Hands down best explanation that I have seen so far in youtube for Monte Hall Puzzle! This puzzle has been bothering me on and off for years! Thank you!
worse most complicated explanation possible of an easy puzzle. just ask what is the odds that the car is behind one of the other two doors? now one card is revealed not to be the card so what are the odds the remaining card is the car? exactly 2 out of 3 same as the odds that the two cards had the car in one of t hem.
easiest way to understand the problem is to do it yourself with 3 objects, for example 2 red pens (goats) and a blue pen (car). First round: Pick the 1st red pen, remove the 2nd red pen (like the host does when he opens the door to a goat he knows is there), switch and you end up with the blue pen. Second round: Pick the 2nd red pen, remove the 1st red pen, switch and you end up with the blue pen. Third round: pick the blue pen, remove any of the 2 red pens, switch and you get the other red pen. So if you switch you get the blue pen (car) 2/3 or the times. Now do the same thing without switching and you get the blue pen 1/3 of the times. It's even more easy to figure it out if you do this with 19 red pens and 1 blue pen, removing 18 red pens before doing the switch.
Hi! I have watched and read several explanations of the Monty Hall problem. Yours is the one that helped me to actually see and understand what is going on, and the logic employed. Thank you so very much for taking the time to post this explanation of the problem!! You are an excellent teacher!
Best explanation regarding the Monty Hall problem I've seen so far. The key being the host will always open a door with a goat behind (making the probability of the car being shown by the host 0) thus inverting the probabilities.
Thank you for this! I've known the right strategy on this forever, but never fully understood the math behind it. This was beautifully done...thank you!
Thank you so much!! So many of these videos keep on explaining the question and the answer over and over again without actually explaining HOW it works.
Best explanation of probability of two events Thanks - I was confused before watching. Now clear a. Expecting Event1 AND Event2 Multiply the two probabilities b. Expecting Event1 OR Event2 Add the two probabilities Combine above with condition
Your explanation is wonderful. I see many videos but they can't explained properly or may be they copy the language of other video but your expalination is 100 times better than other people Thank you. Sorry far bad english
Very well done! My intuition kept telling me 50/50, but her math shows my intuition can't always be trusted! The fact that the host will never open the car door changes the numbers. I finally get it! Thank you!!!
You are wrong! Her explanation is correct. Mathematically there is only a 50/50 chance of winning. The first door was always without a car. You gain no extra information from it. That means that event doesn't count.
Does The Mathematical Probability Tree Work In Relationships . Should You Stay Or Switch ? Do Goats Turn Into Frogs ? Is There A Way To Establish A Base - Line Or Constant Or Do The Variables Make This A More Complex Problem ?
Before you ask about switching doors, the complete rules of the problem need to be presented. You can't just inject new rules after you've asked the question. Monty knows whats behind each door and Monty will always reveal a goat are essential facts which need to be stated prior to asking the contestant whether he or she wants to switch doors.
why. The question of whether to switch doors happens after he shows the goat? Even if it was random and monty didnt know there was a goat there, the fact is, its still a goat there and the chance of it being the car is now added to the other card that you did not pick. If it was a car there, the game would be over
@@ScreamingEagleFTW Mate, the options to make are only four: - STAY with your door and win the car - STAY with your door and lose the car - SWITCH and win the car - SWITCH and lose the car FOUR options and two paths to win, and two to lose. Everything else is just a distraction. It is a 50/50 chance. You don't win any extra information by eliminating one of the doors, because a door without the car was always going to be eliminated from the game and everybody knew that from the start.
It was fantastic . But the thing is had I assumed 1st door to be the car then also I have to switch right ? But after all ultimately it is 50-50 chances of getting a car . So how can that be solved ?
It was never explained to me that the host wasn't opening a door at random. I was under the assumption that you could lose right after choosing if they guessed the car door. Nobody I argued with ever brought this up though, guess they didn't know either and just knew the answer.
@@anthonyng3705 1) It makes the game more fun for the contestant and the audience, and longer lasting. If you picked a door and the host just said, You win, or You lose, that would be too fast and far more boring. 2) Psychologically, I think it makes the contestant feel their chances of having picked the winning door increased. "I had only a 1 in 3 chance originally of choosing the winning door; the winning door was probably one of those two that I didn't pick. But look, I now know that one of the doors I didn't pick was a losing door! There were two doors I didn't choose that could be the car, but now there is only 1 door I didn't choose that could be the car! I am closer to winning. Of the two remaining doors, 1 has the car and 1 doesn't. Now I am just as likely to win as lose"
Nice explanation. However, you are assuming that Monty always gives the player the option to switch. The way I see it, once the player has chosen, Monty has four choices - reveal the car (unlikely, as you've noted), reveal the pink goat, reveal the brown goat, or shut up and let the player take the prize they've chosen. Monty's behavior depends upon such things as whether he wants to give away more cars in an attempt to increase his ratings, or give away fewer cars because the corporate bean counters don't see things that way. Or maybe Monty has some other strategy... So, how goes the analysis when we allow Monty all four options? How much can his motivation shift the odds?
Oops... Actually, after the player has chosen a door, Monty only has three choices - opening either one of the other two doors (his range of choices depends on what the player has chosen), or shutting up. Nonetheless, it seems that Monty's motivation is a significant factor.
@@zevfarkas5120 You'd be somewhat correct if you were talking about the show. There, he didn't always give a switch option nor revealed one door - but he always knew which one had the car. But on the mathematical problem, it is stated that he always will open one door, and it's always a goat.
@5:25 after, say, the Gp is revealed, don't you need to eliminate the option from every decision point? IOW, you are left with tree C or Gb, and since you know the host acted on our decision and under its constraints, then isn't better to not switch, because our actions influenced him less when we picked incorrectly?
@klaus7443 you really think I'm trolling ( that is, responding just to increase responses)? I am genuinely interested in this because of the implied game theory. Again, look at the what the OP calls a probability tree. The OP changes the Host after concluding that revealing the car would end the game, but then doesn't follow through with changing the trees after a goat is revealed. Why?
"after, say, the Gp is revealed, don't you need to eliminate the option from every decision point?" Wow... what an expert in probability calculation you are?! You think you can change parts of other branches of probability tree because of action in one of the branches?! " then isn't better to not switch, because our actions influenced him less when we picked incorrectly?" Of course not, since, by staying, only first branch will end by player winning, while, by switching, the other two branches ends with player winning, so switching is better.
@@Reaction1s Yes you're trolling. You could have tried the problem yourself with three cards as a host to see how the problem actually works. You're not even close to understanding it pusnutz.
@@max5250 No, not one of the branches. All three branches are effected by the random choice made when picking the door, and the other branches are then effected: IOW it eliminates one branch while forcing Monty's choice in the only one left.
Okay I've been on this hunt for hours now. I just want to understand this realistically. I've gone to a random number generator online and had it choose from 1-3 and hit the button 100 times and recorded the data in power point and used that number as the door the car would be behind. I came up with 53 1's, 19 2's, and 28 3's. Just to make things simple I always chose number 1 as my choice and kept swapping like said and lost 47 times. Is this because of a set rule? Can it be that my sample size wasn't big enough? Or is it simply because even though it's a 2/3 chance doesn't mean I can't land in the 1/3 more often ? This question maybe stupid but I'm just trying the figure this out. Thanks
"I came up with 53 1's, 19 2's, and 28 3's" "Just to make things simple I always chose number 1 as my choice and kept swapping like said and lost 47 times." So of the three doors the car was behind Door1 53 times in 100 tries. Of course that door if picked would win only 47 times by switching (and NOT losing 47 times as you said). If you would have chosen Door2 instead you would have won 81 times by switching, and Door3 would have won 72 times. If 100 players picked Door 1 and switched, another 100 picked Door 2 and switched and same for Door 3 they would have won 47+81+72= 200 cars among the 300 players, twice as many as compared to if they all stayed. On average a random number generator show have produced results of the car being behind each door closer to 33, 33, and 34. Switching in that case would win 67 times for each of Doors1 and 2, and 66 times for Door3. Switching wins by having the player and the host both picking a door with a goat. The probability of the player to pick one is 2/3 and for the host it's 1. So the chances to win by switching is 2/3x1=2/3.
