The key to the Monty Hall problem is "the reveal". The host HAS TO show you a wrong door... But he doesn't choose which door that is until AFTER you've made your choice. The original odds (1/3 v 2/3) never changed. Ex, let's say prize is in door #2... -If you pick door 1, he HAS TO show you #3 - Switch and you Win. -If you pick #2, he can show you EITHER door #1 or door #3 - Switch and you Lose -If you pick door #3, he HAS TO show you #1 - Switch and you Win.
@@minrityreprt6302no, you have 1/3 chances if you keep your original pick and 2/3 if you switch. imagine the same game with 1000 doors where the host opens 998 doors which he KNOWS to be wrong AFTER you picked a random one in 1000. you’re obviously gonna switch. it’s intuitive. now in this case it’s 1/1000 versus 999/1000 so it’s not the same but the math is the same and, in the 3 doors case, it’ll lead you to 1/3 vs 2/3
If the point of the Monty Hall problem is to take into account possible deception of the host, then there is NO math. It is only understanding the host's stratagem: He guessed the $1m curtain, use doubt to get him to pick the banana
@@nathanlawson313 Yes, if you know beforehand that the host will offer the switch whether you pick the right curtain or not. But what if the host only offers the switch if you pick the right curtain?
The keys to understanding The Monty Hall Problem are realizing that the host knows which curtain hides the winner, and that when he opens a curtain, he'll always choose to reveal one of the losers (otherwise it'd ruin the game, right?) When you choose one of the three curtains at the beginning, you are also deciding which two curtains the host can choose from when it's his turn to open one. If you pick the winner initally (a 1 in 3 chance), he'll have two losers to pick from, and if you pick one of the losers (a 2 in 3 chance), he'll have one loser and one winner to pick from (being forced to pick the loser as stated above). So when he opens one of the remaining curtains and reveals the loser, there's a 2 in 3 chance he was forced to do so (because the other curtain that you didn't chose initially is the winner), meaning 2 in 3 times you chose a loser at the beginning. So yeah, you should switch. EDIT: I love the pedants insisting on applying real game show logic to a math problem: I get it, Monty could have not followed the rules. I can only imagine you in grade school: "ACKSHUALLY MRS SMITH, if Ralphie had 5 Pennies and gave 2 to Suzie, he would have 4 left because he likely palmed one!"
great explanation. I was just thinking that since he is not going to show you the winner, and he is not going to show the one you picked, so seems that there was a good chance that he was forced to show the one he did. so I was on the right track, but your explanation clearly explained the why while I was only circling around the edges of the reason! nicely done.
This. This explains it. Thank you. Now I understand why the probabilities change the way they do…too bad my stats teacher couldn’t have used this in class…
I actually kind of like the way Mythbusters explained the Monty Hall problem. If you consider your door is a 1/3 chance and the other two doors are a combined 2/3 chance, the fact that one of those doors is opened doesn't change the fact that they are a combined 2/3 chance.
Wait a minute. Combined they have a 2/3 chance. Break them up and they don't each retain a 2/3 chance of winning! The pairing is arbitrary. Option 1. Say you have two picks. The first pick has a 1/3 chance of being right. If you were wrong, your second pick would have a 50% chance of being right. Regardless of which one you pick. In total you had a 2/3 chance of winning. The doors could have been 2 red and 1 blue door or 2 blue and 1 red door. It doesn't matter how they are paired. Option 2. You can pick one door or pick a pair of doors. Naturally you would pick a pair of doors (say #2 & #3) because that would also give you a 2/3 chance of winning. Then the host says surprise, door three just disappears, it's gone. So does the 1/3 chance that that door being the winner. I don't see why the 1/3 chance that door three had would transfer to the remaining door in your pair, any more than it would transfer to door number 1. There are no transfer of odds.
@@djtjm7000 It actually does matter because the door that is shown the contestant is *always* a loser door. If it were random, you might be right, but it's not. And that's why this math trick is deceptive. People act like the last two doors are a 50/50 chance because they act like the elimination of the first door is random *but it's not*. If you picked a winning door, the eliminated door will be a loser because both other doors are losers. But if you picked a losing door, the remaining door that doesn't get opened *is always the winning door*. Given there is a 2/3 chance you picked a losing door to begin with, there is a 2/3 chance the remaining door is the winning door, because the door that gets opened is not random.
@@djtjm7000 I tend to agree with you. When one door is removed, the odds remaining turn into 50/50 that I picked correctly the first time versus the remaining door. I would argue that the 1/3 that got removed gets split into the two remaining doors, since we don't know which of them is the winning door. It is kind of like flipping a penny. You flip it 9 times and it comes heads (unlikely but possible, presuming 50/50 change heads/tails). The next flip still has a 50/50 chance of coming heads.
This video hit close to home. Being color-blind in the green-red spectrum, I can't count the number of times that I have attempted to explain what the world looks like to me. People are often surprised to find that to me the world looks perfectly normal for it is the only way that I have ever seen it. I was told that certain color was green and therefore I call it green. The odds are my green differs from your green. The best way to explain how my vision varies is to show its' impact. For instance, showing up for a dress-green uniform inspection in the military wearing blue suit pants (yes, I really did that.)
The questions posed in this in regards colorblindness I totally understand my best friend has proposed the prospective situation. What his of green perception to my green there's no way to confirm or deny that.
The trouble with explaining colorblindness to people is that Certain Shade blend together there are certain shades of green very light look yellow to me are things like moss which are red sort of green throwing some Ferns and I can't find my golf discs
@@khironkinney1667 I have know idea what you wrote. It is extremely hard to understand. I would recommend either editing it with correct grammar and punctuation. Otherwise if English isn’t your first language then just write in the language your best in.
The Monty Hall problem was not clearly explained in the first go. It's absolutely critical to say the host then reveals a door that ISN'T the winner based on his knowledge of which door has the winning prize. If he just shows door 2 when you choose 1 and door 3 when you choose 2 and door 1 when you choose 3 (or any other just guess) regardless of what has the prize or not then your odds don't change. But he has a chance of just revealing the winner that way and then you don't get a choice to swap anyhow. It's using that extra information that changes the odds. He must know what the winning door is.
I think most people understand that option 2, revealed by the host, will never...can never be the prize. No one thinks the host doesn't know where everything is. Your option 1 is probably wrong, option 2 is always wrong, therefore option 3 is probably right.
With the Monty Hall one there's a simple way to look at it. When you first pick you are choosing one out of three, which gives you a 33.3% chance. When Monty shows you whats behind a curtain it removes that 33.3% from the guess. If you stick with your original pick then you stay at the 33.3%. If you switch though, you get the 33.3% from the curtain you switched to plus the 33.3% from the one he revealed.
This man probably does not understand the Monty Hall problem or he would initially explained that the host does not open a random door but one of the doors that does not hide the million or the door you chose
@@bjorneriksson6480 Are you suggesting that the man doing the video does not understand? If that is your assertion, I am disagreeing with you. Of course he knows the Monty door is by design. The host knows where the prize is. It is presented as a random door because the contests does not know.
Opening the”random” door was proven to increase suspense, captivate the audience, and stall for a commercial break. In the 3 door case it’s still a high chance of being wrong, and most people don’t understand how the probability equation would work. But the guy doing the video is from the “Today I found out” and he’s awesome.
The Monty Hall problem is that your choice of curtain is random - the hosts choice to show you is NOT random. The host never shows you your curtain OR the curtain with the $1,000,000. It's the combination of non-random and random variables that causes most of the confusion.
Yeah, it especially causes confusion if those assumptions you're stating are not even mentioned. Why do I have to go along with these assumptions then?
@@insignificantfool8592 It doesn't actually depend on those factors though. By choosing a door, you yourself are fixing a condition. The fact that one of the doors doesn't contain the goat that was chosen while you were holding a door means the odds are now 2:3 for the last goat. This is irrespective of the fact the host will never open your door or the goat's door.
@@Demmrir many people think so, but it's not true. Imagine the host really doesn't want the contestant to win the car (maybe he has to pay for the prizes himself and is low on cash). What exactly prevents him from opening the contestant's door if he chose a goat door and only offer the switch when he chose the car door?
The tendency to stick with one's original choice is also reinforced by the contestant assuming that the host knows he or she chose the $1 million and is intentionally trying to throw him or her off. The other assumption being that if you had chosen wrong, he wouldn't give you an opportunity to change your mind.
Thanks to @StorkBomb7417 below, I remembered the proper explanation to the Monty Hall problem (which is vastly different to what Simon gave). It boils down to two scenarios: Scenario A - you picked the right door (1/3 chance): if you switch here to the other door you lose. Scenario B - you picked the wrong door, Monty opens the only other wrong door (2/3 chance): if you switch your door here, you are guaranteed to get the right door as it’s the only other door except the one you initially chose. Since Scenario B happens 2/3rds of the time, it’s a statistically better strategy to presume you’re in scenario B and switch your choice.
Imagine the Monty hall problem starting with 100 doors, only one having the car behind it. You pick a door and then the host opens 98 doors (all but the one you picked and another door you didn't pick) that have a goat behind it and then asks you if you want to switch.
Because he's let us know that while we may not know how to calculate the velocity of a baseball, our brains are fully aware of how the baseball is moving.
It's not a neutron bomb. It's a thermonuclear weapon, aka the H-bomb. The neutron bomb was a low yeald, high radiation weapon meant to kill with minimal damage to the area, mostly meant to stop Soviet tanks in Europe during the cold war.
After trying to contradict your Monty Python odds, I concede. You're correct. You have 2/3 chance of picking the wrong door. So it's better to choose the other door.
I still dont think that is true. When you are given the chance to either keep your current choice or choose another, you are actually just rechoosing based on what is still unknown. Just because they say "keep your current choice" does not mean that you are keeping your initial probability. Please respond, because there might be a high probability that I'm missing something :P
@Wuddigot Simplest way to think about it. You know how the game works. You know that you will be shown a losing door after you choose a door. Beginning of the game you have a 1 in 3 chance of picking the winning door. A 2 in 3 chance of picking the wrong door. Would you agree 2 in 3 odds are better than 1 in 3? So you are most likely to choose a wrong door. So knowing how the game works, you pick your first door knowing you're more likely to pick one of the 2 out of 3 that are losers. Because of this, when a door is removed, you pick the other. Definitely not a fool proof way to win Just gives you better odds. You gotta look at the odds of picking the wrong door to determine the best odds of picking the correct door.
@@Wuddigot I just reread you comment. Keeping the first door keeps your original 1 in 3 odds. So keeping your first choice does keep your original probability. Gotta bring the odds of choosing one of the wrong doors into the equation. 2 in 3, which is more likely. Start of game you're thinking "I have the best chance of picking a wrong door" (2 in 3) So after you pick a door (odds are its a loser) you know one of the other two doors are more likely to be the winning door. Right? Since one of them doors are removed from the game, you pick the other. Giving you 2 in 3 odds of it being the winning door.
@@al1383 That just isnt true though. The odds of you choosing the wrong door is now less because there are less wrong doors that you would choose from. You will not choose a door you know is wrong, and therefor the conditions (and probability) have changed. Am I hearing the initial problem incorrectly or something? guy chooses door, before result is revealed he is told one of the wrong doors and given the choice to choose again... Is that it? he now still does not know which of the doors is right, but he does know that one of the two remaining doors is wrong. If I am not missing a detail, this is an iterative problem, as in each completion of the action loop starting at the beginning with a different set of conditions.
The Monty Hall is easier to understand when you try to replicate the experiment in code. winningDoor = random number between 1-3 initialChoice = random number between 1-3 if winningDoor = initialChoice, stayingWins otherwise, switchingWins So there’s a 1/3 chance the player’s initial pick was the winning door. So the rest of the chance (2/3) has to be that switching would win, because the wrong door was already revealed.
I once thought I'd prove the right answer to the Monty Hall problem to myself once and for all by writing a simulation and running it a bunch of times to see how the probabilities landed -- but once I wrote it, I didn't have to run it at all. The answer jumped out at me from the code while writing the step where the host opens a door. I ended up with 3 cases to account for: the one where the contestant picked the right answer -- in which case it didn't matter which other door I picked -- and the two cases where they picked the wrong door, in which case there was no choice about which door to open because the rules allowed for only one of them. In only one of those scenarios would switching be the wrong choice; in the other two it would be right.
You can't put the two stage MH problem into code, since we don't know what the host will decide to do. How do you model the host's decision as to whether to open the player's door or whether to offer a switch?
@@insignificantfool8592 the host knows which door is the winning door. If you picked the winning door, then the eliminated door is irrelevant. If you didn't then there's only one door that can be eliminated. So basically your code is - identify which 2 doors are left, if the first isn't the winning one, eliminate it, if it is, eliminate the other non-selected one.
You mention Richard Hamming in the "Maths may not be real segment" - he's well known in computer science circles for the development of error detection and correction algorithms, the descendants of which are used in computers & data communications to this day. Wikipedia has an article on these, named for him: en.wikipedia.org/wiki/Hamming_code
Another way to look at the Monty Hall problem is that it's effectively the same as giving you the option of sticking with your original choice or changing to selecting both other doors.
Exactly. It doesn't matter when you decide to switch, the probability locks in when you choose the door. 1/3 you're right and 2/3 chance you're wrong. Switching is basically just saying: I think it's not the one I picked first.
Great explanation of the Monty Hall problem. I think it's so confusing because it _isn't_ a pure maths problem. The host doesn't pick a door at random, but people intuitively assume it's random chance.
It doesn't matter what the host chooses to reveal if it's a fair game, though. In the video's example, you still either chose the $1M or the banana, it's not like the banana was what was revealed. The remaining 2 curtains becomes that psychological poison chalice dilemma with the host at that point. It would only be unfair odds if the host can cheat by changing what prize was behind the chosen curtain when asking if you're sure.
Simon's explanation is wrong. He said: "He [the host] reveals that concealed behind curtain number 3 is a thousand dollars", but he did not establish that the host always reveals a curtain. That is actually a huge difference, because without this additional information, the problem is unsolvable. The host could always reveal a curtain, he could reveal a curtain because you guessed right and it is too early for someone to win the main prince. Many versions of the Monty Hall problem make this mistake.
The host knows which curtain hides the million, and he will never open this curtain, and he will never reveal what's behind the curtain you chose, so there are three scenarios : 1 : If you chose the Million - he can reveal what's behind either of the other curtains at random - if you change you lose 2 : If you chose the Thousand - he can *only* reveal the Banana - if you change you win 3 : if you chose the Banana - he can *only* reveal the thousand - if you change you win He has knowledge and restricted choices, so what is revealed does change the odds ..
@@davidioanhedges The problem with your idea is there will never be a 2nd curtain reveal out of 3. If the 1st one revealed shows $1000, what's the 2nd reveal going to show? If it reveals the $1M, you choose the revealed $1M. If it reveals the banana, you choose the only unrevealed curtain remaining since it will be the $1M.
"Math is perfect, real life is not" a very beautiful quote the represents the reality. Here is an example: I studied electronics and since I love playing bass, I decided to make an amplifier from scratch. I even got specialized equipment like an oscilloscope to get the best idea of what's going on. Here is what I found out: the equations that are supposed to give you the results of your circuit are simply not correct. Of course some *DO* give you the results you should expect after doing the math but things were acting very weird to the point I gave up because they made no sense and I didn't know how to proceed to complete it. Multisim, a program that allows digital simulation of a circuit would give me errors all the time because it, just like me, just couldn't comprehend wtf is going on there. The circuit outputted about 3% of what I wanted and since it wasn't any major project or something, I simply gave up and left it for when I have the will to complete it, at some point in the future, hopefully. Before you blame me, I am confident that the circuit was correct. Only faulty parts could have been the reason for this but I just bought them and they were brand new, so, I don't know.
The circuit output 3% of what you expected? Well that is you :) Circuit theory is only an approximation of circuit behaviour. V=IR, I=C dv/dt, V = L di/dt, KVL, KCL, etc are all only valid given assumptions about the size of the circuit, no interference from outside, and the speed of stimulus. They don't even describe the physical processes that lead to those quasi-steady state conditions described by those equations. A similar thing is going on with mathematical models describing semiconductor behaviour. They are simplified approximations whose equations are only considered valid when corresponding assumptions are valid. A single device can have multiple equations modelling behaviour and you can make the model more or less accurate at the expense of complexity. The models that humans use to design with are broadly simplified and will get you in the right ballpark. Computer simulation can use more complicated models and get you closer still. But this is before manufacturing variability is factored in, not to mention environmental factors that you may not be aware of or choose to ignore. Long story short, designs are broadly correct if your assumptions are correct but building something in the environment it is intended to operate in is always necessary to make sure things operate as intended and something(s) hasn't been overlooked. Getting an output 3% of expected is a significant error someplace. If everyone's designs were 97% off, it would be a fluke that anything worked. Electronic design would be more like witchcraft and the models used would be essentially worthless.
I think that the main thing that people who explain the Monte Hall problem never really pin down into simple words is that Monte Hall would NEVER open the door to the million dollar prize on purpose. That is critical to the problem and I've never seen anyone explain the problem and mention that. Anyone who has seen the show automatically knows this critical detail.
I don't understand wy the words "game show" don't automatically infer this to people. In what game show in the world is the big prize given away at random?
I think the funniest example of a Nash Equilibrium is when two people approach each other while walking around a corner. Both frighten each other and for whatever reason both step to the left and both step to the right at the same exact time. This is actually the opposite of the equilibrium. The correct method would be to converse with the other participant and say "Go to the right" or just grab them by the shoulders and do a 180 degree spin with them so neither can get confused. My method is that whenever this happens to me my first and only instinct, instead of getting stuck in the dumb left right left right loop of laughter and embarrassment, is to stand directly upright and perfectly still like a deer in the headlights. It freaks the other person out which makes them go around me faster, but also leaves no room for guesswork and they know that my intention is to let them go around me in whichever direction they want.
Not sure where it falls, but my default is to stand still and let the other person move around me. Saves a bit of time, but I lose out on the two-step shuffle dance moves.
My theory is most people will step to the dominant side. That being the right hand side as most people are right handed. So my question is are you ambidextrous???
This is a Canadian stalemate. Both wants to move for the other and not look rude, so we spend hours just going back and forth saying "you first, no you first, no I insist you first". And "sorry. No I'm sorry."
If you're going in the opposite direction, it works fine when both step to the right or to the left. It only fails if one person steps to the right and the other steps to the left.
I just grab them by the scruff of the neck and beat the living daylights out of them, then I turn around and walk back the way I came from, it's far from ideal but who's perfect right
The way I'd look at it is that much of mathematics is a toolkit for manipulating information. But not in arbitrary ways; the manipulations have to follow various properties like being self consistent and following deductive logic. It's not surprising that tools for deriving information -- essentially augmenting our intuitive reasoning -- would be useful. And then, further to this, the question of whether maths will always be useful comes down to whether our universe has properties like self consistency. Well, if it doesn't, then all bets are off.
The monty hall problem can be better described as you pick door 1. The others are grouped together. [1] [2, 3]. The fact that door 3 gets revealed does not change that grouping. Visually grouping [2, 3] seems to help explain it better for most people.
