# The Monty Hall Problem - where does our intuition fail us?

From Wikipedia :

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

The answer is, of course, yes - but it's incredibly un-inituitive. What misunderstanding do most people have about probability that leads to us scratching our heads -- or better put; what general rule can we take away from this puzzle to better train our intuition in the future?

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No, it is not true that the answer is, of course, yes (see en.wikipedia.org/wiki/…), as the problem is underspecified and different interpretations can give strikingly different results. However, for arguably the simplest solution the answer is yes. –  Piotr Migdal Feb 28 '12 at 19:11
I already supplied an answer one year ago. But as I reread the final question, I wonder: do we actually want to 'train our intuition'? Does that even make sense? –  Henk Langeveld Aug 1 '12 at 22:40

Consider two simple variations of the problem:

1. No doors are opened for the contestant. The host offers no help in picking a door. In this case it is obvious that the odds of picking the correct door are 1/3.
2. Before the contestant is asked to venture a guess, the host opens a door and reveals a goat. After the host reveals a goat, the contestant has to pick the car from the two remaining doors. In this case it is obvious that the odds of picking the correct door is 1/2.

For a contestant to know the probability of his door choice being correct, he has to know how many positive outcomes are available to him and divide that number by the amount of possible outcomes. Because of the two simple cases outlined above, it is very natural to think of all the possible outcomes available as the number of doors to choose from, and the amount of positive outcomes as the number of doors that conceal a car. Given this intuitive assumption, even if the host opens a door to reveal a goat after the contestant makes a guess, the probability of either door containing a car remains 1/2.

In reality, probability recognizes a set of possible outcomes larger than the three doors and it recognizes a set of positive outcomes that is larger than the singular door with the car. In the correct analysis of the problem, the host provides the contestant with new information making a new question to be addressed: what is the probability that my original guess is such that the new information provided by the host is sufficient to inform me of the correct door? In answering this question, the set of positive outcomes and the set of possible outcomes are not tangible doors and cars but rather abstract arrangements of the goats and car. The three possible outcomes are the three possible arrangements of two goats and one car behind three doors. The two positive outcomes are the two possible arrangements where the first guess of the contestant is false. In each of these two arrangements, the information given by the host (one of the two remaining doors is empty) is sufficient for the contestant to determine the door that conceals the car.

In summation:

We have a tendency to look for a simple mapping between physical manifestations of our choices (the doors and the cars) and the number of possible outcomes and desired outcomes in a question of probability. This works fine in cases where no new information is provided to the contestant. However, if the contestant is provided with more information (ie one of the doors you didn't choose is certainly not a car), this mapping breaks down and the correct question to be asked is found to be more abstract.

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I find that people find the solution more intuitive if you change it to 100 doors, close first, second, to 98 doors. Similarly for 50 doors, etc.

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This is the way I explain things, fully agree ! –  robin girard Aug 9 '10 at 7:24
ditto. I usually put it in terms of 52 cards, and the goal is to find the ace of spades. –  shabbychef Sep 12 '10 at 4:08

Our aim is to eliminate both goats. We do this by marking one goat ourselves. The quizmaster is then forced to choose between revealing the car or the other goat. Revealing the car is out of the question, so the quizmaster will reveal and eliminate the one goat we did not know about. We then switch to the remaining door, thereby eliminating the goat we marked with our first choice, and get the car.

This strategy only fails if we do not mark a goat, but the car instead. But that is unlikely: there are two goats and only one car.

So we have a chance of 2 in 3 to win the car.

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Nice explanation. Doesn't explain people's cognitive failings, but +1 anyway. –  user87 Aug 1 '10 at 3:17
I believe we as humans are hardwired to prefer those representations of a problem/challenge that matches its chronology. The Monty Hall problem is always presented as a story, in chronological order. This hampers our ability to reframe the challenge. –  Henk Langeveld Jun 6 '12 at 7:46
I love this answer, it is so clever. –  CharlieK Nov 24 at 14:51

I'd modify what Graham Cookson said slightly. I think the really crucial thing that people overlook is not their first choice, but the host's choice, and the assumption that the host made sure not to reveal the car.

