Goats and Doors (originally posted on uCoz)

Today's issue of Parade has a 30th anniversary Marilyn vos Savant article that among other things revisits the Monty Hall problem (also known as two goats, three doors) that caused an alarming amount of controversy, including furor from a lot of from people who really ought to have known better. "Of the letters from the general public, 92% are against my answer, and and of the letters from universities, 65% are against my answer", she wrote in a follow-up, and many of these letters were condescending if not outright rude and insulting. (archived link)

For anyone unfamiliar with it, here's the basic premise of the problem:

You're on a game show where you are presented with three closed doors. One, chosen at random, has a car behind it, while the other two have goats. You pick a door, trying to find the car. After you have picked, the host opens one door, always selecting one of the two that you did not choose, and always selecting one that has a goat behind it (since the host knows where the car is). You then are given the option of changing your selection from your original one to the other remaining unopened door. What do the odds say you should do?

It should be obvious enough that your initial chance of picking the correct door is 1/3, since there are three doors, each equally likely to hide the car, and you can only pick one. Let's label the doors A, B, and C, with A arbitrarily being the door that you picked. This leaves these three, equally likely, cases:

• A) Door A has the car. Doors B and C have goats.
• B) Door B has the car. Doors A and C have goats.
• C) Door C has the car. Doors A and B have goats.

At this point, the host opens a door, revealing a goat. Opening the door didn't change anything. The goats and car didn't move around. So now there are only two doors left, one with the car and one with a goat. Either door could be correct, so it doesn't really matter whether you switch or not, right? There's a huge problem with that assumption, though.

There are still three equally likely cases.

• A) Door A, which you picked, still has the car. The host can open either of B or C; it doesn't really matter. If you switch, you lose.
• B) Door B still has the car. The host won't reveal your choice or the car, so must open door C. Switching gives you door B and the win.
• C) Door C still has the car. The host won't reveal your choice or the car, so must open door B. Switching gives you door C and the win.

It's easy to assume that, with two doors to pick from, you have a 50-50 chance either way. And that would be true, if you picked a door only after the host opened one. But you didn't! That's the trick. Your initial selection has a 1 in three 3 chance of being correct, period. Nothing the host does after you pick can alter that.

The host knows something you don't, though, namely where the car is. The host never opens the door you picked, so not opening that door tells you nothing about what might be behind it. However, opening a door does affect the odds for the remaining unopened door, since the host never opens the door with the car. In two out of three cases, the unchosen closed door remains closed precisely because it hides the car.

The confusion comes from incorrectly conflating cases B and C by thinking of the situation as "the car is either behind the door I picked or behind the other unopened door". The thing is, there's only one way it can be behind the door you picked, but two different ways it can be behind the other door. It doesn't matter that you can express it as "either one or the other"; the probabilities simply aren't equal.

(revised 2022-12-13 to correct the initial cases and for clarity)