The most direct and irrefutable explanation is that the design of the game is such that if you switch, you will inevitably end up with the opposite of what you picked the first chance. In other words: a) If you initial pick was the car, you'll end up with a goat by switching. b) If your initial pick was a goat, you'll end up with the car. This is because it is not possible to pick goat #1 and end up with goat #2. Since there are two goats and one car, the odds are 2/3 that you will initially pick a goat, and 1/3 that you will initially pick the car. Switching therefore means that there is a 2/3 chance of winning the car and 1/3 chance of getting the goat.
+Nin: Yes, it is not necessary to haul out the math for this problem. Simplicity: On that original selection you have a 1/3rd chance to pick the car and a 2/3rd chance to pick a goat. Obviously, (if you always switch), every time your original choice was a goat, you will end up getting the car
En este víudeo se explica cómo utilizar una simulación para calcular probabilidades y resolver el problema de Monty Hall: th-cam.com/video/ZjGUamzJt08/w-d-xo.html
@@5:11 full stop. I'm OK up to here. However after the host reveals a goat, don't you then you have to remove all of those choices from the tree as you just did because the car was never going to be revealed? Also, isn't this a reveal of a hidden rule that the host will always show you an incorrect door? Now, we don't know why the host revealed what was revealed, but we know we 1/3 picked the other goat or 1/3 car, and we know that the host had to either pick from a 50/50 or always pick the goat he showed, then wouldn't that mean that our original decision influenced his decision either 1/6th( probability we picked correct) or 1/9 (probability we picked incorrect), then wouldn't it be better to not switch because our original choice made the host change more often?
"However after the host reveals a goat, don't you then you have to remove all of those choices from the tree as you just did because the car was never going to be revealed?' Learn how probability tree is read. "Also, isn't this a reveal of a hidden rule that the host will always show you an incorrect door? " There are no hidden rules, and there is no mentioning of a host opening the door with a car in wording of the problem.
@max5250 thanks for all of your replies. I setup the experiment: 3 doors (brown grocery bags), 1 car( black chess pawn) and two goats(white chess pawn), I randomized the placement of car/goat, and randomized the selection of door. My thought was correct, but my math and conclusion were wrong. Here are the correct answers: the probability that Monty is effected by our choice of door is 1/3, the probability Monty is not(and can pick either of the other doors) is 1/6. Therfore, we should switch because Monty must choose the other door more often. As a side note, you DO eliminate( or weight differently) trees based off of Monty''s actions. The probability changes to 1/3 vs 1/6, which means the ratio is the same as if it were what the host in the OP calculated, however there are games were that difference does influence the outcomes. Meaning, on contestant's next turn to play, they should "weight" their options carefully. Also, in almost every game there are hidden rules. Again, thanks for responding.
@@Reaction1s "The probability changes to 1/3 vs 1/6, " Probability of the door host left closed for you to switch, to hold a car behind, is 66.6% from the moment player picked 1 out of 3 door, since one of TWO doors holds a car TWICE as likely as ONE door alone. The only difference is that, whenever player picks the door with a car, which happens with 33.3% chance, host has choice of 2 doors with goats, to decide which one to open.
@@Reaction1s When someone place 1 car behind 3 doors, the following statements are valid: 1. One door holds a car with 33.3% chance. 2. One of two doors (any two doors of these three doors) holds a car with 66.6% chance. 3. One of three doors holds a car with 100% chance. This is elementary probability.
It's a conditional probability tree. Every door has 1/3 chance of the prize. One door/goat is eliminated by the host, who is not going to reveal the car until the final selection. If you unknowingly picked the car ( which occurs 1/3 of the time ) and switch, you lose ( as you switch to a goat ). However if you unknowingly picked a goat (which occurs 2/3 of the time) and one door/goat is eliminated, then when you switch you win ( as you switch to the car ). Thus 2/3 of the time switching wins and 1/13 of the time switching loses.
@@JoeyBlogs007 "If you do nothing you have 0% chance of winning the car." That's s brilliant conclusion. Bet it took you hours just to do the probability tree.
5:17 -- _"After the host opens the door, you still have a 1:3"_ -- This is where the logic breaks down. This is IN NO WAY true. Once the host opens a door, that door is ELIMINATED from the equation, therefore you have a 1 in *_TWO,_* not a 1 in 3. This is a massive failure.
You are wrong. The 1/3 vs 2/3 does not come from still including the revealed door. It comes because from the times that the host reveals the door that this time he opened, twice of them will be games in which the switching door has the car, not yours. That occurs because when your door has a goat, the host only has one possible door to remove: which has the only other goat. Instead, when your door has the car, the host is free to reveal any of the other two, we never know which in advance, they are equally likely for us, because both would have goats in that case. That means that the games in which you start picking the car door (that in total are 1/3) are divided in two halves of 1/6 each according to which door is revealed later, and therefore when he reveals one we must discard the half in which he would have opened the other. For example, if you start picking door 1 and he opens door 3, only two possibilities remain: 1) Door 1 happened to have the car (yours), which made the host have to take a decision about what of the others reveal, and this time he preferred #3 instead of #2. They are two conditions that must have been fulfilled: 1/3 * 1/2 = 1/6 2) Door 2 happened to have the car, which forced the host to open door 3 and not any other. 1/3 * 1 = 1/3 The only way to have kept the entire original 1/3 of door 1 is if we were sure that as long as the correct door is #1, the host will reveal #3 and not #2, but nothing in the rules states that. But those probabilities 1/6 and 1/3 were with respect of the total original cases. They are the only possibilities now, so they must sum 1. Applying rule of three you get that at this point case 1) is 1/3 likely, in which staying wins, and case 2) is 2/3 likely, in which switching wins. If you had started picking door 2, the opposite would have occurred: The host would be able to reveal #1 or #3 when the correct is yours, but would be forced to only reveal #3 when the correct is #1.
@@RonaldABG _"You are wrong."_ -- That's very arrogant of you, but let's examine. _"The 1/3 vs 2/3 does not come from still including the revealed door. It comes because from the times that the host reveals the door that this time he opened, twice of them will be games in which the switching door has the car, not yours."_ -- So, the reason that the door the participant chose doesn't benefit at all from the door the host opens is ONLY that the participant chose it in the first place, right? There are no other factors that determine how the probability is divided, correct?
@@RonaldABG Are you not responding because you see the trap coming? If you had any intellectual integrity, you'd admit that here on the thread, but I wouldn't worry. Most people are really cowards that can't admit when they are wrong, so I'm sure society won't hold it against you.
