In case you were wondering "how big is n before the probability of collision becomes remotely possible, slightly possible, or even likely?"
Given a fixed probability of collision p, the formula to calculate n is: n = 0.5 + sqrt( ( 0.25 + 2*l(1-p)/l((d-1)/d) ) ) (That's just the same equation as before, solved for n) p=0.000001 n=4.8*10^35 ~= 2^118 p=0.00001 n=1.5*10^36 ~= 2^120 p=0.0001 n=4.8*10^36 ~= 2^122 p=0.001 n=1.5*10^37 ~= 2^123 p=0.01 n=4.8*10^37 ~= 2^125 p=0.1 n=1.5*10^38 ~= 2^127 p=0.5 n=4.0*10^38 ~= 2^128 p=0.9 n=7.3*10^38 ~= 2^129 p=0.99 n=1.0*10^39 ~= 2^130 p=0.999 n=1.3*10^39 ~= 2^130 Recall that 2^256 ~= 1.15*10^77 Something somewhere says the n for "expected" collision happens around when the exponent is halved... Half of 77 is around 38, which is supported above. Half of 256 is 128, which is supported above. So as n is reduced exponentially from the "expected" point, the probability of collision exponentially approaches 0. You'll notice for n larger than the "expected" point, the probability even more dramatically approaches 1. I cannot get my poor little computer to compute anything higher than 5.1*10^39, no matter how many 9's I put on there. At this point, it becomes exponentially near impossible to avoid a collision. _______________________________________________ zfs-discuss mailing list zfs-discuss@opensolaris.org http://mail.opensolaris.org/mailman/listinfo/zfs-discuss
