I was wrong. I lost a couple hours sleep last night realizing that I was wrong. I will say that the Wikipedia article not only didn’t convince me, it totally confounded me. I won’t bother to test my theory; I’m now convinced that, in two-thirds of all scenarios for Guest A, Guest A will gain by switching doors when that switch is offered. Even if that does give me a headache.

Oh, and Seth: Thanks for continuing to push at this. Even the thickest skulls, like mine on occasion, can occasionally be penetrated.

Note that, to reach those who read the first post but are unlikely to return, the new text at the top of this post also appears as a separate post.

Information possessed != “mental state”

If a stagehand slipped the person the location of the prize beforehand, that would be information, not “mental state”. Suppose that slipped information was not a certainty, but a probability – e.g “There are two different assistants who set up the car prize, Goofus and Gallant. If Goofus does it, it’ll always be on the left side, since Goofus is lazy and won’t push it further than necessary. Gallant is conscientious and will put the car anywhere on stage. But I don’t know who is on duty tonight”. That’s still useful even if not definitive.

“That is, either the car is behind door X or door Y.” – are you saying this means it has a 50-50 chance of being behind door X or door Y? That’s not the same thing.

“At that point, assuming a fair setup, the odds are 50% in each case”

But that’s the point – the setup is not 50/50, and Guest A knows that, while Guest B doesn’t (note, but Guest B still has 50/50 chance of guessing right, even if the setup is NOT 50/50 chance for each door).

“If they’re 50/50 for Guest B, they must be 50/50 for Guest A”

Umm, why? This is quite false. If Guest A knows that the prize is almost always in one door, say due to the way the stagehands use it more often because it’s easier to get the car into it, it’s still a 50/50 chance for Guest B to pick the right door (not knowing that), but Guest A has a much higher chance.

That is, the previous game has given Guest A information that the setup for the current game is not 50/50 for each door.

Seth: You’re saying that the mental state of the person making the choice affects the results?

Guest B’s knowledge does not, as far as I can tell, affect the actual situation at the point at which she must choose: That is, either the car is behind door X or door Y.

At that point, assuming a fair setup, the odds are 50% in each case. If they’re 50/50 for Guest B, they must be 50/50 for Guest A…even as Guest A knows that another 999,998 doors were discarded during the process. It’s a new game at the point where “change your choice?” is offered; I don’t believe any of the previous odds are, at that point, relevant. Or, rather, I don’t think they are: it’s not a matter of belief.

The current state in this problem is always: There was some number of doors N greater than two (the number doesn’t matter). The guest chose one door. The host revealed some number N-2 of non-car doors. At that point, the guest has a choice: Which of two doors to choose? The odds, at that point, are 50/50 for both doors. The remainder (N-2) of the doors are no longer part of the game.

Look at it this way: Suppose the game is viewed as follows – Monty says “There are a million doors. Pick one. This divides all the doors into two sets, Monty’s set of 999,999 doors, and Guest’s set of 1 door”. Now, would you argue that because there are two sets, and the prize is either in one set or the other, it has a 50% chance of being in either set? (see joke above).

Monty then gives information about the wrong doors in his set. Did the OVERALL PROBABILITY of the prize being in his *set* of doors change, because he gave information about *specific* wrong doors within it?

That is, Monty’s set was originally 999,999 doors. Using his knowledge of which were wrong, he threw out 999,998 of the wrong ones. Did that change the probability of the prize being within his original set of 999,999 doors?

Guest B doesn’t know this. Guest A does.

Suppose Guest B is invited onstage for this million-door unveiling, at the point that all but two doors have been opened. He’s told, correctly, that there’s a car behind one door. Would we agree that he has a one-in-two chance of choosing the right door?

Effectively, that’s what “Do you want to switch doors?” does for Guest A: It abandons the previous game and begins a new one, one in which there are two doors, one of which has a car behind it. The guest has a one-in-two chance of choosing the right door. The other 999,998 doors are simply irrelevant at this point: Those affect the odds for the other game, the game that’s no longer in play.

I don’t believe that the position of the car can be affected by the number of doors-without-a-car that are opened. That simple.

Why do you believe this (50-50 chance) is true? Forgive me, it sounds like the old joke “All events have a 50/50 probability of happening, because either they happen, or they do not”.

As to the final thought experiment, it becomes a 50/50 proposition, because 999,998 doors are irrelevant. No matter what door you pick, it won’t be one of the 999,998 that he opens.

And I agree that I may be “mathematically wrong” or whatever. As for simulations and other testing, the only way that would matter is if it amounted to: I choose one. All but one of the “wrong” choices are removed. Now, I don’t honestly believe that whether I retain the one of two that I originally chose or switch to the other of two that hasn’t yet been revealed will matter–that each one has a 50/50 chance of being the right one.