Quite a long time ago (as the Internet goes) a big debate raged online about a probability problem. Recently Sasha Volokh posted an analysis which goes in an unusual direction. The argument results from unstated assumptions that people read into the problem.

The problem is this, as Volokh phrases it:

you’re on a game show, there are three doors, and there’s a car behind one door. You choose door 1. The host, Monty, opens a door which (1) is different than the door you chose and (2) has no car behind it. So let’s say he reveals that door 2 is empty. Now he offers you a choice: Should you switch to door 3?

The obvious answer (to me, anyway) is that the only knowledge you have is that there is no car behind door #2, but there is one behind either #1 or #3. Thus, the probability for each of the doors is equal, and it doesn’t matter whether you switch or not.

But no, we’re told, the odds are 2/3 that the car is behind a door which is not #1, and that isn’t changed by the opening of the second door. The door you didn’t choose acquires all this probability by osmosis, so it now has a probability of 2/3.

The problem is that this depends on an assumption which isn’t stated in the problem. The most reasonable assumption (to me) is that the host picks a door at random from the two remaining ones. The people concluding that the odds switch are assuming that the host always chooses a door which does not have a car. We aren’t told which strategy he followed, only that he opened a door without a car in the immediate case. If the host avoided the car, he’s giving you additional information about the remaining door. If not, he’s only telling you that the door he opened doesn’t have a car, and the odds are equal between the remaining two, regardless of which you picked initially.

Volokh brings out another unstated assumption: that if the car isn’t behind either of the remaining doors, the host will choose randomly. He explores the consequences of a non-random choice. However, he keeps the assumption that the host has peeked and picked a door which doesn’t have the car.

Enough of probability theory. Let’s write a little program to test it. I’ve put it on my website. As it turns out, when I upload it as a .php file, it actually runs it, showing the numbers for 1000 games played without switching and 1000 with switching, with the host choosing at random and with the host avoiding the door with the car. This is convenient, so I’ll leave it, but I’ve also made it available with a .txt extension for easier downloading.

I know, a “Monty” game should be written in Python. Sorry, I know PHP better. If you want to translate it into Python, feel free. In any case, the results show that if the host chooses randomly, your odds don’t change, but if the host always avoids the car, the odds for the remaining door double. It depends entirely on the assumptions in an incompletely stated problem.

The argument which I hear most often for the doubling of the odds is that Marilyn vos Savant said they double, and Marilyn vos Savant is the smartest person in the world (her very name says so), so how can you disagree with her? This is simply the argument from authority. Even the smartest person in the world — as if there were a definitive ranking of intelligence — stands or falls by the validity of her arguments.