I think he must have explained it wrong, because the way he showed it still comes out to a 50/50 choice, by my calculations. Step 1. Pick between three. Step 2. Throw that choice in the garbage because it doesn't matter. Step 3. Pick between two.
@@johndaniel7161 the odds don't change because an incorrect choice is revealed. That is why grouping visually helps explain. [1] [2, 3], odds of 1 are 33.3%, odds of 2 or 3 are 66.7% because they are grouped. Revealing 3 is a goat does not change the odds, 1 remains 33% and the group is 66% even though one in the group is invalid.
@@johndaniel7161 I should add that the host knows which is the winning choice and which are the non-winning choices. When the host reveals a door in the grouped set they intentionally and knowingly reveal a non-winning choice (which actually matters a bit).
@@bretmcdanel8595 It technically does not matter if the host revealed a winning choice, if the format was still "Do you want the best prize from among the doors you did not pick?", it would just make a bad gameshow - because that's what the host is *actually* asking, just phrased to hide it, and why the grouping helps. You're not being given a chance to switch within two doors, you're being given the chance to pick the best from two thirds of the doors, just you already know one of them is a dud.
@@iskierka8399 But that's kind of the point. The host can't do that, because it's against the rules of the game. When you're playing blackjack, the dealer cannot refuse to deal you a card, because then the game cannot commence. Similarly, the game host cannot reveal the winning door, because then the game is over. It's because the rules are set in stone that this is a mathematics problem instead of a sociology experiment.
With the Monty Hall one there's a simple way to look at it. When you first pick you are choosing one out of three, which gives you a 33.3% chance. When Monty shows you whats behind a curtain it removes that 33.3% from the guess. If you stick with your original pick then you stay at the 33.3%. If you switch though, you get the 33.3% from the curtain you switched to plus the 33.3% from the one he revealed.
In The Monty Hall Problem, if you imagine the steps in the game taking place in a different order, the probabilities become clearer to see. Imagine that the host asks you to pick one of three curtains. After you pick one she simply says you can have the prize behind the curtain you picked, or the best prize that lies behind the other two curtains. I think most people would intuitively pick the “best prize behind two curtains” over the “prize behind one curtain”. Considering this in the context of the actual order of things in the game, the host knows where the big prize is and will never open that curtain. After the host opens that curtain, when the player chooses to switch it seems like they are choosing the prize behind one of two remaining curtains, but actually they are still choosing the “the best prize that lies behind the other two curtains”. I like when the problem is extrapolated to the 100 curtains. It becomes easier to see. Will you take the prize behind the curtain you picked or the best prize that lies behind the other 99 curtains?
Excellent explanation. I've always stuck with "yeah nah it's now 50/50", even after watching the video, but 10 minutes ago something happened in my brain and something made sense but I didn't know why. This explained it well.
I always explain the Monty Hall problem by saying "I'm thinking of a person". When they guess I say "I'll give you a clue, it's either the person you picked, or it's Abraham Lincoln". Pretty similar to your 100 curtain example.
By the way, you explained the answer before you did the explanation. The probability of your choice being correct is 33% (ok, 33 1/3 -- let's just run wth it). Tha means that the chances of it being one of the other two is 66%. Then he shows you which one of the two it is not. But the probability of it being one of the other two (the one the host didn't show you) is still 66%. I knew there was a simpler explanation than all the conditional probability waffle I've read.
The thing about the Schrödinger's cat experiment is that it's completely wrong. The cat is alive **or** dead before you open the box to observe the cat. Why? Because the cat is an observer. At no point is the cat alive and dead at the same time. Yes, it's a thought experiment, but that's how my thoughts work.
a crucial piece of information was left out of the 3-door problem. It is only true IF THE HOST OF THE GAME SHOW KNOWS BEHIND WHICH DOOR THE TOP PRIZE IS (as other commenters have mentioned). If the host doesn't know (and if the host may inadvertently open a door which conceals the top prize) then the chances/probability don't change. This is easier to appreciate in the 100 door example. If the host doesn't know where the top prize is it would be a 1 to 100 chance that he or she would open all the doors except the one where the prize lies.- i.e. the same chance as the game show player has that their choice is correct.
Exactly. That one bit of information changes the whole game. I just feel sorry for the Noble Prize winning physicists who got the answer wrong because he just assumed it was a completely random game.
You are right, and what’s worse is that not that the crucial piece of information was left out inadvertently. Is that they didn’t understand its implications and their whole reasoning about the answer is wrong, as evidenced by the way he proposes to experiment with cards, and pick one at random to reveal from the 2 that are not the chosen one.
Exactly, you've introduced a known into the game. Which is the field where conmen and magicians work. Simple force a choice. Don't ask how I know this 🙂
Everyone always states the Monty Hall problem incorrectly. You're leaving out critical information. You have to tell us that (A) The host is fully aware of what's behind each curtain at the start, (B) the host always chooses to reveal a curtain, and (C) when he reveals one of the other two curtains, he always does so, using his prior knowledge, in a way to avoid revealing the grand prize. Also, the Monty Hall problem doesn't qualify as a "mathematical discovery."
Many of the people putting forward the "switching is better!" idea are doing so, directly or indirectly, because of the article written by Marilyn vos Savant, in response to the letter written by Craig F. Whitaker. She had a whole web page devoted to this one problem, though you can now only find it in archives. You know what she wrote? "So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose." These are her words. She straight up says she's using circular reasoning. The answer does not define the conditions for getting from the question to the (correct) answer. The *question* defines the conditions for getting from the question to the correct answer. If the answer is defining the conditions to get to the answer, that's circular reasoning.
Yep I joined the comments to state this also. The monty hall problem is indeed counter intuitive, and it's not helped in this video by not only not explaining that it's the way monty chooses a curtain to reveal that's significant, but not even mentioning it in the description of the puzzle.
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"You're leaving out critical information." Exactly! And even worse, the experiment in the video is slightly different to the original MHP. As there are three different prizes, it means there are more strategies the host can follow. Depending on what strategy, the probabilities changes. For example, the host could follow the strategy "reveal the door with the lowest prize". If that is the case, switching doors in the video example is only 50%. Another example that shows that knowing the strategy is important, is to imagine the guest is a monkey. For the monkey, the big prize is the banana. Should the monkey also always switch doors to improve the change to get the highest prize?
@@hughobyrne2588 If it were circular reasoning it wouldn't need explaining. But it does, it's counter intuitive to most people (I'll admit, like 99% of people, I didn't get this puzzle when I first heard it)
I think an intuitive way to understand the Monty Hall problem is this: The chances that your initial pick is correct are low (1 in 3). So when Monty reveals an empty door and you have the option to pick the other one, you're basically being asked, "What are the chances that you blindly picked the winner on the first try?" They're not good, so it's likely the one remaining is the winner, and so you should switch
Or, Monty doesn't want to give away a million bucks, so he shows you a sure thing after you picked the best door, hoping you'll switch and save him $999,000.
The problem with that intuitive way to understand the problem is it's not actually understanding the problem. If the empty door is arbitrarily revealed, there's no advantage in switching. The door not containing the prize isn't the only thing that factors into why it's advantageous to switch.
@@strifera yes, of course, this is assuming knowledge of how the show always works, which is fair to assume in this case. That's part of the stated problem, and it's an event that is repeated every time on the show. The host always knowingly picks an empty door to reveal
Your introductory example for the Monty Hall Problem is flawed. You left out a couple vital pieces of information. After making your pick, Monty is going to 1. Show you a door that does NOT have the million, and 2. Do so knowingly. The way you described him revealing the door with $1k makes it seem like the reveal is random, and not intentional. The reason we know that switching is better once he reveals is that we know he is going to reveal a non million door. But by using two different worse prizes, him revealing the 1k (even if he is revealing the 1k intentionally) then that means that he is also revealing a non-banana door just as much as he's revealing a non- million door. I also suggest using an example where both goat prizes are the same, for charity. Such as... um... actual goats. Which is actually what is used in the original problem.
No, it is not. See, if the presenter opens a curtain where the $1M is not hidden, the 2 choices (your original choice and the optional to change your mind) became dependent choices, meaning your first choice influences the outcome of the second one. That changes the probabilities. Otherwise it’d be 50-50.
@@miikayak Actually you kind of got it right. The presenter takes out a non $1M option. Here is how it works. The example in the video is a bit confusing. Let’s say that there are 3 curtains with 2 of them hiding a banana and one the $1M. Your first choice has a 66.6% chance to be a banana. When the presenter takes out 1 banana that you did not choose, he leaves you with a curtain with a banana and one with $1M. So if you chose the banana the first time and then change to the other curtain you’ll get the $1M. Since the probability that you chose the banana was 66.6% at the start, switching results in a higher probability to win the money then not to. The key information is that the presenter opens a curtain where a banana is, and not a random one.
@@fudhater8592 It wasn't a question, it was a statement.... in a video about statements. Thank you for trying to sound smart at my expense. Better luck next time.
@@fudhater8592 You're like a smartass, _hold-the-smart._ If you eat enough fiber to dislodge your head from your rectum, you'll realize that in order for the Monty Hall problem to work, the host has to know where the prize is. IE, it's not random. Do try to keep up, dear.
6:10 The prisoner's dilemma! I can see where it doesn't match up perfectly, but influencing the outcome of the other with the decision of one when both must decide, that part lines up. The first one I knew because of VSauce2 - Kevin does a great job at explaining the brain-melting probability math with /more math/ except it works well. Y'all did a good job here with these, especially making the point that the game show host is the factor that changes the probability.
for people who don’t understand monty hall: just picture in your head the same game with a 1000 doors and 1 prize. and the host opens 998, which he KNOWS to be losing, of the 999 remaining after your pick. wouldn’t you swap? on now remember that even though the gain in chances is way more marginal in the 3 doors case, the math is still the same. so if it 1/1000 vs 999/1000 in the 1000 doors case, it has to be 1/3 vs 2/3 in the 3 doors case. pretty intuitive.
The real key to understanding the Monty Hall Problem is that _the prize does not move._ Starting out, there are X doors (where X = 3 in the classic problem), each with a prize probability of 1/X. All the doors are in one group and the probability of the prize being behind a door in that group is 1. However as soon as you choose a door, you have split the doors into two groups: we will call them C (door you chose) and R (the doors you rejected). There is only one door in group C, so the probability of the prize being in C is 1/X. There are X - 1 doors in R, so the probability of the prize being in R is (X-1)/X. No matter which doors in R the host opens the probabilities of the two groups remain as they were, 1/X for C and (X-1)/X for R, because _the prize does not move._ Eventually there will be only one door left unopened in group R. What is the probability that that door hides the prize?
That's a useful explanation, thanks. Maybe you can answer a follow-up question or two please? First question: Let's use Simon's 100 curtain example, but instead of the host opening curtains that they choose, a number of curtains are revealed but randomly so. The host is just there for show but doesn't actually choose anything. Does this affect the outcome? Second question: Let's say that after I pick a curtain, the host reveals 98 curtains (either randomly or selectively, I don't know if/what difference it makes), then I switch to the other curtain, as offered by the host, and then a friend of mine walks onto the stage. However, my friend hasn't watched the show and isn't told which curtain I originally picked and isn't told if I switched or not. He just sees 98 opened banana curtains and 2 closed curtains. Then he's told to pick one. Intuitively I'd say my friend has a 50/50 chance of picking correctly, but how can it be that he and I seemingly have the same choice but might have different probabilities of outcomes? Thanks in advance.
@@charles7866-o1l 1. It only changes the outcome if the prize is accidentally revealed. If the Random Curtain Opening Device opens only one curtain, and it is a banana, you should still switch to one of the other 98 closed curtains. Your probability of winning will increase from 0.01 to 0.01010204... If the RCOD opens two curtains with bananas, switching to one of the other 97 closed curtains increases your probability of winning to 0.01020619..., and so on. 2. Your intuition is correct. If your friend walks in, has no information of what has happened ("what's with all the bananas?"), and is asked to choose a closed curtain, they have a 0.5 probability of winning. The two of you have the same choice, but you have _more information._ Your friend only knows that there are two curtains and one prize. You know that the prize was somewhere behind one of 100 curtains. When you chose your curtain you split the curtains into two groups: your one curtain, with a 1/100 probability of winning, and the other 99, each with a 1/100 probability but _collectively_ with a 99/100 probability of concealing the prize. As curtains are opened that 99/100 becomes concentrated onto fewer and fewer curtains until only one remains in the group you did not choose. With no information, your friend's choice is an _independent trial_ and I think this is where a lot of people get tripped up by this problem. They think that your second choice, to stick or switch, is an independent trial when it is not. For it to be an independent trial for you, you would have to lose the information you had already been given. The only way that could be done is if after the reveal of all the bananas the prize was somehow randomly redistributed between the two remaining closed curtains. Here is another example of the problem, a simple card game with a standard deck of cards. The goal is to get the ace of spades, if you have it when all the cards have been turned over, you win some money. The deck is shuffled and spread out face down. You are asked to pick a card, any card. The card you pick is placed just aside, still face down, you do not get to look at it. Then the dealer flips over the remaining cards, one at a time for dramatic effect. If it isn't the ace of spades the card is tossed in a garbage can under the dealer's table. The dealer gets through all the cards until just two remain: your card, and the dealer's last card. Let's now consider three scenarios. In the first scenario, you are given the option to stick with your card or switch to the dealer's last card. You know now that the correct choice is to switch: your card has a 1/52 probability of being the ace of spades, the dealer's card has a 51/52 probability. In the second scenario your friend walks up to join the game. They have no idea what has happened so far - not even which card is "yours" - they just know the goal is to find the ace of spades and that one of the two cards on the table is it. From their perspective, with the information they have, the probabilities are equal. In the third scenario, instead of your friend coming by, the dealer shuffles the last two cards (in such a way that you cannot tell which is which), places them back on the table, then asks you if you want to stick with "your" card, or switch. Your card isn't really "your" card anymore, is it? The prize has been randomly redistributed. You have lost all the information you had. Your choice has become an independent trial. Now you are just like your friend: the probabilities of either card being the ace of spades are equal.
@@charles7866-o1l YES. Exactly. Most explanations of the MHP assert that somehow when an option is eliminated from a set (a door opened in this case) that its initial "share" of probability is passed only to some sub groups and not the others. This is arbitrary and problematic. If this is the case, I want to include MY door in the group with the opened door, and have MY door receive special treatment.
@@JonMartinYXD Thankk you very much. I think I understand the concept better now :). Took me an hour or so in bed pondering it and varying the thought experiments, but I think I got it!
@@khemkaslehrling3840 The reason that your door doesn't get special treatment is that when you make your initial choice you've split the groups of doors into two groups: the one you chose and the ones you rejected. If something happens to one of the doors in the rejected group (like the door being opened) then that affects the rejected group, not the one you chose, which is why the "success probability" (SP) of your chosen door doesn't change from its initial position. However, in a way, your door is included: if (as discussed in my reply to Jon Martin) you have a friend who walks in after the door in the rejected group is opened, and they don't know which door you chose, and they see the opened door and understand the game concept, then they'll know that the door opening has just now changed the SP on all the doors (for them but not for you), meaning your door is receiving the treatment of a change of SP, which you wanted. It might help to know that your advantage in information over your friend still puts you in a better position. Let's say there are 10 doors. You picked one. The SP of the right door being in your chosen group (a group of one) is 1/10=0.10, and 9/10=0.90 for the rejected group. A door in the rejected group is then opened, meaning the SP of the 8 remaining closed doors in the rejected group is now 0.90/8=0.1125 Your friend then walks in and just sees 9 closed doors, 1 open door, and doesn't know which door you chose. They ignore the open door, so their SP is 1/9=0.1111111111 and this is less than your SP.
What helps me understand the Monty Hall problem is looking at it like "I have a 1/3 chance to pick the correct door on the first try and a 2/3 chance to have picked the wrong door." The 100 door example really helps too, such an extreme case makes it a lot more clear.
@@SigFigNewton Or if they know it, they don't see that they selective revelation creates a sampling bias. I am not proud, I admit that it took me a long time to understand it.
By which you mean, I think, that if Monty Hall always opens 98 doors, none of which have $ 1,000,000, then it is more intuitively obvious that you should switch.
You could also just list all the possible situations you can end up in after you have chosen your door and the host has revealed a door: Door in [] is chosen, door in || is revealed by the host, L looser door and W winner door. Assuming the host always opens a looser door, we get these cases: 1. [L] |L| W 2. [L] W |L| 3a. [W] |L| L 3b. [W] L |L| Now, we can say that 3a and 3b are the same and count them as one single case, because the order of the remaining doors (the L's) doesn't matter in this case. This leaves us with 3 cases. 2 of which switching would lead to you winning and one which wouldn't. This is how you can reach the 2/3 probability of winning by switching. Edit: Another way too look at it, is that you have 2/3 probability to chose one of the L's at first. And whenever you have chosen one of those L's, you're in case 1 or 2. Thus, switching wins in 2/3 of all cases.
I like how Simon says asking the question "is my red the same as your red" is something we probably all did in childhood meanwhile here I was at like 26 while bingeing on math and science youtube videos that I finally had the thought that we could all be seeing different colors but just calling them the same names.
When I was a kid, most of my friends thought I was crazy when I brought up this question... I was always happy being that dude, and now my bubble was popped 20yrs after the fact; when I find out it was just my friends that were...................... Lmao 😂🤣😂
What helps with the Monty Hall problem is that while your choice is random, Monty's in NOT random. When you pick, you have a 1 in 3 chance of picking the prize door. When Monty opens a door, he has a 0 in 3 chance of picking the prize door. Therefor the remaining door has the remaining 2 in 3 chance of being the prize. :)
It doesn't matter what you pick initially because it's ignored and a new scenario is presented. The host's reveal is never random. It must and will always be a goat or the game is ruined. The player always and only makes a real choice between one of two options, a car and a goat.
@@lrvogt1257 That's not true. Try the scenario with 100 doors. Your first pick is 1% chance of success. Then Monty opens 98 doors with goats. Now there are two doors left, Your door (1%) and Monty's door (99%). It is NOT 50-50 ;)
Actually Relativity IS important to your drive to work if you use satellite navigation. The operators of GPS have to take Special and General Relativity into account for it to remain accurate otherwise it would drift because of time dilation.
The math section of this video is amazingly satisfying. I was literally going through the motions of all of the things that were predicted by math that became real in the real life-like the location of Neptune Neptune or Einstein predicting that gravity could bend light those were all things predicted mathematically that came to be reality. And then Simon goes on to say math is perfect but reality is not. If you take that statement into account and also the previous statement that we may not be understanding it because we represented in symbols that only we created and we understand on this planet, I've come to a conclusion that either there's a mathematical hump that we're not seeing yet or we're living in a simulation. Even with that being said science was always progressing I remember when Hubble was launched I remember the Challenger accident I remember so many things that have changed as when I was in middle school and high school regarding science and thinking that a lot of those things that I was learning about happened 50 to 100 years ago they're possibly could be an aspect of reality similar to how to we express mathematics that will make the universe more comprehensible. I think the real key is thinking outside of the box. People with scientific minds and engineering minds always have to know why. And that why has led to some amazing discoveries. So why is math so perfect in a reality that's not? 🤯🤯🤯 (This message was brought to you buy an educated stoner who was high while watching this video)
A help for the Monty Hall problem: 1) Regardless of which curtain you initially select, the probability that the million bucks is behind one or the other curtain is ⅔. 2) No matter which curtain you initially select, Monty can _always_ open one of the other two curtains, showing you a worthless item behind it because he knows where the big money is. 3) When he opens one of the other curtains, he shifts the entire ⅔ probability from two curtains to the one other curtain he didn't open. 4) After he opens one curtain you have the choice of two curtains: your original choice (⅓ probability) and the one other curtain Monty didn't open (⅔ probability). By changing your selection you therefore double your chance of getting rich.
