Author: Rolf Tueschen
Date: 12:40:13 09/26/02
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On September 26, 2002 at 13:57:27, Sune Fischer wrote: >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. That is correct.And you are still not at thebottom of the problem as it was posed by Mr. Whitaker. I will comment on the strange happening here in answering your other question. You "solved" problems that didn't exist. And you left open problem that were the reason for the topic here. Rolf Tueschen
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