A former college roommate called me with a probability question based on Let’s Make a Deal (yes, my friends are also geeks). The problem is commonly refered to as “The Monty Hall Problem” (after the host of Let’s Make a Deal) and has a clear solution.

My friend posed the question as follows:

Phase 1: Three doors are on stage and Monty asks you to guess which one the prize is behind. You select a door, say door #1. Monty opens an empty door that you did not pick, say door #2. Monty then asks if you would like to switch to door #3 or stay with door #1. Should you switch?

Phase 2: Suppose three people from the audience play the game at once and all three have to pick different doors. Suppose further that Monty reveals door #3 is empty and that player is eliminated. Should the people guessing door #1 and door #2 switch?

Phase 3: Suppose there are three people playing the game simultaneously, but they are unaware of each other’s presence. Should players switch doors?

Okay, so no one actually talks like that and I am paraphrasing my friend. But the gist of the question remain the same.

Good.

So the answer to the Phase 1 Monty Hall problem is that you should switch. When you initially select the door, you have a 1 in 3 chance of guessing correctly. That is, there is a 2 in 3 chance that the prize lurks behind one of the other two doors. Monty then removes one of the doors. Removing the door tells you nothing about the accuracy of your guess, since the rules of the game stipulate that he will remove one of the two remaining doors. So your initial guess still has a 1 in 3 chance of being correct. However, the removal of the door DOES tell you something about the other two doors. Namely, it eliminates one of the options, so you can now pinpoint that 2 in 3 chance behind only one door. This line of reasoning suggests that switchers will be correct 66.7% of the time and stayers will be correct only 33.3% of the time.

Being a good empiricist, I decided to simulate 100,000 games of “Let’s Make a Deal.” The results are unambiguous. (Note: You can play the game by hand here or here. If you’re too lazy to play 100,000 times yourself, this site will speed up the process. Email me if you would like a Stata do-file accomplishing the same feat.)

Play Stay (i.e., do NOT switch)

 Result Number Percent LOSE 66,739 66.7% WIN 33,261 33.3%

Play Switch

 Result Number Percent LOSE 33,261 33.3% WIN 66,739 66.7%

Well, what about Phase 2 where there are three contestants? It would seem absurd for the remaining two contestants to swap doors, right? Of course it would be absurd. The rules of the game are not the same as in Phase 1. When one of the contestants can get eliminated, then the removal of a losing door is informative. Your door could have been removed, but it wasn’t. So your door went from having a 33% chance of being correct, to a 50% chance of being correct. Switching wouldn’t hurt, but it also wouldn’t help.

And the answer to Phase 3 is the same as to Phase 2 because the other two players are a red herring. All the action is in the rules and whether or not the door you pick can be eliminated by Monty. The epistemological and alternative universe type set up makes me suspect that my friend overheard the conversation of a philosophy graduate student.

Can’t trust those philosophers.