By what magic does the share of probability of the opened curtain transfer to only one of the other curtains, and not be split across both? Why not then just declare that the curtain you chose is grouped with the curtain the host opens, giving it "the entire probability of the opened curtain" and making your door 2/3rds? This is the fallacy of this Monty Hall thing.
@@khemkaslehrling3840 Think of it this way; 3 items behind 3 curtains, 1 prize & 2 duds (worthless), no matter which door you choose the host can reveal a dud. Lets go through all 3 choices, Scenario #1 lets say you pick the prize curtain then the host reveals a dud curtain, if you switch you get another dud (1/3). Scenario #2 lets say you choose a dud curtain, the host can only show you the remaining dud so if you switch you pick the prize (1/3). Scenario #3 you choose the other dud, yet again the host can only choose a dud, so if you switch you pick the prize curtain (1/3). So in closing in only 1 out of 3 scenarios do you lose by switching, but if you chose to switch you'd be right 66.7% (2/3) of the time.
Another help to those who aren't yet convinced: when Monty opens one door, think of it as him opening *ALL* the other doors except the one you chose and one other one.
@@khemkaslehrling3840 You choose one door out of 3. If I offer you to let you switch for the two remaining doors, would you take the deal? Of course you would, because your odds would be 2/3. That's what's happening here, except that the host is opening one of the two doors first. But you are still effectively getting to choose 2 doors by switching.
Your explanation of the Nash Equilibrium is spot on. Every time the traffic lights outside my work break down, there's a nail biting display of idiocy which always results in someone crashing their car.
Savvy and slightly sarcastic gamblers have a saying, "Don't bet on anything that talks". Well, that's that's the problem with the Monty Hall Problem. Rather than strictly being a math or statistics problem, it confuses things by including Monty, without explaining anything about his consistency or motives.
The majority of people seem not to really understand the problem, and it shows by the way they put it. It order for the logic to work, this has to be the standard format of the game. ALWAYS the participant selects one curtain, ALWAYS Monty opens a curtain that doesn't contain the prize, ALWAYS the participant has the choice to switch.
@@ntomata0002 Oh no. Here we go again. I can't tell you how many hours I've wasted trying to clarify the ambiguities still inherent in the problem. I have no intention of going through all that again, because 99+% of it falls on deaf ears. I referred to not betting on anything that talks for a reason. Because humans easily and often project other potential motives, rendering this a pointless game to discuss. It certainly is not the straightforward reliable simple statistics game you suggest. Why? Because we have no idea what Monty's motives are, just when he makes this offer or does not. He said so himself, that he has total control, and that sometimes he would prefer to make the offer to switch when he KNOWS they have already chosen the car. I see no valid reason we should accept that he has no such motives. The problem would be a lot clearer if no Monty were involved! Then it might be a math and statistics problem, unencumbered by ambiguity. Just because the problem does not suggest any of this is no reason we must assume no ambiguities exist. Incidentally, Marilyn herself admitted there were ambiguities in her original presentation, because she tried to deflect away from this by saying that MOST of the thousands of letters she received did not mention the ambiguities. The entire failure of this famous problem (from her column) goes back to her failure to present it as a purely statistical problem.
Okay I'm gonna be that asshole 😂 your explanation of the Monty Hall problem left something out. It's crucial to state that the game show host has decided in advance to open a curtain that does not have the million dollars behind it. In this case, switching is the superior strategy. By contrast, if the host randomly opened one of the two remaining curtains, and you saw that that curtain did not hide a million dollars, there is no advantage in switching. It's 50/50. You can test this yourself at home with a deck of cards. Take an Ace, King and Queen. Your objective is to find the Ace. Setup 1: you pick a card, then instruct a friend to look at the other two cards, and remove one which is not an Ace. From the two remaining unrevealed cards, you will now find that switching finds you the Ace 2/3 of the time. Setup 2: you pick a card, then reveal one of the other two cards. If it's an Ace, game over. If it's not, continue. You'll now find that it makes no difference whether you switch or not; you'll pick the Ace half the time, regardless of your strategy, in the long run.
It depends on where you are assigning the odds. If you are assigning the odds before the host makes the choice, then you are talking about different game rules. With the altered rules, 2/3 of the time the car will remain behind the unselected doors. At that point the host has a 1/2 of randomly ending the game. However, if the car was not chosen and the game continues, it doesn't matter whether it was random or with knowledge. For that instance of the game, the 2/3 probability still exists with the remaining set, which has been reduced down to one door. With this altered set of rules there are different probabilities before the host chooses though. There's a 1/3 probability the game ends at the point of the host making a choice. 2/3 of the time, there's then a 2/3 chance to still win. Therefore, there's a 4/9 overall. But again if we go to the point at which the host has already revealed a goat, it is 2/3 probability in that moment for the contest with a choice switch.
@@jasonalquiza noooo in the altered game (setup 2) there's a 1/3 chance the game ends immediately, a 1/3 chance your first pick is correct, and a 1/3 chance the box which isn't picked or removed will be correct. That's all before the host opens a box. The conditional probability in a game that continues - when the removed box does not end the game - is now 50/50 stick/switch. The intuition that it's a 50/50 choice is incorrect in setup 1, because the choice of removal made by the host is constrained in 2/3 of games played. But it's correct in setup 2.
It was clearly stated that the host revealed $1,000. This is not by accident. It just adds another level to the drama. Is $1,000 enough or is it worth risking for $1M. But the real point is that it's not the big prize. So what are the odds that switching gives you a better chance of winning the big prize? The first choice has a 1/3 chance of being $1M. That means the other two together have a 2/3 chance of being $1M. The host then reveals which one is $1,000 Therefore, the remaining one now has a 2/3 chance of being $1M.
@@lrvogt1257 sure, but there's a difference between "the host chose a random box (out of the two which weren't the contestant's first choice) that happened to contain $1000" and "the host chose one of the two boxes with the stipulation that they were not allowed to choose a $1m box, and it happened to contain $1000". It's this constraint that has an effect on the subsequent probabilistic assessment of where the $1m lies.
My favourite way to explain the Monty Hall problem is: Your chance of guessing the million dollar door when you make your initial choice is one out of three. Therefore in two out of three cases, the prize is not behind your door, but behind one of the other doors. When the host shows you which of the other two doors does not hide the prize, those two thirds become the one remaining door. In ONE out of three cases you will already have chosen the million dollar door. In that case the game show host shows you one out of two empty doors, and switching over to another door will make you lose. However in TWO out of three cases you will initially have chosen an empty door. Then the gameshow host opens another empty door. In these two out of three cases, the remaining door is the million dollar door.
Baloney, two doors do not become one. The game changes from one where you have a 1-in-3 chance to a game where you have a 1-in-2 chance. You do not improve you odds in the second game by changing your selection. Math major here and I'm calling you a BS artist.
You can think about the probability of prizes problem as your original choice keeps the same probability as when you selected, but the probability of the other doors gets added together and placed on the door not revealed.
The confusion becomes, why doesn't that same probability exist for the door selected first? Why does revealing one door increase the probability of only the door not selected?
@@dudeinoakland The key to understanding the problem is that _the prize does not move._ Starting out, there are X doors (where X = 3 in the classic problem), each with a prize probability of 1/X. All the doors are in one group and the probability of the prize being behind a door in that group is 1. However as soon as you choose a door, you have split the doors into two groups: we will call them C (door you chose) and R (the doors you rejected). There is only one door in group C, so the probability of the prize being in C is 1/X. There are X - 1 doors in R, so the probability of the prize being in R is (X - 1)/X. No matter which doors in R the host opens the probabilities of the two groups remain as they were, 1/X for C and (X-1)/X for R, because _the prize does not move._ Eventually there will be only one door left unopened in group R so that door must have the same probability of having the prize as group R does (and always had).
@@dudeinoaklandyou literally can switch from your initial one door to both the other doors and Monty also shows you, which of the other two doors is a loss.
The part that Simon leaves out of the puzzle that needs to be given to make the game work is that Monty will always open a curtain that does NOT reveal the million dollars.
@@twotubefamily9323 No, I have a science degree and went all through calculus. The point is that the "chances" are calculated by telling us that after selecting a door and one of the other doors is exposed, should we change our selection to the other door? BUT there is nothing in this set of "rules" that says the second door MIGHT expose the million dollars, and nothing in the rules that we cannot select the opened door showing the million dollars. THAT change the odds dramatically. But we ASSUME the rules are that the million dollars cannot be exposed at the second door opening and that we can't select the open door.
@@twotubefamily9323 blaster is correct. This is yet another sloppy/incorrect formulated Monty Hall problem. The host does not "cheekily decides" to open another door. In classic Monty Hall, the host opens another door knowing that it is not the big price. And he would have done this regardless which door you initially choose. THEN you will have a 2/3 chance to win if you switch. For example, if the host "cheekily decides" to open one of the other doors randomly, discovers it to not be the price and therefore offer you to switch. THEN you will still have 50/50 chance between the two remaining doors.
@sverkeren Respectfully, no! When you pick the first curtain, it has 1/3 chance of being correct. Meanwhile... *The other two curtains together* have 2/3 chance of holding the jackpot. The host removing the non-jackpot curtain of these two is just a detail - that 'side of the equation' is still 2/3 .
@@Michael_Arnold Respectfully, yes! In classic Monty Hall, if you first choose wrong (with 2/3 chance), then the host will practically show you where the price is by AVOIDING that door. If the host does not actively avoid the winning door, then you don't get that probability boost. Instead that probability get evenly distributed on the remaining doors. What could have happened matters.
The Monty Hall puzzle was presented incorrectly here, as it nearly always is. The fact that Monty offers you another choice is not sufficient. He may have done that specifically to throw you off. The math only works if he is guaranteed to give you that offer.
One could argue that it is not presented incorrectly, because the most famous presentation of the problem (the letter to Parade that vos Savant answered) also doesn't say the host is guaranteed to give you that offer.
@@insignificantfool8592 In that case, the solution offered is incorrect. The only way that solution works is if the host is guaranteed to offer you another door.
The "Monty Hall Problem" came about because of a question to Marylin Vos Savant. Somebody asked if it was better to switch or keep their original choice, though with one prize, and 2 with essentially nothing, and she proceeded to say it's best to change it. She was inundated with mail about the whole thing, including from mathematics professors who said she was wrong. (Many of them sent a second letter asking her not to publish their first letters, or keep their names out of it, after figuring out they were wrong.) Mrs. Savant followed up and even had a bunch of kids create an experiment where they run multiple tests using the same parameters, and the results all supported what she had said. She gave a similar example to the one above, with a hundred curtains. Since she was the one who originated this and gave the correct answer which confused the "experts" at first, her name should be included here.
Her answer is to be expected from the mind of a person conditioned to answering IQ questions. The correct answer is indeed 50/50 and not ⅓ to ⅔ . Interestingly Tibees does a video where she answers an IQ question as might a person with an IQ of 300 by coming up with a polynomial equation to generate the pattern presented for the trend given for which the next step in the series needs to be given . Her polynomial equation yields the original starting point instead of an obvious intuitive next number for the series.
Have you *read* what vos Savant wrote? She had a web page about the whole thing, though it seems you can now only find it in archives. In her own words, "So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose". The answer defines the conditions for answering the question? No. The question defines the conditions for answering the question. If the answer is defining the conditions for reaching the answer, it's circular reasoning.
@@hughobyrne2588 Wrong. In order to always open a losing door, whoever opens that door has to know which is a loosing door. But again, the video above shows the answer is correct, and a lot of grade schools participated in an experiment where it worked. There are even web pages that demonstrate this. I shared one link, but apparently that post vanished. I did 200 tests where I always changed, and hit real close to a 66% win rate.
@@philip5940 Not sure what happened to my response, possibly removed because it had a link. But the link was to where you could test it yourself. There are a few web pages set up to test this, and I ran 200 samples, always choosing to switch, and won real close to 66%. Also watch the video I am responding to. They say it works.
The Monty Hall problem only rewards changing your guess if it's assumed that the host knows where the million dollars is and will always open another curtain. If that condition isn't met, then the remaing two curtains each have s 50:50 chance of hiding the million dollars (and in one game out of every three, the million dollars will be behind the curtain the host opens, so you're choosing between the thiusand dollars and the banana, and you have a 50:50 chance of that)
I’m confused. If the host does not know, the only difference would be that they would sometimes accidentally reveal the big prize. And then the contestant would surely switch. So whichever way you think about it it’s always favorable to switch, but of course that does not mean you will always be correct. It only increases your chance of winning.
@@lenaforsgren the host reveals what's behind one door, and the contestant can choose to switch to the remaining unopened door. So if the host opens a door and the prize is there, the game is over and the contestant has lost.
@@Jaccayumitty If that were the rule, the switching strategy wouldn't even be favorable from the start. The outset was what I was thinking of. By the way, I realize now that if the host indeed has opened a door randomly AND it turned out to NOT be the car, the advantage of switching is actually lost. No disadvantage, though, either. I've never thought about that before, it's in my opinion less intuitive, since we're suddenly not talking about all the original possible outcomes.
The Monty Hall problem purely mathematically sure it make sense, but it’s not a purely mathematical decision. Deception is a very strong human trait and psychology of someone who didn’t have it figured out as Monty did would usually be trying to get you to change your answer because you had the right one picked. Just like in poker when people try to look strong when they have weak hands and are bluffing and try to look weak with strong hands to get paid. It’s basic human psychology more than math in this situation.
The believers of the "switching is better" solution will tell you that Monty will always offer the switch. In that case there is no room for deception. But as you say, this is counter everything that makes up human Interaktion, so obviously people tend to interpret it in a way that it was Montys decision and not a forced play. In that case, staying with your door is indeed not a bad idea.
Astonishingly, for every day life most of us probably learn all the math we're ever going to need by about grade 3 or 4. (Or, college in Florida) Throw in a bit of percentages and knowing how to average that comes in handy sometimes, and in some backwards countries fractions which I think are taught a little later, and that's about it. Which is not to say learning beyond add subtract multiply and divide is a waste of time. Depending on what you end up doing, you may need algebra or geometry or calculus, but the vast majority of us don't. Where we bump into it in day to day life, it's done for us.
On my first day of second year algebra in the tenth grade, my math teacher drew a 4 on the blackboard, then said, "See this four? It's not real." My fifteen year old brain was blown.🤯
The confusion comes in when forget that all of math comes down to counting. It's just that what is being counted has become more and more complex. The symbol "4" isn't real, but is used to mark that "there are 4 things". "1/4" means "1 thing out of 4 things", "4/2" is "4 things separated into 2 groups", and so on. The numbers themselves are not real, they're just labels for convenience that represent what IS real.
@@Vaeldarg Symbols are actually real, they represent real things. When we say they aren't real we mean they aren't concrete, but right now I'm writing letters that convey to your mind real things.
@@Vaeldarg Concepts are real things, so language and symbols are real. An idea of something is real. Real things aren't just material, what is invisible is also real. When someone says maths isn't real because it's just a symbol it's denying the spiritual reality of ideas, which are also real.
Simon: let me show you two mathematical ways you may be wrong about the world and how to better understand it. also Simon: Hehe just kidding, it's all a lie thanks Simon for destroying my mathematically perfect world today.
Instead of three, imagine there 10 doors. You pick one first, chance of picking right is 1/10. (The door is not revealed yet as that could end the show.) The chance of the prize behind the remaining doors is 9/10. The host opens 8 out of the remaining 9 doors, none of the 8 would be the prize. (The host knows where the prize is and revealing the prize would end the show.) Now the host basically eliminated the 8 wrong choices out of the 9 doors. The chance of the one remaining door the host did not open being the prize is very high (90%). The host basically just show us the answer and you should switch to that one remaining door.
I think this is an easy visual demonstration of the Monty Hall Problem odds. -Take 3 playing cards, The Ace of Spades is the winner. The other two are red cards. -Lay them face down (For this demonstration they can be face up since you are the host so you know which is which.) -Put a coin on each card which represents a 1 in 3 chance of winning. -Separate 1 set from the other two. -Turn over a red card from the remaining two while moving the coin off of that red card and putting it on the last remaining card. -Your original choice has 1 coin and the remaining card has 2. -Always switch.
I feel like the assumptions on the Monty Hall problem are not properly described in such videos. Most people assume that choosing a door and the proposal from the host are random. Which is not the case. The assumption should be that the host always has a calculated motive on his proposal.
@@lenaforsgren "If the host would choose randomly they would sometimes reveal the big prize and then the contestant would of course switch without hesitation…" The point is that if the host did reveal a non-prize randomly there would be no advantage in switching.
@@lenaforsgren "The point is that if the host did reveal a non-prize randomly there would be no advantage in switching." "That’s incorrect." It IS correct!!
@@lenaforsgren "Not relevant, however. Your initial chance of getting it right is 1/3, so the likelihood that your initial guess was wrong is bigger and you should therefore switch." Initially yes. But if the host acts randomly, when they reveal the non-prize door, the probability of having guessed wrong is reduced. "If the host would choose randomly they would sometimes reveal the big prize and then the contestant would of course switch without hesitation…" They don't though. The problem defines that they always reveal a non-prize door. Which means the player still is uncertain, and the host is not acting randomly.
@@lenaforsgren It does matter. If they acted randomly, the two selections by the player are two separate random experiments. The first selection gives the player 1/3 to win. And the second gives him 1/2 since the open door has 0 probability to win and is not in the population of possible winners. The initial 2/3 possibility to win if they changed no longer exists because of the revealing of the non-winner door. If the host does not act randomly, the three acts/selections are tied into a single experiment. What the host selects depends on what the player selected (always has to reveal a non-winner) and the second player selection depends on what the host selects to open. We no longer have plain random experiments. We are working with conditional probabilities.
The science is called Mathmatics. One side shortens it to Maths the other to Math. This also changed over the years which one was the more accepted one where.
Thank you for adding the 100 curtain option to the Monty Hall explanation. That finally made me understand this after hearing it at least 15 million times and not getting it.😊
This is an imprecise formulation of the Monty Hall problem. It is imperative to state that the host knows what is hidden behind each curtain and that he has no choice whether to give you a preview or not.
@@scottparker1741 explain then. op said the host must have complete knowledge of what door has what prize and must always offer the swap in order for the problem to have this solution. why is this wrong? idk maybe you know the solution is the same even when one or both of these conditions aren’t satisfied? i’ve always wondered how you solve it in those cases and i have sort of an idea but i’m not sure so please tell me if you know cause i’m curious
I get it, the Monty Hall Problem is from getting more information. You picked first of 100 squares randomly, no info. You want the million $ square, Then you see 98 bananas and one square not revealed. That's probably the $. You picked a random banana before probably. Simple. I love it.
