If you have a 1/3 probability of winning each time, and out of 100 trials you won 76, the probability of that happening is:
(1/3)^76*(2/3)^24*(100 nCr 76) Where 100 nCr 76 is 100 choose 76. I think that's correct, but I might be wrong, but of course, that number is definitely really small. Let's compete the probability that someone can win 76 or more times. Sum(i=76 to 100): (1/3)^i*(2/3)^(100-i)*(100 nCr i) Using Mathematica, I got 3*10^(-18). With that type of probability, I'd risk my testicles. And Leyan's too. Tyson Mao MSC #631 California Institute of Technology On Dec 28, 2005, at 7:11 PM, Stefan Pochmann wrote: > P.S. Is somebody capable of computing the odds for winning those 76 > out of 100? ------------------------ Yahoo! Groups Sponsor --------------------~--> Get fast access to your favorite Yahoo! Groups. Make Yahoo! your home page http://us.click.yahoo.com/dpRU5A/wUILAA/yQLSAA/MXMplB/TM --------------------------------------------------------------------~-> Yahoo! Groups Links <*> To visit your group on the web, go to: http://groups.yahoo.com/group/speedsolvingrubikscube/ <*> To unsubscribe from this group, send an email to: [EMAIL PROTECTED] <*> Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/
