I just confirmed these numbers. Thought I would be first but you beat me 
to it :) I'll bet my testicles too.

/Gustav

Tyson Mao skrev:

>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?
>>    
>>
>
>
>
>
> 
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