I think it's important to emphasize the fact that the people disclosing what's behind curtain 3 KNOW it isn't the grand prize when they open it, and have chosen it for that very reason. When I've heard this explained before, I had the impression they were simply revealing one of the others at random, which might include the grand prize. That makes all the difference.
The Monty Hall problem (and this one is a considerable variation from the original - where are the goats) is, as is often the case, inadequately described. Once extremely important point is the the host must always open one of the other two curtains. It is not a matter of deciding to do it on the day, but they have to do it each and every time and, what's more, you have to know they are going to do it. Otherwise the host might be doing it selectively to mislead you. Also, of course, the host must know where each prize is, and they must never open either the competitor's original curtain or the winning curtain. Once all the conditions are clearly explained then it's simple; you change your decision and you double the chance of winning the prize from 1 in 3 to 2 in three. It is, incidentally, not a mathematical discovery at all. It's a bit of basic probability theory, just with a (usually) incompletely explained scenario.
Try modeling all the scenarios. Imagine a game with 3 doors, and 1 car and 2 goats. Assume the contestant chooses door 1. There are now 4 scenarios and two options each, resulting in eight outcomes. And the split of win/lose for stay/switch is 50/50. 1. Car / Goat / Goat - host shows you door 2 - switch lose / stay win 2. Car / Goat / Goat - host shows you door 3 - switch lose / stay win 3. Goat / Car / Goat - host shows you door 3 - switch win / stay lose 4. Goat / Goat / Car - host shows you door 2 - switch win / stay lose
@@khemkaslehrling3840 I know perfectly well the scenarios, and worked all this out long ago and even tested my mathematics with a simulation, just to make sure. However, the point you completely miss is that the Monty Hall problem is often not fully specified. The extra conditions are a) the host must know what is behind each door b) the host must only open another door with the lesser prize c) most importantly of all, the host must do this every single time Without (a) & (b) we are in a different scenario as either could lead to the main prize being revealed. If (c) is not met, then the host might only decide to open another door when the competitor has already picked the main prize. Of they might only do so if the competitor chose a lesser prize, or maybe some mixture of the two. The host might act to help, or to hinder. Like many probabilistic problems, it's essential to be very precise or it is easy to be lead astray.
@@TheEulerID I am forever mystified that people hear the words "game show" and do not immediately understand that the host will never reveal the prize. Thus, the game show host must know what is behind the doors. And with this it is inferred thhat the host always never reveals the prize. Perhaps some people really don't understand how game shows work?
The Monty Hall problem has a simple explanation that is easy to demonstrate. Consider every possible outcome. Let's say money is behind door 1 and bananas are behind 2 and 3. If you choose to not swap your door, the 3 outcomes are: Select door 1, stay with door 1, WIN Select door 2, stay with door 2, LOSE Select door 3, stay with door 3, LOSE If you choose to swap, the possible outcomes are: Select 1, he opens door 2 or 3, you swap, LOSE Select 2, he opens door 3, you swap to 1, WIN Select 3, he opens door 2, you swap to 1, WIN It's obvious then if you stay with your selection, 1 in 3 outcomes win. If you choose to swap, 2 in 3 outcomes win.
@@Stubbari that's true, but as everybody knows, the host won't reveal the big prize and will also never reveal your door. So, obviously, when contestant picks the thousand dollar door, the host will show the banana. It can't be any other way...
@@insignificantfool8592 Yes and no. I really think this is a poor example of the MHP because there are 3 different items rather than 2. Also, $1000 seems like a prize to me so there are 2 prizes and 1 non-prize.
The way I rational the Monty Hall problem is this: The first probability is still collapsing... If you don't modify your choice, the original 1/3 probability continues to collapse like normal... If you do modify your choice you are really getting 2 doors instead of 1... one picked by the host, and one by you.
You still get the same odds. All doors are 1/3. So the host reveals one... you still have 50% chance with the remaining doors. You gain nothing in terms of odds, you only get to see what was behind one door that you did not choose anyway. There's no 66% chance in this scenario, ever.
@@lgmediapcsalon9440 No. The key to understanding the problem is that _the prize does not move._ Starting out, there are X doors (where X = 3 in the classic problem), each with a prize probability of 1/X. All the doors are in one group and the probability of the prize being behind a door in that group is 1. However as soon as you choose a door, you have split the doors into two groups: we will call them C (door you chose) and R (the doors you rejected). There is only one door in group C, so the probability of the prize being in C is 1/X. There are X - 1 doors in R, so the probability of the prize being in R is (X - 1)/X. No matter which doors in R the host opens the probabilities of the two groups remain as they were, 1/X for C and (X-1)/X for R, because _the prize does not move._ Eventually there will be only one door left unopened in group R. What is the probability that that door hides the prize?
@@lgmediapcsalon9440 The only way the odds can "reset" is if the prize can move after Monty opens a door. Let's put it another way. What if before Monty asked you to choose a door, he asked you if you would like to choose one door or two doors? You'd be crazy to restrict yourself to just one door, right? Choosing one door gives you a 1/3 chance of winning, choosing two gives you a 2/3 chance of winning. Well that is exactly what Monty is offering you after he opens a door. Three doors: A, B, and C. From your perspective, A has a 1/3 chance of hiding the prize, B has a 1/3 chance of hiding the prize, and C has a 1/3 chance of hiding the prize. You choose door A. You have a 1/3 chance of winning. Monty opens door B or C to reveal a goat (the goat is not the prize). Monty now offers you the chance to stick or switch. This is equivalent to you previously being given the option of choosing door A or choosing _both B and C._ Since you are the one determining if the choice is between A or B+C, B or A+C, and C or A+B, and since the prize is already in place, Monty cannot change the probabilities. If you choose only one door you have a 1/3 chance of winning. If you choose two doors, by switching, you have a 2/3 chance of winning.
Velocity is the wrong word anyway. Velocity is a vector with a direction component. People on the opposite sides of the planet fall in opposite directions to each other. The word he's looking for is "speed", and even then (like you pointed out) he's still wrong. Much like his re-telling of the Monty Hall Problem was crucially inaccurate.
This, like so many explanations of the Monty Hall problem are wrong. This fun mathematical trick ONLY works if the game show host deliberately shows you an incorrect curtain. If he reveals at random, then you have a 1/3 to see the million dollars, and that is where your other 33% is coming from.
Correct, the host knows where the car is so the curtain he reveals is not at random. But this should go without saying. He’s not just going to reveal the car because then there’s no game. So of course he knows where it is.
I never understood how the Monty Hall problem worked until I saw the Parade Magazine chart solution, and the importance of the "intervention" of the Monty Hall person. The key idea is that the person revealing what is behind one of the curtains follows the rule that they cannot show you the prize. They have to show you the other bad choice. They cannot show you what is behind the curtain you picked either. This is the key idea that changes the odds of winning if you switch your choice. There are three possible outcomes. If you keep your choice, only in one scenario you are right, other two choices are wrong. But in the other two scenarios you are wrong in your first choice. That means that two out of three times you have eliminated the lesser choice. Monty then shows you the other curtain, and it cannot be the prize. Monty has now, two out of three times, shown (or proved) to you which curtain has the prize. It is not the one he showed you, and two out of three times it is not the one you picked. So two out of three times it is the other curtain. If you can switch you will therefore win two out of three times! Your odds of winning when switching are 66%.
In what game show in the world is the big prize given away at random. i'm mystified as to why people hear the words "game show" and don't immediately understand the host doesn't just give the prize away at random, but makes a selection that gives the contestant a choice, and thus intrigue for the viewer.
@@barryschwarz Try 3-2-1. In it the contestants were given clues as to what each prize was. There were 5 clues given one at a time, but once 3 were on the table the contestants had to get rid of one. The prizes ranged from the big main prize down to a dustbin booby prize (and yes, it was won on a few occasions). If they could not work out the clues, they would have to pick one at random to get rid of. Either by chance or by not working out the clues properly they could end up rejecting the big main prize. Although not exactly giving the main prize at random, it did have the chance that the main prize would be eliminated, possibly by random chance.
@@cigmorfil4101 No, that doesn't fit the bill. I don't know any game show that would give away a million due to a random choice made by the host. And it the context of a game show, it just doesn't make sense. I think what actually happens is that people make the common intuitive choice that the remaining two doors are a 50/50 chance, and then try to justify their intuition (eve when they figure out it was incorrect) by arguing post-hoc that the host might give away the prize at random (every 3rd game!). You see various iterations of this post-hoc justification. Some people argue that it's 50/50 if a new contestant steps in after the first choice, and doesn't see which door was first chosen. Well, yes, now the choice is 50/50, but why change the format of the game to make that argument? I'm sure there are people who hear the words "game show" and actually imagine that the host might pick a door at random, often revealing the $1,000,000 prize. But I doubt this is the majority of people. And the riddle infers the same action each time, it does not infer in any way that the host might reveal the prize... and then ask the contestant to "stick or switch?" No, it doesn't maker sense.
The Monty Hall problem is explained ever simpler than this. Before you started, you had a 2 out of 3 chance of picking either the banana or the thousand bucks right out of the gate, or a 66.67% chance. You only had a 1 in 3 chance of picking the million. However, if he offers you the chance to switch and you take it, you have doubled your chances of winning the million. Here's how. There is a 1 in 3 chance you picked the million, and if did and you switch, you will lose a million bucks. But there was only a 33 & a third% chance of this. If you pick either the banana or the thousand dollars, Monty will always reveal the OTHER lesser prize, so when you switch, 2 out of 3 times you will switch TO the million bucks.
Monty Hall problem originated from Marylin vos Savant. She originally faced quite a backlash even among University Professors. And we see who was right.
I think it's fair to say the originator of the problem is the person who asked the problem, Craig F. Whitaker. He wrote a letter to vos Savant while she was writing a regular column in Parade magazine. In Whitaker's letter, as in the question as phrased at the beginning of the video, there is no indication that a compulsion is placed on the game show host to always reveal a door with the booby prize, a critical component in the chain of logic which leads to the conclusion "it's better to switch".
Well done summary and so quick. This way of learning is a secret that most will never get to hear. I hope all watchers give it a shot, especially if you’re still in school and have time to think.
Another way to consider the Monty Hall problem is not to think you have a 1/3 chance of getting it right on your first guess, but that you have a 2/3 chance of getting it WRONG. So the 2/3 chance of getting it right is now with the other two doors, initially split between them. And by removing one of those doors from the equation, that 2/3 correct chance is now with the remaining door.
That is not correct. You would have a 2/3 chance of picking a wrong door regardless as to the host's knowledge. In that case if he opened one of his two doors that reveals a non-prize there would be no advantage in switching.
But by removing one door and making that door 100% wrong then it is a 50/50 it may be one of the remaining doors. 6ou can come u0 with some pseudo intellectual nonsense explaining how the myths shows that no to be true but in real world application its 50/50 not 66/33
@@williamdoherty7824 Create an experiment to test it. You can do it with playing cards and a friend who knows the cards in the hole. But you need to repeat the experiment a number of times to get an approaching probability.
@@klaus7443how can you start a sentence correctly sayin your initial chance is 2/3 to be wrong and then ending it with the conclusion that going away from that choice gives you no advantage? Weird.
Great video! Re: Part 3 - It always made me think about how the simplest visible shape we can conceive of is a circle. But our math & ten-based number system can't describe the relationship between the diameter and the circumference without an undefinable number (pi). Always thought that maybe we went down a slightly wrong rabbit hole right from the beginning
2:17 sorry my friend but you are wrong. You did not at any point stipulate that the host MUST reveal any particular prize. the host could have revealed the million dollars. there is a 1/3 chance that you could have picked the banana or the million dollars. if you'd stipulated that the host must reveal either the 1000 or the banana, then you MAY have a point, but you did not.
I was thinking the same thing. If you are trying to get the bananas, and the host showed you the $1000, so you switched, is your chances of getting the banana higher? Or course not. This only works if 2 of the 3 curtains have the same thing begun it.
I know you posted this 3 months ago - but you're logic on this is pretty common, but it's wrong. The accepted premise is that the host will reveal 1 of the 2 non selected doors before giving you the decision to switch or not. When you originally pick your door, you have a 1/3 chance in picking right. There is a 2/3 chance you chose wrong though, right? So would you trade your one door probability for the two doors you didn't choose? Of course you would. Well essentially that's what you're doing. But simply the host is revealing that one of his doors doesn't have the prize. But, we already knew one of them was empty, so revealing which one is empty doesn't actually change the probability that one of his doors is more likely to have the prize. Hope that makes sense
The simple explanation of the Monty Hall Problem: At the beginning you have a 1/3 chance of picking the prize. After the host opens a door you still have that 1/3 chance of having picked correctly. Therefore the other closed door must have a 2/3 chance as the prize had to lie behind one of the two closed doors, so you should switch.
I look back in time to Sagan, et al and how much fun I had reading about science. Listening to your posts has reignited my interest in science. I have Van Nostrems mathematical book.
Just when I thought I had a handle on 1 + 1 = 2 you come along and say hold on might not be real. I always loved the Monty Hall problem. Great video thumbs up.
The best way I've found to think of the Monte Hall problem is to make the question a little different from "What are the probabilities?" Imagine instead that person A *always* switches curtains and person B *always* keeps the same one. Person B will win if they picked the right curtain the first time, and person A will win if they picked the *wrong* curtain the first time (since they'll definitely switch to the right one). Who would you rather be? This ends up matching the probabilities explained in the video: person A picks the wrong curtain at first about 2/3 of the time, and will thus win after switching on those 2/3 of the games. Person B picks the right curtain at first about 1/3 of the time, and will thus win without switching those 1/3 of the games.
There are other variables that haven't been considered. 1. Am I a monkey or other great ape that would enjoy a banana over a monetary prize? 2. Am I starving at the moment? 3. Based on the current inflation trend could a million dollars at the point in time discussed in this video buy a single banana?
Awesome video. Missed opportunity when you pointed at the light and said, "Just like I can't be sure my blue is your blue." Imagine if you said red or another colour.
monty hall problem. when there are 3 doors, each one is 33% to be correct, which mean whichever one you pick is 66% to be wrong. then 1 door gets eliminated, leaving only 2, which is actually just smoke and mirrors. obviously , the host is never gonna show you the correct door. he's always gonna show you an incorrect door, which actually make that one irrelevant. your initial odds never change. the door you picked is still 66% to be wrong and 33% to be correct, but since you are now being given the opportunity to switch to the other remaining door, then switching means you are 66% to be correct 33% wrong..
My dad used to always say there's 3 types of people in this world.. Those that can count and those that can't.... Very important life lesson I think 🤔
Yours too? That's 2 now let's see if we can get a third person whose dad didnt say that and see who can count.
All 3 of my dads would say that
There are 10 types of people in the world. Those who understand binary and those who don’t.
There's 1 types of people, those who know binary and those who don't.
That's deeper than it seems
The key to the Monty Hall problem is "the reveal". The host HAS TO show you a wrong door... But he doesn't choose which door that is until AFTER you've made your choice. The original odds (1/3 v 2/3) never changed.
Ex, let's say prize is in door #2...
-If you pick door 1, he HAS TO show you #3 - Switch and you Win.
-If you pick #2, he can show you EITHER door #1 or door #3 - Switch and you Lose
-If you pick door #3, he HAS TO show you #1 - Switch and you Win.
Still only have a 1/2 chance. Monty will open the curtain if you chose the right door.
@@minrityreprt6302no, you have 1/3 chances if you keep your original pick and 2/3 if you switch. imagine the same game with 1000 doors where the host opens 998 doors which he KNOWS to be wrong AFTER you picked a random one in 1000. you’re obviously gonna switch. it’s intuitive. now in this case it’s 1/1000 versus 999/1000 so it’s not the same but the math is the same and, in the 3 doors case, it’ll lead you to 1/3 vs 2/3
If the point of the Monty Hall problem is to take into account possible deception of the host, then there is NO math. It is only understanding the host's stratagem: He guessed the $1m curtain, use doubt to get him to pick the banana
@@iamcomcy Nope. It's already been proven mathematically - if you switch, you'll win 66% of the time.
@@nathanlawson313 Yes, if you know beforehand that the host will offer the switch whether you pick the right curtain or not. But what if the host only offers the switch if you pick the right curtain?
The keys to understanding The Monty Hall Problem are realizing that the host knows which curtain hides the winner, and that when he opens a curtain, he'll always choose to reveal one of the losers (otherwise it'd ruin the game, right?)
When you choose one of the three curtains at the beginning, you are also deciding which two curtains the host can choose from when it's his turn to open one. If you pick the winner initally (a 1 in 3 chance), he'll have two losers to pick from, and if you pick one of the losers (a 2 in 3 chance), he'll have one loser and one winner to pick from (being forced to pick the loser as stated above).
So when he opens one of the remaining curtains and reveals the loser, there's a 2 in 3 chance he was forced to do so (because the other curtain that you didn't chose initially is the winner), meaning 2 in 3 times you chose a loser at the beginning. So yeah, you should switch.
EDIT: I love the pedants insisting on applying real game show logic to a math problem: I get it, Monty could have not followed the rules. I can only imagine you in grade school: "ACKSHUALLY MRS SMITH, if Ralphie had 5 Pennies and gave 2 to Suzie, he would have 4 left because he likely palmed one!"
Ah! Now it makes sense! Thanks:)
Exactly! Really well explained
great explanation. I was just thinking that since he is not going to show you the winner, and he is not going to show the one you picked, so seems that there was a good chance that he was forced to show the one he did.
so I was on the right track, but your explanation clearly explained the why while I was only circling around the edges of the reason!
nicely done.
Wonderful explanation, it makes complete sense now. Thanks!
This. This explains it. Thank you. Now I understand why the probabilities change the way they do…too bad my stats teacher couldn’t have used this in class…
I actually kind of like the way Mythbusters explained the Monty Hall problem. If you consider your door is a 1/3 chance and the other two doors are a combined 2/3 chance, the fact that one of those doors is opened doesn't change the fact that they are a combined 2/3 chance.
Wait a minute. Combined they have a 2/3 chance. Break them up and they don't each retain a 2/3 chance of winning! The pairing is arbitrary.
Option 1. Say you have two picks. The first pick has a 1/3 chance of being right. If you were wrong, your second pick would have a 50% chance of being right. Regardless of which one you pick. In total you had a 2/3 chance of winning. The doors could have been 2 red and 1 blue door or 2 blue and 1 red door. It doesn't matter how they are paired.
Option 2. You can pick one door or pick a pair of doors. Naturally you would pick a pair of doors (say #2 & #3) because that would also give you a 2/3 chance of winning. Then the host says surprise, door three just disappears, it's gone. So does the 1/3 chance that that door being the winner. I don't see why the 1/3 chance that door three had would transfer to the remaining door in your pair, any more than it would transfer to door number 1. There are no transfer of odds.
