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Let’s make a deal: Revisiting the Monty Hall problem

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"Charles Sanders Peirce once observed that in no other branch of mathematics is it so easy for experts to blunder as in probability theory."

Thus began an article in the October 1959 Scientific American by the celebrated math columnist Martin Gardner. In fact, as John Allen Paulos observed in last January’s issue ("Animal Instincts" [Advances]), humans can sometimes be even worse than pigeons at evaluating probabilities.

Paulos, a mathematician at Temple University in Philadelphia, was describing a notoriously tricky problem known as the Monty Hall paradox. A problem so tricky, in fact, that scores of professional mathematicians and statisticians have stumbled on it over the years. Many have repeatedly failed to grasp it even after they were shown the correct solution.

According to an article by New York Times reporter John Tierney that appeared on July 21, 1991, after a writer called Marilyn vos Savant described the Monty Hall problem—and its uncanny solution—in a magazine the year before, she received something like 10,000 letters, most of them claiming they could prove her wrong. "The most vehement criticism," Tierney wrote, "has come from mathematicians and scientists, who have alternated between gloating at her (‘You are the goat!’) and lamenting the nation’s innumeracy."

Sure enough, after Paulos mentioned the Monty Hall problem in Scientific American, many readers (though nothing in the order of 10,000) wrote to complain that he had gotten everything wrong, or simply to confess their befuddlement.

"Paulos shows a strange lack of understanding of basic conditional probability," wrote one reader, "and as a result his article is nonsense." The reader added that Paulos’s blunder shook his trust in the magazine. "What are your procedures for evaluating submitted papers?" he wrote. This reader was a retired statistics professor.

So we thought it might be worthwhile to try and clarify things a bit. What is this Monty Hall business, and what’s so complicated about it?

The Monty Hall problem was introduced in 1975 by an American statistician as a test study in the theory of probabilities inspired by Monty Hall’s quiz show "Let’s Make a Deal." (Scholars have observed that the Monty Hall problem was mathematically identical to a problem proposed by French mathematician Joseph Bertrand in 1889—as well as to one, called the three-prisoner game, introduced by Gardner in his 1959 piece; more on that later.) Let’s hear the game’s description from Paulos:

A guest on the show has to choose among three doors, behind one of which is a prize. The guest states his choice, and the host opens one of the two remaining closed doors, always being careful that it is one behind which there is no prize. Should the guest switch to the remaining closed door? Most people choose to stay with their original choice, which is wrong—switching would increase their chance of winning from 1/3 to 2/3. (There is a 1/3 chance that the guest’s original pick was correct, and that does not change.) Even after playing the game many times, which would afford ample opportunity to observe that switching doubles the chances of winning, most people in a recent study switched only 2/3 of the time. Pigeons did better. After a few tries, the birds learn to switch every time.

But wait a minute, you say: after Monty opens the door, there are only two options left. The odds then must be 50-50, or 1/2, for each, so that changing choice of door makes no difference.

To understand what’s going on, we must first make some assumptions, because as it is, the problem’s formulation is ambiguous. So, we shall assume that Monty knows where the car is, and that after the player picks one door he always opens one of the remaining two. Moreover, if the player’s first choice was a door hiding a goat, then Monty always opens the door that hid the other goat; but if the player picked the car, Monty picks randomly between the other two doors, both of which hide a goat.

So imagine you are the player. You take your pick: we’ll call it door 1. One-third of the time, this will be the door with the car, and the remaining 2/3 of the time (66.666… percent) it will be one with a goat. You don’t know what you’ve picked, so you should formulate a strategy that will maximize your overall odds of winning.

Let’s say you picked a goat-hiding door. Monty now opens the other goat-hiding door—call that door 2—and asks you if you want to stick to door 1 or switch to door 3 (which is the one hiding the car). Obviously, in this case by switching you’ll win. But remember, this situation happens 2/3 of the times.

The remaining 1/3 of the time, if you switch, you lose, regardless of which door Monty opens next. But if you adopt the strategy of switching always, no matter what, you’re guaranteed to win 2/3 of the time.

Seems easy enough, doesn’t it? If however you happen to know a little bit of probability theory and you pull out your paper and pencil and start calculating, you might start to doubt this conclusion, as one statistically savvy reader did.

(Warning: this post gets a bit more mathy from here on.)

The reader analyzed the problem using conditional probability, which enables you to answer questions of the type "what are the odds of event A happening given that event B has happened?" The conventional notation for the probability of an event A is P(A), and the notation for "probability of A given B" is P(A | B). The formula to calculate the latter is:

P(A | B) = P(both A and B) / P(B)

The reader wrote:

Let A be the event that the prize is behind door 1 (the initially chosen door), and let B be the event that the prize is not behind door 2 (the door that has been opened). Here, A implies B, so P(both A and B) = P(A) = 1/3, while P(B) = 2/3. Thus P(A | B) = (1/3) / (2/3) = 1/2. Contrary to the claim of Prof. Paulos, nothing is gained by switching from door 1 to door 3. Prof. Paulos is mistaken when he says that P(A | B) = P(A) = 1/3.

What is wrong with this reasoning? It seems utterly plausible and in fact it gave me a headache for about an hour. But it is flawed.

The probability of Monty opening one door or the other changes depending on your initial choice as a player. If you picked a door hiding a goat, Monty has no choice: he is forced to open the door hiding the other goat. If, however, you picked the door hiding the car, Monty has to toss a coin (or some such) before he decides which door to open. But in either case, Monty will open a door that does not hide the prize. Thus, the "event that the prize is not behind door 2 (the door that has been opened)" happens with certainty, meaning P(B) = 1.

