Author: Sune Fischer
Date: 14:53:24 12/06/01
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On December 06, 2001 at 15:21:17, J. Wesley Cleveland wrote: >This problem is like the problem "How many people does it take before >it is probable that two have the same birthday ?". The answer, which >many people find suprising is 23. To calculate this, calculate the >probability p, that two people have different birthdays = 364/365. >Then calculate how many pairs of people n, you need before this is less >than 1/2, p^n <.5. Then find the number of people g, which taken two at >a time is >= n, g = n*(n-1)/2. > >The same method tells you how many different positions you can have >before it is likely that two will have the same hash key. > >32 bits 77163 >48 bits 1.97536627683E+7 >64 bits 5.05693754118E+9 > > >Thanks to Cliff Leitch for providing a high precision freeware calculator. Exactly, which is why 32 bits will only work for pawn hashing. Hyatt claims to be getting a few collisions a second, this must mean he is doing a lot more more than 77163 pawn positions and I doubt that, so something is not right here. However nobody wants to count the number of pawn positions, so I only have my own numbers to go by, and I get ~10000 in a blitz game. Can anybody confirm that number, pleazzzze? -S.
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