Author: Uri Blass
Date: 07:26:54 09/23/01
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On September 23, 2001 at 09:53:47, Janosch Zwerensky wrote: > >>I use the estimate 1/2^64+2/2^64+3/2^64+... 2^32/2^64~=1/2 > >P.S.: A relatively good efficiently computable lower bound for the probability >of getting two equal hash-keys from N randomly generated positions should be >(for N that are small compared to 2^64) something like > >1-(1-1/2^64)^(N*(N-1)/2) according to my calculations. > >Regards, >Janosch. My estimate was for the average number of hash collisions and not for the probability for hash collision. your formula when N=2^32 should give something that is close to (e-1)/e when e is close to 2.72(e=lim n->infinite (1+1/n)^n) Uri
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