Thus, when we apply the formula, we get P(A | B) = (1/3) / (1) = 1/3, not 1/2. The probability that the car is behind door 3 is now 2/3, which means you had better switch.

The Monty Hall paradox is mathematically equivalent to a "wonderfully confusing little problem involving three prisoners and a warden," the one that Gardner introduced in 1959. Here is Gardner:

Three men—A, B and C—were in separate cells under sentence of death when the governor decided to pardon one of them. He wrote their names on three slips of paper, shook the slips in a hat, drew out one of them and telephoned the warden, requesting that the name of the lucky man be kept secret for several days. Rumor of this reached prisoner A. When the warden made his morning rounds, A tried to persuade the warden to tell him who had been pardoned. The warden refused.

"Then tell me," said A, "the name of one of the others who will be executed. If B is to be pardoned, give me C’s name. If C is to be pardoned, give me B’s name. And if I’m to be pardoned, flip a coin to decide whether to name B or C."

"But if you see me flip the coin," replied the wary warden, "you’ll know that you’re the one pardoned. And if you see that I don’t flip a coin, you’ll know it’s either you or the person I don’t name."

"Then don’t tell me now," said A. "Tell me tomorrow morning." The warden, who knew nothing about probability theory, thought it over that night and decided that if he followed the procedure suggested by A, it would give A no help whatever in estimating his survival chances. So next morning he told A that B was going to be executed.

After the warden left, A smiled to himself at the warden’s stupidity. There were now only two equally probable elements in what mathematicians like to call the "sample space" of the problem. Either C would be pardoned or himself, so by all the laws of conditional probability, his chances of survival had gone up from 1/3 to 1/2.

The warden did not know that A could communicate with C, in an adjacent cell, by tapping in code on a water pipe. This A proceeded to do, explaining to C exactly what he had said to the warden and what the warden had said to him. C was equally overjoyed with the news because he figured, by the same reasoning used by A, that his own survival chances had also risen to 1/2.

Did the two men reason correctly? If not, how should each calculate his chances of being pardoned?

Gardner saved the answer for his next column.

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  1. 1. pwfisher 3:54 am 04/16/2011

    Monty is "suggesting" a pick for you. Consider these examples:

    You pick a lottery ticket with the numbers [1,2,3,4,5,6]. Monty then tells you the winning ticket is either your choice, [1,2,3,4,5,6], or [4,8,15,16,23,42].

    There are 1,000 doors. You pick door #1. Monty then tells you that the winning door is either your choice, door #1, or door #472.

    Monty’s choice is quite revealing.

    The principle is the same, whether he has eliminated 1 door or 998. Monty has given you information, so your second choice is not a blind coin toss.

    If there are 1,000 doors, switching will be the right move 999 times out of 1,000. If there are 3 doors, switching will be the right move 2 times out of 3.

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  2. 2. tharriss 8:54 am 04/16/2011

    Thanks! I really enjoyed this article and the comments.

    It is interesting to see how smart people with a rock solid rule can be dead wrong because their assumptions are off (ie, assuming that the odds involved in the 2nd choice can/should be made without regard for information gained from prior events).

    Very illuminating!

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  3. 3. Flash604 2:04 pm 04/16/2011

    I think the important point is that whenever someone attempts to prove that the odds are 50/50, their math has assumed that Monty revealed a random door, and thus there is now a 50-50 chance for the remaining doors. And yes, if Monty revealed a random door, then when asked if you want to switch it would not make a difference. But at the same time, if Monty revealed a random door, then 1/3 of the time he would reveal the car and the game becomes pointless. (And no, Monty’s chances of randomly revealing a car are not 50-50 even though he only has 2 doors to choose from.)

    But Monty does not always reveal a random door. His job is to use his knowledge of the choices to always present you with a goat. Therefore, 2/3′s of the time his choice has already been made for him by the contestant.

    Remember that probability theory assumes random variables. The math the 50-50 supporters present assumes everything has happened randomly. Since Monty does not act randomly, you cannot put his actions into the formula as a random event.

    The example of "what if it was 100 doors instead of 3, and then 98 are revealed to you with the host making sure a car is never revealed" is an excellent further explanation of the correct answer.

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  4. 4. jtdwyer 2:43 pm 04/16/2011

    IMO, statistics is the mathematical study of unknown conditions. Information analysis is the study of conditions about which some information is available.

    Statisticians may call me ignorant if they wish, but if Monte Hall always reveals a door with a goat behind it as stated, no additional information is revealed about one’s initial choice.

    The problem state has been altered such that there are now only two doors from which to chose, one of which hides a car. IMO, regardless of any formal mathematical interpretation, the actual solution is now a 1/2 choice.

    I suggest that the mathematical interpretation is the product of a formal methodology which does not allow any restatement of the original problem, requiring all choices be considered in the context of the original 3 doors.

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  5. 5. DiscomBob 2:52 pm 04/16/2011

    It may help to think of it like this- assign the door you initially choose to pool A and the remaining doors to pool B. With 3 initial doors the chance the prize is in pool A is 1/3 while the chance it is in pool B is 2/3. Eliminating a door from pool B that does not have a prize behind it does not change the probability that the prize is in pool B.

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  6. 6. Flash604 3:03 pm 04/16/2011

    As I just stated above you, a 50-50 chance requires "CHANCE", which means complete randomness. But Monte did not act randomly.

    At first the contestant is choosing between 3 completely random doors, so his probability is 1 in 3.