The fact that the host won't show you if you win ruins that
@@djtjm7000 It actually does matter because the door that is shown the contestant is *always* a loser door. If it were random, you might be right, but it's not. And that's why this math trick is deceptive. People act like the last two doors are a 50/50 chance because they act like the elimination of the first door is random *but it's not*.
If you picked a winning door, the eliminated door will be a loser because both other doors are losers. But if you picked a losing door, the remaining door that doesn't get opened *is always the winning door*. Given there is a 2/3 chance you picked a losing door to begin with, there is a 2/3 chance the remaining door is the winning door, because the door that gets opened is not random.
@@djtjm7000 I tend to agree with you. When one door is removed, the odds remaining turn into 50/50 that I picked correctly the first time versus the remaining door. I would argue that the 1/3 that got removed gets split into the two remaining doors, since we don't know which of them is the winning door. It is kind of like flipping a penny. You flip it 9 times and it comes heads (unlikely but possible, presuming 50/50 change heads/tails). The next flip still has a 50/50 chance of coming heads.
You explained it better than Simon.
This video hit close to home. Being color-blind in the green-red spectrum, I can't count the number of times that I have attempted to explain what the world looks like to me. People are often surprised to find that to me the world looks perfectly normal for it is the only way that I have ever seen it. I was told that certain color was green and therefore I call it green. The odds are my green differs from your green. The best way to explain how my vision varies is to show its' impact. For instance, showing up for a dress-green uniform inspection in the military wearing blue suit pants (yes, I really did that.)
The questions posed in this in regards colorblindness I totally understand my best friend has proposed the prospective situation. What his of green perception to my green there's no way to confirm or deny that.
The trouble with explaining colorblindness to people is that Certain Shade blend together there are certain shades of green very light look yellow to me are things like moss which are red sort of green throwing some Ferns and I can't find my golf discs
That happened to a co-worker of mine, too.
There are women who have 4 different cones..
@@khironkinney1667 I have know idea what you wrote. It is extremely hard to understand. I would recommend either editing it with correct grammar and punctuation. Otherwise if English isn’t your first language then just write in the language your best in.
The Monty Hall problem was not clearly explained in the first go. It's absolutely critical to say the host then reveals a door that ISN'T the winner based on his knowledge of which door has the winning prize. If he just shows door 2 when you choose 1 and door 3 when you choose 2 and door 1 when you choose 3 (or any other just guess) regardless of what has the prize or not then your odds don't change. But he has a chance of just revealing the winner that way and then you don't get a choice to swap anyhow. It's using that extra information that changes the odds. He must know what the winning door is.
I think most people understand that option 2, revealed by the host, will never...can never be the prize. No one thinks the host doesn't know where everything is.
Your option 1 is probably wrong, option 2 is always wrong, therefore option 3 is probably right.
With the Monty Hall one there's a simple way to look at it. When you first pick you are choosing one out of three, which gives you a 33.3% chance. When Monty shows you whats behind a curtain it removes that 33.3% from the guess. If you stick with your original pick then you stay at the 33.3%. If you switch though, you get the 33.3% from the curtain you switched to plus the 33.3% from the one he revealed.
Oh! Thanks for the clarification that makes a lot more sense now!
Probably one of my favorite SP videos so far, great Job from Simon and Team. I'd love to see more content on rather complicated topics like this.
Yepp, please!
Yes!
My cognitive dissonance isn’t allowing me to accept some of this 😔
0:45 - Chapter 1 - The monty hall problem
4:30 - Chapter 2 - The nash equilibrium
11:30 - Chapter 3 - Math might not be real
This man probably does not understand the Monty Hall problem or he would initially explained that the host does not open a random door but one of the doors that does not hide the million or the door you chose
He does understand. From the contestants view it's random. Obviously if you selected wrong, he isn't going to open the million dollar door.
@@pingamalinga that makes no sense, come again
@@bjorneriksson6480 Are you suggesting that the man doing the video does not understand? If that is your assertion, I am disagreeing with you. Of course he knows the Monty door is by design. The host knows where the prize is. It is presented as a random door because the contests does not know.
@@pingamalingabut it being random or not changes the qhole situation
Opening the”random” door was proven to increase suspense, captivate the audience, and stall for a commercial break. In the 3 door case it’s still a high chance of being wrong, and most people don’t understand how the probability equation would work. But the guy doing the video is from the “Today I found out” and he’s awesome.
The Monty Hall problem is that your choice of curtain is random - the hosts choice to show you is NOT random. The host never shows you your curtain OR the curtain with the $1,000,000. It's the combination of non-random and random variables that causes most of the confusion.
Yeah, it especially causes confusion if those assumptions you're stating are not even mentioned. Why do I have to go along with these assumptions then?
I agree... When Monty opens a curtain he is revealing information.
@@insignificantfool8592 It doesn't actually depend on those factors though. By choosing a door, you yourself are fixing a condition. The fact that one of the doors doesn't contain the goat that was chosen while you were holding a door means the odds are now 2:3 for the last goat. This is irrespective of the fact the host will never open your door or the goat's door.
@@Demmrir many people think so, but it's not true. Imagine the host really doesn't want the contestant to win the car (maybe he has to pay for the prizes himself and is low on cash). What exactly prevents him from opening the contestant's door if he chose a goat door and only offer the switch when he chose the car door?
The tendency to stick with one's original choice is also reinforced by the contestant assuming that the host knows he or she chose the $1 million and is intentionally trying to throw him or her off. The other assumption being that if you had chosen wrong, he wouldn't give you an opportunity to change your mind.
Thanks to @StorkBomb7417 below, I remembered the proper explanation to the Monty Hall problem (which is vastly different to what Simon gave).
It boils down to two scenarios:
Scenario A - you picked the right door (1/3 chance): if you switch here to the other door you lose.
Scenario B - you picked the wrong door, Monty opens the only other wrong door (2/3 chance): if you switch your door here, you are guaranteed to get the right door as it’s the only other door except the one you initially chose.
Since Scenario B happens 2/3rds of the time, it’s a statistically better strategy to presume you’re in scenario B and switch your choice.
Imagine the Monty hall problem starting with 100 doors, only one having the car behind it. You pick a door and then the host opens 98 doors (all but the one you picked and another door you didn't pick) that have a goat behind it and then asks you if you want to switch.
This was a great presentation, Simon. I also quite enjoy this format so much. Thanks for all you do : )
And thank you for watching :)
I gotta be honest. This is one of my favorite videos you've ever made, Simon. Not really sure why but it's very interesting.
Because he's let us know that while we may not know how to calculate the velocity of a baseball, our brains are fully aware of how the baseball is moving.
Best SideProjects I've seen in a while. Not that your other SPs weren't good, just that this one was amazing.
Thanks, Simon!
Agreed. This was very solid.
There’s two kinds of people in the world.
1. Those that can extrapolate information from incomplete data
2.
It's not a neutron bomb. It's a thermonuclear weapon, aka the H-bomb. The neutron bomb was a low yeald, high radiation weapon meant to kill with minimal damage to the area, mostly meant to stop Soviet tanks in Europe during the cold war.
It's amazing how quickly this information has faded into the past. It just shows how young the host is and how poor our education system is.
After trying to contradict your Monty Python odds, I concede. You're correct.
You have 2/3 chance of picking the wrong door. So it's better to choose the other door.
I still dont think that is true. When you are given the chance to either keep your current choice or choose another, you are actually just rechoosing based on what is still unknown. Just because they say "keep your current choice" does not mean that you are keeping your initial probability.
Please respond, because there might be a high probability that I'm missing something :P
@Wuddigot Simplest way to think about it.
You know how the game works. You know that you will be shown a losing door after you choose a door.
Beginning of the game you have a 1 in 3 chance of picking the winning door. A 2 in 3 chance of picking the wrong door.
Would you agree 2 in 3 odds are better than 1 in 3? So you are most likely to choose a wrong door.
So knowing how the game works, you pick your first door knowing you're more likely to pick one of the 2 out of 3 that are losers. Because of this, when a door is removed, you pick the other.
Definitely not a fool proof way to win
Just gives you better odds.
You gotta look at the odds of picking the wrong door to determine the best odds of picking the correct door.
@@Wuddigot I just reread you comment.
Keeping the first door keeps your original 1 in 3 odds. So keeping your first choice does keep your original probability.
Gotta bring the odds of choosing one of the wrong doors into the equation. 2 in 3, which is more likely.
Start of game you're thinking "I have the best chance of picking a wrong door" (2 in 3)
So after you pick a door (odds are its a loser) you know one of the other two doors are more likely to be the winning door. Right?
Since one of them doors are removed from the game, you pick the other. Giving you 2 in 3 odds of it being the winning door.
@@al1383 That just isnt true though. The odds of you choosing the wrong door is now less because there are less wrong doors that you would choose from. You will not choose a door you know is wrong, and therefor the conditions (and probability) have changed.
Am I hearing the initial problem incorrectly or something? guy chooses door, before result is revealed he is told one of the wrong doors and given the choice to choose again... Is that it? he now still does not know which of the doors is right, but he does know that one of the two remaining doors is wrong.
If I am not missing a detail, this is an iterative problem, as in each completion of the action loop starting at the beginning with a different set of conditions.
@@Wuddigot you going to get me racking my brain on this again! 😃
To tired today. I'll respond later
The Monty Hall is easier to understand when you try to replicate the experiment in code.
winningDoor = random number between 1-3
initialChoice = random number between 1-3
if winningDoor = initialChoice, stayingWins
otherwise, switchingWins
So there’s a 1/3 chance the player’s initial pick was the winning door. So the rest of the chance (2/3) has to be that switching would win, because the wrong door was already revealed.
What, Simon actually watched a movie?
We need to start making a list of the movies he's actually admitted to watching
new channel for him: CinemaGraphics
@@alyssinwilliams4570 lol
That can't be more than the third time I've heard that from Simon. Although thousands of times it's wtf? Lol
All 3of them!
yes
I once thought I'd prove the right answer to the Monty Hall problem to myself once and for all by writing a simulation and running it a bunch of times to see how the probabilities landed -- but once I wrote it, I didn't have to run it at all. The answer jumped out at me from the code while writing the step where the host opens a door. I ended up with 3 cases to account for: the one where the contestant picked the right answer -- in which case it didn't matter which other door I picked -- and the two cases where they picked the wrong door, in which case there was no choice about which door to open because the rules allowed for only one of them. In only one of those scenarios would switching be the wrong choice; in the other two it would be right.
Same here. I also thought it was wrong when I first heard it, wrote a computer simulation, and had figured out the same thing before I even hit run.
You can't put the two stage MH problem into code, since we don't know what the host will decide to do. How do you model the host's decision as to whether to open the player's door or whether to offer a switch?
@@insignificantfool8592 the host knows which door is the winning door. If you picked the winning door, then the eliminated door is irrelevant. If you didn't then there's only one door that can be eliminated. So basically your code is - identify which 2 doors are left, if the first isn't the winning one, eliminate it, if it is, eliminate the other non-selected one.
@Jonathan Newman that would be assuming the host always has to offer the switch. That was not mentioned and thus shouldn't be assumed.
@@insignificantfool8592 the premise of the problem is that the switch is always offered.
You mention Richard Hamming in the "Maths may not be real segment" - he's well known in computer science circles for the development of error detection and correction algorithms, the descendants of which are used in computers & data communications to this day. Wikipedia has an article on these, named for him:
en.wikipedia.org/wiki/Hamming_code
I use the Hamming Distance in anomaly detection frameworks. Uncanny how it works!
I read your comment as "You mention Richard Hammond..." and was subsequently a little confused.
Another way to look at the Monty Hall problem is that it's effectively the same as giving you the option of sticking with your original choice or changing to selecting both other doors.
My favourite explanation as well. Or in the case of 100 doors, the choice is keep your 1 door or choose all of the 99 other doors.
Totally agree
That’s a great insight
Alright, I can live with that. But people want to believe in their luck so that's where my problem was.
Exactly. It doesn't matter when you decide to switch, the probability locks in when you choose the door. 1/3 you're right and 2/3 chance you're wrong. Switching is basically just saying: I think it's not the one I picked first.
Great explanation of the Monty Hall problem. I think it's so confusing because it _isn't_ a pure maths problem. The host doesn't pick a door at random, but people intuitively assume it's random chance.
It doesn't matter what the host chooses to reveal if it's a fair game, though. In the video's example, you still either chose the $1M or the banana, it's not like the banana was what was revealed. The remaining 2 curtains becomes that psychological poison chalice dilemma with the host at that point. It would only be unfair odds if the host can cheat by changing what prize was behind the chosen curtain when asking if you're sure.
Simon's explanation is wrong. He said: "He [the host] reveals that concealed behind curtain number 3 is a thousand dollars", but he did not establish that the host always reveals a curtain. That is actually a huge difference, because without this additional information, the problem is unsolvable. The host could always reveal a curtain, he could reveal a curtain because you guessed right and it is too early for someone to win the main prince. Many versions of the Monty Hall problem make this mistake.
The host knows which curtain hides the million, and he will never open this curtain, and he will never reveal what's behind the curtain you chose, so there are three scenarios :
1 : If you chose the Million - he can reveal what's behind either of the other curtains at random - if you change you lose
2 : If you chose the Thousand - he can *only* reveal the Banana - if you change you win
3 : if you chose the Banana - he can *only* reveal the thousand - if you change you win
He has knowledge and restricted choices, so what is revealed does change the odds ..
@@davidioanhedges The problem with your idea is there will never be a 2nd curtain reveal out of 3. If the 1st one revealed shows $1000, what's the 2nd reveal going to show? If it reveals the $1M, you choose the revealed $1M. If it reveals the banana, you choose the only unrevealed curtain remaining since it will be the $1M.
@@davidioanhedges Precisely.
"Math is perfect, real life is not" a very beautiful quote the represents the reality. Here is an example: I studied electronics and since I love playing bass, I decided to make an amplifier from scratch. I even got specialized equipment like an oscilloscope to get the best idea of what's going on. Here is what I found out: the equations that are supposed to give you the results of your circuit are simply not correct. Of course some *DO* give you the results you should expect after doing the math but things were acting very weird to the point I gave up because they made no sense and I didn't know how to proceed to complete it. Multisim, a program that allows digital simulation of a circuit would give me errors all the time because it, just like me, just couldn't comprehend wtf is going on there. The circuit outputted about 3% of what I wanted and since it wasn't any major project or something, I simply gave up and left it for when I have the will to complete it, at some point in the future, hopefully.
Before you blame me, I am confident that the circuit was correct. Only faulty parts could have been the reason for this but I just bought them and they were brand new, so, I don't know.
whelp, atleast you got your name right.
The circuit output 3% of what you expected? Well that is you :)
Circuit theory is only an approximation of circuit behaviour. V=IR, I=C dv/dt, V = L di/dt, KVL, KCL, etc are all only valid given assumptions about the size of the circuit, no interference from outside, and the speed of stimulus. They don't even describe the physical processes that lead to those quasi-steady state conditions described by those equations.
A similar thing is going on with mathematical models describing semiconductor behaviour. They are simplified approximations whose equations are only considered valid when corresponding assumptions are valid. A single device can have multiple equations modelling behaviour and you can make the model more or less accurate at the expense of complexity.
The models that humans use to design with are broadly simplified and will get you in the right ballpark. Computer simulation can use more complicated models and get you closer still. But this is before manufacturing variability is factored in, not to mention environmental factors that you may not be aware of or choose to ignore.
Long story short, designs are broadly correct if your assumptions are correct but building something in the environment it is intended to operate in is always necessary to make sure things operate as intended and something(s) hasn't been overlooked. Getting an output 3% of expected is a significant error someplace. If everyone's designs were 97% off, it would be a fluke that anything worked. Electronic design would be more like witchcraft and the models used would be essentially worthless.
This video is gold and I plan to show it again and again in my classes. (HS maths teacher) Your work continues to be so very good and awesome.
Your gold, my red.
I think that the main thing that people who explain the Monte Hall problem never really pin down into simple words is that Monte Hall would NEVER open the door to the million dollar prize on purpose.
That is critical to the problem and I've never seen anyone explain the problem and mention that. Anyone who has seen the show automatically knows this critical detail.
Oh? I've read numerous examples just today, where people point out that Monte would never open the big prize door.
I don't understand wy the words "game show" don't automatically infer this to people. In what game show in the world is the big prize given away at random?
It's always stated that the second step is to reveal a non-prize. It's never an IF.
@@lrvogt1257 Yep. In every version of this I've seen, the clear inference is that the host always opens a goat door.
@@barryschwarz : Right, but it's more than an inference. It is the second step of the game.
I think the funniest example of a Nash Equilibrium is when two people approach each other while walking around a corner. Both frighten each other and for whatever reason both step to the left and both step to the right at the same exact time. This is actually the opposite of the equilibrium. The correct method would be to converse with the other participant and say "Go to the right" or just grab them by the shoulders and do a 180 degree spin with them so neither can get confused. My method is that whenever this happens to me my first and only instinct, instead of getting stuck in the dumb left right left right loop of laughter and embarrassment, is to stand directly upright and perfectly still like a deer in the headlights. It freaks the other person out which makes them go around me faster, but also leaves no room for guesswork and they know that my intention is to let them go around me in whichever direction they want.
Not sure where it falls, but my default is to stand still and let the other person move around me. Saves a bit of time, but I lose out on the two-step shuffle dance moves.
My theory is most people will step to the dominant side. That being the right hand side as most people are right handed. So my question is are you ambidextrous???
This is a Canadian stalemate. Both wants to move for the other and not look rude, so we spend hours just going back and forth saying "you first, no you first, no I insist you first". And "sorry. No I'm sorry."
If you're going in the opposite direction, it works fine when both step to the right or to the left. It only fails if one person steps to the right and the other steps to the left.
I just grab them by the scruff of the neck and beat the living daylights out of them, then I turn around and walk back the way I came from, it's far from ideal but who's perfect right
Watching this after a 12 hour shift and being told maths might not be real, I’m too tired my brain hurts
You probably shouldn't watch videos on the channels Stand-Up Maths or Dr Brian Keatong after a 12 hr shift then either lol
The way I'd look at it is that much of mathematics is a toolkit for manipulating information. But not in arbitrary ways; the manipulations have to follow various properties like being self consistent and following deductive logic. It's not surprising that tools for deriving information -- essentially augmenting our intuitive reasoning -- would be useful.
And then, further to this, the question of whether maths will always be useful comes down to whether our universe has properties like self consistency. Well, if it doesn't, then all bets are off.
The monty hall problem can be better described as you pick door 1. The others are grouped together. [1] [2, 3]. The fact that door 3 gets revealed does not change that grouping. Visually grouping [2, 3] seems to help explain it better for most people.
I think he must have explained it wrong, because the way he showed it still comes out to a 50/50 choice, by my calculations.
Step 1. Pick between three.
Step 2. Throw that choice in the garbage because it doesn't matter.
Step 3. Pick between two.
@@johndaniel7161 the odds don't change because an incorrect choice is revealed. That is why grouping visually helps explain.
[1] [2, 3], odds of 1 are 33.3%, odds of 2 or 3 are 66.7% because they are grouped. Revealing 3 is a goat does not change the odds, 1 remains 33% and the group is 66% even though one in the group is invalid.