@@dienekes4364 Sorry man, but you are talking about being arrogant? Just listen to yourself being toxic to someone who is right. If you don't understand it, then it doesn't necessarily mean that it's wrong: the part where most people don't understand it is, that the host can only eliminate the door behind which is a goat
@@nikmrn _"Sorry man, but you are talking about being arrogant?"_ -- Can't imagine why you'd be sorry about that, but okay, I accept your appology. _"Just listen to yourself being toxic to someone who is right."_ -- It's not my fault these people are too stupid to understand basic logic or probability. _"If you don't understand it, then it doesn't necessarily mean that it's wrong:"_ -- That is correct. But I DO understand it, which is why I can explain EXACTLY why it is wrong. _"the part where most people don't understand it is, that the host can only eliminate the door behind which is a goat"_ -- Yes, and I have explained several times now, that the ONLY thing this does is push the game to round 2 and completely invalidates any "probability" in the first round. People like you keep bringing this up, but the problem is that you can't explain how this impacts the probability. If you can actually explain this, then feel free. But until you do, it is meaningless.
The chance is 50:50, as the actual game begins only after the host opens one of the door. prior to opening of the it was 33.33 % probability. but as you eliminate the one of the door, the 33.33% probability of the eliminated dor is shared between the other two doors , making each doors probability 50.(33.33+16.67). the way the problem is narrated makes you feel that probability never changed once one of the door opened. while it actually changes the probability. on the calculation provided it shows three scenario for each choice.while in reality the scenario is only two. Take the 1st scenario. if you choose door and stay. the host has option to open the door onlu once and not twice( either door 2 or door 3). so the scenario can only be two.
It is very likely (more probable) you choose a got first. If you choose car first, you will lost by switching, but that will happen two times less then choosing a goat.
yes and the odds of you choosing a goat are 66% so you probably did choose a goat. Thats why mathematically if you switch you will win the car 6 out of 9 times.
Is just logic 2/3 chance of u losing. But if u switched then 2/3 chance of using winning. 1/3 chance of u winning. But if u switch only 1/3 chance of u losing
@3:21 the host has 100% chance of makeing the right move because the host has complete information. A correct choice influenced the options the has to 50/50. Whereas, a incorrect choice always influences the host.
i guess what bothers me is how the smartest woman saw this intuitively and the esteemed math professor, who eventually conceded, really refused that she was right. THAT is puzzling.
That actually made a lot of sense. Thank you so much. I was skeptical when you started saying 1/2s but then you switch it up correctly. I believe that true understanding of something comes when you understand the fundamentals of it.
You are making this WAY too difficult by examining this "probability tree". Obviously if you choose a door and stick to your choice you will win 1 out of 3 times over the long term. Doesn't matter what might happen to those other doors later in the game, you will win 33.3% of the games. (Do we really need to test this obvious fact???) You have chosen a door, and you will get the car if it is under that one door. But if you switch, you get what is under the other 2 doors. 1 out of 3 chance, or 2 out of 3 -- your choice. Its not a math problem its a logic problem. Another way to look at it is that you had a 33.3% chance to have a car under your door, and a 66.7% chance of a goat. Its not overly difficult to see that if a goat was under that first door you chose, if you switch, you will always get the car. 33.3% if you don't switch and 66.7% chance if you do switch.
@@klaus7443 "Lol" Yes it is a logic problem, but most logic problems will contain some maths. Here is an answer from someone who dealt with this problem from before you were ever in diapers. And note, it is from the "rainman" so it cannot be bettered. (-; We are presented with this wellknown problem. The question is asked, "Do we get the best chance at the prize by staying with our original pick, or by switching" The answer: "We get the best chance to win the prize by switching" Examination/explanation: We are presented with three doors with 1 prize under 1 of the doors, at random. We pick a door. Our door has a 1 out of 3 chance to have the prize behind it, just as all the doors do. Now notice something. At this point picking that door can divide those 3 doors into 2 relevant/useful sets: #1: A set of 1 door, the one we just picked. That set has a 1/3 chance to have the car in it., and,#2, a set containing the other 2 doors. Notice, that the set containing 2 doors has exactly a 2/3 chance to have the car in it. Those 2 facts exist, they exist whether we do nothing afterward, whether we do something afterward. They exist whatever you might do or not do afterward. (Open all doors, open NO doors, whatever!). They exist if there are goats, they exist if there are no goats. Goats are totally immaterial to this simple fact. No acts afterward change these facts. Thats it! In being allowed to switch, we get the car EVERY TIME it is in the larger set. THE REASON WHY SWITCHING GIVES YOU THE CAR EVERY TIME IT IS IN THE 2 DOOR SET: The set only has 2 doors in it. Monty opened 1 EMPTY door in that set. Obviously it cannot be behind that door. There is only 1 other door in that set. WHEN THE CAR IS IN THAT SET, YOU WILL ALWAYS GET THE CAR IF YOU SWITCH TO THAT SET Over the long haul, the small set will have the car in it 33.333...% of the time, and the 2 door set will have a car in it 66.666...% of the time. How simple! Yet the way it is presented it stumps a large percentage of people in the beginning. The reason is that they simply cannot visualize that upon opening the empty door, and inviting you to switch, Monty gives you the opportunity to get car every time it is in the larger set. And it is in that set 66.66...% of the trials
@@37rainman Yeah I know. If the probability of contestant picking a goat is 2/3 and for the host it's 1, then the probability they both do is 2/3x1=2/3.
The music in the beginning and the end belongs to me, yet I do not get any credit or fee. This intelligent lady was not honest enough to buy my music. She had to steal it.
With all respect, the reason the game show analogy does not work with Bayes theorem is that the door Monty reveals is taken out of the game and therefore the probabilities altogether. The remaining possibilities have actually, not just apparently changed. It is an example of very educated folks getting tricked by the math. I demonstrated the fallacy with my 9 year old nephew whose argued with me as he believes all he reads on the internet. We used 3 cards and repeated the test 50 times holding his card in every test, the result was as expected near 50% (actually 27 times) he won when holding rather than switching. MY advise to the author of this video is get some cards and test it before being sure of your math.
I made a simulation program that can run the game as many times as I want; for example, 1,000,000. The proportions tend to get closer to 1/3 for staying and 2/3 for switching when more trials are done. Your case was a coincidence. The point you are not getting is you win by staying exactly the same times as if you had only one opportunity selecting from three doors (1/3). This happens because the host is forced by the rules to reveal a goat after you have selected in all the games, and he can do it always because he knows the positions and despite what you caught there was always going to be at least one goat available to show in the other two doors. So once it is revealed it does not say anything about your door, you already knew there was a goat hidden in at least one of the other two. You are confused with if the host revealed the goat randomly. In that case there wouldn't be reasons to think one of the closed doors is more likely than the other but in this case he is forced to leave the prize door closed, and since it is not your door the majority of times, then it is the other he didn't reveal.
You shouldn't do this experiment and then just calculate the result at the end. You should calculate the results after each iteration. What you will notice the more and more trials you do is that the results are not going to stay 50/50 over time. The more trials you perform the closer you will come to a 1/3 v 2/3 result. Also I assume that you knew where each goat and the car were and only revealed goats when you did your experiment right? Because if you didn't then you would not have gotten the right result.
+Peaceful: You are a hoot. If you actually cannot visualize with your own mind that if you select one door from 3 doors, that the possibility is 33%, there is little help for you. If you must do a physical experiment to determine such a mindlessly simple aspect of this problem, then, actually do one. Gee, do the test 100 times, or 500. You will have your answer. You demonstrated to your nephew that you cannot visualize the most simple of probability issues, and that is all. He is still scratching his head over you and your difficulties
WAIT WAIT WAIT... 2:38 So he would open the door with the car, EVER? I don't think that is accurate. I'm going to try and find a video of that... I bet I can't find one.
Bravo....I was starting to feel stupid....makes sense once you think there's a 2/3 chance you chose a goat on the first pick. Then he opens a goat. Odds say switch. Simplified.....2 out of three times you'll chose a goat on your first choice. Monty opens a goat door.