    But after Monte eliminates a door using prior knowledge and making an informed choice, then the two doors the contestant has to choose from the second time are no longer completely random choices.

    Think of it this way, Monte is not asking "Pick randomly from these two doors", but rather he is asking "Do you want to switch?"

    The original door chosen remains at a 1 in 3 chance. The door that Monte offers as a switch has a 2 in 3 chance, because Monte was given a 2 in 3 chance of having the car behind his 2 doors and that has not changed because no random selection has occurred since that point.

    Look up the theory of probability, you will find it relies on randomness throughout. If Monte acted randomly in his choice, then the contestants final choice would have a 1 in 2 probability. HOWEVER, if Monte was acting randomly, then 1 in 3 times he would reveal the car himself and the contestant would never get to make a final choice. That never happens, thus Monte is not acting randomly, thus probability does not apply at this point.

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  7. 7. jtdwyer 3:32 pm 04/16/2011

    I understand your better explanation in the context of probability theory, which assumes no prior knowledge – thanks.

    IMO, the 1/3 probability originally assigned to the revealed door that you transfer to the unselected door should actually be distributed to both remaining doors.

    That Monte revealed a door with a goat provides no information about the unselected door that it does not also provide for the initial selection – they both remain as equally viable car choices.

    IMO, the transferral of the revealed door’s probability to only the unselected door is an artifact of mathematical methodology.

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  8. 8. jaycubed 7:19 pm 04/16/2011

    I think a clear way to understand the nature of the problem is to rephrase the choice.

    You are playing Let’s Make A Deal and are offered the choice of what’s behind either door #1, #2 or #3.

    You choose any door. Your chance of picking the correct door (say a car as the prize) is 1/3 (a lucky guess). Your chance of picking the wrong door (say a goat as the prize) is 2/3 (an unlucky guess).

    Monty Hall the offers you a choice of switching your guess, but instead of showing you what is behind one of the other doors and offering you the remaining door, he phrases the offer as this, "You may switch your choice to whichever of the remaining doors has the best prize behind it."

    This is in essence what Monty Hall does. He has hidden knowledge of what is behind all doors so will always pick a door with the goat prize. The chance of picking the door with the preferred prize the first time is 1/3. The chance of picking the wrong door is 2/3.

    Switching doors makes it twice as likely to pick the door with the preferred prize because the odds of picking the preferred door correctly or incorrectly never change even though one door is opened after your initial choice.

    There is NEVER a 50/50 chance at any time in the game. The only chance is 1/3 (the lucky guess) vs. 2/3 (the unlucky guess). The probability of guessing wrong the first time is 2/3. This never changes.

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  9. 9. jtdwyer 12:21 am 04/17/2011

    That’s an entirely different problem. The original problem statement is:

    "The guest states his choice, and the host opens one of the two remaining closed doors…"

    Your analysis does not address the original problem statement.

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  10. 10. @dcastelvecchi 11:12 am 04/17/2011

    Hans: I think the problem here is that you are treating this as a simple, cast-your-dice probability problem. It is not: the game has two steps, and it is probabilistically different depending on what happens at the first step.

    Here’s why: if the player first picks the door with the car (by a random choice that assigns 1/3 probability to each possibility), then Monty has to randomly pick another door. (This is part of the rules of the game, which are supposed to be fair and known to the player.) Thus there is a second random event, which has probability 1/2 for each possibility.

    But if the player’s first choice is a door hiding a goat, then Monty’s behavior is determinitic: He has to pick the other door with a goat.

    You write "If Monty opens a goat door no matter whether the contestant chose a car or a goat door…" But that reasoning ignores the fact that Monty’s behavior _depends_ on what door the player first picked.

    Because the game evolves differently depending on the player’s first choice, the rigorous way to analyze the game is with a decision tree.

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  11. 11. jtdwyer 12:50 pm 04/17/2011

    I agree, except that Monte’s secret information is not determinable by any other observer or participant. Monte’s choice provides no information about the contents of either remaining door – they are both equivalent.

    In a decision tree, the revealed door branch is simply trimmed off.

    In any case, Monte’s revelation of a door with a goat simply eliminates that option without contributing any additional information to the game analysis concerning either remaining door. The new problem state is a choice of two doors, one of which hides a car.

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  12. 12. jtdwyer 12:56 pm 04/17/2011

    I challenge anyone to demonstrate in a rigorously controlled experiment that changing one’s choice produces a success rate of more than 50%.

    Consider that perhaps some goat lover might want to select the remaining door with a goat. Is the probability theory solution to _not_ change his initial choice? I bet the result is always 50% when either changing or not changing the initial selection. Good luck!

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  13. 13. Flash604 3:24 pm 04/17/2011

    Why would we? The math and logic has already proved that changing your choice gives you a 2/3 chance of success. Despite that you cling to "It just makes sense to me that it’s 50-50." I can understand if you can’t see the logic, but please don’t think that no one proving the math through experimentation vindicates your viewpoint. When the math proves our viewpoint but it just seems wrong to you despite logical arguments, then it is not incumbent upon us to prove our supported view but rather it is incumbent upon you to do the the controlled experiment to prove your unsupported viewpoint.

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  14. 14. jtdwyer 4:57 pm 04/17/2011

    I agree in principle – I would do so if I were 20 years younger and physically capable of doing so – that’s why it came to mind. Experimental resolution would provide a definitive resolution.

    Unfortunately it is you who can’t see the logic as you cling to your faith in mathematical doctrine. What real justification is there for increasing the probability that one of the identically unrevealed doors contains the car? How did Monte reveal any information about what only he knows – which door hides the car?