@@johndaniel7161 I should add that the host knows which is the winning choice and which are the non-winning choices. When the host reveals a door in the grouped set they intentionally and knowingly reveal a non-winning choice (which actually matters a bit).
@@bretmcdanel8595 It technically does not matter if the host revealed a winning choice, if the format was still "Do you want the best prize from among the doors you did not pick?", it would just make a bad gameshow - because that's what the host is *actually* asking, just phrased to hide it, and why the grouping helps. You're not being given a chance to switch within two doors, you're being given the chance to pick the best from two thirds of the doors, just you already know one of them is a dud.
@@iskierka8399 But that's kind of the point. The host can't do that, because it's against the rules of the game. When you're playing blackjack, the dealer cannot refuse to deal you a card, because then the game cannot commence. Similarly, the game host cannot reveal the winning door, because then the game is over. It's because the rules are set in stone that this is a mathematics problem instead of a sociology experiment.
With the Monty Hall one there's a simple way to look at it. When you first pick you are choosing one out of three, which gives you a 33.3% chance. When Monty shows you whats behind a curtain it removes that 33.3% from the guess. If you stick with your original pick then you stay at the 33.3%. If you switch though, you get the 33.3% from the curtain you switched to plus the 33.3% from the one he revealed.
In The Monty Hall Problem, if you imagine the steps in the game taking place in a different order, the probabilities become clearer to see. Imagine that the host asks you to pick one of three curtains. After you pick one she simply says you can have the prize behind the curtain you picked, or the best prize that lies behind the other two curtains. I think most people would intuitively pick the “best prize behind two curtains” over the “prize behind one curtain”.
Considering this in the context of the actual order of things in the game, the host knows where the big prize is and will never open that curtain. After the host opens that curtain, when the player chooses to switch it seems like they are choosing the prize behind one of two remaining curtains, but actually they are still choosing the “the best prize that lies behind the other two curtains”.
I like when the problem is extrapolated to the 100 curtains. It becomes easier to see. Will you take the prize behind the curtain you picked or the best prize that lies behind the other 99 curtains?
Excellent explanation
You did the best explanation but I'm still confused 😆
Excellent explanation. I've always stuck with "yeah nah it's now 50/50", even after watching the video, but 10 minutes ago something happened in my brain and something made sense but I didn't know why. This explained it well.
Pretty bad explanation here
What if you happen to have picked the right door anyway? The host will be forced to pick crappier doors, and in that case switching would be bad.
I always explain the Monty Hall problem by saying "I'm thinking of a person". When they guess I say "I'll give you a clue, it's either the person you picked, or it's Abraham Lincoln". Pretty similar to your 100 curtain example.
By the way, you explained the answer before you did the explanation. The probability of your choice being correct is 33% (ok, 33 1/3 -- let's just run wth it). Tha means that the chances of it being one of the other two is 66%. Then he shows you which one of the two it is not. But the probability of it being one of the other two (the one the host didn't show you) is still 66%.
I knew there was a simpler explanation than all the conditional probability waffle I've read.
The thing about the Schrödinger's cat experiment is that it's completely wrong. The cat is alive **or** dead before you open the box to observe the cat. Why? Because the cat is an observer. At no point is the cat alive and dead at the same time.
Yes, it's a thought experiment, but that's how my thoughts work.
A valid point and one that highlights the weakness of the anthropocentric perspective that science is conducted in.
a crucial piece of information was left out of the 3-door problem. It is only true IF THE HOST OF THE GAME SHOW KNOWS BEHIND WHICH DOOR THE TOP PRIZE IS (as other commenters have mentioned). If the host doesn't know (and if the host may inadvertently open a door which conceals the top prize) then the chances/probability don't change. This is easier to appreciate in the 100 door example. If the host doesn't know where the top prize is it would be a 1 to 100 chance that he or she would open all the doors except the one where the prize lies.- i.e. the same chance as the game show player has that their choice is correct.
Exactly. That one bit of information changes the whole game. I just feel sorry for the Noble Prize winning physicists who got the answer wrong because he just assumed it was a completely random game.
You are right, and what’s worse is that not that the crucial piece of information was left out inadvertently. Is that they didn’t understand its implications and their whole reasoning about the answer is wrong, as evidenced by the way he proposes to experiment with cards, and pick one at random to reveal from the 2 that are not the chosen one.
Exactly, you've introduced a known into the game. Which is the field where conmen and magicians work. Simple force a choice. Don't ask how I know this 🙂
Everyone always states the Monty Hall problem incorrectly. You're leaving out critical information. You have to tell us that (A) The host is fully aware of what's behind each curtain at the start, (B) the host always chooses to reveal a curtain, and (C) when he reveals one of the other two curtains, he always does so, using his prior knowledge, in a way to avoid revealing the grand prize. Also, the Monty Hall problem doesn't qualify as a "mathematical discovery."
Many of the people putting forward the "switching is better!" idea are doing so, directly or indirectly, because of the article written by Marilyn vos Savant, in response to the letter written by Craig F. Whitaker. She had a whole web page devoted to this one problem, though you can now only find it in archives. You know what she wrote?
"So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose."
These are her words. She straight up says she's using circular reasoning. The answer does not define the conditions for getting from the question to the (correct) answer. The *question* defines the conditions for getting from the question to the correct answer. If the answer is defining the conditions to get to the answer, that's circular reasoning.
Yep I joined the comments to state this also. The monty hall problem is indeed counter intuitive, and it's not helped in this video by not only not explaining that it's the way monty chooses a curtain to reveal that's significant, but not even mentioning it in the description of the puzzle.
"You're leaving out critical information."
Exactly! And even worse, the experiment in the video is slightly different to the original MHP. As there are three different prizes, it means there are more strategies the host can follow.
Depending on what strategy, the probabilities changes.
For example, the host could follow the strategy "reveal the door with the lowest prize". If that is the case, switching doors in the video example is only 50%.
Another example that shows that knowing the strategy is important, is to imagine the guest is a monkey. For the monkey, the big prize is the banana. Should the monkey also always switch doors to improve the change to get the highest prize?
@@hughobyrne2588 If it were circular reasoning it wouldn't need explaining. But it does, it's counter intuitive to most people (I'll admit, like 99% of people, I didn't get this puzzle when I first heard it)
I think an intuitive way to understand the Monty Hall problem is this: The chances that your initial pick is correct are low (1 in 3). So when Monty reveals an empty door and you have the option to pick the other one, you're basically being asked, "What are the chances that you blindly picked the winner on the first try?" They're not good, so it's likely the one remaining is the winner, and so you should switch
Or, Monty doesn't want to give away a million bucks, so he shows you a sure thing after you picked the best door, hoping you'll switch and save him $999,000.
Now, this example I understand! Thanks
The problem with that intuitive way to understand the problem is it's not actually understanding the problem. If the empty door is arbitrarily revealed, there's no advantage in switching. The door not containing the prize isn't the only thing that factors into why it's advantageous to switch.
But the 2/3 probability of the second door is also not good
@@strifera yes, of course, this is assuming knowledge of how the show always works, which is fair to assume in this case. That's part of the stated problem, and it's an event that is repeated every time on the show. The host always knowingly picks an empty door to reveal
Your introductory example for the Monty Hall Problem is flawed. You left out a couple vital pieces of information. After making your pick, Monty is going to
1. Show you a door that does NOT have the million, and
2. Do so knowingly.
The way you described him revealing the door with $1k makes it seem like the reveal is random, and not intentional. The reason we know that switching is better once he reveals is that we know he is going to reveal a non million door. But by using two different worse prizes, him revealing the 1k (even if he is revealing the 1k intentionally) then that means that he is also revealing a non-banana door just as much as he's revealing a non- million door.
I also suggest using an example where both goat prizes are the same, for charity. Such as... um... actual goats. Which is actually what is used in the original problem.
I wish this answer was higher up.
It's always stated that a non-prize IS revealed by the host. There's no IF involved.
@@lrvogt1257agreed. But it wasn't stated in this video, which is a problem.
I'm convinced this whole monty hall problem is just one big psyop to gaslight you into thinking that the odds change.
It's okay, the problem was never meant for everyone to understand.
It's not gaslighting but, you're right that the odds don't change. The contestants just don't understand them.
No, it is not. See, if the presenter opens a curtain where the $1M is not hidden, the 2 choices (your original choice and the optional to change your mind) became dependent choices, meaning your first choice influences the outcome of the second one. That changes the probabilities. Otherwise it’d be 50-50.
@@szaboandras9843 yea because if I rolled a 1 with a dice, now for the next role I have bigger chances of winning if I pick 2-6...
@@miikayak Actually you kind of got it right. The presenter takes out a non $1M option. Here is how it works. The example in the video is a bit confusing.
Let’s say that there are 3 curtains with 2 of them hiding a banana and one the $1M. Your first choice has a 66.6% chance to be a banana. When the presenter takes out 1 banana that you did not choose, he leaves you with a curtain with a banana and one with $1M. So if you chose the banana the first time and then change to the other curtain you’ll get the $1M. Since the probability that you chose the banana was 66.6% at the start, switching results in a higher probability to win the money then not to.
The key information is that the presenter opens a curtain where a banana is, and not a random one.
In the Monty-Hall problem, _the host does _*_not_*_ choose randomly._
... and that's it. That's the one thing that changes *everything.*
And it's irrelevant to the question
@@fudhater8592 It wasn't a question, it was a statement.... in a video about statements.
Thank you for trying to sound smart at my expense. Better luck next time.
@@TROOPERfarcry It's irrelevant to the question of "What are the odds if you stay or switch". Try to keep up
@@fudhater8592 You're like a smartass, _hold-the-smart._ If you eat enough fiber to dislodge your head from your rectum, you'll realize that in order for the Monty Hall problem to work, the host has to know where the prize is. IE, it's not random.
Do try to keep up, dear.
Is math a plural word?
6:10 The prisoner's dilemma! I can see where it doesn't match up perfectly, but influencing the outcome of the other with the decision of one when both must decide, that part lines up. The first one I knew because of VSauce2 - Kevin does a great job at explaining the brain-melting probability math with /more math/ except it works well. Y'all did a good job here with these, especially making the point that the game show host is the factor that changes the probability.
for people who don’t understand monty hall: just picture in your head the same game with a 1000 doors and 1 prize. and the host opens 998, which he KNOWS to be losing, of the 999 remaining after your pick. wouldn’t you swap?
on now remember that even though the gain in chances is way more marginal in the 3 doors case, the math is still the same. so if it 1/1000 vs 999/1000 in the 1000 doors case, it has to be 1/3 vs 2/3 in the 3 doors case. pretty intuitive.
The real key to understanding the Monty Hall Problem is that _the prize does not move._ Starting out, there are X doors (where X = 3 in the classic problem), each with a prize probability of 1/X. All the doors are in one group and the probability of the prize being behind a door in that group is 1. However as soon as you choose a door, you have split the doors into two groups: we will call them C (door you chose) and R (the doors you rejected). There is only one door in group C, so the probability of the prize being in C is 1/X. There are X - 1 doors in R, so the probability of the prize being in R is (X-1)/X. No matter which doors in R the host opens the probabilities of the two groups remain as they were, 1/X for C and (X-1)/X for R, because _the prize does not move._ Eventually there will be only one door left unopened in group R. What is the probability that that door hides the prize?
That's a useful explanation, thanks. Maybe you can answer a follow-up question or two please?
First question:
Let's use Simon's 100 curtain example, but instead of the host opening curtains that they choose, a number of curtains are revealed but randomly so. The host is just there for show but doesn't actually choose anything. Does this affect the outcome?
Second question:
Let's say that after I pick a curtain, the host reveals 98 curtains (either randomly or selectively, I don't know if/what difference it makes), then I switch to the other curtain, as offered by the host, and then a friend of mine walks onto the stage.
However, my friend hasn't watched the show and isn't told which curtain I originally picked and isn't told if I switched or not. He just sees 98 opened banana curtains and 2 closed curtains. Then he's told to pick one.
Intuitively I'd say my friend has a 50/50 chance of picking correctly, but how can it be that he and I seemingly have the same choice but might have different probabilities of outcomes?
Thanks in advance.
@@charles7866-o1l 1. It only changes the outcome if the prize is accidentally revealed. If the Random Curtain Opening Device opens only one curtain, and it is a banana, you should still switch to one of the other 98 closed curtains. Your probability of winning will increase from 0.01 to 0.01010204... If the RCOD opens two curtains with bananas, switching to one of the other 97 closed curtains increases your probability of winning to 0.01020619..., and so on.
2. Your intuition is correct. If your friend walks in, has no information of what has happened ("what's with all the bananas?"), and is asked to choose a closed curtain, they have a 0.5 probability of winning. The two of you have the same choice, but you have _more information._ Your friend only knows that there are two curtains and one prize. You know that the prize was somewhere behind one of 100 curtains. When you chose your curtain you split the curtains into two groups: your one curtain, with a 1/100 probability of winning, and the other 99, each with a 1/100 probability but _collectively_ with a 99/100 probability of concealing the prize. As curtains are opened that 99/100 becomes concentrated onto fewer and fewer curtains until only one remains in the group you did not choose.
With no information, your friend's choice is an _independent trial_ and I think this is where a lot of people get tripped up by this problem. They think that your second choice, to stick or switch, is an independent trial when it is not. For it to be an independent trial for you, you would have to lose the information you had already been given. The only way that could be done is if after the reveal of all the bananas the prize was somehow randomly redistributed between the two remaining closed curtains.
Here is another example of the problem, a simple card game with a standard deck of cards. The goal is to get the ace of spades, if you have it when all the cards have been turned over, you win some money. The deck is shuffled and spread out face down. You are asked to pick a card, any card. The card you pick is placed just aside, still face down, you do not get to look at it. Then the dealer flips over the remaining cards, one at a time for dramatic effect. If it isn't the ace of spades the card is tossed in a garbage can under the dealer's table. The dealer gets through all the cards until just two remain: your card, and the dealer's last card. Let's now consider three scenarios. In the first scenario, you are given the option to stick with your card or switch to the dealer's last card. You know now that the correct choice is to switch: your card has a 1/52 probability of being the ace of spades, the dealer's card has a 51/52 probability. In the second scenario your friend walks up to join the game. They have no idea what has happened so far - not even which card is "yours" - they just know the goal is to find the ace of spades and that one of the two cards on the table is it. From their perspective, with the information they have, the probabilities are equal. In the third scenario, instead of your friend coming by, the dealer shuffles the last two cards (in such a way that you cannot tell which is which), places them back on the table, then asks you if you want to stick with "your" card, or switch. Your card isn't really "your" card anymore, is it? The prize has been randomly redistributed. You have lost all the information you had. Your choice has become an independent trial. Now you are just like your friend: the probabilities of either card being the ace of spades are equal.
@@charles7866-o1l YES. Exactly. Most explanations of the MHP assert that somehow when an option is eliminated from a set (a door opened in this case) that its initial "share" of probability is passed only to some sub groups and not the others. This is arbitrary and problematic. If this is the case, I want to include MY door in the group with the opened door, and have MY door receive special treatment.
@@JonMartinYXD Thankk you very much. I think I understand the concept better now :). Took me an hour or so in bed pondering it and varying the thought experiments, but I think I got it!
@@khemkaslehrling3840 The reason that your door doesn't get special treatment is that when you make your initial choice you've split the groups of doors into two groups: the one you chose and the ones you rejected.
If something happens to one of the doors in the rejected group (like the door being opened) then that affects the rejected group, not the one you chose, which is why the "success probability" (SP) of your chosen door doesn't change from its initial position.
However, in a way, your door is included: if (as discussed in my reply to Jon Martin) you have a friend who walks in after the door in the rejected group is opened, and they don't know which door you chose, and they see the opened door and understand the game concept, then they'll know that the door opening has just now changed the SP on all the doors (for them but not for you), meaning your door is receiving the treatment of a change of SP, which you wanted.
It might help to know that your advantage in information over your friend still puts you in a better position.
Let's say there are 10 doors. You picked one. The SP of the right door being in your chosen group (a group of one) is 1/10=0.10, and 9/10=0.90 for the rejected group.
A door in the rejected group is then opened, meaning the SP of the 8 remaining closed doors in the rejected group is now 0.90/8=0.1125
Your friend then walks in and just sees 9 closed doors, 1 open door, and doesn't know which door you chose. They ignore the open door, so their SP is 1/9=0.1111111111 and this is less than your SP.
What helps me understand the Monty Hall problem is looking at it like "I have a 1/3 chance to pick the correct door on the first try and a 2/3 chance to have picked the wrong door." The 100 door example really helps too, such an extreme case makes it a lot more clear.
Andrew Well, it helps some people.
@@SigFigNewton Or if they know it, they don't see that they selective revelation creates a sampling bias. I am not proud, I admit that it took me a long time to understand it.
I don’t think you understand the Monty problem
By which you mean, I think, that if Monty Hall always opens 98 doors, none of which have $ 1,000,000, then it is more intuitively obvious that you should switch.
You could also just list all the possible situations you can end up in after you have chosen your door and the host has revealed a door:
Door in [] is chosen, door in || is revealed by the host, L looser door and W winner door. Assuming the host always opens a looser door, we get these cases:
1. [L] |L| W
2. [L] W |L|
3a. [W] |L| L
3b. [W] L |L|
Now, we can say that 3a and 3b are the same and count them as one single case, because the order of the remaining doors (the L's) doesn't matter in this case.
This leaves us with 3 cases. 2 of which switching would lead to you winning and one which wouldn't. This is how you can reach the 2/3 probability of winning by switching.
Edit:
Another way too look at it, is that you have 2/3 probability to chose one of the L's at first. And whenever you have chosen one of those L's, you're in case 1 or 2. Thus, switching wins in 2/3 of all cases.
I love the Monty Hall problem. I've won money with that as most people don't understand it.
Most people think it's an even chance. They don't use logic
I like how Simon says asking the question "is my red the same as your red" is something we probably all did in childhood meanwhile here I was at like 26 while bingeing on math and science youtube videos that I finally had the thought that we could all be seeing different colors but just calling them the same names.
When I was a kid, most of my friends thought I was crazy when I brought up this question... I was always happy being that dude, and now my bubble was popped 20yrs after the fact; when I find out it was just my friends that were...................... Lmao 😂🤣😂
What helps with the Monty Hall problem is that while your choice is random, Monty's in NOT random. When you pick, you have a 1 in 3 chance of picking the prize door. When Monty opens a door, he has a 0 in 3 chance of picking the prize door. Therefor the remaining door has the remaining 2 in 3 chance of being the prize. :)
It doesn't matter what you pick initially because it's ignored and a new scenario is presented. The host's reveal is never random. It must and will always be a goat or the game is ruined. The player always and only makes a real choice between one of two options, a car and a goat.
@@lrvogt1257 That's not true. Try the scenario with 100 doors. Your first pick is 1% chance of success. Then Monty opens 98 doors with goats. Now there are two doors left, Your door (1%) and Monty's door (99%). It is NOT 50-50 ;)
Hmm I don’t think you understand the Monty problem
0:07 Bold of you to assume I know anything about math.