I do not understand the math. But I will share with you a possible simple way to explain it: Say you choose Door 1 as your first pick. (You know that your first attempt at choosing the right door will fail 2/3rds of the time because 2 of the 3 doors are losers). Then next thing to happen is that Monty reveals one of the losing doors. In this example, say Monty opens Door 3 to reveal Door 3 is a loser. Remember, your first choice has 2/3rds chance of losing. That means that the one remaining door has only 1/3rd chance of losing. So switch to it. I've never heard anyone explain it this way...I'm curious to know people's thoughts.
Heyling Chan your welcome! I know how great the feeling is to have something explained in a way where it’s finally simple and clear and all makes sense. That being said, I hope I did this correctly. It’s been a long time since I was on this thread, but in Re-reading my answer, seemed good to me.
yeah this is the same thing except rephrased with 2/3rd chance of losing or 1/3rd chance of losing. Although this is correct :) What I did was lets say I pick door 1 which i know has 1/3rd probability of having a car Now monty knows that he cant pick a door that has car in it so he picks lets say door 2 that has goat. This essentially means that the probability of door 2 having car is 0. in other words the door 3 has 2/3rd probability. People get confused over the fact that the car could have been in door 2 as well. Thing is, if monty opened door 3 to have a goat, then door 2 has 2/3rd chances of having a car. And door 1 has 1/3rd chances of having a car Hopefully this is crct xd also been a year why am i replying to this
Except that the choise is not picking one door out of 3 and win the car. The choice is only between staying with your door or switching. TWO DOORS, not three: Stay and win / Stay and lose Switch and win /Switch and lose. Four paths and only two lead to winning. A clear 50/50 chance, not 2/3 anywhere. The illusion here is that the first pick is 100% IRRELEVANT. It doesn't even matter to the game. At the end you win by picking ONE out of TWO.
NOPE. Nobody has ever answered my question. see at 6:00 = the circled 1/3 chances on the RIGHT bottom of the graph are the very same 1/3 chances on the LEFT graph for Gp and Gb. Which means, we go back to initial 1/3 chance for each one of the original options. the GROUPING together the chances 1/3 + 1/3 that occurs in ANY of these videos explanations is THE QUESTION nobody ever answer. WHY GROUPING 1/3 chances of remaining options towards only one specific door , and not the other door??
Yes, the choice was door 1, 2 or 3. Then, after opening door 3, the choice is door 1 or 2. Doors 2 and 3 equal 2/3rds chance of winning only when they're selected both at the same time -- choosing door 1 or door 2 and 3 together. And the Monty Hall problem was never the door selection -- it was always the cash he would offer in lieu of the prize behind the door. So people would reject a bunch of cash in hand and then win a goat.
That happens because when the player picks a door, they are creating two groups of doors... the player's group (which has one door) and Monty's group which has two doors. Since Monty can only open his own doors (and not the player's) the odds of each group don't change unless the whole group is revealed.
Yes, but it's the modified DeLorean from Back to the Future, which is a time machine. You are both correct (well, except for your spelling of DeLorean).
The probability tree is incorrect. The host WILL always choose the door with a goat behind it because he knows which door has the car behind it. That is part of the setup of this whole scenario. Essentially, the odds do not change whether the host opens a goat door before the contestant chooses a door (but does not open it) or immediately following that. No change in outcome from doing it either way. Because the contestant KNOWS the host will reveal one of the goat doors (the host will have one OR two doors available for him to open and he knows what's behind them), the actual probability of him choosing the correct door (with the car) is ALWAYS 50%. It only looks like 30% initially because there are three doors. Thing is... one of the doors with the goat behind it WILL be revealed by the host. That part of this equation is not taken into account when one continues to think of three doors. It is a logical fallacy. From 2:16 you are suggesting a possibility that the host could pick the door with the car, or the goat. No. He will not do that. There is no probability in this equation that he will pick the car door ever, and that is understood.
"The host WILL always choose the door with a goat behind it because he knows which door has the car behind it." Which means the probability of the contestant and the host both picking a door with a goat is 2/3x1=2/3.
"The probability tree is incorrect. The host WILL always choose the door with a goat behind it because he knows which door has the car behind it." Nope, you just need to watch more of the video. She prunes out the branches where the host shows the car.
Hands down best explanation that I have seen so far in youtube for Monte Hall Puzzle!
This puzzle has been bothering me on and off for years!
Thank you!
Its not good to walk around and bother urself over math. You don't need anxiety to do math.
@@scottwarren4998 lmao
worse most complicated explanation possible of an easy puzzle. just ask what is the odds that the car is behind one of the other two doors? now one card is revealed not to be the card so what are the odds the remaining card is the car? exactly 2 out of 3 same as the odds that the two cards had the car in one of t hem.
easiest way to understand the problem is to do it yourself with 3 objects, for example 2 red pens (goats) and a blue pen (car). First round: Pick the 1st red pen, remove the 2nd red pen (like the host does when he opens the door to a goat he knows is there), switch and you end up with the blue pen. Second round: Pick the 2nd red pen, remove the 1st red pen, switch and you end up with the blue pen. Third round: pick the blue pen, remove any of the 2 red pens, switch and you get the other red pen. So if you switch you get the blue pen (car) 2/3 or the times. Now do the same thing without switching and you get the blue pen 1/3 of the times. It's even more easy to figure it out if you do this with 19 red pens and 1 blue pen, removing 18 red pens before doing the switch.
the fact that she says "the host won't open the car door" helps a lot :-)
Yes whole problem relies on that piece of information she did a good job explaining it.
Yeah, the idea that it's a show and the host does what it takes to keep it running is what usually laks in explaining the reasoning behind it all.
Hi! I have watched and read several explanations of the Monty Hall problem. Yours is the one that helped me to actually see and understand what is going on, and the logic employed. Thank you so very much for taking the time to post this explanation of the problem!! You are an excellent teacher!
You explained this a lot better than my stats prof..... Thanks!
Best explanation regarding the Monty Hall problem I've seen so far. The key being the host will always open a door with a goat behind (making the probability of the car being shown by the host 0) thus inverting the probabilities.
Thank you for this! I've known the right strategy on this forever, but never fully understood the math behind it. This was beautifully done...thank you!
This is the best video about the the Monty Hall problem on TH-cam. Time saver, very simple, very clear, very well done. THANK YOU
This is the best explanation for the visual learners. Thank a lot!
Thank you so much!! So many of these videos keep on explaining the question and the answer over and over again without actually explaining HOW it works.
Yes and thee was I with my Bayes Theorem ....forgetting that the the game was rigged!
Best explanation of probability of two events Thanks - I was confused before watching. Now clear
a. Expecting Event1 AND Event2 Multiply the two probabilities
b. Expecting Event1 OR Event2 Add the two probabilities
Combine above with condition
THANK YOU AFTER WATCHING 3 VIDEOS ON THIS, THIS ONE WAS THE MOST EASY TO UNDERSTAND!!
Your explanation is wonderful.
I see many videos but they can't explained properly or may be they copy the language of other video but your expalination is 100 times better than other people
Thank you.
Sorry far bad english
OK I got lost at 5:03. I thought the probabilty would be zero and 1, right?
The probability for step 2 is 0 and 1. But that is multiplied by the probability of step 1, which is 1/3.
1/3 * 0 = 0
1/3 * 1 = 1/3
This is the first time that this problem has made sense to me , I've been arguing my ass off, incorrectly I have to say. Thank you
Good for you to admit your error.