    I particularly object to your putting words in my mouth with falsified quotations. Perhaps you can’t understand my explanation, but I do have 30+ years planning and resolving operational problems in the world’s largest transaction processing systems – I think my explanation is clearly stated. If you don’t understand, question my reasoning and I’ll be happy to explain further.

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  15. 15. djlewis 7:22 pm 04/17/2011

    Heh heh… the part I love about discussion of this problem, is the absolute, morally and mathematically righteous certainty that so many people display in favor of the wrong — but seductively simple — answer. Human foibles on parade!

    Actually, it’s scary. We see this kind of righteous certainty about bald mistakes operating in the political realm all the time, and right now, it’s killing us — literally! I wonder what correlation we’d see if everybody opining on this had to declare their political ideology.

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  16. 16. jtdwyer 8:12 pm 04/17/2011

    After further research, I concede my argument. The following is the explanation that I can best understand, in case it can help others:

    Of the three doors, each has a 1/3 probability of being a winner.

    When the player selects a door, it has a 1/3 probability of wining; the other two doors have a combined 2/3 probability.

    Monte determines one door of the two remaining that definitely has zero probability of winning and reveals it, leaving the unselected door representing the combined 2/3 probability of being the winner.

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  17. 17. djlewis 8:27 pm 04/17/2011

    For those still doubting, try this "experiment" (in thought, or for real if you still doubt it)…

    Say you always pick door 1 (may as well, right, since you know nothing) and adopt the policy to always switch after Monty shows you a goat. What happens?

    Let’s run 300 trials. For 100 of the 300 trials, the car is behind door 1 (your initial choice), 100 behind door 2, 100 behind door 3. That’s about what you’d get if the car is placed randomly.

    Now what happens in the 100 times the car is behind door 1, your initial choice. Monty will show you door 2 or 3 — doesn’t matter — you’ll switch to the one he didn’t show you — a goat — and you lose all 100 times. Oh, too bad!

    B-u-t in the 100 trials with the car behind door 2, Monty HAS to show you door 3 and your policy of always switching wins all 100 times! Likewise if the car is behind door 3, Monty has to show you door 2, you switch to 3 and always win. So 100 more times.

    Adding the results from those 300 trials up, you lose 100 times and win 200 times — your odds are clearly 2/3 using the switching policy, and consequently, only 1/3 with the sticking policy. Which policy would YOU use?

    Note, in the 200 times that you won, Monty did reveal a lot, in effect he gave you a total hint. But in the 100 times you lost, he revealed nothing. Just because you don’t know when he is revealing anything or not doesn’t change the calculation.

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  18. 18. djlewis 8:36 pm 04/17/2011

    @jtdwyer — good way of looking at it!

    Also, try my "experiment". It kind of factors out the probability issue, which is the tricky part to understand, in favor of a concrete number of trials. That’s what probability is about, under the surface, and it can help conceptually to return to the underlying "experiment".

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  19. 19. david winter 9:22 pm 04/17/2011

    Done, here’s the code ( and <a href="">here's the result.</a>

    Now, what do I win?

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  20. 20. david winter 9:24 pm 04/17/2011

    Sorry for the ascii garble, is there a way to link in sci-am comments? (And JT, sorry I didn’t catch the comment in which you’d converted ;)

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  21. 21. DavCo 11:10 pm 04/17/2011

    For those puzzled by the problem:

    Consider a variation of the Monty Hall problem:

    There are three doors: A, B and C

    Monty says there is a new car behind one door and one goat behind each of the other doors.

    You are allowed to choose two doors to open. Monty always opens one the doors a contestant chooses and demonstrates that behind it is a goat.

    He does this for you. And offers you the choice of the two remaining doors.

    Is there any benefit for you to switch?

    The answer is no. The other door you have chosen has a probability of 2/3 of success — just has your original choice had given you a probability of 2/3 of success.

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  22. 22. jmblock2 12:45 am 04/18/2011

    For those who say probability has no memory: If you have 10 marbles in a paper bag. 5 red and 5 green. You draw one out a red. What is the probability of drawing another red? The two events are not independent, just as the two events of Monty Hall are not independent. By pretending they are you are failing the puzzle, that is the point of having the puzzle to begin with.

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  23. 23. jmblock2 12:49 am 04/18/2011

    And for those saying ‘do the experiment’ to show it’s 50/50… do the experiment yourself! There is even code available for you to do it (available in 40+ languages for your convenience).

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  24. 24. jtdwyer 5:24 am 04/18/2011

    Thanks – you win the goat head cheese!

    Unfortunately, I don’t know of any automagic solution for the character set translation issue… Copying text between applications should not be such a probabilistic affair!

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  25. 25. jtdwyer 5:28 am 04/18/2011

    Thanks! And you thought I was too old to learn!

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  26. 26. jtdwyer 9:22 am 04/18/2011

    Say I have a bag with 2 goats and a car. Monte peeks in and pulls out a goat. What is the probability of drawing a car? I just have to ask…

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  27. 27. jason.dworkin 10:37 am 04/18/2011


    The difficulty people have in understanding this concept is due to the game description having 3 doors to begin with as opposed to more doors which is irrelevant to the logic and the fact that the difference between 1/3 odds or 2/3 or 1/2 being very `similar’ conceptually.
    The game description could read:

    “There are X number of doors behind which only one has a prize. All the rest have nothing. A contestant chooses one door he thinks the prize is behind and the game host then removes all other doors except for 2 – the contestants chosen door and one other door with the prize being behind one of the 2 remaining doors”

    So now the concept is easier to visualise. "There are ONE MILLION doors with the prize behind only one. The contestant chooses one door out of a million that he guesses the prize may be behind. The host then removes all doors except for the chosen one and one other. The prize is behind one of the two remaining doors." Now it’s easy to say that the there’s DEFINATELY NOT a 50/50 chance the prize is behind one or the other. `Obviously’ it’s behind the card the contestant didn’t choose because he picked that card out of 1 000 000 others and the host has to keep that card plus the winning card except in 1/1 000 000 times if the contestant chose the winning card.

