose when you do select the car on the first try. Leaving you with a 66% chance to win if you always switch.
Way too logical. And a good way to think about it
But having a chance and actually proving you are correct is two different things. The car could have been behind door #1 and then everyone using their math ability to choose #2 would lose. Sometimes you need luck too. That is how nature works. You cannot predict anything definately. You can only guess.
@@edwardspencer9397 Sure this game depends on luck. But the chances to win are higher.
@@edwardspencer9397 that's why it's called probability. Out of three doors you will win one in three times if you stay with your original door. Out of 100 Doors you will win one in 100 times if you stick with your original door. It would be ridiculous to not switch unless you didn't want to win
@@edwardspencer9397 yes you need 'luck', but by always switching you will be 'lucky' 2 out of 3 times. Whereas, if you do not switch, you will be 'lucky' 1/3 times. Therefore, by switching you increase your 'luckiness'. Considering this, you may be less fortunate if you merely guessed at what to do.
The explaination is quite simple :
Only one car and TWO goats.
So :
1/3 : you pick the car, then you must stay to win.
2/3 : you pick goat 1 or goat 2, then you must switch to win.
Saying that the final choice is a 50/50 chance means that your initial choice is 1/2 to pick the car and 1/2 to pick a goat, which is mathematically wrong.
Then : ALWAYS SWITCH.
CQFD as we say in France :)
This is the problem i have with it. In math if you take something away it is removed from the equation. Yet here it stays with the equation. Not very math like.
@@kevinriffey9970 You are confused because you don't understand where does the 1/3 vs 2/3 comes from. You think that it is because they are still keeping the original 1/3 of the door that was removed. That's not true; it does not come from that. That 1/3 was in fact removed. The thing is that also half of the 1/3 of your original door was removed, and that's why it is now half as likely as the other closed door.
Remember that the host must reveal a losing option from the two that you did not pick. That means that if yours is already a losing one, he is only left one possible losing option to remove, but if yours is the winner, he can reveal any of the other two, making it uncertain which he will take in that case.
So, the original 1/3 chances of your door, that is #1 in the example of the video, are actually divided on two halves of 1/6 each depending on which of the others is revealed then, #2 or #3, so once #3 was revealed you must discard the 1/6 in which the revealed one would have been #2.
Therefore, your door #1 is only left with 1/6 chance, while #2 remains with its entire 1/3, as we are sure that if the correct were #2 the host would have opened #3.
Now, since the total probabilities must add up 1 = 100%, you must scale those chances 1/6 and 1/3. Applying rule of three, you get that the original 1/6 represents 1/3 now, and the original 1/3 represents 2/3 now.
@@kevinriffey9970 Continuing with my previous example, the confusion comes because the final proportions end up being the same as in the beginning, which gives the impression that the information was not changed, but that is not the case. It is just that when you make the new calculation, the proportions happen to be the same.
If you think this is strange, think about a soccer match. We know that each team starts with 11 players, so each has 50% of the total. Now suppose that during the match, a player from each team is ejected from the game. Each team is left with 10 players, so if you calculate the current proportion, you get that each team still has 50% of the total.
That does not mean that they are still counting the players that were ejected, it is just that the proportions with the remaining ones happen to be the same.
@@RonaldABG No I'm saying she is using a hypothetical of a hypothetical. For example, try using her logic if you fix the car to 1 door and make the game show host open only 1 door as the question is worded. Most people see the question as a real life scenario in which there is only 1 attempt and a set door.
@@RonaldABG If you look at the possibilities listed its a 50/50 at 3:21 She is just picking a subset of possibilities to validate the switch. In reality its a 50/50 in a statistics world that does not exist its different. In reality there is only 1 possibility which is a 50/50. Fix the car to 1 door like it would be in real life and its only a 50/50 which is what most logical people assume the question is asking.
I love that the professors were humble enough to publicly retract their first responses
Whether their retractions were praiseworthy depends on why they did so.
They probably wouldn't have written them to a man in the first place
Nah! They were more worried about how they would be received if they did not. The humble would not rebuke someone in public, as in a letter to the editor!
@@Elle-ht3km There was an extreme amount of sexism back then, and there still is a shocking amount now, but I don't think all of them were because she was a woman. At least a couple acknowledged that she still had higher IQ than them, and thought she simply made an obvious mistake. Even the cleverest people can do incredibly dumb things sometimes. I'm sure even Marilyn oversaw something really dumb in her private life at some point; and they probably thought it was one of those, but in public. They did grew heated to make sure she "corrects" the mistake and thought it was so obvious that it would severely undermine the population's understanding of probability. It was the exact opposite, but I can definitely see why they'd think it was a mistake.
At least one of those letters was definitely sexist though, and knowing those times, probably a lot more. There's probably a lot of enchancement on it too, after having been told she's more intelligent than you are and then given actual proof of that statement. Thinking someone make a dumb mistake when you're both sexist and jealous would probably compound and get you extremely heated up. And then after she proves you wrong, and you get personal proof that you're dumber, you'd either get more embarassed and angry or you'd be embarassed and start respecting her.
#MeTOO
The easiest way to understand this: The probability of you choosing the right door on the first try is only 1/3, but choosing a goat is 2/3. There is a higher chance that the door you first chose has a goat behind it than a car. And since door C has already been revealed to contain a goat, door B receives the rest of the probability of getting a car, which is a 2/3 chance (because there was only a 1/3 chance it was in door A).
I can understand the confusion. There are three possible interpretations:
1. You randomly choose one out of three doors and stay - 1/3 chance of winning every time
2. You randomly choose to switch or stay - 1/2 chance of winning every time
3. You always switch - 2/3 chance of winning every time
And the funny thing is, they are all correct, which is what makes this problem interesting and controversial.
Great explanation. Thank you.
I still can't get the "correct" answer to sink in...
When the goat is revealed, you're starting all over, but instead of 3 doors you only have with a 50/50 chance. Why must the calculation continued to have anything to do with 3 doors when it gets reduced to 2 doors and 2 choices? Is there a reason why a person is forced to choose from the original odds after they change to different odds?
Non sense
That's math. But if you take in the human element, the show host can know you have the correct door and tempt you away.
4:10 NOW I get it. Two out of every three times you make the wrong choice so switching will lead to the right choice two out of three times.
Weirdly that does make sense.
Exactly. The thing people don't seem to understand is that the host must open one of the two wrong doors. He cannot open the door with the prize in it (this defeats the point of the game) or your door (which would render the decision obvious). If the host were randomly opening doors (and could open your door or the one with the prize in it), then this wouldn't work.
@@user-wu4bo1hz3p what if not randomly opening door? Just not opening door that is yours and.prize ?
@@yuquoint6633 Huh? That's what he's doing, which is why it works.
@@user-wu4bo1hz3p Ah finally i get it, thanks for explanation.
The reason it is not 50/50 is because the host ALWAYS opens a LOSING door. He does not pick the door at random - it's always a losing door he exposes - this changes everything. Therefore you will be correct 2/3rd of the time if you change your pick.
"The reason it is not 50/50 is because the host ALWAYS opens a LOSING door."
So what was the contestant's probability in the host's FIRST game that he did?
@@klaus7443 If you keep your original pick - you have a 1/3 chance of being right. If you switch after host purposely opens a losing door - it's 2/3rd chance of being right.
@@markmiller3210
I know that. Now answer the question that I asked you.
The best way to understand this is to imagine doing it many times. If you employ the strategy of always staying with your original decision, you'll only win a third of the time.
They did it on Mythbusters. Switching works.
@@archwaldo Anyone who can write macros is a mythbuster here xD
It's not hard to understand the pyramids !! By vibrating within the earth's "B" field they could communicate as well as travel intergalactically. My goal is to educate others. Just for clarification. The center of a magnet is by far the most powerful. AC current Radio frequency signals could not travel without the "B" field. The oscilloscope never shows the "B" field but it is there. otherwise it would show a short circuit as the two polarities collide. The "B" field pushes everything forward. Things you can do to see the "B" field:1> vibrate the "B" field of a magnet near a trickle of water and watch the water repel the same.2> Vibrate the "B" field on plastic or glass and watch the item lose weight!! You must understand, don't try this on iron or metal; it will distort the magnetic field . 3> Vibrate on Granite rock and it will become weak. and You can cut and shape ONLY with copper tools. AGAIN pay attention to the magnetic field you can't use iron tools for this action. . 4> I have not proven this part yet, but you should be able at the correct Two frequencies to levitate non metallic objects. FYI I think that's why most UFO's don't show on radar they are not metallic.
I have come to believe we are all operating on the magnetic "B" field. The entire universe is 100% magnetic, we are an algorithm of the same.
Think about this : The pyramids were communication and transportation device's of yesteryear: When you install such a huge mass within the "B" field of earth they intercommunicate via the "B" field. FYI I have teaching credentials and much much more. Want some Proof, Check out Coral Castle in Florida on TH-cam, all his generators were set up in the repel mode to maximize the "B" field into the earth FACT!! The stones at Stonehenge and KT were part of the "B" field system, The stones would vibrate at frequencies that would stimulate the crops and many other things. Those were the true power plants of yesteryear. WB6HUN/1958
I remember this problem from quite awhile ago. I was able to get it by the method used in the middle which involved the hundred doors, in which case one’s intuition works properly. It’s interesting that if you think of the given case with only two remaining doors, it’s easy for your intuition to be wrong, but in the hundred door case, anyone would switch.
In probability problems our intuition is often wrong, so if you generalize the problem by considering what happens as you increase the numbers and look for a pattern, that’s often helpful.
