Author: Michael Yee
Date: 10:05:11 10/03/03
Go up one level in this thread
On October 03, 2003 at 10:05:06, Tord Romstad wrote:
>On October 02, 2003 at 15:49:13, Oliver Roese wrote:
>
>>On October 02, 2003 at 05:21:45, Tord Romstad wrote:
>>
>>>I have just finished a 100-game test match between a version of Gothmog with
>>>the Botvinnik-Markoff extension and an identical version without the extension.
>>>The version with the new extension added won by 56.5-43.5.
>>>
>>>Not enough data to make any definite conclusions, of course, but it certainly
>>>looks interesting.
>>>
>>>Tord
>>
>>Ignoring draws, a match over 100 games between 2 programs of equal strength,
>>should end with a score of the first programm being in the interval [44:56] in
>>about 80.66% of all cases (computed with R, btw).
>>
>>That confirms your conclusion.
>
>Thanks.
>
>After 200 games, the score was 111.5-89.5. Could you do a similar
>calculation with these numbers (or better yet, teach me how to do it
>myself)?
>
>Tord
Hi Tord,
Computing 80.66% goes as follows:
P(k wins by A) = (100 "choose" k) * (0.5)^i * (1 - 0.5)^(100 - i)
where (100 "choose" k) is the number of combinations. This comes from viewing k
as a binomial random variable with probability of success on each trial = 1/2.
This simplifies a bit, and summing over k = 44 to 56 gives:
(0.5)^100 * sum("100 choose k", k = 44 to 56) = 0.806652...
Using similar expressions, I guess you could compute probability that program A
would score >= 111 in a 200 game match. For the record, I think it comes to
6.86%
Michael
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