## The Monty Hall problem

The Monty Hall problem is inspired by an American television game show. There are three doors, and behind one of them, the host of the show, Monty, hides a car. Each of the other two doors hides a goat.

The contestant is asked to pick a door, so that if she finds the car, she wins the game (and the car). Since there are three doors, chances are $1/3$ that she picked the door behind which is the car. But Monty doesn't open yet the door, but he opens one of the remaining doors, revealing a goat. He then asks the contestant either to keep her original choice, or to switch to the other unopened door. The problem is, what should the contestant do?

The first instinct of anybody may be to think that since there are only two remaining doors, it doesn't matter if you switch the door or not, because the chances are $1/2$ in both ways. However, Marilyn vos Savant explained that if the contestant switches the doors, the chances are $2/3$. while if she doesn't switch them, the chances are $1/3$. This is counterintuitive, and the legend says that not even Paul Erdős understood it. You can find on Wikipedia some solutions of this puzzle.

## An equivalent puzzle

I will present another, simpler puzzle, and show that it is equivalent to the Monty Hall problem.

Consider again three doors, one hiding a car. The contestant is asked to pick either one of the three doors, or two of them. What is the best choice?

Obviously, the contestant should better choose two doors, rather than one. Since if she thinks that the car is behind door number three, choosing also door number one will only double the chances to win.

But how is this related to the Monty Hall problem? Well, it is, because if you play the Monty Hall problem, you can pick two doors, but then tell Monty, you just tell you picked the remaining one. When Monty asks if you want to switch, then you switch to the other two doors, and since one is already open, you choose the remaining one. This means that choosing a door and switching is equivalent to choosing the other two doors.

So the Monty Hall problem is actually equivalent to having to choose one or two doors. Not switching is equivalent to choosing one door, and switching is equivalent to choosing two doors. So switching gives indeed probability $2/3$.

## 3 comments:

I did not believe the story that established scientists did not get the correct explanation until it happened to me. I will not name names, but I found it impossible once to explain the correct solution to some well established scientist. No matter what argument I gave I could not get the point across.

You will similarly get resistance from people who accept this explanation, when they are told that it is wrong. Paul Erdős accepted the correct answer when shown a simulation of it. What he did that was unusual, was to accept that correct solution without being shown what was wrong with his own. And note that just getting the right answer does not make a solution correct.

Make a slight change to the original problem: Monty Hall always opens the highest-numbered door that he can. 1/3 of the time he won't be able to, so he will open the lower-numbered of the two you didn't pick. If you see that happen, knowing his favoritism, then you will know where the car is with 100% certainty. Your original door has a 0% chance to win.

The point isn't that Monty has such a bias, or that if he does you know about it. The point is that how he chooses between the two doors can change the probability of your original door from 1/3 to something else. Both solutions above (Marilyn vos Savant's, and Cristi Stoica's) are wrong even though they get the right answer, because they are based on the incorrect assertion that your original probability can't change. People refuse to accept both, whether or not they realize why, because they are based on an assertion and not a proof.

The correct solution is that if you originally pick a goat, Monty Hall has no choice about what door to open. But half (assuming no biases) of the time you start with the car, Monty would have opened a different door than he did. Making you door half as likely to have the car, once a goat is revealed, as the third door.

JeffJo,

You say about Erdős "What he did that was unusual, was to accept that correct solution without being shown what was wrong with his own." I think an answer explaining why you should change the door also explains pretty well why the fifty-fifty answer is wrong, so I doubt that what you said about Erdős is true. About the rest of your comment, if you "Make a slight change to the original problem", then of course the solution have to be changed too :). Suppose that Monty always opens the smallest-numbered door. Then your solution will be wrong 100% of cases. Of course there are other ways he could choose what door to open, ways that can be guessed after a number of trials too. To each of Monty's possible strategies favoring the highest-numbered doors, there is one favoring the lower-numbered doors with the same probability. So Monty's possible strategies balance each other, just as if his choices would be random. When Monty's choice is random, or even when he has a strategy like the one you described, but you have no information about it, vos Savant's solution is correct, and this was all about. And the simulation shown to Erdős was assuming Monty's choice is random too. Your (mistaken) argument can be generalized to any problem of probabilities. Just bias consistently the data, and you will think that the solution offered by the theory of probabilities to the problem is wrong! But the same theory of probabilities also knows how to deal with biased data.

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