What if you didn’t pick any door and I eliminated all but two doors from the 100 and told you to pick. Would switching between them change your odds? I don’t think so.
Doh … with the 100 door example boths doors have a 1/100 chance of having the car, the same odds. After 98 are eliminated two remaining doors each have a 1/2 chance of having the car…the SAME odds relative odds. The absolute addos of both doors changes but the relative odds of those two doors remain unchanged. The premis is simply wrong and the big joke is that it is just a con trick like the hare and the tortoise, which clearly never catches up. Also some people might actually prefer the goat, for a lifetime supply of goat milk. The woman was taking the piss and most people are too dumb to spot the trick, including the narrator.
The problem is that people confuse what the question is based on. It is based on knowns and unknowns. as soon as a door is known it is removed from the equation AS AN UKNOWN because it is no longer an unknown door. to keep the door in the equation you have to PRETEND you don't know what is there which throws people off. so you have 3 doors and open one so now you have 2 doors and a goat standing there NOT 3 doors as is the mistake here. from 100 unknown doors if you open 98 you then have 2 doors of unknowns and 98 'objects or nothing' the 98 'things' are NO LONGER DOORS. Why the people are insisting on keeping the *known* doors in the math as though they are *unknown* doors is the real stupidness here .. lol as soon as an unknown door becomes known it isn't unknown anymore and should be removed entirely or if you pretend it is still unknown then you should also pretend it was never opened. lmfao i think people miss the fact that the room with the door only exists because it's unknown and should disappear from being unknown when it IS known. how can you possibly include the third door as an unknown in calculating odds WHEN IT IS STATED THAT IT WAS TRANSFORMED INTO A KNOWN.
@@spivvo it seems few people see the root of the issue here. after the door is discovered people are still including it in the math as an 'unknown'. If the door has been opened it is no longer a door so how can you possibly include it. you have to PRETEND it's still a door even though it's been destroyed by becoming a 'known' i would like to talk to the people who said she was wrong then said she was right.. lol
@@jaybefaulky4902 You are not using the information given by the host. This is a conditional probability problem and as such is mathematically and logically solved with the doors closed. There are three ways it could have been worded, two of them with the doors closed. The author of this puzzle chose the wording to include an open door only for the purpose of fooling the most readers. If you read the rules carefully you would know where a goat is even if it wasn't revealed. Having said that the door that must have a goat is useless to both the contestant and the host. To show it's of no value the host could have given that whole door to the contestant and his chances of winning by staying is still 1/3+0=1/3.
It’s one of those problems, which one can understand better if one writes down the possibilities of each choice and calculate the probability of switching.
The constant in each game is the host opening the door with a goat in the second step.
Game 1 and variations
Door 1 chosen by the player
1 Car 2 Goat 3 Goat (2nd or 3rd door reveals the goat) - Switching makes the player lose the game
1 Goat 2 Car 3 Goat (3rd door reveals the goat) - Switching makes the player win the game
1 Goat 2 Goat 3 Car (2nd door reveals the goat) - Switching makes the player win the game
Switching wins 2 out of 3 games. In other words, there is a 66.67% chance of winning when you switch the doors.
Game 2 and variations
Door 2 chosen by the player
1 Car 2 Goat 3 Goat (3rd door reveals the goat) - Switching makes the player win the game
1 Goat 2 Car 3 Goat (1st or 3rd door reveals the goat) - Switching makes the player lose the game
1 Goat 2 Goat 3 Car (1st door reveals the goat) - Switching makes the player win the game
Switching wins 2 out of 3 games. In other words, there is a 66.67% chance of winning when you switch the doors.
Game 3 and variations
Door 3 chosen by the player
1 Car 2 Goat 3 Goat (2nd door reveals the goat) - Switching makes the player win the game
1 Goat 2 Car 3 Goat (1st door reveals the goat) - Switching makes the player win the game
1 Goat 2 Goat 3 Car (1st or 2nd door reveals the goat) - Switching makes the player lose the game
Switching wins 2 out of 3 games. In other words, there is a 66.67% chance of winning when you switch the doors.
There are nine total games possible (technically 12, but that’s irrelevant to the problem). Switching will win you 6, and not switching will win you 3. Switching 2/3 (66.7%) Not switching 1/3 (33.33%)
What we’re really calculating here is not the probability of what is behind the door, but whether switching makes a win or not.
The or is what makes this statistical assumption wrong. There are 12 distinct possible outcomes. you can't put an or on the ones that make the player lose and treat the ones that make the player win as separate. for each variation, there are 2 that will win and 2 that will loose. 50/50. They even show it at the 3:23 in the video that its equal while saying that it's not. hilarious.
There seems to be an error, because in the example in the video door 1 was chosen (your GAME 1) and exactly door 3 was shown. That only leaves the first to lines/options as remaining possibilities, otherwise door 2 had been opened to reveal the goat there and still hiding the car.
Do you anderstand, that you cannot "switch" here and count that, because "2nd door reveals the goat" did not happen. There are only options left that could reveal the goat behind door 3, and this is not goat, goat, car (with the information of a goat behind door 3, this is eliminated. Likewise, showing a goat behind door 2, completely eliminates goat, car, goat - and so on).
God how we hated Statistics 101, just to get our degrees in Sociology. YOU PROVING MY "theory" MATH MATH MATH...
Ah, maybe YOU can answer my question then ? I'd heard of her before, but "as a mathematical expert ?!"
I'd like to see the BRAINY DOSE VIDEO MAKERS get rid of their mechanical narrators and DO AN INTERVIEW WITH Marilyn..wouldn' tyou ?
Smart? Intuitive? Knowledgeable ? Verbally expressive of her thoughts ? The list could go on, yes ?
Seemed to me "they" were sort of picking on her for not using the "math stats" as they did ?
Oh, of course; now let's hear it for the "Intelligence of Creative Thinking!?"
It seems THERE ARE different "kinds" and "types" of IQ " Tests." Experience and Education , two possibly different types ?
So we should be also asking WHICH IQ Test did Marilyn excel in, or on ?
MISSING FROM the video; Does this lady write and speak in both German and Italian ?I've always felt there is an extreme indication of high intellect in regard to peoples' abilities TO express themselves in foreign languages ?Seems there is a certain "type" or "kind" of logic it seems in learning to "relate " foreign language to one's own ability to speak and write in their native language ? It appears Marilyn 's " (by assumption?) that Marilyn had TWO "Native languages" yes ?
Her opinion of "public schooling" holds great merit. I remember a question required to be asked on a high school test , was "Who were the Phoenicians ?" The ABCD Answers included the answer " Venetian" . Most admitted later that they all misunderstood the word Phoenicians because they were all more "familiar" with Venetian Blinds, than historical terms of peoples and places! (Ha Welcome to American World History 101-we (all) need to repeat that one!?) Which btw lead to my last question (for you or Marilyn, ha ?) Is the inability to "spell" properly (in any language/especially ones native language ) indicate ignorance ?Duh...As a retired teacher, I submit I've become dependent on the Google Gargoyles ' offers for correction, which often just doesn't exist.
The robots tell me I've misspelled something, yet/while, offering no options with which TO correct it.
Good at Questions; Slow at the answers. Anyone ?
"I Ain't no Middleman"
Fred Gold & Lynda Faye
Copyrighted 2016 by LyndaFayeSmusic@gmail.com or Yahoo, if censored for using the word " God" too often?
@@jameslinmd
1/3x1/2=1/6 Pick Car, host shows Goat A
1/3x1/2=1/6 Pick Car, host shows Goat B
1/3 Pick Goat A, host shows Goat B
1/3 Pick Goat B, host shows Goat A
Probability of winning by staying 1/3
Probability of winning by switching 2/3
This is the second time your faulty theory has been debunked.
You can also include all the cases where the host opens a door with the car and the player loses - it makes no difference to the question.
Wasn't Bayes theorem backing this fact the whole time?
Yup.
I am amazed at this! I consider myself a smart person, but I'm not a genius. This made perfect sense after listening to the explanation. I never would've come to that conclusion. The one with 100 doors was clear as day. It's HIGHLY unlikely that you would choose the correct door out of 100 possible choices. After they eliminate 98 doors, it pretty much tells you which door has the car. You would've had to be extremely lucky (or unlucky, as it turns out) to select the correct door out of 100. Brilliant!
You got it
Great. Now note that if you select door #1, switching offers no advantage at all unless the host opens door #2 and there's a goat behind it. Then, if you don't switch, you lose.
Do you understand why this is?
@@klocke5247
Still desperately sticking to one out of 3 equally probable states?!
Poor K Lockeless...
@@klocke5247 In the case of 3 doors, it gives you a very slight advantage if you switch. If you look at the scenarios in this image (3:20) The top three show that you would win 2 out of 3 times if you switch. The bottom three show that you would lose 2 out of 3 times if you stay.
In the case of 100 doors, your odds are 99/100 if you switch. For 3 doors, it's 66.6/100 if you switch. It seems like it shouldn't be that high of a probability with 3 doors. I guess 66.6/33.3 is a lot closer to 50/50. I wouldn't be as confident switching with just 3 doors.
@@BillGraper thank you Bill
I've loved listening to this lady for as long as she's written. She is always inspiring. And she's funny!
We Russians knew of several men and women with IQs higher than 235 to 239 and you boast about her? If I were you I keep them as a group of natural security assets. But one of them, a woman said this, IQ IS NOT EVERYTHING! You may have a higher IQ but still remain a complete utter boop if you did not receive the required proper education to make use of it. In fact, she says, proper educational environments can practically increase the IQ of EVERYBODY, man, womand, and child if we are willing to have the political will to do it!
