Winning percentage and centipawns

[metax]

Has anyone an idea for a function that approximates the winning percentage for a given centipawn value of a position (and vice versa)? I only need an approximate value.

[Zach Wegner]

http://chessprogramming.wikispaces.com/ ... 2C+and+ELO

[gingell]

Take a look at:

https://chessprogramming.wikispaces.com ... 2C+and+ELO

I did some follow up on the same idea. You can see my write up at:

http://sourceforge.net/apps/wordpress/c ... tatistics/

and

http://sourceforge.net/apps/wordpress/c ... eneration/

[jwes]

If you are referring to material advantage the links above will help. If you are referring to positional values in general, then the results will be highly engine dependent, and difficult to determine.

[metax]

As I only need approximations, the links above are sufficient.

[hgm]

This is similar to the results I obtained for material-imbalance self-play of Fairy-Max, and an experiment I read about with pawn-odds games for Rybka 1-min games. The latter found ~74% score advantage, while I found 68%. But In my measurements I averaged out the white advantage, while the Rybka results where the classical pawn+move advantage. So subtracting a 4% white advantage would bring the results quite close. An 68% score is closer to 125 Elo advantage, but note that in the reported results by Pradu a Pawn was weighted a bit low: usually we do not consider a Knight to be worth 400 cP.

Note that GMs hardly ever give away a Pawn without any compensation, so analyzing GM games is bound to underestimate the efect of pawn odds. Starting from quasi-symmetric material-imbalance positions, like I always do, should not suffer from this.

So on a scale where B=N=325 (Kaufman values), 100 would be worth about 125 Elo or 68% score.

[Martin Brown]

http://chessprogramming.wikispaces.com/ ... 2C+and+ELO

That is an interesting graph. Odd that they show a model fitted for K=4 when the tails would be better represented with K=5. Is there some empirical reason not to vary K for a best fit solution?

Regards,
Martin Brown

[Pradu]

http://chessprogramming.wikispaces.com/ ... 2C+and+ELO

That is an interesting graph. Odd that they show a model fitted for K=4 when the tails would be better represented with K=5. Is there some empirical reason not to vary K for a best fit solution?

Regards,
Martin Brown

K=4 was chosen so that it would be very simple to relate ELO to centipawns: 1ELO = 1cp. You could use other models to try to fit this curve better:

http://www.pradu.us/old/Nov27_2008/Buzz ... ercent.xls

I'd say matters most is the nearly linear region between -5 to +5 pawns where K=4 fits very well. An evaluation function that has units of ELOs should produce a value close to the same evaluation function in units of cps. Therefore, it should be ok to obtain evaluation terms in units of ELOs from a database of games or whatever else and combine it with other evaluation terms which have intuitive hand-crafted values in centipawns.

[Pradu]

This is similar to the results I obtained for material-imbalance self-play of Fairy-Max, and an experiment I read about with pawn-odds games for Rybka 1-min games. The latter found ~74% score advantage, while I found 68%. But In my measurements I averaged out the white advantage, while the Rybka results where the classical pawn+move advantage. So subtracting a 4% white advantage would bring the results quite close. An 68% score is closer to 125 Elo advantage, but note that in the reported results by Pradu a Pawn was weighted a bit low: usually we do not consider a Knight to be worth 400 cP.

Note that GMs hardly ever give away a Pawn without any compensation, so analyzing GM games is bound to underestimate the efect of pawn odds. Starting from quasi-symmetric material-imbalance positions, like I always do, should not suffer from this.

Computer games were used, not GM games; so the effect should not be as prominent, but will still be there.

So on a scale where B=N=325 (Kaufman values), 100 would be worth about 125 Elo or 68% score.

Some strong programs like Fruit use pawn-normalized values of 4 in the opening and it worked better, atleast for Buzz, than Kaufman values.

[Don]

Has anyone an idea for a function that approximates the winning percentage for a given centipawn value of a position (and vice versa)? I only need an approximate value.

1 / (1 + exp(-v))

where v is the evaluation of the position.

The formula gives you 56% winning percentage if you are up 1/4 pawn, about 62% if you are up 1/2 pawn and if you are up 3.0 pawns it gives you about 95% winning probability.

You can fine tune this by scaling V accordingly.

Of course you realize that in reality this doesn't quite make sense. You probability of winning is going to be based on how strong/weak your opponent is in relation to you and other factors. But in practice with equal players this is pretty reasonable.