Author: Sune Fischer
Date: 10:57:27 09/26/02
Go up one level in this thread
On September 25, 2002 at 17:33:00, Roy Eassa wrote: > >OK, I'll take my shot at wording the explanation: > > >"If n is the starting number of doors... > >The door you choose always has (from your perspective) a 1/n chance of being the >winner. All the unopened doors COMBINED (assuming there's at least one) always >have a (n-1)/n chance. If Monty opens one of those doors that he KNOWS is not a >winner, that does not change the fact that all the (remaining) unopened doors >combined have a (n-1)/n chance. But since there are now fewer of them among >which to divide it, each INDIVIDUAL unopened door's chance is (from your >perspective) higher than before. > >Thus in the 3-door example, your door has a 1/3 chance and the one remaining >unopened door has a 2/3 chance of being the winner." Yes, this is the correct solution. The funny thing is, that he must KNOW where the car is. If he doesn't KNOW and still opens a door to a goat, there is no reason to change your choice of door. In that case ALL the remaining doors simply get 1/(n-1) chance. Quite an interesting exercise :) -S.
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