Ideal komi and winning odds, based on pro games.
- Ootakamoku
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Re: Ideal komi and winning odds, based on pro games.
To get around the problem. I looked for pairs of games, where same two players had played against each other, within a years timespan, 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, 53.26% both won by same player, 54.23% won by black.
6.5 komi, 56.74% both won by same player, 49.88% won by black.
7.5 komi, 51.15% both won by same player, 48.06% won by black.
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 : 6.8143
multiplier per point : 0.1148
strength difference (5.5) : 3.9187
strength difference (6.5) : 5.7777
strength difference (7.5) : 2.2413
these values ends up predicting..
5.5 komi, 53.19% both won by same player, 53.83% won by black.
6.5 komi, 56.77% both won by same player, 50.79% won by black.
7.5 komi, 51.21% both won by same player, 47.71% won by black.
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.
I then counted how often the same player won both games, and how often black won games.
5.5 komi, 53.26% both won by same player, 54.23% won by black.
6.5 komi, 56.74% both won by same player, 49.88% won by black.
7.5 komi, 51.15% both won by same player, 48.06% won by black.
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 : 6.8143
multiplier per point : 0.1148
strength difference (5.5) : 3.9187
strength difference (6.5) : 5.7777
strength difference (7.5) : 2.2413
these values ends up predicting..
5.5 komi, 53.19% both won by same player, 53.83% won by black.
6.5 komi, 56.77% both won by same player, 50.79% won by black.
7.5 komi, 51.21% both won by same player, 47.71% won by black.
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.
- Ootakamoku
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Re: Ideal komi and winning odds, based on pro games.
jeromie wrote:kivi wrote: So what's the point of the exponential model, or the significance of the value of 6.75?
It sounds like the exponential model was chosen as a nod to other ranking systems. I suppose knowing the "real" value of a continuous Komi might be interesting in some situations (e.g. Predicting the value of each handicap stone in high level games could be used to select an appropriate reverse komi), but I think this is mostly an exercise undertaken because math is fun.
The ideal komi is a byproduct. What I find interesting, is being able to calculate the odds a player wins if he is 1 stone stronger, or two stones stronger, etc. Or in reverse, determine the strength difference between two players, based on their win ratio over multiple games.
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Re: Ideal komi and winning odds, based on pro games.
Further discovery, ELO based on exponential was only better estimation, when ideal komi wasn't calculated correctly. Ie when games were assumed to be fair (say 5.5 komi) while it infact favored one of the players, in this case exponential curve produced better results. But when we have a more precise measure of komi, logistic curve actually produces better results. Why does this have to be so complicated 
- HermanHiddema
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Re: Ideal komi and winning odds, based on pro games.
You don't mention your sample sizes, but it looks like you've got a huge case of false precision going on. Unless you specifically give error bars, you should only give the number of significant figures that is warranted by the data, and I think it is unlikely that your data is accurate to within 0.01%
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Re: Ideal komi and winning odds, based on pro games.
HermanHiddema wrote:You don't mention your sample sizes, but it looks like you've got a huge case of false precision going on. Unless you specifically give error bars, you should only give the number of significant figures that is warranted by the data, and I think it is unlikely that your data is accurate to within 0.01%
The first post, which I had updated to reflect the most recent results also contains sample sizes. Forgot to include it in the last post.
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Re: Ideal komi and winning odds, based on pro games.
Latest discovery. With pass stone and white always getting last move, area scoring and territory scoring produces equal results for all practical purposes. However, since chinese rules dont use pass stone, this ends up favoring black by half a point, as roughly 50% of the time, black gets the last move instead of white. This means that chinese 7.5 komi is in fact 7.0 komi when compared to territory scoring. Accounting for this I redid the calculation, getting theory and practice to match better.
Not only did exponential curve match by far better than either logistic or normal, which was my expectation, especially for handicap. But the eventual total error was much smaller too.
Ideal komi : 6.516
Multiplier per point : 0.1737
There by if two equal strength pros played against each other, with 0.5 komi and 2 handicap, white would have 5% chance of winning, which matches intuitive expectations much better.
0.5^( 1 + ( 1.5 * 2 * 6.156 - 0.5 ) * 0.1737 ) = 0.0504
The 1.5 in this formula comes from the fact that black gets 2 handicap stones but white gets first move (which is worth half a stone).
Not only did exponential curve match by far better than either logistic or normal, which was my expectation, especially for handicap. But the eventual total error was much smaller too.
Ideal komi : 6.516
Multiplier per point : 0.1737
There by if two equal strength pros played against each other, with 0.5 komi and 2 handicap, white would have 5% chance of winning, which matches intuitive expectations much better.
0.5^( 1 + ( 1.5 * 2 * 6.156 - 0.5 ) * 0.1737 ) = 0.0504
The 1.5 in this formula comes from the fact that black gets 2 handicap stones but white gets first move (which is worth half a stone).
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Re: Ideal komi and winning odds, based on pro games.
Pass stones are actually irrelevant to all that. Pass stones do not impact area scoring at all, since they are prisoners and prisoners are not counted under area scoring.
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Re: Ideal komi and winning odds, based on pro games.
HermanHiddema wrote:Pass stones are actually irrelevant to all that. Pass stones do not impact area scoring at all, since they are prisoners and prisoners are not counted under area scoring.
http://senseis.xmp.net/?EquivalenceScoringExample
http://senseis.xmp.net/?LogicalProofOfT ... reaScoring
In area scoring, black has by default half point advantage. If white fills last dame, the result is equal to territory scoring. If black fills last dame, the result equals to territory scoring plus 1 point for black. Since (and I checked) its equally likely either player gets the last dame, the result is that chinese rules favor black by half a point more than komi would imply, hence chinese rules 7.5 komi is in practice more like 7.0 komi.
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Re: Ideal komi and winning odds, based on pro games.
Ootakamoku wrote:HermanHiddema wrote:Pass stones are actually irrelevant to all that. Pass stones do not impact area scoring at all, since they are prisoners and prisoners are not counted under area scoring.
http://senseis.xmp.net/?EquivalenceScoringExample
http://senseis.xmp.net/?LogicalProofOfT ... reaScoring
In area scoring, black has by default half point advantage. If white fills last dame, the result is equal to territory scoring. If black fills last dame, the result equals to territory scoring plus 1 point for black. Since (and I checked) its equally likely either player gets the last dame, the result is that chinese rules favor black by half a point more than komi would imply, hence chinese rules 7.5 komi is in practice more like 7.0 komi.
Yes, exactly. This is a property of area scoring, and is not related to pass stones.
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Re: Ideal komi and winning odds, based on pro games.
HermanHiddema wrote:Yes, exactly. This is a property of area scoring, and is not related to pass stones.
Yes its property of area scoring that it favors black by half a point on average, compared to territory scoring. I was merely trying to convey how the use of pass stone + white moves last (which affects territory scoring), is one way to ensure area scoring and territory scoring gives same result.
And yes, it was incorrect of me to say how chinese rules dont use pass stone, should be how japanese rules dont use pass stone.