Author: Sune Fischer
Date: 05:41:51 09/28/02
Go up one level in this thread
On September 28, 2002 at 07:49:50, Rolf Tueschen wrote: >On September 28, 2002 at 05:53:07, Omid David wrote: > >>We have to understand what the "switch" operation does here: If our first choice >>was correct, it will turn it into false; and if our first choice was false, it >>will turn into correct. So in order to eventually succeed, we have to initially >>choose the correct door in "stay" method, and the *wrong* door in "switch" >>method. >> >>Thus, the operation "switch" merely "negates" the probability, i.e. if the >>initial probability was x%, it will now become (100-x)%. >> >>For n doors, the success probability of "stay" will be (100/n)%, and the success >>probability of "switch" will be (100 - 100/n)%. >> >>Examples: >> >>In the case of 3 doors, the initial probability was 33% (x=33), so after the >>"switch" operation, it will become 67% (100-33=67). > >This is mathematical nonsense. No definition here. See my answer to Bruce in >CTF. > >Rolf Tueschen I think it is time to give in, actually it isn't very complicated at all: Assume the hosts always opens to a goat (by luck or not): You first pick a car (1/3), you switch and pick a car (0%) You first pick a goat (2/3), you switch and pick a car (100%) So by always switching you get 66.66% chance of a win. With N>3 the odds gets even better. How the host is supposed to "always" pick a goat by shear luck is a different question, in fact it seems contradictory to you assumption about luck, but if you only run 1 experiment you don't have worry about those questions. -S. > >> >>For 10,000 doors, the success probability of "switch" will be 99.99%, while >>"stay" will only have a 0.01% probability of success. >> >>In Marilyn vos Savant's example with 777,777 doors, the probability of success >>with "switch" will be 99.99987%. >> >>Omid.
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