He does and I can promise you that.
Actually Relativity IS important to your drive to work if you use satellite navigation. The operators of GPS have to take Special and General Relativity into account for it to remain accurate otherwise it would drift because of time dilation.
The math section of this video is amazingly satisfying. I was literally going through the motions of all of the things that were predicted by math
that became real in the real life-like the location of Neptune Neptune or Einstein predicting that gravity could bend light those were all things predicted mathematically that came to be reality. And then Simon goes on to say math is perfect but reality is not. If you take that statement into account and also the previous statement that we may not be understanding it because we represented in symbols that only we created and we understand on this planet, I've come to a conclusion that either there's a mathematical hump that we're not seeing yet or we're living in a simulation. Even with that being said science was always progressing I remember when Hubble was launched I remember the Challenger accident I remember so many things that have changed as when I was in middle school and high school regarding science and thinking that a lot of those things that I was learning about happened 50 to 100 years ago they're possibly could be an aspect of reality similar to how to we express mathematics that will make the universe more comprehensible. I think the real key is thinking outside of the box. People with scientific minds and engineering minds always have to know why. And that why has led to some amazing discoveries.
So why is math so perfect in a reality that's not? 🤯🤯🤯
(This message was brought to you buy an educated stoner who was high while watching this video)
A help for the Monty Hall problem: 1) Regardless of which curtain you initially select, the probability that the million bucks is behind one or the other curtain is ⅔. 2) No matter which curtain you initially select, Monty can _always_ open one of the other two curtains, showing you a worthless item behind it because he knows where the big money is. 3) When he opens one of the other curtains, he shifts the entire ⅔ probability from two curtains to the one other curtain he didn't open. 4) After he opens one curtain you have the choice of two curtains: your original choice (⅓ probability) and the one other curtain Monty didn't open (⅔ probability).
By changing your selection you therefore double your chance of getting rich.
By what magic does the share of probability of the opened curtain transfer to only one of the other curtains, and not be split across both? Why not then just declare that the curtain you chose is grouped with the curtain the host opens, giving it "the entire probability of the opened curtain" and making your door 2/3rds? This is the fallacy of this Monty Hall thing.
@@khemkaslehrling3840 Think of it this way; 3 items behind 3 curtains, 1 prize & 2 duds (worthless), no matter which door you choose the host can reveal a dud. Lets go through all 3 choices, Scenario #1 lets say you pick the prize curtain then the host reveals a dud curtain, if you switch you get another dud (1/3). Scenario #2 lets say you choose a dud curtain, the host can only show you the remaining dud so if you switch you pick the prize (1/3). Scenario #3 you choose the other dud, yet again the host can only choose a dud, so if you switch you pick the prize curtain (1/3). So in closing in only 1 out of 3 scenarios do you lose by switching, but if you chose to switch you'd be right 66.7% (2/3) of the time.
@@justblaze6802 that explanation just gave me the light bulb moment, thank you
Another help to those who aren't yet convinced: when Monty opens one door, think of it as him opening *ALL* the other doors except the one you chose and one other one.
@@khemkaslehrling3840 You choose one door out of 3. If I offer you to let you switch for the two remaining doors, would you take the deal?
Of course you would, because your odds would be 2/3. That's what's happening here, except that the host is opening one of the two doors first.
But you are still effectively getting to choose 2 doors by switching.
Your explanation of the Nash Equilibrium is spot on. Every time the traffic lights outside my work break down, there's a nail biting display of idiocy which always results in someone crashing their car.
Savvy and slightly sarcastic gamblers have a saying, "Don't bet on anything that talks". Well, that's that's the problem with the Monty Hall Problem. Rather than strictly being a math or statistics problem, it confuses things by including Monty, without explaining anything about his consistency or motives.
The majority of people seem not to really understand the problem, and it shows by the way they put it. It order for the logic to work, this has to be the standard format of the game. ALWAYS the participant selects one curtain, ALWAYS Monty opens a curtain that doesn't contain the prize, ALWAYS the participant has the choice to switch.
@@ntomata0002 Oh no. Here we go again. I can't tell you how many hours I've wasted trying to clarify the ambiguities still inherent in the problem. I have no intention of going through all that again, because 99+% of it falls on deaf ears. I referred to not betting on anything that talks for a reason. Because humans easily and often project other potential motives, rendering this a pointless game to discuss. It certainly is not the straightforward reliable simple statistics game you suggest. Why? Because we have no idea what Monty's motives are, just when he makes this offer or does not. He said so himself, that he has total control, and that sometimes he would prefer to make the offer to switch when he KNOWS they have already chosen the car. I see no valid reason we should accept that he has no such motives. The problem would be a lot clearer if no Monty were involved! Then it might be a math and statistics problem, unencumbered by ambiguity. Just because the problem does not suggest any of this is no reason we must assume no ambiguities exist. Incidentally, Marilyn herself admitted there were ambiguities in her original presentation, because she tried to deflect away from this by saying that MOST of the thousands of letters she received did not mention the ambiguities. The entire failure of this famous problem (from her column) goes back to her failure to present it as a purely statistical problem.
Okay I'm gonna be that asshole 😂 your explanation of the Monty Hall problem left something out. It's crucial to state that the game show host has decided in advance to open a curtain that does not have the million dollars behind it. In this case, switching is the superior strategy.
By contrast, if the host randomly opened one of the two remaining curtains, and you saw that that curtain did not hide a million dollars, there is no advantage in switching. It's 50/50.
You can test this yourself at home with a deck of cards. Take an Ace, King and Queen. Your objective is to find the Ace.
Setup 1: you pick a card, then instruct a friend to look at the other two cards, and remove one which is not an Ace. From the two remaining unrevealed cards, you will now find that switching finds you the Ace 2/3 of the time.
Setup 2: you pick a card, then reveal one of the other two cards. If it's an Ace, game over. If it's not, continue. You'll now find that it makes no difference whether you switch or not; you'll pick the Ace half the time, regardless of your strategy, in the long run.
It depends on where you are assigning the odds. If you are assigning the odds before the host makes the choice, then you are talking about different game rules.
With the altered rules, 2/3 of the time the car will remain behind the unselected doors. At that point the host has a 1/2 of randomly ending the game. However, if the car was not chosen and the game continues, it doesn't matter whether it was random or with knowledge. For that instance of the game, the 2/3 probability still exists with the remaining set, which has been reduced down to one door.
With this altered set of rules there are different probabilities before the host chooses though. There's a 1/3 probability the game ends at the point of the host making a choice. 2/3 of the time, there's then a 2/3 chance to still win. Therefore, there's a 4/9 overall. But again if we go to the point at which the host has already revealed a goat, it is 2/3 probability in that moment for the contest with a choice switch.
@@jasonalquiza noooo in the altered game (setup 2) there's a 1/3 chance the game ends immediately, a 1/3 chance your first pick is correct, and a 1/3 chance the box which isn't picked or removed will be correct. That's all before the host opens a box.
The conditional probability in a game that continues - when the removed box does not end the game - is now 50/50 stick/switch.
The intuition that it's a 50/50 choice is incorrect in setup 1, because the choice of removal made by the host is constrained in 2/3 of games played. But it's correct in setup 2.
It was clearly stated that the host revealed $1,000. This is not by accident. It just adds another level to the drama. Is $1,000 enough or is it worth risking for $1M. But the real point is that it's not the big prize. So what are the odds that switching gives you a better chance of winning the big prize?
The first choice has a 1/3 chance of being $1M.
That means the other two together have a 2/3 chance of being $1M.
The host then reveals which one is $1,000
Therefore, the remaining one now has a 2/3 chance of being $1M.
@@lrvogt1257 sure, but there's a difference between "the host chose a random box (out of the two which weren't the contestant's first choice) that happened to contain $1000" and "the host chose one of the two boxes with the stipulation that they were not allowed to choose a $1m box, and it happened to contain $1000". It's this constraint that has an effect on the subsequent probabilistic assessment of where the $1m lies.
If math were not real, my Grade Point Average would have been much higher.
My favourite way to explain the Monty Hall problem is: Your chance of guessing the million dollar door when you make your initial choice is one out of three. Therefore in two out of three cases, the prize is not behind your door, but behind one of the other doors. When the host shows you which of the other two doors does not hide the prize, those two thirds become the one remaining door.
In ONE out of three cases you will already have chosen the million dollar door. In that case the game show host shows you one out of two empty doors, and switching over to another door will make you lose.
However in TWO out of three cases you will initially have chosen an empty door. Then the gameshow host opens another empty door. In these two out of three cases, the remaining door is the million dollar door.
In the description here, we don't even know what happens when the player picks the door that hides 1000 $.
@@insignificantfool8592 I guess we can assume banana = $1000 = empty door when compared to $1Mn
@@priyamd4759 the number of things we're supposed to "assume", just to arrive at the sought after "unintuitive" result is amazing.
Baloney, two doors do not become one. The game changes from one where you have a 1-in-3 chance to a game where you have a 1-in-2 chance. You do not improve you odds in the second game by changing your selection. Math major here and I'm calling you a BS artist.
@@sailingelectricgitana1286 you failed your major and did it on a public forum
You can think about the probability of prizes problem as your original choice keeps the same probability as when you selected, but the probability of the other doors gets added together and placed on the door not revealed.
The confusion becomes, why doesn't that same probability exist for the door selected first? Why does revealing one door increase the probability of only the door not selected?
@@dudeinoakland It doesn't. Whoever thinks that revealing one door gives you better chances if you change your mind is delusional
@@dudeinoakland The key to understanding the problem is that _the prize does not move._ Starting out, there are X doors (where X = 3 in the classic problem), each with a prize probability of 1/X. All the doors are in one group and the probability of the prize being behind a door in that group is 1. However as soon as you choose a door, you have split the doors into two groups: we will call them C (door you chose) and R (the doors you rejected). There is only one door in group C, so the probability of the prize being in C is 1/X. There are X - 1 doors in R, so the probability of the prize being in R is (X - 1)/X. No matter which doors in R the host opens the probabilities of the two groups remain as they were, 1/X for C and (X-1)/X for R, because _the prize does not move._ Eventually there will be only one door left unopened in group R so that door must have the same probability of having the prize as group R does (and always had).
@@dudeinoaklandyou literally can switch from your initial one door to both the other doors and Monty also shows you, which of the other two doors is a loss.
The part that Simon leaves out of the puzzle that needs to be given to make the game work is that Monty will always open a curtain that does NOT reveal the million dollars.
I'm guessing math is tricky for you
@@twotubefamily9323 No, I have a science degree and went all through calculus. The point is that the "chances" are calculated by telling us that after selecting a door and one of the other doors is exposed, should we change our selection to the other door? BUT there is nothing in this set of "rules" that says the second door MIGHT expose the million dollars, and nothing in the rules that we cannot select the opened door showing the million dollars. THAT change the odds dramatically. But we ASSUME the rules are that the million dollars cannot be exposed at the second door opening and that we can't select the open door.
@@twotubefamily9323
blaster is correct. This is yet another sloppy/incorrect formulated Monty Hall problem. The host does not "cheekily decides" to open another door.
In classic Monty Hall, the host opens another door knowing that it is not the big price. And he would have done this regardless which door you initially choose. THEN you will have a 2/3 chance to win if you switch.
For example, if the host "cheekily decides" to open one of the other doors randomly, discovers it to not be the price and therefore offer you to switch. THEN you will still have 50/50 chance between the two remaining doors.
@sverkeren Respectfully, no!
When you pick the first curtain, it has 1/3 chance of being correct.
Meanwhile...
*The other two curtains together* have 2/3 chance of holding the jackpot.
The host removing the non-jackpot curtain of these two is just a detail - that 'side of the equation' is still 2/3 .
@@Michael_Arnold Respectfully, yes!
In classic Monty Hall, if you first choose wrong (with 2/3 chance), then the host will practically show you where the price is by AVOIDING that door.
If the host does not actively avoid the winning door, then you don't get that probability boost. Instead that probability get evenly distributed on the remaining doors. What could have happened matters.
The Monty Hall puzzle was presented incorrectly here, as it nearly always is. The fact that Monty offers you another choice is not sufficient. He may have done that specifically to throw you off. The math only works if he is guaranteed to give you that offer.
One could argue that it is not presented incorrectly, because the most famous presentation of the problem (the letter to Parade that vos Savant answered) also doesn't say the host is guaranteed to give you that offer.
@@insignificantfool8592 In that case, the solution offered is incorrect. The only way that solution works is if the host is guaranteed to offer you another door.
@@Rick_Cavallaro I would agree.
The "Monty Hall Problem" came about because of a question to Marylin Vos Savant. Somebody asked if it was better to switch or keep their original choice, though with one prize, and 2 with essentially nothing, and she proceeded to say it's best to change it. She was inundated with mail about the whole thing, including from mathematics professors who said she was wrong. (Many of them sent a second letter asking her not to publish their first letters, or keep their names out of it, after figuring out they were wrong.)
Mrs. Savant followed up and even had a bunch of kids create an experiment where they run multiple tests using the same parameters, and the results all supported what she had said.
She gave a similar example to the one above, with a hundred curtains.
Since she was the one who originated this and gave the correct answer which confused the "experts" at first, her name should be included here.
Her answer is to be expected from the mind of a person conditioned to answering IQ questions. The correct answer is indeed 50/50 and not ⅓ to ⅔ . Interestingly Tibees does a video where she answers an IQ question as might a person with an IQ of 300 by coming up with a polynomial equation to generate the pattern presented for the trend given for which the next step in the series needs to be given . Her polynomial equation yields the original starting point instead of an obvious intuitive next number for the series.
@@philip5940
It is not 50/50.
Why would it be 50/50?
Have you *read* what vos Savant wrote? She had a web page about the whole thing, though it seems you can now only find it in archives. In her own words, "So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose".
The answer defines the conditions for answering the question? No. The question defines the conditions for answering the question. If the answer is defining the conditions for reaching the answer, it's circular reasoning.
@@hughobyrne2588
Wrong. In order to always open a losing door, whoever opens that door has to know which is a loosing door.
But again, the video above shows the answer is correct, and a lot of grade schools participated in an experiment where it worked. There are even web pages that demonstrate this. I shared one link, but apparently that post vanished. I did 200 tests where I always changed, and hit real close to a 66% win rate.
@@philip5940
Not sure what happened to my response, possibly removed because it had a link. But the link was to where you could test it yourself. There are a few web pages set up to test this, and I ran 200 samples, always choosing to switch, and won real close to 66%.
Also watch the video I am responding to. They say it works.
The Monty Hall problem only rewards changing your guess if it's assumed that the host knows where the million dollars is and will always open another curtain. If that condition isn't met, then the remaing two curtains each have s 50:50 chance of hiding the million dollars (and in one game out of every three, the million dollars will be behind the curtain the host opens, so you're choosing between the thiusand dollars and the banana, and you have a 50:50 chance of that)
I’m confused. If the host does not know, the only difference would be that they would sometimes accidentally reveal the big prize. And then the contestant would surely switch. So whichever way you think about it it’s always favorable to switch, but of course that does not mean you will always be correct. It only increases your chance of winning.
@@lenaforsgren the host reveals what's behind one door, and the contestant can choose to switch to the remaining unopened door. So if the host opens a door and the prize is there, the game is over and the contestant has lost.
@@Jaccayumitty If that were the rule, the switching strategy wouldn't even be favorable from the start. The outset was what I was thinking of. By the way, I realize now that if the host indeed has opened a door randomly AND it turned out to NOT be the car, the advantage of switching is actually lost. No disadvantage, though, either. I've never thought about that before, it's in my opinion less intuitive, since we're suddenly not talking about all the original possible outcomes.
Amazing as always, I love maths, but I don't need maths to Love Simon's videos. Thanks guys.
👍
The Monty Hall problem purely mathematically sure it make sense, but it’s not a purely mathematical decision. Deception is a very strong human trait and psychology of someone who didn’t have it figured out as Monty did would usually be trying to get you to change your answer because you had the right one picked.
Just like in poker when people try to look strong when they have weak hands and are bluffing and try to look weak with strong hands to get paid. It’s basic human psychology more than math in this situation.
The believers of the "switching is better" solution will tell you that Monty will always offer the switch. In that case there is no room for deception. But as you say, this is counter everything that makes up human Interaktion, so obviously people tend to interpret it in a way that it was Montys decision and not a forced play. In that case, staying with your door is indeed not a bad idea.
@@insignificantfool8592
IF: Everyone will tell you that Monty will never kill you.
This video showed me that it's okay that I never did well in maths, since it doesn't matter much in the day to day of my life.
Astonishingly, for every day life most of us probably learn all the math we're ever going to need by about grade 3 or 4. (Or, college in Florida) Throw in a bit of percentages and knowing how to average that comes in handy sometimes, and in some backwards countries fractions which I think are taught a little later, and that's about it. Which is not to say learning beyond add subtract multiply and divide is a waste of time. Depending on what you end up doing, you may need algebra or geometry or calculus, but the vast majority of us don't. Where we bump into it in day to day life, it's done for us.
On my first day of second year algebra in the tenth grade, my math teacher drew a 4 on the blackboard, then said, "See this four? It's not real."
My fifteen year old brain was blown.🤯
The confusion comes in when forget that all of math comes down to counting. It's just that what is being counted has become more and more complex. The symbol "4" isn't real, but is used to mark that "there are 4 things". "1/4" means "1 thing out of 4 things", "4/2" is "4 things separated into 2 groups", and so on. The numbers themselves are not real, they're just labels for convenience that represent what IS real.
@@Vaeldarg Symbols are actually real, they represent real things. When we say they aren't real we mean they aren't concrete, but right now I'm writing letters that convey to your mind real things.
@@konroh2 Where were you going with this?
@@Vaeldarg Concepts are real things, so language and symbols are real. An idea of something is real. Real things aren't just material, what is invisible is also real. When someone says maths isn't real because it's just a symbol it's denying the spiritual reality of ideas, which are also real.
@@konroh2 "the spiritual reality of ideas" is easy to deny, because no it isn't real.
Simon: let me show you two mathematical ways you may be wrong about the world and how to better understand it.
also Simon: Hehe just kidding, it's all a lie
thanks Simon for destroying my mathematically perfect world today.
Instead of three, imagine there 10 doors.
You pick one first, chance of picking right is 1/10. (The door is not revealed yet as that could end the show.)
The chance of the prize behind the remaining doors is 9/10.
The host opens 8 out of the remaining 9 doors, none of the 8 would be the prize. (The host knows where the prize is and revealing the prize would end the show.)
Now the host basically eliminated the 8 wrong choices out of the 9 doors. The chance of the one remaining door the host did not open being the prize is very high (90%).
The host basically just show us the answer and you should switch to that one remaining door.
I cannot express how much I love and hate this at the same time. I wasn't ready for the whole maths is perfect but not real bit.
The "maths might not be real" theory actually messed with my brain
I think this is an easy visual demonstration of the Monty Hall Problem odds.
-Take 3 playing cards, The Ace of Spades is the winner. The other two are red cards.
-Lay them face down (For this demonstration they can be face up since you are the host so you know which is which.)
