You can't just look at the mean. If you take a histogram and look at the 
distribution of scores, you'll see a Gaussian-like bump in the middle, but 
also huge tails where only one color was left. You can calculate the histogram 
once and then use it to derive the win rate for different Komis. (That would 
save time.) But you can't use the mean for that purpose.

- Dave Hillis


-----Original Message-----
From: Nuno Milheiro <nunomilhe...@gmail.com>
To: computer-go <computer-go@computer-go.org>
Sent: Thu, 8 Jan 2009 2:14 pm
Subject: Re: [computer-go] Black/White winning rates with random playout?


Given random play, komi value does not change play, so we could see what is the 
mean score (no komi) instead of playing games at different komis. But in this 
case we should not see those 2 exceptions.
Or else I'm wrong somewhere on my assumption.


2009/1/8 Rémi Coulom <remi.cou...@univ-lille3.fr>


Isaac Deutsch wrote:

I ran some tests with 500k games each and came to this result:

with komi 0.5, white has 47.5 winn. perc.
with komi 1.5, white has 50.7 winn. perc.
with komi 2.5, white has 50.9 winn. perc.
with komi 3.5, white has 54.0 winn. perc.
with komi 4.5, white has 53.8 winn. perc.  <---------- ?
with komi 5.5, white has 57.3 winn. perc.
with komi 6.5, white has 57.1 winn. perc. <--------- ?
with komi 7.5, white has 60.3 winn. perc.
with komi 8.5, white has 60.3 winn. perc.
with komi 9.5, white has 63.1 winn. perc.
with komi 10.5, white has 63.1 winn. perc.
with komi 1
1.5, white has 65.9 winn. perc.

The percentage is mostly according to komi, with 2 exceptions.
 



If your playouts don't score seki, then komi = 2n + 1.5 and komi = 2n + 2.5 are 
equivalent. So you should get the same winn. perc. for 1.5 and 2.5. Ditto for 
3.5 and 4.5, etc.

That's because White Score + Black Score = 81, so Black Score - WhiteScore = 81 
- 2 * WhiteScore is always odd.

Rémi 





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