The confusion people have, is that the door being opened by the host is NOT RANDOM. The host opening the door KNOWS it's NOT the winning door. If the host opened the door randomly then the odds wouldn't change.
Right, and the random opened door might accidentally reveal the car.
Not really. Because if the car was behind door 1 and you chose door 1, then the host could have opened either of the other doors at random. In that case, switching would have caused you to lose. However, that doesn't change the answer to the question. The question simply stipulated if your chances would be better for you to change doors, not if changing doors would guarantee you the win. And the answer is still yes, that technically, logically, and mathematically, switching, even in that specific circumstance, would have increased your chances of winning. For example, if you look at the example the narrator gives with 100 doors. If you just so happen to pick the door with the car as your first choice, and they eliminated 98 of the remaining doors, that other door still would have a higher chance of being the correct one, except for the fact that you had the dumb luck to happen to pick the one door that wouldn't have helped you.
@@TalkingHands308 The host does nothing randomly, he knows the winning door.
@@TalkingHands308 The host doesn't randomly reveal the car if you picked a goat. That's what they meant. This makes your odds after switching 2/3 instead of 1/2 or 1/3. You will still lose if you switch after originally picking the car.
@@CarMoves you've missed the point of my comment...
3:33 Yes, because the host would not open a winning door as in game #3 with door #3 and would have therefore opened door #2, since she had already chosen door #1. And switching to #3 in game 3 is a winner.
This is by far the best explanation of the Monty Hall problem I've encountered, thank you for making this video.
I am so happy about it too. I knew that you could easily write it out but the idea to do it with 100 doors blew my mind. Then without writing it out, it will make more sense to many people.
Probability is a coin does it fall heads or tails?? and may be one in two thousand flips it lands on its edge ?? play roulette on red or green?? sit and wait for a run of 7 red?? then bet on green? logic says it as a good chance of falling on a red??? Yet really every time is a new happening and as a chance of red or green or may be the bankers zero's You just got to feel lucky
It's not an explanation. It's a false spin .
The speed of Light is a very insignificant item. Has little to do with the universe we live in. The pyramids were used for not only instantaneous communications but also for instantaneous physical transportation to other galaxy's. E=MC2 has absolutely no relevance to anything!! Wake up please!! We live in a 100%magnetic universe! The center of a magnet will show you anti gravity, It also will show you how to repel water, we will no longer have a problem going to extreme depths. Do some basic experimentation using the center of a magnet it will blow your mind. However, Keep in mind the magnetic field you cannot use any iron anywhere near, it would disrupt the center field of the magnet. As you know I refer to the center of a magnetic force the "B" field. You can tap glass plastic anything other then iron and it will loose weight. That is exactly why they used copper tools to build the Pyramid's
Please review "CORAL CASTLE Florida on TH-cam> What he did was transmit the "B"field into the ground via his generator that was set up with 25 "V" magnets five deep, all in the repel mode to maximize the "B"field energy about his property and he also had a horizontal stone similar to Stone Henge on his property. Don't waste your time on E=mc2 it has no relevance to anything!! If you take a granite rock just like those used on the pyramids and "VIBRATE the "B" field into same at the correct frequency it will become easy to cut with a copper tool. FACT!!!!
Please get the tools you need to learn the facts.!! Thanks
The Monty Hall problem is flawed. The question poses two completely different scenarios 1 being "If you choose a door #1 and the Host opens 'another door'" ...then it proceeds to add information thus changing the question..."If you choose door #` and then opens the third door" So there's nothing mystical or profound about this quesiton, it's simply flawed. Yes the player might have an advantage with the first scenario as seen by Marylin's Graph, but the Second question, once that third door is opened the problem negates the graph because she's showing the third door not beign opened as a variable xD
If I remember correctly, the way the problem was originally worded in the newspaper didn't have one key condition - the host being required to open a non-winning door. It only said the host opened one of the remaining doors. So her answer was actually wrong for the specific problem, and it really was 50/50. She gave the right answer to the wrong problem.
@@weevil601
Right, because the Host isn't going to open a door with the PRIZE behind it....eyeroll....
@@weevil601 "It doesn't matter if he simply guessed or if he knew where the goat was."
It does matter. If he didn't know where the car was and opened another door that has a goat then it's 50/50.
@@weevil601 If the host does not where the car is and randomly opens another door with a goat then here are all the cases, each of them with the same 1/6 probability.
Pick Car, host shows Goat A
Pick Car, host shows Goat B
Pick Goat A, host shows Goat B
Pick Goat B, host shows Goat A
Probability of winning by staying and switching is equal, 50/50.
@@weevil601 It's 50/50 and the mathematics proves it.
@@weevil601 Even if you perform 99 trials the car would still be randomly placed behind the host's door 33 times, behind the contestant's door 33 times, and behind the door no one picked 33 times.
The answer to the "Monte Hall Question" was known well before Marilyn was asked it.
But the clicks! We must write the clickbait for the clicks!!!
This comment is gold.
@1978ajax No she achieved the Mensa 228 IQ test in a supervised manner (they are always monitored tests). However, she spent a lot of time actually studying IQ tests and sitting them privately which makes you much more versed in the problems and questioning style and can help achieve much higher scores, than just being thrown an IQ test out of the blue as most people are when tested at school etc. You must still read and comprehend the questions however, as they may slip in updates / variance or even completely new problems.
She almost certainly has a photographic / Eidetic memory and can memorize anything.
She writes and publishes books. So for all the massive IQ she has, she hasn't really invented a space warp drive or anything, which a lot of people perhaps expect. That's not how people's intelligence and lives work really.
@1978ajax She's in my Psych text from over 22 years ago... her maiden name was Marilyn Mach ( descended from Ernst Mach, the Physicist ).
yes same as people knew that earth is center of universe. :D ..... To avoid being burned at the stake, the 69-year-old was forced to renounce his belief in a heliocentric model of the universe. And so I see you, Trying to burn those with different oppinion :)
Marilyn has been one of my idols since I was 17... that was 50 years ago. I used to read her column in the paper every Sunday and loved it! She's a national treasure I rank right up there with Dr. Thomas Sowell and John Stoffel.
By reading the comments I thought of my own way of understanding the problem. You have higher chance to pick the wrong door in first question. So when the second question arrives, you have higher chance of winning by changing doors because it is dependent on the chance of you picking the wrong door, and since you have higher chance of picking the wrong door, you have higher chance of winning if you switch doors.
This is it exactly.
You understand it right. Since the winning door is 2/3 likely to be among the 2 doors you didn't pick, and the host eliminates one of them, you should switch and double your chances (1/3→2/3)
That's a good way to think about it.
The professional mathematicians who pointed out her error are correct and vos savant is wrong . She may perhaps be confusing the odds that relate to coin tosses .
Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans.
Ah, I get it now. You pick a door:
Outcome 1: The prize is in the door you picked. So switching would lose the game.
Outcome 2: The prize is in the 2nd door you did not pick. And since monty hall only removes the goat door, switching would win the game.
Outcome 3: The prize is in the 3rd door you did not pick. And since monty hall only removes the goat door, switching would win the game.
Switching has 1/3 chance of losing and 2/3 chance of winning!
this actually does it for me, at first i thought monty hall problem is only logical to switch when you have more than 3 doors
@@tactic2569 well if the total num of doors is 3 + x, with x being more doors added. If monty hall removes 1+x amount of goat doors,
Then the chance of winning is 2+x/3+x and losing is 1/3+x.
Meaning it would be even more logical to switch if more doors are added and monty removes more goat doors.
It still makes absolutely no sense to me. There are two doors after you remove the 1.. q of the two doors has the prize.. I don't get it lol
I got it a different way-- in the 100 door example, you pick a door randomly, then 98 wrong doors are removed. Your door can't be touched, but the right door also can't be touched. Your door was originally chosen at random, so switching doors would be better, since you have more information to make the choice with.
Although this method probably wouldn't work for any other problem and I just didn't know any other way to rationalize it..
Here's an extra thing:
I also thought, "hey, the probabilities just merged into one door! That's cool." Without really understanding why it happened.. :/
@@joeyeddy9344 even if probability of something is 99%, it's still possible that 1% is the final answer.
I remember when she presented this puzzle in the newspaper. I was one of those who was skeptical, so I set up a quicky computer program to simulate it over many trials to see what would happen. I was surprised to find that she was absolutely right! lol
I cannot believe it.
@@cvn6555 Why can't you believe it?
What about intention in this equation. The people at Let's Make A Deal know where the winning door is, so their intention could be to help or to harm. Is that not a variable.
@@joewhite4170 Yes, that is why they only open a door that does NOT have a car behind it when they open one of the doors. ;-)
I just laid it out like at 3:24. It took 2 minutes, and made it apparent why it's better to switch in a concrete and intuitive way.
ok i never got this. i thought it fell under the gamblers fallacy, that the first choice is irrelevant to the second choice. but after thinking about the hundred doors, i get it now. thanks for the video
It's a very easy problem actually. There is a (1/3) chance you pick the right door. In that case you are wrong if you switch. However, there is a (2/3) chance you pick the wrong door and the presenter opens up another wrong door for you. In that case there is a 100 percent chance you pick the right door when you switch. 0+(2/3)*1 is 2/3. I think half the schooled population could figure this out as well.
If the presenter opened up another wrong door you are implying a 3rd chance for the 100% pick. Or more than half of course could figure this out mathematically. There's always someone that says "it's a very easy problem actually". There's only two chances offered and there is no 100% chance.