In fact, when I discuss this problem in a class, I present it in part as a case study in being clear on your assumptions. It is to your advantage to switch if the host is making sure only to reveal a goat. On the other hand, if the host picked randomly between doors 2 and 3, and happened to reveal a goat, then there is no advantage to switching.

(Of course, the practical upshot is that if you don't know the host's strategy, you should switch anyway.)

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I have to admit that, even being a convinced Bayesian, having read several treatments of the subject (popular science ones, in particular Mlodinow’s, and text books) as well as understanding the underlying statistics, this result surprised me. Now, it’s easy to see that it’s in fact true – both by systematically enumerating all possible scenarios or by simulating (I did both). But surprising nonetheless. –  Konrad Rudolph Jun 15 '12 at 13:03

This doesn't give a general rule, but I think that one reason why it's a challenging puzzle is that our intuition doesn't handle conditional probability very well. There are plenty of other probability puzzles that play on the same phenomenon. Since I'm linking to my blog, here's a post specifically on Monty Hall.

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The answer is not, "of course YES!" The correct answer is, "I don't know, can you be more specific?"

The only reason why you think it is correct, is because Marliyn vos Savant said so. Her original answer to the question (although the question was widely know before her) appeared in Parade magazine on September 9, 1990. she wrote that the "correct" answer to this question was to switch doors, because switching doors gave you a higher probability of winning the car (2/3 instead of 1/3). She got lots of responses from Mathematics PhDs and other intelligent people that said she was wrong (although many of them were incorrect too).

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? — Craig F. Whitaker Columbia, Maryland

I have bolded the important part of this logic question. What is ambiguous in that statement is:

Does Monty Hall always open a door? (What would it be to your advantage to switch doors if he only opened a losing door when you picked a winning door? Answer: No)

Does Monty Hall always open a losing door? (The question specifies that he knows where the car is, and this particular time he showed a goat behind one. What would your chances be if he randomly opened a door? i.e. The Monty Fall question or what if sometimes he chooses to show winning doors.)

Does Monty Hall always open a door you did not pick?

The basics of this logic puzzle have been repeated more than once, and many times they aren't specified well enough to give the "correct" answer of 2/3.

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male? — Stephen I. Geller, Pasadena, California

Did the fellow look at both dogs before responding "Yes," or did he pick up a random dog and discovered it was a male and then responded "Yes."

Say that a woman and a man (who are unrelated) each has two children. We know that at least one of the woman's children is a boy and that the man's oldest child is a boy. Can you explain why the chances that the woman has two boys do not equal the chances that the man has two boys? My algebra teacher insists that the probability is greater that the man has two boys, but I think the chances may be the same. What do you think?

How do we know that the women has at least one boy? Did we look over the fence one day, and see one of them? (Answer: 50%, same as man)

The question has even tripped up our very own Jeff Atwood. He posed this question:

Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?

Jeff goes on to argue that it was a simple question, asked in simple language and brushes aside the objections of some that say that the question is incorrectly worded if you want the answer to be 2/3.

More importantly though, is why the woman volunteered the information. If she was speaking the way normal people do, when some one says "one of them is a girl," inevitably the other is a boy. If we are to assume this is a logic question, with the intent of tripping us up, we should ask that the question is more clearly defined. Did the woman volunteer the sex of one of her children, randomly selected, or is she talking about the set of her two children.

It is clear that the question is poorly worded, but people don't realize it. When similar questions are asked, where the odds are much greater to switch, people either realize that it must be a trick (and question the motive of the host), or get the "correct" answer of switching as in the one hundred doors question. This is further supported by the fact that doctors when asked about the likelihood of a woman having a particular disease after testing positive (they need to determine if she has the disease, or it is a false positive), they are better at arriving at the correct answer, depending upon how the question is phrased. There is a wonderful TED Talk that half way through covers this very case.

He described the probabilities associated with a breast cancer test: 1% of women tested have the disease, and the test is 90 percent accurate, with a 9% false positive rate. With all that information, what do you tell a woman who tests positive about the likelihood they have the disease?

If it helps, here’s the same question phrased another way:

100 out of 10,000 women at age forty who participate in routine screening have breast cancer. 90 of every 100 women with breast cancer will get a positive mammography. 891 out of  9,900 women without breast cancer will also get a positive mammography. If 10,000 women in this age group undergo a routine screening, about what percentage of women with positive mammographies will actually have breast cancer?