    Now you apply that logic to 10 doors or 5 doors and you can more easily see it’s the same and therefore for three doors also.

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  28. 28. jason.dworkin 10:40 am 04/18/2011


    The difficulty people have in understanding this concept is due to the game description having 3 doors to begin as opposed to more with which is irrelevant and that the difference between 1/3 odds or 2/3 or 1/2 being very `similar’ conceptually.

    The game description could read:

    “There are X number of doors behind which only one has a prize. All the rest have nothing. A contestant chooses one door he thinks the prize is behind and the game host then removes all other doors except for 2 – the contestants chosen door and one other door with the prize being behind one of the 2 remaining doors”
    So now the concept is easier to visualise. There are ONE MILLION doors with the prize behind only one.

    " The contestant chooses one door out of a million that he guesses the prize may be behind. The host then removes all doors except for the chosen one and one other. The prize is behind one of the two remaining doors." Now it’s easy to say that the there’s DEFINATELY NOT a 50/50 chance the prize is behind one or the other. `Obviously’ it’s behind the card the contestant didn’t choose because he picked that card out of 1 000 000 others and the host has to keep that card plus the winning card except in 1/1 000 000 times if the contestant chose the winning card.
    Now you apply that logic to 10 doors or 5 doors and you can more easily see it’s the same and therefore for three doors also.

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  29. 29. djlewis 5:18 pm 04/18/2011

    @daveco — Yup, you are right — in the modified game (pick two doors, Monty opens one of them) the switch strategy wins 1/3 of the time. To "prove" this, again use the 300 trial experiment.

    You may as well pick doors 1 and 2, since you know nothing.

    If the car is behind door 3 (100 of the 300 trials), switching wins. If the car is behind door 1 or door 2 (200 of the 300 trials) switching loses. So, you win 100/300 times with the switching strategy, 200/300 times if you resolve to stick with the door you initially picked but Monty did not open.

    This kind of analysis often works and is much less subject to fallacious thinking because it avoids subjective probability. Instead, it goes back to the fundamental mathematical definition of probability as a fraction or count of trials (or, as later mathematicians realized the fractional area of a region, suitably generalized to sparse "regions" and multiple dimensions.)

    Read the book "The Unfinished Game" by Keith Devlin about the invention of modern probability by Pascal and Fermat in the 17th century. Prior to that, subjective probability was all there was, and fallacious reasoning was the norm.

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  30. 30. jtdwyer 10:07 pm 04/18/2011

    Yes, I that does make it clearer, since there is a 1/1M probability that the initial selection holds the car.

    The confusion is caused by the final selection between two doors, one of which definitely holds the car.

    In isolation, that final selection would be a 1/2 proposition, except the door initially selected still had that 1/1M probability of holding the car – giving the unselected door the 999,999/1,000,000 probability of holding the car. It’s nearly a sure thing!

    Very good – Thanks!

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  31. 31. gwmckenzie 12:23 pm 04/19/2011

    Make it simple: take nine glasses and arrange them in groups of three (representing the three doors). In group 1, put a penny under glass #1; in group 2 under glass #2; in group 3 under glass #3. You now have all three possible combinations of doors and "prizes". Now as a "contestant" choose, say, glass #1. Now, as "Monte" turn over one of the two other glasses, choosing in each case a glass that has no prize under it. For group 1 you can turn over either glass #2 or glass #3, for group 2 you turn over glass #3, and for groupe 3 you turn over glass #2. The solution to the puzzle is now glaringly simple: In only one of the three cases (the first group of three) does "door #1" – your initial choice – have the penny under it. In the remaining two cases (groups 2 and 3) the glass you did not turn over has the prize under it. In other words, for two of the three possible combinations, you would "win" by switching.

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  32. 32. gwmckenzie 12:30 pm 04/19/2011

    That’s a description of "sampling without replacement" vs "sampling with replacement". If you replace the red marble before you make the second draw (sampling with replacement) the odd are the same for the second draw (no memory). When you flip a coin that’s also sampling with replacement – the "head" is still there for the second flip, even if it came up the first time.

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  33. 33. Mark Pine 9:53 pm 04/19/2011

    cuda7051 is completely correct. Thanks to her or him for presenting this simple way to understand the difficult Monty Hall problem. It’s now clear why the contestant should switch doors.

    In cuda’s version there are 100 doors and the contestant picks one. It’s no problem to understand that there is only a 1% chance the pick was correct and a 99% chance the prize is behind one of the other doors.

    After the contestant chooses, the omniscient (and, in this case, very helpful) MC obligingly opens 98 of the other doors without the prize, leaving one closed. That doesn’t change the probabilities of the situation at all. At the start, there was a 99% chance the prize was behind one of the other doors, and there still is after the MC opens all of them but one. So it’s clear that by switching, the contestant increases the probability of winning from 1% to 99%.

    What the MC has changed are not the probabilities, but the information available to the contestant. After the MC opens 98 doors, the contestant has much more information and is able to make a much better choice.

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  34. 34. btvdan 8:35 pm 04/21/2011

    Fortunately, there aren’t many options… don’t even need to write a program.

