Monty Hall Two

(From 8/14/06 on my old blog)

Read the posting below called “The Monty Hall Problem”

Now suppose that Monty does not know which of the doors is good and which are bad. You randomly pick one of the three doors and then Monty randomly picks one of the others and reveals what’s behind it. If he picks the good door, the game is over and you lose. If he picks a bad door, you get the choice of whether to switch or not. Should you switch?

Consider. The selected door is in caps. There are three initial possibilities.
GOOD bad bad
good BAD bad
good bad BAD

There’s a 1/3 probability of selecting the good door. This time, Monty does not know which door is good and which ones are bad. He randomly picks one of the doors you did not select. So for each of the original three cases, there are two possible doors that Monty can reveal. That means that there are now six cases.

For the "GOOD bad bad" case:
1. Monty could reveal the first bad door leaving GOOD and bad.
2. Monty could reveal the second bad door leaving GOOD and bad.

For the "good BAD bad" case:
3. Monty could reveal the good door - you lose.
4. Monty could reveal the bad door leaving good and BAD.

For the "good bad BAD" case:
5. Monty could reveal the good door - you lose.
6. Monty could reveal the bad door leaving good and BAD.


Of the six case, there are four where you don’t automatically lose. In two of them switching is bad and in the other two, switching is good. The probability is equal so it doesn’t matter whether you switch or not.