Actually, switching wins two times out of three! Sticking with your original choice ones one time in three.
For the most part, people don't think much of this problem. Of those who do (because they are good logicians, or mathematicians, or statisticians, or some such) follow Leonard's logic.
>. . . So, whichever door
>you pick originally, they are
>then going to show you
>a door that isn't.
>That being the case, nothing
>has changed. Your odds
>originally were one in three.
> They remain that way
>between either of the two
>remaining doors. (Nothing changed
>since the EmCee is going
>to pick the "solder" door
>under any circumstances.)
>
>However, if this was a "random"
>choice, then the door you
>did not pick is a
>50-50 shot of being solder
>or a model (airplane, I
>assume, not a girl.
>Wouldn't want that.) But,
>then, after going through all
>this, and having now eliminated
>one door, that now makes
>your original pick a 50-50
>shot as well. So,
>the odds remain the same.
> Change if you want.
> Don't change. Shouldn't
>make any difference. (If
>your door had been interchanged
>with the door that was
>not picked before the EmCee
>made his choice, the odds
>would remain the same as
>they are now when it
>wasn't in the "group of
>two". No difference.)
The correct answer can be arrived at with Baysean methods. For us lay people, it easier to see the correct answer by imagining you are the MC instead of the contestant. The MC has no doubt as to the correct solution! Imagine standing being behind the doors with full view of all three items.
What the MC sees is that the contestant picks "solder" two times out of three. The MC simply shows the other piece solder. In other words, two times out of three the MC has no choice to make. Two times out of three the model (indeed an airplane, not a girl) is behind the other door. So switching wins two out of three.
When the contestant picks the model first, the MC shows one of the solders. That happens one time in three. Switching, therefore, loses one time in three.