    1 C 2 K – C
    1 G 2 K
    S – C
    1 G 2 K
    S – C

    K = keep, S = switch, C = Car

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  35. 35. btvdan 8:38 pm 04/21/2011

    Fortunately, there aren’t many choices…

    K = keep, S = switch, C = car

    1 C 2 K – C
    1 G 2 K
    S – C
    1 G 2 K
    S – C

    favors switching 2 to 1. Though, it doesn’t seem that way retorically.

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  36. 36. btvdan 9:04 pm 04/21/2011

    OK formatting got deleted… let’s try once more for the sake of my vanity.

    Choice 1 (pick a door)
    Door 1(Car); Door 2(Goat); Door 3(Goat)
    Choice 2 (Keep your choice or Switch)
    … the result of the 3 previous paths is:
    Switch = Loose; Switch = Win; Swith = Win

    It’s strange, but it’s true. Your three original choices provide 3 sub-jars. However, we know that in 2 jars there must be keys to a car. It does’t matter how we consider it… the mechanics of the experiment create the possible outcomes. They favor switch 2 to 1… it’s obvious when it’s countable, but it’s not intuitive in an algebraic sense. At least, it isn’t to me.

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  37. 37. hoamingin 2:28 am 04/23/2011

    Fascinating to watch the ebb and flow of comments. I cheered jtdwyer and although the argument for switching looks convincing, my brain is still not convinced.

    Shade1974 (way back at #3) commented "I have heard we evolved to not understand risk individually so that we can act without hesitation as a collective." What comes up for me is that the human brain did not evolve to use numbers or probabilities, but to put trust in intuitive approximations. External conditions worked out the probabilities of different behaviours by eliminating individuals whose behaviours did not enable them to survive in the prevailing conditions.

    For the first 6my of human evolution, numbers did not exist. Humans evolved speech about 200,000ya, but recent studies of hunter gatherer tribes in South America long cut off from contacts with other societies found that they had no words for numbers. Their counting was 1, 2, many. Franz Boas found the same among Inuit a century earlier. Numbers may have originated to record commercial transactions in complex societies that developed over the last 6,000 years.

    As far as we know, humans are the most numerately competent species on the planet, yet numbers (and probability) did not come naturally to them. That suggests that no species is able to calculate probabilities, yet the assumptions of biology require them to do so. When pressed on this issue, Richard Dawkins wrote that "the idea of animals behaving as if calculating odds without really doing so is fundamental to an understanding of the whole of sociobiology."

    In fact, Dawkins’ comment applies to the whole of biology, the assumptions of which are based on Natural Selection. Biology may well be the only science in which pretence is necessary to understand its basic concepts.

    To arrive at Natural Selection, Darwin decided that external conditions had no direct effect. What caused evolution were variations that gave individuals improved internal qualities that favoured them in competitive struggle that resulted in extermination of less improved individuals. Species adapt to specific conditions not because of the effect of those conditions, but because their improved qualities favour them in those conditions. Read that in more detail in Ch. 3 of Origin.

    How about a few of you mathematicians apply probability to that argument.

    Like Monty Hall, Darwin’s explanation is intuitively plausible and it requires careful reasoning to find the logical gaps.

    David Bainbridge

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  38. 38. jtdwyer 10:43 am 04/24/2011

    I do like your relating the evolution of intuitive evaluations to the mathematics of probabilities. While in the million to one case, where ‘Monte’ removes 999,998 goats from the equation, I’d have to switch, in the three door case I might still prefer to rely on intuition.

    If I were a committed and intuitive gambler and the proposition were put to me to bring the 3 door successful initial selection up from to 1/3 to 1/2 over 100 attempts, I might wager my $1,000 against a new Corvette, for example. I have had some success at blackjack even though I can’t count cards…

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  39. 39. hoamingin 11:29 pm 04/24/2011


    I agree(d).

    Some people get turned on by formulae. Formulae leave me cold – no, worse – when I did statistics (I called it sadistics) as part of an MBA, it was the only subject that caused me physical pain. Watching symbols with little bars and hats being written on the board had my brain sending pain down my back and telling my body to get up and walk out. Sadistics was a compulsory subject, so I took the pain and finished it.

    I prefer to prove it by logic. It occurs to me that the difficulty comes from the focus on the guest’s odds of winning. Let me look at it from Monty’s perspective and the odds of losing. (Like the Wason Selection Task, the proof is clearer from the negative, rather than the positive)

    We know that the initial odds are prize = 1 in 3, not-prize = 2 in 3. We also know that the second stage will include the prize door and a not-prize door and Monty will open the other not-prize door.

    Knowing that one door is the prize door and the other a not-prize means that the odds are 50/50, but we are not looking at present odds, we are looking at the probability of the initial choice. The probability that the choice was not-prize was 2 in 3. If the choice was one of those not-prize doors, Monty had to open the other not-prize door, so opening one or other not-prize door does not change the probability of that initial choice being not-prize. In effect, opening the door simply concentrates the probability of not-prize into one of the remaining doors and the issue becomes the probability of the initial choice being prize or not-prize.

    A further complication is that we know from cognitive science that the human brain tends to anchor itself on its preconceptions, so we should not trust an inclination to stick with the initial choice. Flipping a coin will give us an unbiased 50/50 chance, but the 2 in 3 probability of our initial choice being not-prize tells us that we increase our chances by switching.

    I initially believed that this only applied to all situations above 3 doors, but when I looked from the counter view, it fell into place for me, so that is why I added the (d) in the first line.

