a message dated 5/10/05 8:04:49 PM Pacific Daylight Time, drd@xxxxxxxxxxxxxxxxx writes:
> One the other hand no matter how many copies of the same program participate,
> it doesn't really change the odds of wining.
Am I in the twilight zone?
Assume program A has a chance P of winning against program B. Now let A play against N copies of B. A will play N games and win P*N of them. Each copy of B will win on average (N-1)/2+(1-P) games. P*N > (N-1)/2+(1-P)=0.5*N+0.5-P for any N >1 and P > 0.5.
The only problem is the fluctuation. P*N - 0.5*N - 0.5 + P = (P-0.5)*(N+1). The probability for a copy of B to win (P-0.5)*(N+1) games more than its average can be calculated as
exp(-(P-0.5)*(N+1)*(P-0.5)*(N+1)/0.5/N)/0.5/sqrt(2*pi).
For P=0.7 and N = 10 it equals to 0.303.
But as N increases, this probability goes to zero.
Daniel liu
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