@@videoshomepage oh yeah , I wast even close on the concept lol
Very well done! My intuition kept telling me 50/50, but her math shows my intuition can't always be trusted! The fact that the host will never open the car door changes the numbers. I finally get it! Thank you!!!
But she shows the game is rigged !
You are wrong! Her explanation is correct.
Mathematically there is only a 50/50 chance of winning.
The first door was always without a car. You gain no extra information from it. That means that event doesn't count.
This is the best explanation that I've watch so far.. thumbs up.
So Does That Work With Intimate Relationships ? / Explain -
Does The Mathematical Probability
Tree Work In Relationships . Should You Stay Or Switch ?
Do Goats Turn Into Frogs ?
Is There A Way To Establish
A Base - Line Or Constant
Or Do The Variables Make This A More Complex Problem ?
This is an elegant, entertaining, and humorous explanation.
Before you ask about switching doors, the complete rules of the problem need to be presented. You can't just inject new rules after you've asked the question. Monty knows whats behind each door and Monty will always reveal a goat are essential facts which need to be stated prior to asking the contestant whether he or she wants to switch doors.
why. The question of whether to switch doors happens after he shows the goat? Even if it was random and monty didnt know there was a goat there, the fact is, its still a goat there and the chance of it being the car is now added to the other card that you did not pick. If it was a car there, the game would be over
@@ScreamingEagleFTW Mate, the options to make are only four:
- STAY with your door and win the car
- STAY with your door and lose the car
- SWITCH and win the car
- SWITCH and lose the car
FOUR options and two paths to win, and two to lose.
Everything else is just a distraction. It is a 50/50 chance.
You don't win any extra information by eliminating one of the doors, because a door without the car was always going to be eliminated from the game and everybody knew that from the start.
Beautifully explained with methodology learned in probability classes rather than "intuition".
It was fantastic . But the thing is had I assumed 1st door to be the car then also I have to switch right ? But after all ultimately it is 50-50 chances of getting a car . So how can that be solved ?
Problem is Monty Hall never offered the chance to change doors. He offered money or the door.
"But the thing is had I assumed 1st door to be the car"
Why would yo make that assumption?
It was never explained to me that the host wasn't opening a door at random. I was under the assumption that you could lose right after choosing if they guessed the car door. Nobody I argued with ever brought this up though, guess they didn't know either and just knew the answer.
I agree. If the host is purposefully choosing a door with a goat, the problem works. If he is picking randomly, the problem does not work.
Yes. The host is not picking the car so is eliminating a goat option. Probability tree illustrates it nicely.
Hey guy, please help this. My question is " What is the purpose of the host opening another goat?"
@@anthonyng3705 1) It makes the game more fun for the contestant and the audience, and longer lasting. If you picked a door and the host just said, You win, or You lose, that would be too fast and far more boring.
2) Psychologically, I think it makes the contestant feel their chances of having picked the winning door increased. "I had only a 1 in 3 chance originally of choosing the winning door; the winning door was probably one of those two that I didn't pick. But look, I now know that one of the doors I didn't pick was a losing door! There were two doors I didn't choose that could be the car, but now there is only 1 door I didn't choose that could be the car! I am closer to winning. Of the two remaining doors, 1 has the car and 1 doesn't. Now I am just as likely to win as lose"
Excellent demonstration of probability theory behind Monty Hall puzzle.
best explanation for monty hall problem
Nice explanation. However, you are assuming that Monty always gives the player the option to switch. The way I see it, once the player has chosen, Monty has four choices - reveal the car (unlikely, as you've noted), reveal the pink goat, reveal the brown goat, or shut up and let the player take the prize they've chosen. Monty's behavior depends upon such things as whether he wants to give away more cars in an attempt to increase his ratings, or give away fewer cars because the corporate bean counters don't see things that way. Or maybe Monty has some other strategy...
So, how goes the analysis when we allow Monty all four options? How much can his motivation shift the odds?
Oops... Actually, after the player has chosen a door, Monty only has three choices - opening either one of the other two doors (his range of choices depends on what the player has chosen), or shutting up. Nonetheless, it seems that Monty's motivation is a significant factor.
Monty has 1 option. He *MUST* open a door with a Goat behind it.
@@zevfarkas5120 You'd be somewhat correct if you were talking about the show. There, he didn't always give a switch option nor revealed one door - but he always knew which one had the car. But on the mathematical problem, it is stated that he always will open one door, and it's always a goat.
@@kabzebrowski AFAIK, there are many different statements of the problem.
@@zevfarkas5120 But only 1 of the ones you proposed is the one mentioned in the video.
Think this is the first way I’ve seen it explained and actually understood it!
Very nice and clear explanation, thank you.
The best explanations I found after so much search
Best explanation
@5:25 after, say, the Gp is revealed, don't you need to eliminate the option from every decision point?
IOW, you are left with tree C or Gb, and since you know the host acted on our decision and under its constraints, then isn't better to not switch, because our actions influenced him less when we picked incorrectly?
Quit trolling. There was a 2/3 chance the host revealed the only goat he had and left the car.
@klaus7443 you really think I'm trolling ( that is, responding just to increase responses)?
I am genuinely interested in this because of the implied game theory.
Again, look at the what the OP calls a probability tree. The OP changes the Host after concluding that revealing the car would end the game, but then doesn't follow through with changing the trees after a goat is revealed. Why?
"after, say, the Gp is revealed, don't you need to eliminate the option from every decision point?"
Wow... what an expert in probability calculation you are?!
You think you can change parts of other branches of probability tree because of action in one of the branches?!
" then isn't better to not switch, because our actions influenced him less when we picked incorrectly?"
Of course not, since, by staying, only first branch will end by player winning, while, by switching, the other two branches ends with player winning, so switching is better.
@@Reaction1s
Yes you're trolling. You could have tried the problem yourself with three cards as a host to see how the problem actually works. You're not even close to understanding it pusnutz.
@@max5250 No, not one of the branches. All three branches are effected by the random choice made when picking the door, and the other branches are then effected: IOW it eliminates one branch while forcing Monty's choice in the only one left.
Best explanation I've ever seen on this problem
Thank you! Best explanation of this problem I’ve come across!
The best explanation I found on the internet 👍🏽💯
Excellent explanation !
Best explanation for the problem on internet
By far the best explanation !
Okay I've been on this hunt for hours now. I just want to understand this realistically. I've gone to a random number generator online and had it choose from 1-3 and hit the button 100 times and recorded the data in power point and used that number as the door the car would be behind. I came up with 53 1's, 19 2's, and 28 3's. Just to make things simple I always chose number 1 as my choice and kept swapping like said and lost 47 times.
Is this because of a set rule? Can it be that my sample size wasn't big enough? Or is it simply because even though it's a 2/3 chance doesn't mean I can't land in the 1/3 more often ?
This question maybe stupid but I'm just trying the figure this out. Thanks
"I came up with 53 1's, 19 2's, and 28 3's"
"Just to make things simple I always chose number 1 as my choice and kept swapping like said and lost 47 times."
So of the three doors the car was behind Door1 53 times in 100 tries. Of course that door if picked would win only 47 times by switching (and NOT losing 47 times as you said). If you would have chosen Door2 instead you would have won 81 times by switching, and Door3 would have won 72 times. If 100 players picked Door 1 and switched, another 100 picked Door 2 and switched and same for Door 3 they would have won 47+81+72= 200 cars among the 300 players, twice as many as compared to if they all stayed.
On average a random number generator show have produced results of the car being behind each door closer to 33, 33, and 34. Switching in that case would win 67 times for each of Doors1 and 2, and 66 times for Door3.