Are you not shown the goat?
But when you stay with the first door, you are also playing the new odds. So this debate is irrelevant. I'd expect more from this smartest person in the world not to realize this.
@@FathomlessJoy no the odds don’t change at all if you stick with the first door you choose, that’s like asking someone to pick a number between 1 and a million trying to guess the number you’re thinking of and then revealing the 999998 wrong numbers leaving two numbers left and he says ”It’s a 50% chance i choose the right number initially!” when in fact it’s still only a one in a million chance like before.
@@omnipotentpotato2436 You are late to the game, 🤡
Another way to try to grap it with intuition is to think about it the other way around.
Initially the Chance to pick a goat is 2 out of 3 and to pick the car is 1 out of 3
So the chance to pick the wrong door is 2:3. At this time the host has only ONE door left to open, as YOU already picked the one with a goat.
Therefore by switching the door you CONTAIN this 2:3 probability. In 2 out of 3 cases you picked the wrong door and the only one left is the one with the car as the host ALWAYS opens one with a goat.
Or out of three shots you wouldn't put the goat side by side, also understanding how the conditioned brain works.
Unfortunately it's not clear how it's not 50%. Even if I picked the right door the first time (say door 2), the host would have picked the wrong door (say door 3) and offered me door 1 or 2. The same applies if I'd picked door 1. So to me I don't see how the hosts actions increases the probability of the unpicked door having the car.
@@nitinullas it's not 50% because there is a much deeper intelligence at work, you have the social norm mathematical equation that would give door one the better chance. 😉 and then you have the social norm that only sees what's right in front of them. Think deeper or overthinking makes for a much wider probability of intelligence.
Hint: shallow thinkers of black and white only see 50% chance.
@@nitinullas It is quite clear.
1/3 times you pick the car. This means your odds of winning by staying are 1/3. By staying you only win when you initially chose the car.
2/3 times you pick the goat. This means your odds of winning by switching are 2/3. By switching you win by initially choosing a goat and switching to a car.
@@nitinullas Well, the probability does not change. That is part of the reason. And the host itself is not free to do what he wants - he is restricted in his actions as he is only allowed to open a door with a goat. If he would open doors freely and the doors would contain like envelopes with a key or a "you lost" note the probability would not change.
Think about it this way. Your chance to pick the correct door is one out of three. In this case the chance that the car is behind one of the other two doors is two out of three.
This does not change when the host opens a door. The chance you chose the correct door is still one out of three and the chance that the car is behind one of the other two doors is still two out of three BUT the host opened a door with a goat as he is not allowed to open the door with the car. Therefore the chance that the door you did not chose contains the car is two out of three.
The point is. You chose door A. Car can be behind A or B or C.
If it is behind A the host can either open door B or C. If you switch, you lose.
If the car is behind door B the host can ONLY open door C. If you switch to B you win.
If the car is behind door C the host can ONLY open door B. If you switch to C you win.
Even in this small example you can already see that you win in two of the three cases if you switch.
The trick, for me, to intuitively understand that she was right is the realization that, when you first picked a door, the odds were against you. You only had a 1 in 3 chance of picking the correct door - probabilistically speaking, you were more likely to choose the wrong door than you were to choose the correct door and so, when more information is received, narrowing your ALTERNATE choices down to only one, you should probably switch (because you were more likely wrong than right the first time). That's it, that's the whole thing (the probability of your first door pick being incorrect).
...
Not convinced yet? Here's the real kicker, the thing that "bends the odds" in your simple choice of two doors at the end. Because you have three doors to choose from up front, in only 1 out of 3 play outs of this game (probablistically speaking) are you going to be in the scenario where you picked the door with the prize and it doesn't matter at all what door the game host opens because they both are going to have goats behind them. The other 2 out of 3 times, you're going to be in the scenario where you didn't pick the correct door at first, and so it ABSOLUTELY MATTERS what door the game hosts leaves closed. 2 out 3 times, what the game host does provides new and relevant information, 2 out of three times the door they leave closed will be the door with the prize. This is the real world world explanation for why Savant's math is probablistically right.
It took me longer than I care to admit to understand all of this. Intuitively, the problem as first stated, is hard because I think there is something inherently "real" to the belief that you have a 50% chance of being right or wrong at the final choice. What's "real" about that instinct is YOUR LEVEL OF "REAL" KNOWLEDGE when the choice has been narrowed down to only two doors. At that moment, you do not know what actually happened when you picked your first door. You don't know if you beat the odds or if you didn't. Also, the fact that you can now choose a new door kind of changes everything (you can argue it sets up a "new coin flip" and a new probability). As always, probability is best observed at scale. When you only get one shot at this game of three doors and a single prize (not 100 chances for 100 prizes in 300 doors, or as the video suggests 1 prize in 100 doors), you're, seemingly, in the same position as someone who has just flipped heads on a coin 9 times in a row. Regardless of those previous outcomes, what happens next is a 50/50 chance.
All that said, she is still absolutely right and the math is sound. Your first choice of door was made at an odds of 1/3 (not 1/2). It, probablistically speaking, is slightly less likely to have the prize than the door that survived elimination by the game host.
"Your first choice of door was made at an odds of 1/3 (not 1/2). It, probablistically speaking, is slightly less likely to have the prize than the door that survived elimination by the game host."
I figured this was a very easy thing to work out (?). First choice most likely wrong, one correct choice left, better to switch. Logical no?
@@SicketMog No, not logical to me at all. You pick the first door with a one in three chance...the second door is also sitting there with a one in three chance. The third door is opened and there is a goat behind it so, with no information to go on with the other two doors, they are equally apt to have a car. I don't get why it is door two that gets the extra kick, you have no information on either door, you can just as easily say door number one gets the kick.
@@mercurysmith563 When you first picked with no information you had only a 33% chance of being right. You are most likely wrong on your first guess; THAT's the information you have to go on.
Yes you COULD have picked the correct door from the beginning, could happen (if so you were ironically unlucky), but there's a smaller chance of that having happened (33% to 66% chance being wrong). It's about improving the odds not wether or not you ACTUALLY win.
To me this is super logical, it was when I first heard of the problem as a dumb kid too (without decent math), but maybe it isn't? 🤷♂
O
@mercurysmith563
Maybe it would help if you actually "played the game" at home with 3 doors in your residence. Under this scenario always make Door 1 your first pick. Run through the 6 scenarios and compare how often you win if you switch to how often you win if you don't. You will find that you will win twice if you switch and only once if you don't. If those aren't your results - you have a flaw in the way you've created the scenario.
If switching should be good you need to take the wrong door first. Only 1 in 3 are the car. So 2/3 are wrong. I understod this when I was 16.
My other posts were wrong . I finally found an explanation that made me see that switching is better. When you first choose you have a one in three chance of picking the car, 2 in 3 chance of picking a goat. If you picked a goat door the host opens the other goat door meaning that no matter which goat door you picked you win by switching. Picking the car door first and switching is the only way you lose by switching. This works in all three possible scenarios. Thanks to Bai Su Zhen for the helpful post. It's really just logic but you have to find the right way to look at it.
i finally manage to understood this problem thanks to you.
is this how you interpreted it?
"i think most people that are like me, got confused by this, is because nobody state the underlying rule.
- that the host would always open the goat door.
through this rule, i manage to understood that the only way to loose is by picking the car first, then switching to the remaining goat door after the reveal, which is 33% compared to picking the goat door first which is 66%, have the host eliminate the other goat door and then switching to the car door.
now if you didn't switch at all, then it would remain 33% each (as if making the choice without taking note of the underlying rule). which is why the video tries to explain that information affects the probability of your choices."
So if it was "if you can switch to a door that hide a different thing from your door whatever it is", I think everyone would get it right. It is mesmerising to think about
tell me what formula you used to pick the first door you do not know what is behind any door. The host knows were all the goats are you do not . If you pick 1 door out of 3 doors you have 33.inf % chance to be right. If the host shows you were one goat is he has removed that door from the equation. You now have 2 doors to pick not the same equation you had 1 choice with 3 doors n ow you have 1 choice with 2 doors 50% chance.
Isn’t that implying the host will only reveal a goat door if you chose a goat door initially?
The host is to open a goat door regardless if you chose a car door or a goat door. So how does one have a higher percentage considering they are unaware if they chose the goat and the host will open a goat door regardless.
@@emptyempty4238 Because you had a 2/3 chance of picking a goat originally, and if you did, switching guarantees a car. (and if you didn't, switching guarantees a goat)
The way I always thought about the problem was this: choosing the right door out of three, you’ll be wrong two times out of three so, two times out of three, you’d better better off switching to one of the other doors. If neither of the other doors is opened, you have no idea which one to switch to. The host takes care of that problem and shows you which other door it _isn’t_ - if you’ve chosen Door 1, he’ll open Door 2 if the right door is Door 3 and Door 3 if the right door is Door 2. So in those two-thirds of the time when you’ve chosen the wrong door, you’re taken care of and you’re better off switching. (And, of course, if Door 1 is the right one, he’ll open _either_ Door 2 or Door 3 and you’re better off _not_ switching but you have no way of knowing that.)
If the correct door is the one you've already chosen, the host could choose either of the other two.
Exactly (you have no way of knowing) which renders this whole exercise pointless. When you pick your initial door and the host removes one with a goat, your chances are now 1/2. Since no new information was presented. You can consider the the one the host removed was never there in the 1st place.
@@DMZRPG “When you pick your initial door and the host removes one with a goat, your chances are now 1/2. Since no new information was presented.”