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(+1) This is a cogent reply, well worth the read. It clearly explains how and why people can so emphatically defend different answers. Thank you! –  whuber Jun 21 '12 at 4:35

I agree that students find this problem very difficult. The typical response I get is that after you've been shown a goat there's a 50:50 chance of getting the car so why does it matter? Students seem to divorce their first choice from the decision they're now being asked to make i.e. they view these two actions as independent. I then remind them that they were twice as likely to have chosen the wrong door initially hence why they're better off switching.

In recent years I've started actually playing the game in glass and it helps students to understand the problem much better. I use three cardboard toilet roll "middles" and in two of them are paper clips and in the third is a £5 note.

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I believe that it is more a question of logic than a difficulty with probability that makes the Monty Hall solution surprising. Consider this description of the problem.

You will decide at home, before going to the TV show, if you are going to switch doors or stick with your first choice, whatever happens during the show. That is, you choose between strategies "Stay" or "Switch" before you play the game. There is no uncertainty involved in this choice of strategy.

Let's understand the differences between the two strategies. We will not talk about probabilities yet.

Under strategy "Stay" you win if and only if your first choice is the "good" door. On the other hand, under strategy "Switch" you win if and only if your first choice is a "bad" door. Please, think about these for a minute, specially the second case. Notice that we didn't talk about probabilities yet. It is just a logical equivalence.

Now, if you agree with that, supposing that you initially assigned probability $1/3$ for the prize being behind each door, it is clear that under strategy "Stay" your probability of winning is $1/3$ (it is the probability of choosing the "good" door). But under strategy "Switch" your probability of winning is $2/3$ (it is the probability of choosing a "bad" door). And that is why strategy "Switch" is better.

P.S. In 1990, Prof. Larry Denenberg sent a letter to TV show host Monty Hall asking for his permission to use in a book his name in the description of the well known three doors problem we've been discussing in this topic.

Here is an image of part of Monty's reply to that letter, where we can read:

"as I see it, it wouldn't make any difference after the player has selected Door A, and having been shown Door C - why should he then attempt to switch to Door B?"

Therefore, we can safely conclude that Monty Hall (the man himself) didn't understand the Monty Hall problem!

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I find this a helpful exercise. As an argument, though, it is unconvincing because it relies on an unstated assumption: namely, that Mr. Hall will even offer an opportunity to switch and, if he does, that his choice is independent of yours. For example, if Mr. Hall happened to learn that you intended to switch (and he wished to minimize his losses), he might elect to open a door only if switching would cause you to lose! In this case, your chance of losing becomes 100%. –  whuber Feb 27 '12 at 21:20
An interesting variant on the problem. I am not surprised that Monty Hall would be fooled also. Also I don't know eactly where the problem originated. Marilyn vos Savant got it from someone else. Also although there were three doors to pick from for what was called "The deal of the day" Monte did not show what was behind a curtain and then allow them to switch. –  Michael Chernick May 5 '12 at 4:59
Betting games like that where players gave up prizes for other unknown prizes went on throughout the game, In the end for dramatic effect they would show a curtain that wasn't yours and wasn't the big deal but switching was never offered. –  Michael Chernick May 5 '12 at 5:07
Are you sure that the original TV show didn't reveal what was behind one of the "bad" doors, Michael? If so, I see no reason to refer to the three doors problem as the Monty Hall problem. –  Zen May 9 '12 at 4:14

The lesson? Reformulate the question, and search for a strategy instead of looking at the situation. Turn the thing on its head, work backwards...

People are generally bad at working with chance. Animals typically fare better, once they discover that either A or B gives a higher payout on average; they stick to the choice with the better average. (don't have a reference ready - sorry.)

The first thing people are tempted to do when seeing a 80/20 distribution, is to spread their choices to match the pay-out: 80% on the better choice, and 20% on the other. This will result in a pay-out of 68%.

Again, there is a valid scenario for people to choose such a strategy: If the odds shift over time, there's a good reason for sending out a probe and try the choice with the lower chance of success.

An important part of mathematical statistics actually studies the behaviour of processes to determine whether they are random or not.

