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RE: [computer-go] future KGS Computer Go Tournaments - two sections?
Title: Message
You
are assuming a full round robin. The variance is much higher in a typical
Swiss tournament. If N is 15 and the 16 programs play in a 4 round single
elimination tournament, the probability of the stronger program winning is
P^4. If P is 0.7, if N is 1, the program has a 70% chance of
winning. If N is 15, the winning probability drops to about 25%. As
N increases the probability of the stronger program winning goes to zero
:)
David
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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