This will be my one and only contribution to this thread,
I have seen the Monty Hall problem too many times.
The situation where switching makes sense is if (as in the
original problem) the announcer actually knows which door
has the prize, and the announcer will always pick a door
without the prize first. Then your analysis is correct.
After the announcer shows you a losing door, your chance
of winning if you switch is 2/3, versus 1/3 if you stay
put.
But if the announcer is an ignorant guesser, then it makes
no difference whether you switch or not. This is the case
that corresponds to most people's intuition - indeed the
value of switching with the usual analysis is because the
announcer affects the information available.
The third case, which chipmunk alluded to, is the case of
an evil announcer who knows the answer and wants you to
fail. In this case you should stick to your guns... | [reply] |

Thank you tilly; there was/is a demo of the "Monty Hall"
problem on the web (way back when that was amazing) and I
argued w/ the prof. who put it up. I never could get a
across the idea that he was
using that assumption in declaring "always change your vote"
as the *right* answer.
*a*
| [reply] |

This is fun... Let's explore the second possibility: the announcer doesn't know which door is the winner. After you pick a door, he shows you what is behind one of the other two doors (with equal probability). If he reveals the prize, you lose. If not, you may switch to the other door.
You switch:
You pick: He shows: Result:
Loser (2/3) Loser (1/2) Win! (1/3)
Loser (2/3) Winner (1/2) Lose (1/3)
Winner (1/3) Loser (1/1) Lose (1/3)
You don't switch:
You pick: He shows: Result:
Loser (2/3) Loser (1/2) Lose (1/3)
Loser (2/3) Winner (1/2) Lose (1/3)
Winner (1/3) Loser (1/1) Win! (1/3)
So, if the announcer doesn't know where the prize is, but just reveals one of the other doors randomly, no matter what you do you have a 1/3 chance of picking the winning door. | [reply] |