Author: Sune Fischer
Date: 13:00:48 09/26/02
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On September 26, 2002 at 15:40:13, Rolf Tueschen wrote: >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 Could you rephrase the question to less than 16 pages then, I don't want to be looking for a needle in a haystack. I simply gave you an answer to whether or you should switch doors. -S.
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