Switching wins by having the player and the host both picking a door with a goat. The probability of the player to pick one is 2/3 and for the host it's 1. So the chances to win by switching is 2/3x1=2/3.
The most direct and irrefutable explanation is that the design of the game is such that if you switch, you will inevitably end up with the opposite of what you picked the first chance. In other words:
a) If you initial pick was the car, you'll end up with a goat by switching.
b) If your initial pick was a goat, you'll end up with the car.
This is because it is not possible to pick goat #1 and end up with goat #2. Since there are two goats and one car, the odds are 2/3 that you will initially pick a goat, and 1/3 that you will initially pick the car. Switching therefore means that there is a 2/3 chance of winning the car and 1/3 chance of getting the goat.
@@derfunkhaus Finally, someone explained it in a way that makes sense to me. Thank you.
+Revik: Did you ever hear of the concept: "garbage in, garbage out"????
+Nin: Yes, it is not necessary to haul out the math for this problem. Simplicity: On that original selection you have a 1/3rd chance to pick the car and a 2/3rd chance to pick a goat. Obviously, (if you always switch), every time your original choice was a goat, you will end up getting the car
Thank you for finally explaining it in a way I can understand. Subscribe worthy
Even my teacher was confusing when he taught this. Thank you. XD
I've listened to 2 other explanations and had just about given up. I love your explanation. I finally get it! Thank you!
En este víudeo se explica cómo utilizar una simulación para calcular probabilidades y resolver el problema de Monty Hall: th-cam.com/video/ZjGUamzJt08/w-d-xo.html
I'm pretty smart but have just never understood the Monty Hall Problem until this moment. Thank you!
@@5:11 full stop. I'm OK up to here. However after the host reveals a goat, don't you then you have to remove all of those choices from the tree as you just did because the car was never going to be revealed? Also, isn't this a reveal of a hidden rule that the host will always show you an incorrect door?
Now, we don't know why the host revealed what was revealed, but we know we 1/3 picked the other goat or 1/3 car, and we know that the host had to either pick from a 50/50 or always pick the goat he showed, then wouldn't that mean that our original decision influenced his decision either 1/6th( probability we picked correct) or 1/9 (probability we picked incorrect), then wouldn't it be better to not switch because our original choice made the host change more often?
"However after the host reveals a goat, don't you then you have to remove all of those choices from the tree as you just did because the car was never going to be revealed?'
Learn how probability tree is read.
"Also, isn't this a reveal of a hidden rule that the host will always show you an incorrect door? "
There are no hidden rules, and there is no mentioning of a host opening the door with a car in wording of the problem.
@max5250 thanks for all of your replies. I setup the experiment: 3 doors (brown grocery bags), 1 car( black chess pawn) and two goats(white chess pawn), I randomized the placement of car/goat, and randomized the selection of door.
My thought was correct, but my math and conclusion were wrong.
Here are the correct answers: the probability that Monty is effected by our choice of door is 1/3, the probability Monty is not(and can pick either of the other doors) is 1/6. Therfore, we should switch because Monty must choose the other door more often.
As a side note, you DO eliminate( or weight differently) trees based off of Monty''s actions. The probability changes to 1/3 vs 1/6, which means the ratio is the same as if it were what the host in the OP calculated, however there are games were that difference does influence the outcomes. Meaning, on contestant's next turn to play, they should "weight" their options carefully.
Also, in almost every game there are hidden rules.
Again, thanks for responding.
@@Reaction1s
"The probability changes to 1/3 vs 1/6, "
Probability of the door host left closed for you to switch, to hold a car behind, is 66.6% from the moment player picked 1 out of 3 door, since one of TWO doors holds a car TWICE as likely as ONE door alone.
The only difference is that, whenever player picks the door with a car, which happens with 33.3% chance, host has choice of 2 doors with goats, to decide which one to open.
@max5250 umm 1/3 is .33, and 1/6 is .16. There is no possibility that there is 2/3 66% after the first selection of the tree.
@@Reaction1s
When someone place 1 car behind 3 doors, the following statements are valid:
1. One door holds a car with 33.3% chance.
2. One of two doors (any two doors of these three doors) holds a car with 66.6% chance.
3. One of three doors holds a car with 100% chance.
This is elementary probability.
This was the only explanation ive understood. THANK YOU
Definitely, the best explanation so far thank you!
Superb explanation, well done!
It's a conditional probability tree. Every door has 1/3 chance of the prize. One door/goat is eliminated by the host, who is not going to reveal the car until the final selection. If you unknowingly picked the car ( which occurs 1/3 of the time ) and switch, you lose ( as you switch to a goat ). However if you unknowingly picked a goat (which occurs 2/3 of the time) and one door/goat is eliminated, then when you switch you win ( as you switch to the car ). Thus 2/3 of the time switching wins and 1/13 of the time switching loses.
If you pick randomly you get a 50% chance of winning the car.
If you do nothing you have 0% chance of winning the car.
@@JoeyBlogs007
"If you do nothing you have 0% chance of winning the car."
That's s brilliant conclusion. Bet it took you hours just to do the probability tree.
ah so it’s basically a 2/3 probability that switching is beneficial
Best explanation that I’ve seen. Thanks!
1. If you swap doors, you will always get other object (ie goat to car, car to goat).
2. Two-thirds of the time you will pick a goat initially.
Exactly. Excellent explanation.
Perfect explanation
Best explanation I have seen. Thanks !
5:17 -- _"After the host opens the door, you still have a 1:3"_ -- This is where the logic breaks down. This is IN NO WAY true. Once the host opens a door, that door is ELIMINATED from the equation, therefore you have a 1 in *_TWO,_* not a 1 in 3. This is a massive failure.
You are wrong. The 1/3 vs 2/3 does not come from still including the revealed door. It comes because from the times that the host reveals the door that this time he opened, twice of them will be games in which the switching door has the car, not yours.
That occurs because when your door has a goat, the host only has one possible door to remove: which has the only other goat. Instead, when your door has the car, the host is free to reveal any of the other two, we never know which in advance, they are equally likely for us, because both would have goats in that case.
That means that the games in which you start picking the car door (that in total are 1/3) are divided in two halves of 1/6 each according to which door is revealed later, and therefore when he reveals one we must discard the half in which he would have opened the other.
For example, if you start picking door 1 and he opens door 3, only two possibilities remain:
1) Door 1 happened to have the car (yours), which made the host have to take a decision about what of the others reveal, and this time he preferred #3 instead of #2. They are two conditions that must have been fulfilled:
1/3 * 1/2 = 1/6
2) Door 2 happened to have the car, which forced the host to open door 3 and not any other.
1/3 * 1 = 1/3
The only way to have kept the entire original 1/3 of door 1 is if we were sure that as long as the correct door is #1, the host will reveal #3 and not #2, but nothing in the rules states that.
But those probabilities 1/6 and 1/3 were with respect of the total original cases. They are the only possibilities now, so they must sum 1. Applying rule of three you get that at this point case 1) is 1/3 likely, in which staying wins, and case 2) is 2/3 likely, in which switching wins.
If you had started picking door 2, the opposite would have occurred: The host would be able to reveal #1 or #3 when the correct is yours, but would be forced to only reveal #3 when the correct is #1.
@@RonaldABG _"You are wrong."_ -- That's very arrogant of you, but let's examine.
_"The 1/3 vs 2/3 does not come from still including the revealed door. It comes because from the times that the host reveals the door that this time he opened, twice of them will be games in which the switching door has the car, not yours."_ -- So, the reason that the door the participant chose doesn't benefit at all from the door the host opens is ONLY that the participant chose it in the first place, right? There are no other factors that determine how the probability is divided, correct?