That’s false. Again, if you choose _any_ door, that initial choice will be incorrect two times out of three on average. If no other door were revealed, you would not know _which_ of the _other_ two doors was the right one. But, by the terms of the problem, the host will _always_ open a door that _isn’t_ the right one so, in those two out of three times, the right door will be the one he _doesn’t_ open and you’re better off switching. So, if you choose, say, Door A and the right door is _either_ Door B or Door C (again, two out of three times), the host will _always_ open the _other_ door (Door C if the right door is B, and Door B if the right door is C). That’s new information. Switching leads to picking the right door in those two out of three times. If you happen to pick the right door initially (one out of three times), the host can (and will) open either of the other doors and switching will result in you choosing the wrong door. But that occurs in, again, only one out of three times on average.
@@DMZRPG thats what Im thinking. The odds reset when one door is eliminated. There are no longer 3 doors, there are only 2 doors. Why do the odds carry on ? In my view it becomes a new probability.
@@ScreamingEagleFTW “The odds reset when one door is eliminated.”
They don’t. Again, when you have three doors, whether you’re dealing with the Monty Hall problem or not, your initial guess is, on average, wrong two out of three times. In those two out three times, if no other doors are opened, you don’t know _which_ of the two doors you didn’t pick is the correct one. In the Monty Hall problem, in those two out of three times, the host, in effect, _tells_ you by opening the _other_ incorrect door-you’ve chosen (incorrect) Door 1; with _either_ Door 2 _or_ Door 3 being correct, the host will open the other one. (By the terms of the problem, he _has to_ because he _has_ to reveal one incorrect door-and, again, two out of three times, you’ll have already chosen the other one.) So, again, if you’ve initially chosen the wrong door, which happens, on average, two out of three times, switching to the _other_ unopened door gives you the right one.
You _don’t know,_ of course, _if_ you’re in the situation where you’ve initially chosen the right door (on average, one out of three times) or in the situation where you’ve initially chosen the wrong one (on average, two out of three times) but the odds remain the same, regardless. (If you knew that-there _could_ be a “foolproof” version of the game where the host opens one of the doors you didn’t choose and then says “And, BTW, your choice was wrong (or right)”-you could win 100% of the time: switch to the _other_ unopened door if you initially chose the wrong one, which would happen, on average, two-thirds of the time, and stick with the one you initially chose if that was the right one, which would happen, on average, one-third of the time.)
I actually read the Parade article. I had 5 classes of math in high school and one chapter was on probability. One thing that stuck in my mind was 'All RANDOM choices must add to 1.' So if I had a 1 in 3 chance the first round, and Monty's choices are not random, then my second choice is actually a 2 in 3 chance.
Yes but you can choose to choose the same door
@@MrSanford65 No, you already chose that door and it had a 1 in 3 chance, not choosing the other door means you still only have a 1 in 3 chance. It only changes to 2 in 3 if you switch doors.
The table is flawed logic because of the labeling of the doors. And that's why it becomes so confusing. Only the first two entries of the table are valid. The rest of the entries are actually just permutations of the first two.
Imagine the doors had no numbers or arrangements and you didn't even know which one you picked, only that you picked one (of three). Now you were told, you thought you picked one of three, but, it's been reduced to one of two. Would you like to change doors?
There is literally no sense in changing.
I'm saying that I look at it this way: Instead of picking door 1 of three...
I pick one door out of three.
This is where the table becomes redundant. The first two entries are the only useful ones.
After studying it for about an hour, I see that I am wrong. I have to say that is very counter intuitive. Thank you for posting... 👍
After the first door is opened, the problem is changed. The remaining doors have a 50% chance of having the car.
But when you chose the first door, it had a 1/3 chance of having the car. So you get to 50% when you choose the other door.
Let's say you buy a lottery (1,000,000 numbers with 1 winning number).
After the result is out (you don't know the result yet), your friend (who know the winning number) say your number and another number. One of them is the winning number.
What is your winning chance?
Matt: I pick the lottery number 000000. My friend revealed that all the other numbers but 415896 are loser numbers.
Matt: There are only 2 options left: My number is the winner number or the other number is the winner number.
Matt: Those 999,998 choices are gone. At this point, my number has a 50% chance to be a winner number.
Matt: When my friend remove 999,998 loser numbers, he makes my winning chance becomes 50%.
Matt: This means my winning chance will be 50% no matter which ticket/number I bought.
Matt: I just found an easy way to win lottery. I am a genius. Hoooraaay!!!
Matt's parents: Hoooraaay!!!
Oh wait. Where the hell is Where the hell is Mat-mz7kk?
@@AraqiusMat from all 4 therory channels retired.
5:11 Yes, information matters. I believe the misconception comes from the idea that the host is working against your interest. Subconsciously you may think he will only open a door in order to confuse you and is only offering you to switch, because he knows you have chosen the car. The way the problem is communicated, you are offered a switch, you are not entitled to one. So game theory kicks in and your brain pushes you to refuse the switch, overriding your base math logic in order to protect you from fraud - a mechanism we have evolved to trust :)
Is more complex than that , psychological game he already wants you to keep number one making you think he's asking you and wants you to switch is obviously asking to switch and pushing you to think you have a winner on one , the obvious game here is to switch
@@makalribera6742 It all depends on how often he is working against you and how often he is working in your favor. You can add that to the equation and figure out how the numbers add up :)
And another level to this that no one has said yet is the unspoken rule of the host not to open the first door if it had a goat. If the host could do that, then the game would reset immediately and it would be a 50-50%. And that's unspoken, just a structure of the show
no, the problem is that it is a hypothetical
and remains such until tested in reality
@@shaft9000 None of that matters. This is a straight mathematical problem, which does have a strange answer thanks to the weird world of probabilities. Mind you, picking door 2 gives you a higher probability of being right, but not a guarantee. There is still a 33% probability it could be behind door 1. But the odds are in your favour with door 2, as she said.
The world always fails to recognize the real heroes.
Always?
She did nothing bro, she wasted her life
Be not misled by megalomanic schizophrenics who in their grandiosity associate themselves with these levels of intelligence based on delusional claims regarding ability, achievement, or test performance. This helps them write books, become famous and make money. Be assured that no one has ever come close to I.Q. 230, that the highest I.Q. to be realistically expected on Earth is about 185, and that only through tentative challenging norms in unexplored score ranges someone will one day qualify.
The real heroes are not those that are born lucky beiy insanely smart. The real heroes are those that use their intelligence (no matter how smart they are) and make the world a better place.
...as you are doing right now. Intelligence without wisdom is like a boat without water. And intelligence without wisdom enables one to outsmart one's common sense. In this case, with MATH. If the car is behind door number one or number two has absolutely nothing to do with a goat being behind door number three, no matter how many math wizards look at the issue. And the 100 door example is a completely different scenario. It doesn't take a high IQ or math wizard to know that. But it does take common sense. GOD's prophet has spoken.
Basically you switch because the door you picked could’ve also been eliminated by the host if you didn’t pick it but the door that remained closed would never, so the unopened door is the best choice
@Don't Fear The Reaper that’s what I said
The better answer is simply that since every door has a 33% chance, thus those 2 other doors have a 67% chance of the car being behind one of them. And if you switch, you will get the prize in every game where the car happens to be under one of those 2 other doors. So switching gives a 67% chance
The real intriguing question is one of why this very simple problem stumps over 90% of people at first glance. But it certainly does
Here’s how ChatGPT explained the Monty Hall Problem:
The Marilyn vos Savant question refers to a famous logic problem she addressed in her "Ask Marilyn" column, specifically the Monty Hall Problem, which generated significant public debate.
The Problem:
You're on a game show with three doors:
1. Behind one door is a car (the prize).
2. Behind the other two doors are goats (not a prize).
You choose a door. The host (Monty Hall), who knows what is behind each door:
1. Opens one of the other two doors, revealing a goat.
2. Asks if you'd like to switch your choice to the remaining unopened door or stick with your original choice.
The question is: Should you switch, stick, or does it not matter?
---
Marilyn's Answer:
She advised always switching, because it increases the probability of winning the car from 1/3 (if you stick) to 2/3 (if you switch).
---
Explanation:
1. When you pick a door initially, the probability of choosing the car is 1/3, and the probability of choosing a goat is 2/3.
2. If you stick with your original choice, you retain the 1/3 chance of winning the car.
3. If you switch, you win the car in the cases where your original choice was a goat (2/3 probability).
Monty's action of revealing a goat doesn't change the initial probabilities but shifts the odds in favor of switching.
---
Controversy:
Many people, including mathematicians, initially disagreed, believing the odds were 50/50 after Monty reveals a goat. However, mathematical simulations and proofs consistently supported Marilyn's explanation.
I was skeptical until I wrote this: There are 3 options:
I select a car, then the host reveals a goat and the remaining door must contain a goat.
I select a goat, then the host reveals another goat and the remaining door must contain a car.
I select a goat, then the host reveals another goat and the remaining door must contain a car.
In 2 of 3 cases the remaining door contains a car... so I should switch to it.
Correct.
She cheated. Door 3 cannot be the car, so games 3 and 6 cannot be indicative of increased probability by switching.
@@polygontower Not after it is shown to be not have a car.
Now do the same, but assume you had selected door 2 on the beginning...
@@kelseysmart9472 the point of the game is that the host is eliminating one of the doors from the equation. From there you are choosing to switch your answer to the remaining door or not. Wether that is door 2 or 3 I do not believe matters.
I think a more intuitive way of understanding this problem is recognising how the question has changed from the first choice when offered the second choice.
