Ideal komi and winning odds, based on pro games.
Posted: Thu Jul 03, 2014 4:49 am
Database with 37380 professional go games.
Prediction performance for elo ranking algorithm (with optimal K), not including first 1/3 of the professional games (which is used exclusively as training data), is 58.0221% when using logistic curve as is the prevailing choice in modern elo implementations. However when using exponential curve the prediction performance increases to 58.0494%. This effectively implies that professional games outcome are based more on the cumulative quality of their moves, and and less on outside factors (than the use of logistic curve would imply).
We can therefor calculate the odds of winning (when below 50%) with 0.5 ^ ( 1 + points * multiplier_per_point ) where each handicap stone is worth 2 * ideal_komi points, and getting to play first move is worth ideal_komi points. So we need to figure out ideal_komi and multiplier_per_point. In order to do that..
I looked for pairs of pro games, where same two players had played against each other, within a years time span, with same komi, with alternating colors.
I then counted how often the same player won both games, and how often black won games.
5.5 komi, 4288 samples, 54.81% both won by same player, 54.13% won by black.
6.5 komi, 8064 samples, 56.29% both won by same player, 50.06% won by black.
7.5 komi, 4672 samples, 50.47% both won by same player, 44.93% won by black.
Its worth noting, that when using pass stone where white always gets last move, territory and area scoring produce equal results. The 7.5 komi with no such rule in place, is equivalent to 7.0 komi when compared to territory scoring.
Obviously, if players were of equal strength and komi was fair, then 50% of the games should be won by same player and 50% won by black.
Anyways, putting these trough a simulated annealing, trying to estimate average strength difference between players (in terms of extra komi he can give), what is ideal komi, and finally what is multiplier per point for win odds assuming exponential curve. Simulated annealing results in..
ideal komi : 6.516
multiplier per point : 0.1737
strength difference in points (5.5) : 3.64
strength difference in points (6.5) : 3.27
strength difference in points (7.5) : 1.31
these values ends up predicting..
5.5 komi, 54.80% both won by same player, 54.08% won by black.
6.5 komi, 56.29% both won by same player, 50.22% won by black.
7.5 komi, 50.94% both won by same player, 47.43% won by black.
So each handicap stone given drops odds of winning by 79.2%.
Interesting sidenote, it seems like pairing in chinese pro games is more evenly matched. Or it could indicate that chinese pro players are in general closer in strength to each other than say japanese players. One possible explanation is bias from which games are recorded.
Prediction performance for elo ranking algorithm (with optimal K), not including first 1/3 of the professional games (which is used exclusively as training data), is 58.0221% when using logistic curve as is the prevailing choice in modern elo implementations. However when using exponential curve the prediction performance increases to 58.0494%. This effectively implies that professional games outcome are based more on the cumulative quality of their moves, and and less on outside factors (than the use of logistic curve would imply).
We can therefor calculate the odds of winning (when below 50%) with 0.5 ^ ( 1 + points * multiplier_per_point ) where each handicap stone is worth 2 * ideal_komi points, and getting to play first move is worth ideal_komi points. So we need to figure out ideal_komi and multiplier_per_point. In order to do that..
I looked for pairs of pro games, where same two players had played against each other, within a years time span, with same komi, with alternating colors.
I then counted how often the same player won both games, and how often black won games.
5.5 komi, 4288 samples, 54.81% both won by same player, 54.13% won by black.
6.5 komi, 8064 samples, 56.29% both won by same player, 50.06% won by black.
7.5 komi, 4672 samples, 50.47% both won by same player, 44.93% won by black.
Its worth noting, that when using pass stone where white always gets last move, territory and area scoring produce equal results. The 7.5 komi with no such rule in place, is equivalent to 7.0 komi when compared to territory scoring.
Obviously, if players were of equal strength and komi was fair, then 50% of the games should be won by same player and 50% won by black.
Anyways, putting these trough a simulated annealing, trying to estimate average strength difference between players (in terms of extra komi he can give), what is ideal komi, and finally what is multiplier per point for win odds assuming exponential curve. Simulated annealing results in..
ideal komi : 6.516
multiplier per point : 0.1737
strength difference in points (5.5) : 3.64
strength difference in points (6.5) : 3.27
strength difference in points (7.5) : 1.31
these values ends up predicting..
5.5 komi, 54.80% both won by same player, 54.08% won by black.
6.5 komi, 56.29% both won by same player, 50.22% won by black.
7.5 komi, 50.94% both won by same player, 47.43% won by black.
So each handicap stone given drops odds of winning by 79.2%.
Interesting sidenote, it seems like pairing in chinese pro games is more evenly matched. Or it could indicate that chinese pro players are in general closer in strength to each other than say japanese players. One possible explanation is bias from which games are recorded.
Original First Post wrote:In the database there are 12012 games with 7.5 komi, 19151 games with 6.5 komi. Black player won 47.977% of the games with 7.5 komi, compared with 50.713% when komi was 6.5. Assuming an exponential curve, we can get the following two equations.
0.5^(1 + ( 7.5-x) * y) = 0.4799
0.5^(1 + ( x- 6.5 ) * y) = ( 1 - 0.50713 )
where x is ideal komi, and y is multiplier per point. solving the equations we end up with ideal komi of 6.75929 and multiplier per point of 0.0799152.
Since komi is compensation for half a move. a full move is worth twice that of komi. Meaning, one handicap stone is worth 13.51858 points.
As an example, based on these results the odds of winning an even game against someone 3 stones stronger would be 5.2%, as per 0.5^( 1 + 3 * 13.51858 * 0.0799152 ) = 0.052...