    If anyone is prepared to take up my challenge on Darwin’s explanation, they will also have to be prepared to look from the negative. The difference between a species before and after a change is defined by what has been removed from the genetic diversity that existed before the change. So what are the probabilities that what removed that diversity were external conditions, or internal competition?

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  40. 40. jtdwyer 3:18 am 04/25/2011

    Yeah, but I used to make my cigarette money in school by flipping quarters…

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  41. 41. SimpleZ 12:13 pm 04/26/2011

    As for the Gardner problem, if analogue to the Monty Hall, i think A had 1/3 chance of being the one who is pardoned. P(not A being pardoned) = 2/3. The ward now must point at one of B or C who will be executed, i e he must point out which one of B or C, who has 2/3 of being pardoned (since the other is then revealed as 100 % doomed). The ward points at B, which means C is the one who has the 2/3 chance of being pardoned. A did not increase his own chance of survival, but he increased C’s chances.

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  42. 42. hoamingin 12:10 am 04/28/2011


    So, based on long practical experience, you trust gut feeling over logic?

    Btw, have you given up the cigarette habit, or did your experimentation lead you to ignore logic on that also?

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  43. 43. jdyf333 12:19 am 05/1/2011

    (I was explaining the Monty Hall Problem to a friend as we were drinking tea in a restaurant. Another customer [a gambler] overheard our conversation and became quite agitated. He was extremely insistent that the guest should stick with his first choice, because the customer said the odds were 50/50 that the first choice was correct. When I very, very politely informed him that the odds were actually more like 33/66, he stormed out of the restaurant, and got in his car. Then he came back in the restaurant in a state of fury and yelled "You are saying I am stupid!!! Why don’t you step outside and tell me I am wrong, that the odds are not 50/50??? I’ll punch you out, you insulting #@*#%!")

    The pigeons are better than people at solving the Monty Hall Problem because, unlike human beings, they have no ego. LSD has been frequently cited as an inspiration by people working in the computer industry because LSD can lessen the ego and thus make it easier to solve problems. People who have not experienced LSD-induced states of lessened ego commonly believe that "self" and "ego" are the same. People who have not tripped sometimes think that if they lose their ego they will also lose consciousness. They tend to be quite frightened by the idea of ego-loss because they think that the less ego one has, the less control one has…

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  44. 44. Bart314 10:40 pm 05/4/2011

    This seemed insane and wrong to me until I eventually realized that the act of Monty opening a door changes the physical odds because he always eliminates one of the goats.

    If you pick the car, he shows a goat and the two remaining doors are one car and one goat. Thus, 50-50.

    If you pick a goat, he shows the other goat and the two remaining doors are one car and one goat. Thus, 50-50.

    Monty’s action physically changes the odds because he ALWAYS shows the other goat AND because he never puts you in the circumstance of having to choose between two goats.

    It is that scenario that everyone seems to overlook. He never eliminates a door without opening it, because the rules of the game say that even if you pick the goat to start with, you get a second chance. In this circumstance he can’t eliminate the car and have you picking between two goats because he would have to open the car door and at that point there would be no point to you choosing between the two goats.

    Statistically speaking, you will pick the goat to start with two thirds of the time. Each time you do he eliminates the other goat and then the odds of choosing randomly make it 50-50. Switching works here every time.

    Similarly, you will pick the car to start with one third of the time. Each time you do he eliminates one of the goats and then the odds of choosing randomly make it 50-50. Switching fails here every time but this scenario happens only one third of the time, as opposed to the immediate above scenario.

    Thus, if I am ever faced with this situation, I will switch for sure.

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  45. 45. jdyf333 5:31 pm 05/5/2011

    "Switching works here every time" is clearly a false statement. "Switching fails here every time" is clearly a false statement.

    You have mistakenly stated that "Statistically speaking, you will pick the goat to start with two thirds of the time. Each time you do he eliminates the other goat and then the odds of choosing randomly make it 50-50. " Actually the odds are approximately 33-66. (Although you are correct to state that if one door has a goat behind it and one door has a car behind it then your CHOICES are 50-50. "Choices", however, are not the same as "odds".)

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  46. 46. RogerInHawaii 10:18 pm 05/20/2011

    There is a much simpler explanation of this issue, and it requires almost no math, just reasoning. Suppose that, instead of revealing the goat behind one of the remaining doors Monty says, "You’ve made your choice, but I’ll let you make a change. You can keep your current choice (one particular door) OR you can change it to BOTH of the other doors, and you’ll win if the prize is behind EITHER of those doors". Everyone will recognize that this increases the person’s chances to 2 out of 3. But notice that it is actually exactly the same as Monty revealing the goat which is behind one of the remaining doors and letting the user switch from his current choice to the other remaining door. Still don’t see it? Suppose that after letting the player switch to selecting BOTH remaining doors (and winning if the prize is behind either one) he THEN opens one of those doors to reveal the goat. Now it’s back to the original description. The player always increases his odds from 1/3 to 2/3 by switching to the other door.

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  47. 47. tulanebarandgrill 12:48 pm 06/24/2011

    So I know this subject has been quiet for a while, but I think I found a really easy way to understand the 2/3 game. Monty eliminates one wrong door after your choice. Forget about the odds of choosing at that point. Think from the viewpoint that you decide from the beginning to switch. Then the game for you is choosing a WRONG DOOR first. If you choose a wrong door as your first choice, which is 2/3 probability as anyone can see, then you will definitely win by switching.

