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Old 11-13-2002, 03:34 PM
BruceZ BruceZ is offline
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Join Date: Sep 2002
Posts: 1,636
Default Re: what\'s the approximation???

Oops, sorry. Make that [(n-1)/n]^n = 1/e for large n. This is the probability of something not happening in 1/p tries if the probability of it happening each try is p. So in this case we have (1325/1326)^1326 = 36.8% = probability of not getting 2 in a row in 1326 tries.
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