@@RonaldABG Are you not responding because you see the trap coming? If you had any intellectual integrity, you'd admit that here on the thread, but I wouldn't worry. Most people are really cowards that can't admit when they are wrong, so I'm sure society won't hold it against you.
@@dienekes4364 Sorry man, but you are talking about being arrogant? Just listen to yourself being toxic to someone who is right. If you don't understand it, then it doesn't necessarily mean that it's wrong: the part where most people don't understand it is, that the host can only eliminate the door behind which is a goat
@@nikmrn _"Sorry man, but you are talking about being arrogant?"_ -- Can't imagine why you'd be sorry about that, but okay, I accept your appology.
_"Just listen to yourself being toxic to someone who is right."_ -- It's not my fault these people are too stupid to understand basic logic or probability.
_"If you don't understand it, then it doesn't necessarily mean that it's wrong:"_ -- That is correct. But I DO understand it, which is why I can explain EXACTLY why it is wrong.
_"the part where most people don't understand it is, that the host can only eliminate the door behind which is a goat"_ -- Yes, and I have explained several times now, that the ONLY thing this does is push the game to round 2 and completely invalidates any "probability" in the first round.
People like you keep bringing this up, but the problem is that you can't explain how this impacts the probability. If you can actually explain this, then feel free. But until you do, it is meaningless.
The chance is 50:50, as the actual game begins only after the host opens one of the door. prior to opening of the it was 33.33 % probability. but as you eliminate the one of the door, the 33.33% probability of the eliminated dor is shared between the other two doors , making each doors probability 50.(33.33+16.67). the way the problem is narrated makes you feel that probability never changed once one of the door opened. while it actually changes the probability. on the calculation provided it shows three scenario for each choice.while in reality the scenario is only two. Take the 1st scenario. if you choose door and stay. the host has option to open the door onlu once and not twice( either door 2 or door 3). so the scenario can only be two.
Maybe you should try to learn about probability first, and then come back, since your reply is pretty embarrassing.
Thank you, I'm satisfied with the answer :-)
this is so clearly explained. Amazing
hey , thank you !! Best explanation ever. and thanks for making the clay goats and gettingthe cards for us. god bless u
Best explanation ever.
I guess it is the fact that there is a higher chance you will choose a goat door 2/3 and then the host will open the non car door
You explained great! So basically, if you choose the car First, you’re screwed & shouldn’t switch? It’s better to choose a goat first
It is very likely (more probable) you choose a got first.
If you choose car first, you will lost by switching, but that will happen two times less then choosing a goat.
yes and the odds of you choosing a goat are 66% so you probably did choose a goat. Thats why mathematically if you switch you will win the car 6 out of 9 times.
Is just logic 2/3 chance of u losing. But if u switched then 2/3 chance of using winning. 1/3 chance of u winning. But if u switch only 1/3 chance of u losing
Thanks a lot... I was struggling to explain the solution to my high school son....
Someone forgot to take his pills again…
@3:21 the host has 100% chance of makeing the right move because the host has complete information.
A correct choice influenced the options the has to 50/50. Whereas, a incorrect choice always influences the host.
Prefect explanation.
Finally I understood it. Thx!
And it finally makes sense. The key is he WONT open the door with the car behind. Kind of ambiguous isn't it
doesnt matter what he does. if he does open the door with the car the game is over.
i guess what bothers me is how the smartest woman saw this intuitively and the esteemed math professor, who eventually conceded, really refused that she was right. THAT is puzzling.
That actually made a lot of sense. Thank you so much. I was skeptical when you started saying 1/2s but then you switch it up correctly. I believe that true understanding of something comes when you understand the fundamentals of it.
You are making this WAY too difficult by examining this "probability tree".
Obviously if you choose a door and stick to your choice you will win 1 out of 3 times over the long term. Doesn't matter what might happen to those other doors later in the game, you will win 33.3% of the games. (Do we really need to test this obvious fact???)
You have chosen a door, and you will get the car if it is under that one door. But if you switch, you get what is under the other 2 doors. 1 out of 3 chance, or 2 out of 3 -- your choice.
Its not a math problem its a logic problem. Another way to look at it is that you had a 33.3% chance to have a car under your door, and a 66.7% chance of a goat. Its not overly difficult to see that if a goat was under that first door you chose, if you switch, you will always get the car. 33.3% if you don't switch and 66.7% chance if you do switch.
"Its not a math problem its a logic problem."
Lol....
@@klaus7443 "Lol" Yes it is a logic problem, but most logic problems will contain some maths.
Here is an answer from someone who dealt with this problem from before you were ever in diapers. And note, it is from the "rainman" so it cannot be bettered. (-;
We are presented with this wellknown problem. The question is asked, "Do we get the best chance at the prize by staying with our original pick, or by switching"
The answer: "We get the best chance to win the prize by switching"
Examination/explanation: We are presented with three doors with 1 prize under 1 of the doors, at random. We pick a door. Our door has a 1 out of 3 chance to have the prize behind it, just as all the doors do.
Now notice something. At this point picking that door can divide those 3 doors into 2 relevant/useful sets: #1: A set of 1 door, the one we just picked. That set has a 1/3 chance to have the car in it., and,#2, a set containing the other 2 doors. Notice, that the set containing 2 doors has exactly a 2/3 chance to have the car in it. Those 2 facts exist, they exist whether we do nothing afterward, whether we do something afterward. They exist whatever you might do or not do afterward. (Open all doors, open NO doors, whatever!). They exist if there are goats, they exist if there are no goats. Goats are totally immaterial to this simple fact. No acts afterward change these facts.
Thats it! In being allowed to switch, we get the car EVERY TIME it is in the larger set.
THE REASON WHY SWITCHING GIVES YOU THE CAR EVERY TIME IT IS IN THE 2 DOOR SET: The set only has 2 doors in it. Monty opened 1 EMPTY door in that set. Obviously it cannot be behind that door. There is only 1 other door in that set. WHEN THE CAR IS IN THAT SET, YOU WILL ALWAYS GET THE CAR IF YOU SWITCH TO THAT SET
Over the long haul, the small set will have the car in it 33.333...% of the time, and the 2 door set will have a car in it 66.666...% of the time.
How simple! Yet the way it is presented it stumps a large percentage of people in the beginning. The reason is that they simply cannot visualize that upon opening the empty door, and inviting you to switch, Monty gives you the opportunity to get car every time it is in the larger set. And it is in that set 66.66...% of the trials
@@37rainman Yeah I know. If the probability of contestant picking a goat is 2/3 and for the host it's 1, then the probability they both do is 2/3x1=2/3.
Great explanation. The cards and car does not really contribute in explaining, the diagram does. :)
Best explanation!
After many explanations …I’ll just switch
If there's no host then i assume it doesn't matter if you switch or not since the probably would be 50%
Correct, the only reason why the better option to switch is, because the host can't eliminate the door with the goat behind it
Well done!
The host can't reveal the car
EXACTLY!
Yes! That's what made it click for me...
The music in the beginning and the end belongs to me, yet I do not get any credit or fee. This intelligent lady was not honest enough to buy my music. She had to steal it.
What is the probability of this happening
@@jakepollen6839 Pretty high because people are dishonest.
Weird. I randomly chose this video and heard my piano music playing!
Not weird
very well explained
please help! WHAT IS THE PURPOSE OF THE HOST WHEN OPENING AN ANOTHER GOAT GATE? I cant figure out why?
to make the game more interesting and provoke his decision by giving him an choice.
tank you for your clear explain. godd job!