The chance of being correct (selecting the car) on the second choice is dependent on being incorrect on the first choice (having selected a either goat). Therefore, the chance of selecting the car in the second choice is the chance of NOT selecting the car in the first choice, which is the chance of selecting a goat on the first choice. This is 2/3.
A very logical explanation. Thank you.
No, you have the choice to keep your door, or pick the other one = 50%.
@@R391s And the odds of your door having the car are 1/3 and the odds of host's door having the car are 2/3.
@@R391s Let’s run through the different scenarios/outcomes (allowing for where the car is and whether you choose to stay or change your answer):
Car behind door 1 and you keep your 1st answer
- Choose Door 1. Shows Door 2 or 3. You stay at 1. Therefore: WIN
- Choose Door 2. Shows Door 3. You stay at 2. Therefore: LOSE
- Choose Door 3. Shows Door 2. You stay at 3. Therefore: LOSE
Car behind door 1 and you change your 1st answer
- Choose Door 1. Shows Door 2 or 3. You change to 3 or 2. Therefore: LOSE
- Choose Door 2. Shows Door 3. You move to 1. Therefore: WIN
- Choose Door 3. Shows Door 2. You move to 1. Therefore: WIN.
The same outcomes will occur if the Car was behind door 2 or door 3.
You may notice that if you stay with you answer you LOSE twice and WIN once. Therefore, your probability remains at 33,33% (as before the door was revealed).
However, if you change your answer you WIN twice and LOSE once. Therefore, your probability of choosing the car goes up to 66,67% when you use the strategy of changing your answer.
Showing the above by force is not ideal, but it illustrates it visually very well (imo). However, I would implore you to consider looking at the Bayesian Statistics proof or running a simulation (with Python for example) if you don’t trust the above at face value.
the chance is rather illusion, if someone else chooses the other door then what happens
It's easy af, by switching you pick 2 doors, the wrong one and the unknown one, that's 2/3 or 66%. If you don't switch you pick only 1 unknown door, that's 1/3 or 33%.
Some people believe it's always 50% but you have to start counting your propability from the very beginning when there were 3 doors, not after one was openend. Because the first choice also matters.
disagree. You flip a coin, neglecting the edge probability, you have either tails or heads. When you flip you always have 50% chance probability of landing either heads or tails. However, flipping 50 heads in a row is extremely unlikely and thus getting a 51st head in a row is, too, extremely unlikely, despite having only 2 outcomes.
Lavant's high IQ is garbage and false or she cheated on her test. IQ test measures several things: verbal, reasoning, memory, processing speed. Yes, those 4 combine into 1 IQ score, still, I would love to see her true published IQ test. Until then I call bullshit.
@@vkaa3k190 Chance for any of the doors to have a car is 1/3. Then the host is opening one door. But host is not allowed to open your door -> your door is a separate group of propabilities -> the door that is revealed transfer its probability to the other door in its separate group...
Or
You make decision to stick with your initail choice(when all 3 doors were unknown) = you are ignoring that extra information provided to you by the host so your chances do not change...
@@vkaa3k190Mate, this problem has been resolved by mathematicians, not just Vos Savant, since the 90s... you're making a right fool of yourself here.
@@fleurdemaiko lots of mathematicians have called out her bs, so you are the one making a n**ger out of yourself. And you don't have to have a PhD in math to figure out the probabilities.
@@vkaa3k190 The fact that it is astronomically unlikely that you might flip 50 heads on a row is irrelevant to the MHP.
But if you did that , the 51st flip has a 1 in 2 chance, The past, in this instance has exactly nothing to do with the future
If you stick to the first choice, the chances are 33%, opening other doors cannot change that fact. If you switch, you get the car if it is behind either of the 2 other doors, which happens 67% of the games.
Huh, interesting. I didn't know about her. She married Jarvik, of artificial heart fame, and once publicly questioned Wiles for relying on hyperbolic geometry for Fermat's last theorem.
The key here is that we know that the host will always open a door with a goat. Because if the host picks one of the doors at random that happens to be a goat, the odds will be 50/50. But as stated, the host will always open a door with a goat and then it’s is better to switch. The easiest ways to see this is that if we pick door #1 and then switch, we will win if the prize was behind door #2 or #3, while by not switching we only win if the prize is behind door #1.
"Another way to look at it is from Monty's perspective:
Monty can only open a door that you didn't pick, and he can only open a door that has a Zonk behind it.
There's a 1 in 3 chance that you picked the car. So there's a 2 in 3 chance that you picked a Zonk. So 2 out of 3 times, Monty has only one door that he's allowed to open. Because one is your Zonk and the other is the car." - @motorsheep
Wrong. Nearly everyone gets this wrong. Let me explain the true flaw in the Monty Hall problem. There's an implicit assumption that few seem to notice, and that's that the host is required to offer a door switch. What if the offer is made only when the host knows that the contestant has chosen the prize? If this is so, then it's best not to switch. Because we don't know the rule the host is using, or the odds of the host adhering to any rule, we cannot say whether it's best to switch or not. Sorry Marilyn, but you blew this one, but you're hardly alone.
@@mrwilkdfly are you familiar with the saying in physics, “let’s assume the cow is a square”? The point is that you are always working within a model that is not reality. Here it is the same, you can bring up any number of valid points that show the model is not a perfect representation of reality, but that is not the point. In the original problem it was the assumption that always a goat door is opened by the host, that creates an interesting problem that many people got wrong as they thought it was 50/50 to switch.
@@mrwilkdfly Your math skills are really poor. If it's not in the question then it's not in the equation either. Math problems are solved with the information that is there, and not from the imagination coming out of your arse.
@@joseywales898 There is no reason to consider a motive one way or the other. So it's not in the equation.
An example showing that changing doors increase your chances:
Door 1: goat
Door 2: goat
Door 3: car
You chose 1, it’s wrong, if you changed you would be on door 3, win.
You chose 2, it’s wrong, if you changed you would be on door 3, win.
You chose 3, it’s correct, if you changed you would be on door 1 or 2, loss.
Here you can see that by keeping the first choice you have 1 in 3 chance, and by changing you have 2 in 3 chance.
Nice
Player chooses Door 3
Door Number 2 is eliminated
Player switches to Door 1 because "sTaTiStIcS"
Player facepalms
What if you pick 3 in the first place? Its a bit confusing. Maybe because I just got introduced to the problem. Maybe I will get some clarity in a few days
@@junaidulislam6170 Since the host knows what is behind all doors he/she would have opened the other door which had a goat behind it. Same result.
Incorrect, he opens one door showing the loser. Your remaining choice is between two doors each with an equal chance of being the winner.
I look at it this way. There are only three possibilities:
1- you pick the car, so the host shows one of the 2 goats - then you should not switch door
2- you pick goat 1, so the host shows goat 2 - you must switch door
3 - you pick goat 2, so the host shows goat 1 - you must switch door
Therefore there is 2 out of 3 chance that switching your door choice will get you the car
@@user-ej9nl1ng9d lol - same principle
@@user-ej9nl1ng9d I know - I was just putting it simple
@@user-ej9nl1ng9dMost of the people who argue about it can't read a table. Many can't even follow John's simple logic. The biggest problem is getting them to realize 50/50 is wrong or they don't even listen to explanations of the right answer.
thank you - that is a good explanation
the key is that *the host always eliminates a losing choice*
the odds that the original choice was correct don't improve but the odds that the remaining non-eliminated door is correct do increase
it is easier to grasp for most people if the door count is say 10 or 20 or some other larger number
if all the *wrong* choices are eliminated except the initial choice (which may or may not be wrong) - plus one other last remaining choice - there is a very high probability that the correct door is the one remaining door (the one not eliminated)
Best explanation
The easiest explanation is: your 1st choice has a probability of 1/3. The revealed door a probability of 0/3. Therefore the final door a probability of 2/3. Is that easy or what?
It goes back to probability. Based on your information, what is the probability that door number one will be the correct one. Probably zero. It also highly depends on your approach too, whether it's thinking of the goats being animals not wanting to make too much noise Near each other, or maybe it's the car, keeping it further away from the goats because goats can be destructive. So keeping the car safe is most important, giving you the answer straight away.
As a french native. "Savant" means someone who is educated and smart, even someone like a professor or a researcher. Or something that is highly intellectual and sophisticated.
The equivalence of "learn" would be "apprendre". Savoir, savant are more an equivalence of "knowing"
There are two ways to say the word learned which have different meanings. One is past tense for learn. The other means knowledgeable. Savant does mean learned but the narrator said the word incorrectly.
Luckily I know nothing at all 🤷♂️
@@Demigodess42 This is twice wrong. First, "Savoir" is to know, not to learn (which is apprendre). Second, there is no "savant" conjugaison of "savoir "(the closest would be its present participle, "sachant"). Savant is always a substantive, which I would translate loosely as scholar. I am also a French native.
@@matthieutricottet You haven't understood the difference between the two pronunciations of 'learned'. If pronounced 'lernd' it is the past tense of learn, but if pronounced 'lern-id' it means educated or knowledgeable.
@@jdb47games OK, now I read again your comment and I understand what you meant. Having heard "learnd" in the video, I did not make the connection with the substantive learned (which actually I knew), and totally misread your remark, sorry about that.
I believe one of the circles of hell in Dante's inferno is trying to explain the Monty Hall problem to people who can't understand it and who smugly insult you for saying the correct answer
A fool takes no pleasure in understanding, but only in expressing his opinion. Proverbs 18:2
Have the person trying to understand the problem play the role of Monty Hall using playing cards or something with someone who will always switch. They should soon realize that Monty Hall will have the car 2/3 of the time.
