Author: Sune Fischer
Date: 05:23:54 09/26/02
Go up one level in this thread
If you didn't understand Marilyn's explanation, then here is another.
You divide the three doors {doorX,doorY,doorZ} into two sets, one set is
A={doorX} and the other
B={doorY,doorZ).
Now you pick doorX, so you pick the smaller set A with outcome 1/3 (we can
always rename the doors if you want to pick doorY or doorZ obviously).
Ok so there is 1/3 chance the stochastic outcome belongs to the set A and 2/3
chance that it belongs to set B, agreed?
Now the host is kind enough to remove one possible outcome from the set B thus
reducing it to 1 element like in A, let's assume he opens doorY, so now
B*={doorZ}
However B and B* must still have the same probability(!) because we know he
opened a door which didn't open to the car, so the car is still in B* with the
same probability.
Why you ask didn't this change the probability of A, so P(A)=P(B*)?
Because there it was 100% certain that the host would open for a goat, we knew
there was a goat behind one of the doors and since the host is all-knowing it is
possible for him to do with 100% certainty.
Okay, now you can choose between sets A and B*, only P(B*)=2*P(A)=2/3, therefore
you should switch.
If the host doesn't know for sure where the car is, and he just happens to open
to a goat by luck, then P(A) increases too(!), because it is more likely that he
would pick a goat if both the doors were goats (ie if you already picked the
right one). So in that case it all pans out and you don't have to switch.
-S.
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