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"Animals typically fare better, once they discover that either A or B gives a higher payout on average". I don't think humans would do worse given access to the same amount of emperical data. A single quiz show contestant, however, plays the game once, not n times. –  Frank Jan 8 '12 at 17:55

One does not need to know about conditional probability or Bayes Theorem to figure out that it is best to switch your answer.

Suppose you initially pick Door 1. Then the probability of Door 1 being a winner is 1/3 and the probability of Doors 2 or 3 being a winner is 2/3. If Door 2 is shown to be a loser by the host's choice then the probabilty that 2 or 3 is a winner is still 2/3. But since Door 2 is a loser, Door 3 must have a 2/3 probability of being a winner.

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@MichaelChernick, have you considered posting these comments as an answer, rather than listing them under a seemingly unrelated answer? –  Macro May 5 '12 at 5:06
@MichaelChernick Please don't abuse comments system this way. If you look for a place for longer statistical anecdotes, consider contributing to the site's blog (i.e. visit the blog chat room). –  mbq May 8 '12 at 20:57
@mbq What are you talking about. Our moderate whuber made a comment about a variant on the Monte Hall problem. I made a simple response back. we did not get into a real chat. i am beginning to think some members abuse the system by chastising members over any little thing they see that they think violates the rules. We can all read the FAQs and learn how to follow the rules ourselves. –  Michael Chernick May 8 '12 at 22:38
@MichaelChernick, no one is asking you to leave. I think we all appreciate what you contribute to the site. The issue was that the comments were, seemingly randomly, located under this unrelated answer. After getting no response to my suggestion that maybe this wasn't the place for them, they were deleted. Re: "But if sharing a lot of useful information related to the subject of the question is abusing the system then there is something wrong with the system.", it seems like you want the site to be something more than it is - it's a Q&A site, not a statistics message board. –  Macro May 9 '12 at 3:34
@MichaelChernick This is not a forum and comment system is only for short comments about the post, period. There was a lot done on SE to redirect the discussions and side notes to a places where they would be easier to follow, notify right people and won't be mixed by collapsing system -- we have chat, meta and blog. Please just use right tools so that your precious content won't get lost in clutter. –  mbq May 9 '12 at 6:16
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I think there are several things going on.

For one, the setup implies more information then the solution takes into account. That it is a game show, and the host is asking us if we want to switch.

If you assume the host does not want the show to spend extra money (which is reasonable), then you would assume he would try to convince you to change if you had the right door.

This is a common sense way of looking at the problem that can confuse people, however I do think the main issue is not understanding how the new choice is different then the first (which is more clear in the 100 door case).

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I'll quote this great article on lesswrong:

The possible hypotheses are Car in Door 1, Car in Door 2, and Car in Door 3; before the game starts, there is no reason to believe any of the three doors is more likely than the others to contain the car, and so each of these hypotheses has prior probability 1/3.

The game begins with our selection of a door. That itself isn't evidence about where the car is, of course -- we're assuming we have no particular information about that, other than that it's behind one of the doors (that's the whole point of the game!). Once we've done that, however, we will then have the opportunity to "run a test" to gain some "experimental data": the host will perform his task of opening a door that is guaranteed to contain a goat. We'll represent the result Host Opens Door 1 by a triangle, the result Host Opens Door 2 by a square, and the result Host Opens Door 3 by a pentagon -- thus carving up our hypothesis space more finely into possibilities such as "Car in Door 1 and Host Opens Door 2" , "Car in Door 1 and Host Opens Door 3", etc:

Before we've made our initial selection of a door, the host is equally likely to open either of the goat-containing doors. Thus, at the beginning of the game, the probability of each hypothesis of the form "Car in Door X and Host Opens Door Y" has a probability of 1/6, as shown. So far, so good; everything is still perfectly correct.

Now we select a door; say we choose Door 2. The host then opens either Door 1 or Door 3, to reveal a goat. Let's suppose he opens Door 1; our diagram now looks like this:

But this shows equal probabilities of the car being behind Door 2 and Door 3!

Did you catch the mistake?

There you go, this is how your intuition fails you.

Check out the correct solution the in the full article. It includes :

• Explanation of the Bayes theorem
• Wrong approach of Monty Hall
• Right approach of Monty Hall
• More problems...
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