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# The Monty Hall problem and counter-intuitive teaching

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The Monty Hall problem is a well-known thought experiment in probability analysis. The problem is fairly simple, but for reasons that aren’t well understood the right answer is sufficiently counter-intuitive that a very large majority of people get it wrong on their first attempt. More interestingly, I’ve found that students often resist the validity of the correct answer, even when the problem is analyzed in some detail. The problem:

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. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?

(Taken from the wiki page if you want to look up the answer).

Here are some strategies I’ve used for explaining the solution to students who resist accepting it.

(1) Redescribing what the offer to switch gives you, i.e., by switching you are in effect choosing two doors instead of one, and thus doubling your odds of success.

(2) Recharacterization via different quantities, i.e., what if there are one hundred doors and 99 goats and a car, and Monty Hall is required to show you 98 goats after you choose a door?

(3) Explicitly working out all the potential iterations, i.e., if you choose Door 1 and there’s a car behind it then X, but if there’s a goat behind it then Y etc.

(4) Empirical testing. Have the student run the experiment and observe the results.

These are listed in descending order of abstraction, and probably not coincidentally ascending order of pedagogical effectiveness. (Occasionally there will be a holdout even after (4). This person is invariably male and almost certainly a future litigator.

Anyway, teaching the problem is a fun way to get students to think about the limits of common sense intuition, which is a much-cited source of wisdom for legal interpretation in general, and statutory interpretation in particular. It’s also a good way to get people to think about how people tend to cling to intuitively correct answers, even in the face of demonstrations that their intuitions are wrong.

Update: Thanks for the comments, and especially to J.W. Hamner’s variation on explanation (1) and Vardibidian’s card trick for explanation (2) — I’m going to use those.

Lemuel Pitkin and Mike Schilling emphasize that it’s crucial that the rules of the game require Hall to reveal a goat after the initial choice, and that without this caveat the situation is different. Just for the heck of it, gaming that out: The contestant doesn’t know what if any post-choice decision rule constrains Hall, or even if Hall knows what’s behind the doors. What should the contestant do?

Possibility (A) Hall doesn’t know what’s behind the doors.

Possibility (B) Hall knows and is indifferent to whether you win or lose.

Possibility (C) Hall knows and wants you to win.

Possibility (D) Hall knows and wants you to lose.

If (A), then the revealing of a goat moves the odds to 50/50 for the remaining doors, and therefore switching neither helps nor hurts.

However, this is where “pure” game theory needs some richer sociological context. In our culture it would be a very strange game show in which the host didn’t know what was behind the doors, and therefore might accidentally reveal the car. So as a practical matter the contestant can probably rule out (A) as an actual possibility. In any case, the sum probabilities created by (B), (C), and (D) remain dispositive, since (A) would leave the contestant indifferent to switching or staying, i.e., whether you estimate the odds of (A) being the case as 1% or 99% makes no difference — the only thing that matters are the odds governing the other possibilities.

Moving right along, if (B) is the case, then as long as you assume he’s not going to choose to show you a car, which given the rules of game shows is a pretty safe assumption, we’re right back to the classic description of the problem, and you double your odds of winning by switching.

If (C) is the case, then deciding whether to switch comes down to your estimate of Hall’s assumptions regarding your mental state. Maybe you’ve chosen the goat, and because he wants you to win he’s giving you additonal information that, if you both understand the probability structure and that he wants you to win, tells you to switch. But here’s a disturbing possibility: maybe you’ve chosen the car and he wants you to win, but he’s showing you a goat precisely because he believes that if he does so he’ll encourage you not to switch, because like most people you’d get the probabilities wrong under the classic assumptions of the game’s rules, and it’s more likely you’ll choose to stay than switch because of endowment effects or sheer stubborness. Remember you don’t know the rules — you don’t know whether he even has to open one of the other doors. The analysis is the same for (D) but reversed.

Ultimately if you don’t know the rules of game, you have to make two separate judgments: what are the probabilities that (B), (C) and (D) are the case, and what are the probabilities within each of those possibilities? Those two estimates then determine whether to switch or stay, since (A) leaves you indifferent.