Host is not allowed to interact with door that was initialy selected that is why its chance for having a car do not change.
So you ending with tree doors one with 1/3 chance(the one selected when no information about one of the doors was known), one with zero chance as the host is allowed to open only the door whithout car and the third door takes the whole propability of the group of doors that host is allowed to interact with => 2/3.
I think there's a simpler way of explaining this puzzle. If you choose the correct door and switch you will be wrong. If you choose the incorrect door and switch you will be right ( because the other wrong door gets eliminated), and you choose the wrong door 2/3 times
Physically actually simulate it.
She WAS wrong. The probability of doors 1 and 2 is IDENTICAL.
@@nnnnnn3647
Yes,
The car cant move based on a choice.
Nothing can increase the chances of the car changing doors
Still a 50/50 chance
@@nnnnnn3647 It's pretty easy to test it for yourself. Get someone to put a matchstick in one of three boxes. Then choose. Do it about a hundred times. Then see how many times you chose the wrong box. It should be about 67 times. So on each of those times if you'd then gone on to play the game, your odds of winning if you stick are only about 33%, but if you switch, your odds will be about 67%. Not 50%, because it's more likely that you will already have made the wrong choice. So 66/33, not 50/50. Your previous decision has already shifted the odds away from an even chance.
@@owenorders5202 the problem is that you only play once and you have 1/3 to choose from.
If the game consisted of 100 draws, the changing door strategy would be better.
Think of the door question this way. When you pick a door, there is a 1 in 3 chance that you are right. That means that there is a 66.66% chance that you are wrong. In other words, you're PROBABLY wrong. When another door is opened to reveal a goat, you now now that it's in your best interest to switch, since you were probably wrong from the beginning. Since your first choice was probably wrong, and you've been shown another door with a goat behind it, that means that switching doors would make you, probably right.
The reason switching is advantageous is due to the host knowing where everything is and not of the original probability of picking the car.
My explanation for the Monty Hall problem: Suppose Monty does NOT open one of the doors you didn't pick, but instead offers that you can swap your door for both the others. Of course you swap your 1/3 chance for a 2/3 chance. Now one of them must have a goat - so what?. Monty just showed you that one, and you take the other.
This sounds interesting. Somebody with an high IQ should read that
Well if the door was not opened essentially yes, but door 3 is opened, no one wants the goat, so 2/3 of a chance, but why would it be a better chance for 2 to be the door? Between 1 and 2 is a 50/50 shot. Both being 2/3 chance
@@anthonyharper4877 Swapping for both of the other doors, will always give you a better chance than staying with just the one - that's beyond question.
The only differences between how the game is normally played and that are:-
1/ You are shown a goat (which must be behind at least one of your two doors)
2/ You don't have to take that goat home, just the other door's content (hopefully the car)
@bertoid as it is stated, the goat is behind door number three, I'm implementing a conclusion that 1 and 2 has equal chances, switching or not does Increase the chances since 3 is not what we would chosen for the car, sure switching and there is the said car, but also the chances of switching and there's not to be.
Clearly the biggest thing people can't wrap their heads around is why the fact that Monty knows which door is right, and will never obviously open that one makes a difference. People are seeing Monty's choice as random (which would make it not any better to switch) when it isn't.
I think I just figured out a unique concise way of explaining this that will be more intuitive for people. If you start out with the wrong door (which is more likely than not), then there is no other wrong door to switch to because the host has opened it.
Actually there's an even more concise way of explaining this - If you start out with the wrong door, you will always switch to the prize :).
@@dan20020 You don't know if you picked the wrong door but it is pretty easy to see that, with two goats and one car, you are twice as likely to pick a goat door as you are the car door with your first pick.
@@dan20020 Sorry, not true. That's like saying a coin flip is not 50/50 unless you do it multiple times. Even in a single event, probabilities matter.
@@dan20020 It is all about the way you approach the problem... stop thinking about the second choice as a new 2-door game, that is not correct and it will be obvious once you understand it.
B and C are grouped together because they are both the host's doors. This game is all about the contestant's first pick splitting the doors into two groups, the contestant's door with 1/3 probability and the host's two doors with 2/3 probability. Would you switch your one door for the host's two doors with both of them still closed?
@@dan20020 look at the 100 door example.
Is the prize more likely to be behind your door? Or the other 99.
With three, is the prize more likely to be behind your door, or the other two?
It comes down to this: One of these three doors has a prize behind it. Do you want to pick one door or two doors?
And if you want to pick two doors but are only given the opportunity to choose one, switch when the host gives you the option after showing the empty door (which is just the empty side of the 2/3 chance).
unless he only asks to switch if you are on the prize door.
@@eymanbaird5847 You don’t know if you have the prize door 🤦🏻♀️
2011CaRe that is a very good way to describe the problem. That should be a very effective way to get stubborn people to change their minds about insisting there’s still a 50-50 chance. However, I think you would still have trouble convincing them that your description is accurate. Many people will bend over backwards to rationalize that what you’re saying isn’t true.
So I would recommend starting with the statement “the host has no choice which door he is going to open.” Then say the host must open a door that has a goat. And then say, if both of the doors that the host may open have a goat, it doesn’t matter which one the host chooses. If the person you were trying to convince doesn’t agree with all of this, then you cannot reason with them. If you both have the same understanding, then you can ask them your rhetorical question.
If I were a college professor, I would give a lecture like this : there is one prize behind one of three doors. The other doors have a ‘goat’. I’m going to let one student ‘own’ one door and I’m going to give both of the other doors to another student. I’m sure you all agree that the student with two doors is twice as likely to get the prize, but won’t necessarily get it. OK, now I’m going to open one of the two doors that the second student controls, revealing that it does not have the prize. Please carefully take note that it doesn’t matter which door the prize is actually behind . There is always at least one goat available for me to choose.
Mr. first student, would you like to change places with the second student now?
What? You made it more confusing.
@@rhysobrien2439 To some, I’m sure
If you have 100 doors the probability of picking the car is 1/100. The host opens 98 doors all goats leaving one closed door asking you if you want to switch. Obviously you want because now you have 99/100 of winning. It reverses the probability. 3 is the least amount of doors you need to for this to work.
I never got the Monty Hall problem, because it seems like your odds go from 1-in-3 to 1-in-2... why switch? It's 50/50, right? So I got some paper and dice out, and using a 6-sided-die, I assigned a roll of a 1 or 2 to door number one, and so on.
-
The point of the dice was to eliminate my own knowledge, because since I was playing both the 'contestant' as well as the 'game-show-host', I always knew where the prize was. Anyway, a little ways into this, it was true that switching improved the odds of winning.... but I still couldn't figure out why. Then I figured it out: it was because the 'game-show-host' *never eliminates the grand-prize* -- so it is a *biased* decision, and *not random.*
-
Anyway, hopefully you guys reading this are able to finally grasp *why* this problem works the way that it does, because it took me a long time and some actual sit-down-and-experiment both to verify it, as well as see why it worked.
It's actually just nunber cranking. The true response would depend on the behaviour of the host. If he always does that, then it makes no difference, if not, and he has a bias towards doing that when you choose right, then you should switch. You can crank numbers on hipothetical number world, but that doesn't correlate to the real world.
@@arthurrosa9403 If the host always opens a door with a goat then there is a 2/3 chance of winning by switching.
It's still 50/50 no matter what. I don't see the big deal here, why is everyone making this hard??
@@klaus7443 That's not true. In the real world. . . The car is in the same door. Assuming switching increases your odds when the car is in the same place hold no sense in the physical world. This is a fun abstract logic exercise, but it holds no real world logic. Just like you can prove different numbers are the same if you crank the math up. But It won't make you Jesus in the real world.
@@arthurrosa9403 Your mathematics is terrible, If the contestant has a 2/3 of picking a door with a goat and the host always picks a door with a goat then there is a 2/3x1=2/3 chance that they both did.
Edit: Adding shorter explanation to the top:
In the mentioned scenario, when you initially pick one from 10 doors, you have a 1/10 or 10% chance of success. None of the variables that are changed are ones that would affect your chances of picking the right door the first time, and unlike the rest of the doors which could be removed, the door you picked had no chance of being removed even if it was wrong. Since the correct door hasn't been randomly shuffled again between the two remaining doors, in order for you to be right the second time by picking the same door you did the first time, you would have had to have been already been right the first time before any doors were removed, and that only had a 10% chance of success.
For those odds to be able to change, it would have to be possible for a different door to become correct than at first, or your door would need equal chance of being removed compared to the rest of the doors, instead of being guaranteed to still be there even when it's wrong; or you'd have to have extra information available to influence your first choice.
If only the correct door and one random door were guaranteed to be there, and your first choice was not guaranteed to stay, then your door would be gone 80% of the time just like the other doors, it would be right 10% of the time, and it would be wrong 10% of the time; then in that scenario, if your door was still there after you picked it the first time, then it would have a 1/2 chance of success, but even that would be 10% of the total, because 80% of the time the door you picked was gone. However, since the door you picked can't be removed, it's still there as an option all of the times that it was the wrong door.
Original comment:
I spent some time pondering this, and at first I still didn't believe it, but then I thought about it. If there were 3 doors, you only have a 1/3 chance of picking the desired door, and even after one door is eliminated, that still doesn't change the fact that you had a 1/3 chance of picking the right door. And you are right that using more doors in an example makes it more obvious.
Your door is no less likely to be unsuccessful than it was before, because your door was ignored only because it is the door you chose, whether it is successful or not. However the other door was only left available either because it contains success, or, on the 1/3 chance that you already picked the right door, then the other door was randomly left available merely to have an extra door available.
