Author: Janosch Zwerensky
Date: 08:04:16 09/23/01
Go up one level in this thread
>I used 1/2^64*(sum i,i=1,2,3,4,5,6,7...2^32-1) and not sum(2^i) and I get >1/2^64*[(2^32)*((2^32)-1)/2] that is almost 1/2 Ok, that's true and that formula doesn't look bad either, and it certainly doesn't look like it contradicts any of the formulas I gave. Sorry for me not having read your post properly. Regards, Janosch.
This page took 0 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.