Her handwriting is nice and she is super cute.
With all respect, the reason the game show analogy does not work with Bayes theorem is that the door Monty reveals is taken out of the game and therefore the probabilities altogether. The remaining possibilities have actually, not just apparently changed. It is an example of very educated folks getting tricked by the math. I demonstrated the fallacy with my 9 year old nephew whose argued with me as he believes all he reads on the internet. We used 3 cards and repeated the test 50 times holding his card in every test, the result was as expected near 50% (actually 27 times) he won when holding rather than switching. MY advise to the author of this video is get some cards and test it before being sure of your math.
I made a simulation program that can run the game as many times as I want; for example, 1,000,000. The proportions tend to get closer to 1/3 for staying and 2/3 for switching when more trials are done. Your case was a coincidence.
The point you are not getting is you win by staying exactly the same times as if you had only one opportunity selecting from three doors (1/3). This happens because the host is forced by the rules to reveal a goat after you have selected in all the games, and he can do it always because he knows the positions and despite what you caught there was always going to be at least one goat available to show in the other two doors. So once it is revealed it does not say anything about your door, you already knew there was a goat hidden in at least one of the other two.
You are confused with if the host revealed the goat randomly. In that case there wouldn't be reasons to think one of the closed doors is more likely than the other but in this case he is forced to leave the prize door closed, and since it is not your door the majority of times, then it is the other he didn't reveal.
You shouldn't do this experiment and then just calculate the result at the end. You should calculate the results after each iteration. What you will notice the more and more trials you do is that the results are not going to stay 50/50 over time. The more trials you perform the closer you will come to a 1/3 v 2/3 result. Also I assume that you knew where each goat and the car were and only revealed goats when you did your experiment right? Because if you didn't then you would not have gotten the right result.
@@RonaldABG in the game ..you only get one chance
+Peaceful: You are a hoot. If you actually cannot visualize with your own mind that if you select one door from 3 doors, that the possibility is 33%, there is little help for you. If you must do a physical experiment to determine such a mindlessly simple aspect of this problem, then, actually do one. Gee, do the test 100 times, or 500. You will have your answer.
You demonstrated to your nephew that you cannot visualize the most simple of probability issues, and that is all. He is still scratching his head over you and your difficulties
@@ronalddump4061 schadenfreude
WAIT WAIT WAIT... 2:38 So he would open the door with the car, EVER? I don't think that is accurate. I'm going to try and find a video of that... I bet I can't find one.
3:54
Thanks was a great explanation
Problem is Monty Hall never offered the chance to change doors or give a reveal.
Thanks a Lot... !¡!
I GET IT. now i get it thank you
Bravo....I was starting to feel stupid....makes sense once you think there's a 2/3 chance you chose a goat on the first pick. Then he opens a goat. Odds say switch.
Simplified.....2 out of three times you'll chose a goat on your first choice.
Monty opens a goat door.
The host must know where the car is...otherwise it is 50/50.
Thanks in a million! I got it.
I do not understand the math. But I will share with you a possible simple way to explain it: Say you choose Door 1 as your first pick. (You know that your first attempt at choosing the right door will fail 2/3rds of the time because 2 of the 3 doors are losers). Then next thing to happen is that Monty reveals one of the losing doors. In this example, say Monty opens Door 3 to reveal Door 3 is a loser. Remember, your first choice has 2/3rds chance of losing. That means that the one remaining door has only 1/3rd chance of losing. So switch to it. I've never heard anyone explain it this way...I'm curious to know people's thoughts.
I FINALLY understood it, thank you!! Everyone on the internet was explaining it the same way, yours is the only different one that made sense to me.
Heyling Chan your welcome! I know how great the feeling is to have something explained in a way where it’s finally simple and clear and all makes sense. That being said, I hope I did this correctly. It’s been a long time since I was on this thread, but in Re-reading my answer, seemed good to me.
yeah this is the same thing except rephrased with 2/3rd chance of losing or 1/3rd chance of losing. Although this is correct :)
What I did was lets say I pick door 1 which i know has 1/3rd probability of having a car
Now monty knows that he cant pick a door that has car in it so he picks lets say door 2 that has goat.
This essentially means that the probability of door 2 having car is 0. in other words the door 3 has 2/3rd probability.
People get confused over the fact that the car could have been in door 2 as well. Thing is, if monty opened door 3 to have a goat, then door 2 has 2/3rd chances of having a car. And door 1 has 1/3rd chances of having a car
Hopefully this is crct xd
also been a year why am i replying to this
Except that the choise is not picking one door out of 3 and win the car.
The choice is only between staying with your door or switching.
TWO DOORS, not three:
Stay and win / Stay and lose
Switch and win /Switch and lose.
Four paths and only two lead to winning.
A clear 50/50 chance, not 2/3 anywhere.
The illusion here is that the first pick is 100% IRRELEVANT. It doesn't even matter to the game. At the end you win by picking ONE out of TWO.
Perfection
Brilliant
Literally feel drunk.
NOPE. Nobody has ever answered my question. see at 6:00 = the circled 1/3 chances on the RIGHT bottom of the graph are the very same 1/3 chances on the LEFT graph for Gp and Gb. Which means, we go back to initial 1/3 chance for each one of the original options. the GROUPING together the chances 1/3 + 1/3 that occurs in ANY of these videos explanations is THE QUESTION nobody ever answer. WHY GROUPING 1/3 chances of remaining options towards only one specific door , and not the other door??
Yes, the choice was door 1, 2 or 3. Then, after opening door 3, the choice is door 1 or 2. Doors 2 and 3 equal 2/3rds chance of winning only when they're selected both at the same time -- choosing door 1 or door 2 and 3 together. And the Monty Hall problem was never the door selection -- it was always the cash he would offer in lieu of the prize behind the door. So people would reject a bunch of cash in hand and then win a goat.
That happens because when the player picks a door, they are creating two groups of doors... the player's group (which has one door) and Monty's group which has two doors. Since Monty can only open his own doors (and not the player's) the odds of each group don't change unless the whole group is revealed.
Thank u.. for explaining the human angle..
It’s a de Lorain.
Yes, but it's the modified DeLorean from Back to the Future, which is a time machine. You are both correct (well, except for your spelling of DeLorean).
this actually helped me tys
The probability tree is incorrect. The host WILL always choose the door with a goat behind it because he knows which door has the car behind it.
That is part of the setup of this whole scenario.
Essentially, the odds do not change whether the host opens a goat door before the contestant chooses a door (but does not open it) or immediately following that. No change in outcome from doing it either way.
Because the contestant KNOWS the host will reveal one of the goat doors (the host will have one OR two doors available for him to open and he knows what's behind them), the actual probability of him choosing the correct door (with the car) is ALWAYS 50%. It only looks like 30% initially because there are three doors.
Thing is... one of the doors with the goat behind it WILL be revealed by the host. That part of this equation is not taken into account when one continues to think of three doors. It is a logical fallacy.
From 2:16 you are suggesting a possibility that the host could pick the door with the car, or the goat.
No. He will not do that. There is no probability in this equation that he will pick the car door ever, and that is understood.
"The host WILL always choose the door with a goat behind it because he knows which door has the car behind it."
Which means the probability of the contestant and the host both picking a door with a goat is 2/3x1=2/3.
"The probability tree is incorrect. The host WILL always choose the door with a goat behind it because he knows which door has the car behind it."
Nope, you just need to watch more of the video. She prunes out the branches where the host shows the car.
The host can't open the car door.