For people who still aren't convinced, think about it from these two different realities of the same scenario:
Example: Without you seeing, a person puts a coin under one cup out of 10. You choose a cup, but instead of revealing it, the person lifts 8 cups that as the dealer they knew were empty, leaving two remaining cups, including the one you had chosen, and they do not move the cups around (because doors don't move) so you still know which one was your first choice, and which one wasn't.
Reality A: The cup you chose has a 1/10 chance of hiding a coin, and the other remaining cup has a 9/10 chance of hiding a coin. This sounds weird to you.
Reality B: Both of the remaining cups have a 1/2 chance of hiding a coin. If this were true, that would mean that the cup you originally chose when there were 10 to choose from, would have the coin, 50% of the time. But revealing the other cups AFTER you made your decision does not increase the odds of your original decision being correct. If you repeat this experiment 100 times, it is highly unlikely that your first pick would be the cup with the coin at least 50 times out of a hundred. See how Reality A makes more sense now?
What if the dealer is just trying to bate you because they know you did pick the right door/cup & they are thinking "Oh no!"
@@amyflana With 3 doors, there is only a 1/3 chance that you originally picked the right door. With 10 doors, there is only a 1/10 chance that you originally picked the right door. With 100 doors, there is only a 1/100 chance that you originally picked the right door. So it is more likely that the right door is the other remaining door.
If the dealer was sabotaging people by only revealing doors when people pick the right door to scare them, then before long people would catch onto that, and it wouldn't work. Any decision-making on their part would only turn into clues for people in the future. In order to not make it obvious, they have to be consistent and reveal doors regardless of whether or not the person picked the right door to begin with.
Sure, there's a chance you picked the right door and then switch to the wrong one. It's just much more likely that you picked the wrong one.
I still don't get it. The probability changes as the number of options changes. As the number of choices decreases, so too does the number of possible outcomes.
@@SeanPAllen Yes, but you made your choice before the amount of options changed. With ten doors, and zero information to go on to help you determine which door hides what you want, you have a 1/10 chance of choosing the correct door. What happens in the future doesn't change what the odds were when you made your decision.
Think about an example with 100 doors. You pick one door, and 98 other doors are revealed, but the door you picked, and one other door, still remain unrevealed. Imagine repeating this experiment 100 times. If you had a 1/2 chance of picking the correct door the first time, that would mean that on average you would have chosen the correct door 50 out of 100 times, despite your choice being blind. Unless you can see the future, the doors being revealed afterwards doesn't change the odds of you originally picking the right door beforehand.
Do an experiment with an object under cups, using as many cups as you have available; they don't even need to be identical. Have another person hide something under one of the cups while you cannot see. Wear noise cancelling headphones or ear plugs if you have to, to eliminate auditory clues, or leave the room. Tell them which cup you chose, then have them reveal all but two cups, being sure to leave out the cup you chose, and the cup with the object (and if these are one in the same, have them still keep one other cup down, for a total of two remaining). Then have them reveal your original choice. Repeat this experiment at least 10 times and see how often you get it right. It will probably end up a fraction close to 1 out of the amount of cups used.
If someone else were to enter the room knowing nothing, not knowing your original choice, and pick from the two remaining cups out of ten, they would have a 50/50 chance of picking the right cup. But that still wouldn't change the fact that the cup you originally chose has a 1/10 chance of being correct, and the other remaining cup has a 9/10 chance of being correct; but the person not knowing this has a 1/2 chance of picking your cup, and a 1/2 chance of picking the other cup.
The thought process leading you to believe it is still a 50/50 chance for you, the person who made the original choice and saw the reveals, is that you think you still have zero information and that new odds are calculated based on the remaining choices. If you truly made a random choice with the two remaining cups, you really would have a 50/50 chance, however you do not actually have zero information and the choice is not random anymore.
thx
I have always enjoyed reading her column and agree with what she said about education vs intelligence.
I know plenty of people who have college degrees and are complete morons outside of their very small realm of study and their profession. I work with engineers who couldn't engineer themselves out of a cardboard box. Studying and passing test doesn't make you a genius or guarantee that you even have an ounce of common sense. You go through their well oiled machine and don't make waves by questioning and you pass. I love that Marilyn was found to be correct. Ha ha ha. You rock girl! and I'm a guy.
These mathematicians who allegedly contradicted Ms. vos Savant didn't even exist.
So you're saying hard work is nothing in front of in born talent
"I work with engineers who couldn't engineer themselves out of a cardboard box" xD xD
I'm amused at the comments . The professional mathematicians are actually correct and vos savant is in error .
@@charlesdickens6706 but the professional mathematicians agree with Savant. The mathematicians that disagreed were proven wrong and now agree with her, because she was correct.
Well yeah, the only case in which you’d lose by switching is if you selected the car to begin with, which would only be 1/3 of the time.
It really is that easy but people will argue about it for _years_ and never understand it!
@@Hank254 .. and if you find someone that chooses to use his energy to understand the problem instead of doubling down, then actually understands what is happening (ie. each door has 1/3 chance because: 3 door, 1 car.. quite simple really) more 50/50 folks come out of the weeds, once one understands and even says thank you, wow, this was great.. 20 more will come out and "debunk" whole conversation of 300 replies above resetting the conversation back to social media standards... sigh
For example my problem is that I actually enjoy explaining things and when someone understands or learns something new it's great.. but.. the manners of stupid people are such that they insult you, call you stupid, ask if you ever graduated from elementary school, this sort of petty things.. I came to conclusion that some people are just beyond salvage.. these videos where they ask geography and other questions from college students.. depressing.. are we really this stupid as a whole? Depressing.. that said, I *never* call someone stupid even if it is plain obvious that they are, it's not because I am polite.. far from it.. the danger is that the discussion is de-railed into direction I am not interested in, so, when they insult me I don't react just keep explaining as clearly as possible and once in a while someone takes their time to read with thought and Bob's your uncle, something nice happens.. if not.. then I will keep explaining until number of comments run out (500) or they don't reply anymore.. just remain calm and rational, keep yappin' guys..
@@n00blamer The number of people that are willing to listen seems to be decreasing, on MHP videos at least, but I suspect it isn't only here. Strange as it might seem, some of those that seem beyond salvage actually snap back when they are ridiculed. I've seen it happen plenty of times in here, sometimes it is the only way to get them to stop trying to explain the 'new two door problem' and actually think. Negative reinforcement is not the best tool, certainly not the first choice, but it can work. It's totally useless for the ones only trolling but I think the reason most of us are here is because we enjoy explaining it to people who are interested in learning it.
She literally did a move whose only logic is used to the fullest in Minesweeper.
And that is the true genius behind it.
We make these mathematically rational decisions when playing minesweeper and don't even realize we are doing it.
She points out how that basic algorithm can apply to a gameshow with only a 1 x 3 row of potentially mined tiles, and everyone looses their minds.
The first tile clicked in minesweeper is the 2nd safest tile on the board, and is never mined.
Making the chosen door a door that reveals other doors' contents.
I remember playing minesweeper. It came down to luck at the last move while the rest were calculated. Brilliant reference.
I was amazing at Minesweeper and yes, I remember using similar logic to determine the best odds of not choosing a mine. I was wrong some of the time using what I thought was 'stellar logic', but most of the time I was correct.
It's good. Typically with Minesweeper I stuck to 100% probabilities. I don't remember many games where guessing was really important.
The kicker seems to be the game operator's knowledge. After I choose, the operator opened one that he knew didn't have the prize. That would mean that he also knew NOT to open the other one that the contestant didn't choose, perhaps because it was the winner.
That, I get
Precisely. The host's knowledge and how they operate is the _key_ to the Monty Hall problem. If the host does not have knowledge of what each door contains, and randomly opens a door other than that which the contest picked, and it just so happens to be a goat, then there's a 50/50 chance that switching wins.
That's what I realised, finally. If I'm on the correct door, then the host has 2 options to open. If I'm on the wrong door, the host only has one option for opening. That's not the whole explanation, but it's part of it.
That is cheating. Serious answer is Math only. There might not be any car at all, only 3 goats. He does not NEED to open all 3, once 2 goats appear he says "you had 2 chances, you lost, contest/game ends here"
@@MuffinsAPlenty what???
No! Even if the moderator doesn’t know and by pure chance picked a door with a goat (that is not your door), you are still better off switching! What are you saying here?
you dont see people holding themselves publicly account like those cited today
Public scrutiny of such a thing was less back then. Yes you would be scrutinised by those in your field and if it was particularly egregious someone in the media might have done it but we have social media these days and people in these fields also tend to become more famous now because there are more people than back in 1980s and thee are more ways for people to acces information through the internet so you also have more armchair experts who have no knowledge, experience, logic, smarts, training think they do know better.
I believe they were upset to see a personal letter to Marilyn become a public strike against their competence.
Today all the US has is what passes for mainstream media ravaged by gems the truth in broadcasting act of 2005 and smith-mundt modernization act of 2012.
Then there’s that ‘87 reject Biden who has had mental illness issues for decades.. Even the DOJ doubts his fitness to speak on basic topics.
It’s awful that during the largest world crisis, Covid-19 neither she nor any mensans or other supposed “high iq” people stepped up to guide or lead during a significant time of need. Lost respect for the lot of them for not speaking out.
This is true. Otherwise the New York Times and Washington Post would be nothing more than retractions from front to back
my thought also, the shame is greater now and so also the shameless - what'r the odds ?? 🤣🤣🤣