The role of probability in solving the Monty Hall problem

[robvalue]

Yes, they agree. If they don't, we're really in trouble.

I have probability 1/3 that I'm in front of the right door. It doesn't matter if monty is a sneaky cunt because I've chosen randomly, he can't psych me out. Take that Derren Brown.

So I have a 2/3 probability I chose wrong.

Now, he must open one of the other two doors. So if I chose wrong initially, I'm now guaranteed to win if I switch. The prize can only be behind the remaining door.

This situation has probability 2/3. The probability I switch and lose only happens when I chose right initially, probability 1/3.

In other words, I'm planning to choose the two doors I don't pick initially. Monty picks one for me, I pick the other. If it's behind either of those, I win. If it's behind the one I pick initially, I lose. As I have 2 chances to win, it's 2/3.

Here's the crucial part: him opening a door gives me no more information about whether I picked right initially, or not. He is making a forced choice (although it's random between the two if I chose right). It looks the same to me, either way.

Now, if monty could also open my door, the game is up. I have no advantage. It's because he can't that I get an instant win 2/3 of the time.