The Monty Hall problem

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.

4 Responses to “The Monty Hall problem”

1. Eyal Mozes Says:

One other factor that complicates the issue: Monty Hall was a real-life game show host; the problem describes the actual game show he hosted (with one major change); and in his actual game show he did in fact always open first a door which does not have a car. So the use of the name “Monty Hall” does imply this assumption, even if it is not explicitly stated.

The one difference, between the actual Monty Hall show and the one described in the problem, is that the real-life Monty Hall did not give contestants the choice of switching the door. The only purpose, of opening first a door that does not have a car, was to increase the tension; Monty followed that by opening the other door that the contestant didn’t pick, followed by the door he picked. So the problem really amounts to: if, on the Monty Hall show, you were offered a choice that contestants are usually not offered, to switch doors after Monty opened the first door, should you do it? The answer is yes.

I agree, of course, that using the name “Monty Hall”, and requiring people who hear the problem to be familiar with the real-life game show in order to understand the problem, is not a substitute for actually stating a problem clearly. In most discussions I’ve heard of the problem, the people insisting that you should switch doors, and denigrating those who dare disagree with Marilyn vos Savant, are the ones who don’t know what they’re talking about; and the ones who say you should not switch doors are correct given the problem as stated.

• Gary McGath Says:

I’m vaguely familiar with the game show, mostly from the term “Monty Hall dungeon” for D&D games where players can pick up huge amounts of treasure with ease. A show based on a simple guessing game sounds dull. The entertainment must have been entirely in the players’ reactions.

• Eyal Mozes Says:

I don’t know that much about the show, but as far as I understand most of the show did involve skill and knowledge on the part of the contestants. Only contestants who succeeded in previous rounds got to the final one, with a 1/3 chance of getting a car based on simple guessing.

2. Capt'n Lisa Says:

Very cool! Thanks for the analysis and the program testing it.