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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