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  48. 48. bob_my 1:53 am 06/26/2011

    in case anyone is actually reading down to this, the 57th comment, here is, I hope, an easy way to explain why the correct answer is to switch:

    a) initial condition: three doors, one prize, two goats.

    b) choose a door at random; no one should disagree that at this point probability of prize is 1/3. At this point, obviously, the probability of the the prize being behind one of the other two doors is 2/3.

    c) Now we propose a slight modification of the game play.

    Suppose that Monty now offers you the opportunity to switch from your first choice of one door, to *both* of the other two doors (i.e., you win if the prize is behind *either* of the other two).

    Under this scenario it is easy to see that you should switch, as you will then win the prize with 2/3 probability.

    d) Under the initial condition, whether your initially chosen door held the prize or not, at least one of the remaining two doors will hide a goat. Monty, knowing the contents of all three doors, can always show you a goat behind one of the two doors not initially chosen.

    I hope that upon reflection it is easy enough to see that Monty opening a goat door and then allowing you to switch to the other of the two doors not initially chosen, is really no different then allowing you to switch to *both* of the other two doors, just as I set out in my modified game play in c), above.

    In simpler words, the offer to switch away from the initially chosen door after one goat has been exposed in the non-chosen doors is equivalent to an offer to switch to *both* of the other two doors, i.e. 2/3 versus 1/3.

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  49. 49. bob_my 2:00 am 06/26/2011

    ha ha. added my comment first, then read through more of the others. I am not the first here to recommend this way of looking at it, for example, RogerInHawaii a few posts back.

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  50. 50. bob_my 3:15 am 06/26/2011

    Finally, for any readers who may still be uncertain, and who may be more inclined to an empirical approach rather than logic, if you have a little facility with spreadsheets it is not difficult to set up a little monte carlo type of simulation. Many years ago when I first encountered this problem I did just that. Doing so might make a amusing and instructive classroom exercise for anyone using this problem in a probably course.

    in the first row, set up a few cells as follows:

    in the first column, generate a random number in the range 0-0.9999… To interpret this as one of three doors, and to make the subsequent calculations simpler, the random number should be multiplied 3 and rounded down ("floor" function) to an integer. This will result in either 0, 1, or 2, representing the three doors.

    in the second column, another random number similarly transformed represents the door initially guessed.

    in the third column, calculate which door Monty opens to exposed a goat. this can be done by adding 1 modulus 3 to the guess, to represent the next door in sequence after the one guessed, then comparing this number to the actual prize location in column A. if the comparison shows no prize in this door, accept it as the exposed goat location, otherwise (by embedding into an "if, then, else" structure), add 2 modulus 3 to the guess and expose that goat instead.

    in the cell in the next column, for the "no switch" case show win or lose as 1 or 0 by comparing the first column to the second in an "if, then, else" structure.

    in the cell over the next column, show win or lose again as 1 or 0, this time for the "switch door" scenario. The door to be switched to can be calculated using a slight modification of the formula used above to select the door for Monty to open. Again we add 1 (or 2) mod 3 to the number representing the door first selected, but this time it is compared to the exposed goat door already calculated in the third column, to avoid a switch to that door. Then this switched-to door result is embedded in another "if, then, else" structure to yield a 1 or 0 result after comparison with the actual prize door in the first column.

    Copy these several formulas down for 2-300 rows, and take the mean at the bottom of the two 1 – 0 win – lose columns. It converges to .333 and .666 pretty fast.

    Aside from showing the answer empirically, the exercise of thinking through the several calculations needed to model the two choices of switch versus no switch actually helps lay bare the logic underlying the correct answer as well.

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  51. 51. jrubin 4:32 pm 10/17/2011

    Start with the prior scenario, except that after Monty Hall shows Person 1 the open door with the goat, Person 1 finds the choice much too perplexing, and leaves the room. Shortly thereafter, Person 2 enters the room, and Monty tells the person that behind one door is a goat, and behind the other is a car, asking him to choose. Person 1′s chance of getting the car by changing his initial choice was 2/3. Person 2′s chance of choosing the car, given the same choice between the same two doors, is 1/2. Does this present a logical inconsistency?

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  52. 52. rockkroc 9:13 am 12/13/2011

    This is a simple case of both sides are right because we are looking at this problem different ways. For the contestant there is only one choice that matters. The fact that you had a 1/3 chance of choosing the right door is irrelevant when it comes to choose whether to stay or switch. You will always choose a door and Monty will always remove a goat.

    Now the game is do I have the right door already or do I not, and, in fact, it is a coin flip. Your initial choice is irrelevant because you had to make it and it doesn’t matter whether you had a higher probability of choosing the goat over the car. Carrying the probability from that first choice over is useless in the context of this game.

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  53. 53. rockkroc 9:22 am 12/13/2011

    Another way of looking at it is if Monty initially removed a goat, then told you to pick between one of the two remaining. No one will contest that this is a 50/50 choice. Simply remember that your first choice is irrelevant and you will see that there is only one choice, between two options.

    Failure to understand the game is the reason people lose hundreds gambling.

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  54. 54. PalmerEldritch 10:12 pm 12/15/2012

    Since the answer to the problem posed is 33/66 and not 50/50 can I take it that that when you say “people lose hundreds gambling” you are talking from personal experience?

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  55. 55. michaelmross 7:03 am 05/21/2014

    You don’t explain what (1) is in the formula you suggest as a proof :

    Thus, when we apply the formula, we get P(A | B) = (1/3) / (1) = 1/3, not 1/2. The probability that the car is behind door 3 is now 2/3, which means you had better switch.

    What is (1)?

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  56. 56. PalmerEldritch 1:52 am 10/2/2014

    Well the mathematical explanation is incorrect since it gives the same answer of 1/3 for P(C|B)

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