-Put a coin on each card which represents a 1 in 3 chance of winning.
-Separate 1 set from the other two.
-Turn over a red card from the remaining two while moving the coin off of that red card and putting it on the last remaining card.
-Your original choice has 1 coin and the remaining card has 2.
-Always switch.
I feel like the assumptions on the Monty Hall problem are not properly described in such videos. Most people assume that choosing a door and the proposal from the host are random. Which is not the case. The assumption should be that the host always has a calculated motive on his proposal.
@@lenaforsgren
"If the host would choose randomly they would sometimes reveal the big prize and then the contestant would of course switch without hesitation…"
The point is that if the host did reveal a non-prize randomly there would be no advantage in switching.
@@lenaforsgren
Again you have shown that you don't know why one door has a probability of 1/3 and the other one has a probability of 2/3.
@@lenaforsgren
"The point is that if the host did reveal a non-prize randomly there would be no advantage in switching."
"That’s incorrect."
It IS correct!!
@@lenaforsgren "Not relevant, however. Your initial chance of getting it right is 1/3, so the likelihood that your initial guess was wrong is bigger and you should therefore switch."
Initially yes. But if the host acts randomly, when they reveal the non-prize door, the probability of having guessed wrong is reduced.
"If the host would choose randomly they would sometimes reveal the big prize and then the contestant would of course switch without hesitation…" They don't though. The problem defines that they always reveal a non-prize door. Which means the player still is uncertain, and the host is not acting randomly.
@@lenaforsgren It does matter. If they acted randomly, the two selections by the player are two separate random experiments. The first selection gives the player 1/3 to win. And the second gives him 1/2 since the open door has 0 probability to win and is not in the population of possible winners. The initial 2/3 possibility to win if they changed no longer exists because of the revealing of the non-winner door. If the host does not act randomly, the three acts/selections are tied into a single experiment. What the host selects depends on what the player selected (always has to reveal a non-winner) and the second player selection depends on what the host selects to open. We no longer have plain random experiments. We are working with conditional probabilities.
Hopefully he will explain why two people that speak the same language call it two separate things.
Maths vs. Math
ooh, I can explain that. Super easy. One of them is wrong. ;)
Math is the subject , Maths are the stuff you do in the subject. i.e. '' There are many maths to do today , lets get mathing''.
The science is called Mathmatics. One side shortens it to Maths the other to Math. This also changed over the years which one was the more accepted one where.
Thanks, great episode! I love maths
Thank you for adding the 100 curtain option to the Monty Hall explanation. That finally made me understand this after hearing it at least 15 million times and not getting it.😊
This is an imprecise formulation of the Monty Hall problem. It is imperative to state that the host knows what is hidden behind each curtain and that he has no choice whether to give you a preview or not.
I don’t think you understand the Monty problem
@@scottparker1741i think you don’t understand it
@@uncopinoI think you don’t understand it
@@scottparker1741 explain then. op said the host must have complete knowledge of what door has what prize and must always offer the swap in order for the problem to have this solution. why is this wrong? idk maybe you know the solution is the same even when one or both of these conditions aren’t satisfied? i’ve always wondered how you solve it in those cases and i have sort of an idea but i’m not sure so please tell me if you know cause i’m curious
I get it, the Monty Hall Problem is from getting more information. You picked first of 100 squares randomly, no info. You want the million $ square, Then you see 98 bananas and one square not revealed. That's probably the $. You picked a random banana before probably. Simple. I love it.
I think it's important to emphasize the fact that the people disclosing what's behind curtain 3 KNOW it isn't the grand prize when they open it, and have chosen it for that very reason. When I've heard this explained before, I had the impression they were simply revealing one of the others at random, which might include the grand prize. That makes all the difference.
The Monty Hall problem (and this one is a considerable variation from the original - where are the goats) is, as is often the case, inadequately described. Once extremely important point is the the host must always open one of the other two curtains. It is not a matter of deciding to do it on the day, but they have to do it each and every time and, what's more, you have to know they are going to do it. Otherwise the host might be doing it selectively to mislead you. Also, of course, the host must know where each prize is, and they must never open either the competitor's original curtain or the winning curtain.
Once all the conditions are clearly explained then it's simple; you change your decision and you double the chance of winning the prize from 1 in 3 to 2 in three.
It is, incidentally, not a mathematical discovery at all. It's a bit of basic probability theory, just with a (usually) incompletely explained scenario.
Try modeling all the scenarios. Imagine a game with 3 doors, and 1 car and 2 goats.
Assume the contestant chooses door 1. There are now 4 scenarios and two options each, resulting in eight outcomes. And the split of win/lose for stay/switch is 50/50.
1. Car / Goat / Goat - host shows you door 2 - switch lose / stay win
2. Car / Goat / Goat - host shows you door 3 - switch lose / stay win
3. Goat / Car / Goat - host shows you door 3 - switch win / stay lose
4. Goat / Goat / Car - host shows you door 2 - switch win / stay lose
@@khemkaslehrling3840 I know perfectly well the scenarios, and worked all this out long ago and even tested my mathematics with a simulation, just to make sure.
However, the point you completely miss is that the Monty Hall problem is often not fully specified. The extra conditions are
a) the host must know what is behind each door
b) the host must only open another door with the lesser prize
c) most importantly of all, the host must do this every single time
Without (a) & (b) we are in a different scenario as either could lead to the main prize being revealed.
If (c) is not met, then the host might only decide to open another door when the competitor has already picked the main prize. Of they might only do so if the competitor chose a lesser prize, or maybe some mixture of the two. The host might act to help, or to hinder.
Like many probabilistic problems, it's essential to be very precise or it is easy to be lead astray.
@@TheEulerID I am forever mystified that people hear the words "game show" and do not immediately understand that the host will never reveal the prize. Thus, the game show host must know what is behind the doors. And with this it is inferred thhat the host always never reveals the prize.
Perhaps some people really don't understand how game shows work?
The Monty Hall problem has a simple explanation that is easy to demonstrate. Consider every possible outcome. Let's say money is behind door 1 and bananas are behind 2 and 3. If you choose to not swap your door, the 3 outcomes are:
Select door 1, stay with door 1, WIN
Select door 2, stay with door 2, LOSE
Select door 3, stay with door 3, LOSE
If you choose to swap, the possible outcomes are:
Select 1, he opens door 2 or 3, you swap, LOSE
Select 2, he opens door 3, you swap to 1, WIN
Select 3, he opens door 2, you swap to 1, WIN
It's obvious then if you stay with your selection, 1 in 3 outcomes win.
If you choose to swap, 2 in 3 outcomes win.
In this version of the MHP there is a banana and $1000 instead of 2 bananas.
@@Stubbari that's true, but as everybody knows, the host won't reveal the big prize and will also never reveal your door. So, obviously, when contestant picks the thousand dollar door, the host will show the banana. It can't be any other way...
@@insignificantfool8592 How do we know that? It's just an assumption you are making. If you choose the $1000 the host might as well open your door.
@@Stubbari haha, now you sound like me, probably to make fun of me?
@@insignificantfool8592 Yes and no. I really think this is a poor example of the MHP because there are 3 different items rather than 2. Also, $1000 seems like a prize to me so there are 2 prizes and 1 non-prize.
The way I rational the Monty Hall problem is this:
The first probability is still collapsing...
If you don't modify your choice, the original 1/3 probability continues to collapse like normal...
If you do modify your choice you are really getting 2 doors instead of 1... one picked by the host, and one by you.
You still get the same odds. All doors are 1/3. So the host reveals one... you still have 50% chance with the remaining doors. You gain nothing in terms of odds, you only get to see what was behind one door that you did not choose anyway. There's no 66% chance in this scenario, ever.
@@lgmediapcsalon9440 No. The key to understanding the problem is that _the prize does not move._ Starting out, there are X doors (where X = 3 in the classic problem), each with a prize probability of 1/X. All the doors are in one group and the probability of the prize being behind a door in that group is 1. However as soon as you choose a door, you have split the doors into two groups: we will call them C (door you chose) and R (the doors you rejected). There is only one door in group C, so the probability of the prize being in C is 1/X. There are X - 1 doors in R, so the probability of the prize being in R is (X - 1)/X. No matter which doors in R the host opens the probabilities of the two groups remain as they were, 1/X for C and (X-1)/X for R, because _the prize does not move._ Eventually there will be only one door left unopened in group R. What is the probability that that door hides the prize?
@@JonMartinYXD Where did I write that the prizes move? If there's 3 doors and one is open, the chance is 1/2.
@@lgmediapcsalon9440 The only way the odds can "reset" is if the prize can move after Monty opens a door.
Let's put it another way. What if before Monty asked you to choose a door, he asked you if you would like to choose one door or two doors? You'd be crazy to restrict yourself to just one door, right? Choosing one door gives you a 1/3 chance of winning, choosing two gives you a 2/3 chance of winning. Well that is exactly what Monty is offering you after he opens a door.
Three doors: A, B, and C. From your perspective, A has a 1/3 chance of hiding the prize, B has a 1/3 chance of hiding the prize, and C has a 1/3 chance of hiding the prize. You choose door A. You have a 1/3 chance of winning. Monty opens door B or C to reveal a goat (the goat is not the prize). Monty now offers you the chance to stick or switch. This is equivalent to you previously being given the option of choosing door A or choosing _both B and C._ Since you are the one determining if the choice is between A or B+C, B or A+C, and C or A+B, and since the prize is already in place, Monty cannot change the probabilities. If you choose only one door you have a 1/3 chance of winning. If you choose two doors, by switching, you have a 2/3 chance of winning.
@@JonMartinYXD You're moving the goal post with your "pick 2 doors". Monty opens one door, two remain. You have 50% chance of winning. End of story.
14:11 everything falls at the same ACCELERATION! Terminal velocities vary widely between objects!
Velocity is the wrong word anyway. Velocity is a vector with a direction component. People on the opposite sides of the planet fall in opposite directions to each other.
The word he's looking for is "speed", and even then (like you pointed out) he's still wrong. Much like his re-telling of the Monty Hall Problem was crucially inaccurate.
This, like so many explanations of the Monty Hall problem are wrong. This fun mathematical trick ONLY works if the game show host deliberately shows you an incorrect curtain. If he reveals at random, then you have a 1/3 to see the million dollars, and that is where your other 33% is coming from.
If he reveals at random it's a different (and rather unsatisfactory) game. "Here's $1m; would you like to switch to that?"
Yeah he made it more complex than it really is by not hitting that point
Correct, the host knows where the car is so the curtain he reveals is not at random. But this should go without saying. He’s not just going to reveal the car because then there’s no game. So of course he knows where it is.
I never understood how the Monty Hall problem worked until I saw the Parade Magazine chart solution, and the importance of the "intervention" of the Monty Hall person. The key idea is that the person revealing what is behind one of the curtains follows the rule that they cannot show you the prize. They have to show you the other bad choice. They cannot show you what is behind the curtain you picked either. This is the key idea that changes the odds of winning if you switch your choice. There are three possible outcomes. If you keep your choice, only in one scenario you are right, other two choices are wrong. But in the other two scenarios you are wrong in your first choice. That means that two out of three times you have eliminated the lesser choice. Monty then shows you the other curtain, and it cannot be the prize. Monty has now, two out of three times, shown (or proved) to you which curtain has the prize. It is not the one he showed you, and two out of three times it is not the one you picked. So two out of three times it is the other curtain. If you can switch you will therefore win two out of three times! Your odds of winning when switching are 66%.
In what game show in the world is the big prize given away at random. i'm mystified as to why people hear the words "game show" and don't immediately understand the host doesn't just give the prize away at random, but makes a selection that gives the contestant a choice, and thus intrigue for the viewer.
@@barryschwarz Try 3-2-1.
In it the contestants were given clues as to what each prize was. There were 5 clues given one at a time, but once 3 were on the table the contestants had to get rid of one. The prizes ranged from the big main prize down to a dustbin booby prize (and yes, it was won on a few occasions). If they could not work out the clues, they would have to pick one at random to get rid of. Either by chance or by not working out the clues properly they could end up rejecting the big main prize.
Although not exactly giving the main prize at random, it did have the chance that the main prize would be eliminated, possibly by random chance.
@@cigmorfil4101 No, that doesn't fit the bill. I don't know any game show that would give away a million due to a random choice made by the host. And it the context of a game show, it just doesn't make sense.
I think what actually happens is that people make the common intuitive choice that the remaining two doors are a 50/50 chance, and then try to justify their intuition (eve when they figure out it was incorrect) by arguing post-hoc that the host might give away the prize at random (every 3rd game!).
You see various iterations of this post-hoc justification. Some people argue that it's 50/50 if a new contestant steps in after the first choice, and doesn't see which door was first chosen. Well, yes, now the choice is 50/50, but why change the format of the game to make that argument?
I'm sure there are people who hear the words "game show" and actually imagine that the host might pick a door at random, often revealing the $1,000,000 prize. But I doubt this is the majority of people. And the riddle infers the same action each time, it does not infer in any way that the host might reveal the prize... and then ask the contestant to "stick or switch?"
No, it doesn't maker sense.
The Monty Hall problem is explained ever simpler than this. Before you started, you had a 2 out of 3 chance of picking either the banana or the thousand bucks right out of the gate, or a 66.67% chance. You only had a 1 in 3 chance of picking the million. However, if he offers you the chance to switch and you take it, you have doubled your chances of winning the million. Here's how. There is a 1 in 3 chance you picked the million, and if did and you switch, you will lose a million bucks. But there was only a 33 & a third% chance of this. If you pick either the banana or the thousand dollars, Monty will always reveal the OTHER lesser prize, so when you switch, 2 out of 3 times you will switch TO the million bucks.
"This problem is pretty easy to understand if you have it explained to you in the best way" and then you proceed to explain it in not the best way.
Monty Hall problem originated from Marylin vos Savant. She originally faced quite a backlash even among University Professors. And we see who was right.
I think it's fair to say the originator of the problem is the person who asked the problem, Craig F. Whitaker. He wrote a letter to vos Savant while she was writing a regular column in Parade magazine.
In Whitaker's letter, as in the question as phrased at the beginning of the video, there is no indication that a compulsion is placed on the game show host to always reveal a door with the booby prize, a critical component in the chain of logic which leads to the conclusion "it's better to switch".
Well done summary and so quick. This way of learning is a secret that most will never get to hear. I hope all watchers give it a shot, especially if you’re still in school and have time to think.
Another way to consider the Monty Hall problem is not to think you have a 1/3 chance of getting it right on your first guess, but that you have a 2/3 chance of getting it WRONG. So the 2/3 chance of getting it right is now with the other two doors, initially split between them. And by removing one of those doors from the equation, that 2/3 correct chance is now with the remaining door.
That is not correct. You would have a 2/3 chance of picking a wrong door regardless as to the host's knowledge. In that case if he opened one of his two doors that reveals a non-prize there would be no advantage in switching.
That is an explanation that makes straightforward sense.
But by removing one door and making that door 100% wrong then it is a 50/50 it may be one of the remaining doors. 6ou can come u0 with some pseudo intellectual nonsense explaining how the myths shows that no to be true but in real world application its 50/50 not 66/33
@@williamdoherty7824 Create an experiment to test it. You can do it with playing cards and a friend who knows the cards in the hole. But you need to repeat the experiment a number of times to get an approaching probability.
@@klaus7443how can you start a sentence correctly sayin your initial chance is 2/3 to be wrong and then ending it with the conclusion that going away from that choice gives you no advantage? Weird.
Great video! Re: Part 3 - It always made me think about how the simplest visible shape we can conceive of is a circle. But our math & ten-based number system can't describe the relationship between the diameter and the circumference without an undefinable number (pi). Always thought that maybe we went down a slightly wrong rabbit hole right from the beginning
2:17 sorry my friend but you are wrong. You did not at any point stipulate that the host MUST reveal any particular prize. the host could have revealed the million dollars. there is a 1/3 chance that you could have picked the banana or the million dollars. if you'd stipulated that the host must reveal either the 1000 or the banana, then you MAY have a point, but you did not.
C: The host could have kill the player.
I was thinking the same thing. If you are trying to get the bananas, and the host showed you the $1000, so you switched, is your chances of getting the banana higher? Or course not. This only works if 2 of the 3 curtains have the same thing begun it.
I know you posted this 3 months ago - but you're logic on this is pretty common, but it's wrong.
The accepted premise is that the host will reveal 1 of the 2 non selected doors before giving you the decision to switch or not.
When you originally pick your door, you have a 1/3 chance in picking right.
There is a 2/3 chance you chose wrong though, right?
So would you trade your one door probability for the two doors you didn't choose? Of course you would.
Well essentially that's what you're doing. But simply the host is revealing that one of his doors doesn't have the prize. But, we already knew one of them was empty, so revealing which one is empty doesn't actually change the probability that one of his doors is more likely to have the prize.
Hope that makes sense
@@edmandolin9310it is indeed higher. I wrote a more detailed response above. Hope it helps.
The simple explanation of the Monty Hall Problem:
At the beginning you have a 1/3 chance of picking the prize.
After the host opens a door you still have that 1/3 chance of having picked correctly. Therefore the other closed door must have a 2/3 chance as the prize had to lie behind one of the two closed doors, so you should switch.
This actually makes sense
I look back in time to Sagan, et al and how much fun I had reading about science. Listening to your posts has reignited my interest in science.
I have Van Nostrems mathematical book.
Just when I thought I had a handle on 1 + 1 = 2 you come along and say hold on might not be real. I always loved the Monty Hall problem. Great video thumbs up.
Simon is trul a one man network with all his shows , writers and editors and I love it and thank you all
The best way I've found to think of the Monte Hall problem is to make the question a little different from "What are the probabilities?"
Imagine instead that person A *always* switches curtains and person B *always* keeps the same one. Person B will win if they picked the right curtain the first time, and person A will win if they picked the *wrong* curtain the first time (since they'll definitely switch to the right one). Who would you rather be?
This ends up matching the probabilities explained in the video: person A picks the wrong curtain at first about 2/3 of the time, and will thus win after switching on those 2/3 of the games. Person B picks the right curtain at first about 1/3 of the time, and will thus win without switching those 1/3 of the games.
There are other variables that haven't been considered. 1. Am I a monkey or other great ape that would enjoy a banana over a monetary prize? 2. Am I starving at the moment? 3. Based on the current inflation trend could a million dollars at the point in time discussed in this video buy a single banana?
You hurt my brain with this one. I love it. If you need me I will be rewatching this video in perpetuity.
Great explanation. And many thanks for that excellent Wigner quote.
Awesome video. Missed opportunity when you pointed at the light and said, "Just like I can't be sure my blue is your blue." Imagine if you said red or another colour.
monty hall problem.
when there are 3 doors, each one is 33% to be correct, which mean whichever one you pick is 66% to be wrong.
then 1 door gets eliminated, leaving only 2, which is actually just smoke and mirrors. obviously , the host is never gonna show you the correct door. he's always gonna show you an incorrect door, which actually make that one irrelevant.
your initial odds never change.
the door you picked is still 66% to be wrong and 33% to be correct, but since you are now being given the opportunity to switch to the other remaining door, then switching means you are 66% to be correct 33% wrong..