The "Let's Make a Deal" Paradox

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Originally posted by hot_pastrami
[b]Well, the problem with the 50/50 intuition is that it assumes that the first choice has no bearing on the second choice. But it does. Consider the two possible scenarios for the initial door selection:

1) You select the door that does contain the prize, which will occur one in three times. Monty can open either of the other two doors at random. His choice is unaffected.

2) You select a door which does not contain the prize, which will occur two in three times. Monty is forced to open the other non-winning door.

Scenario #1 will happen about 33% of the time, and scenario #2 will happen about 66% of the time, right? So, after you make your initial choice, it is safer to bet that #2 happened. So accordingly, it is safe to bet that he was forced to open the other empty door, and therefore it follows that the third door, the one you did not pick, contains the prize. I hope I explained that well, my brain is getting pretty tired.

But... as always, I don't grasp the causal relationship between the two choices.

Let me phrase it this way:

Game #1: Monty shows Joe Contestant three doors. Joe picks Door #1. Monty says "Here's what was behind Door #2," and opens it to reveal a goat. Monty says "Would you like to switch to Door #3?"

Game #2: Monty shows Joe Contestant two closed doors and one open one. The open one contains a goat. (Alternatively, he shows three closed ones, but opens one goat door before Joe makes a choice.) Monty says, "Which one will you choose?"

In my mind, these two games are identical in terms of probability; Game #1 merely has additional non-binding gamesmanship before it's time to choose "for real."

In Game #1, you make an initial choice, but it is in no way binding; the only effect it has (other than suspense for the audience) is to help Monty select which goat door he's going to open. (If you guess the prize door initially, he can open either; if you guess a goat door initially, he will open the other goat door.) Either way, the ambiguity has been removed, and you are now faced with a situation equivalent to Game #2 (two doors that you know nothing substantial about, one prize, one final choice to make).

This doesn't change the fact that there's a 2/3 chance that the prize will be behind one of the two doors you didn't initially choose; that much is simple mathematics. But the situation _after_ Monty opens the door is not the same situation as _before_ Monty opens the door; to claim that the second choice (viewed in a vacuum) isn't a fifty-fifty chance in and of itself is to play clever semantic games with the problem as a whole. That 2/3 chance is tempered by the 100% certainty that one of its two options will be eliminated before the actual choice is made.