Author: Simon Finn
Date: 05:35:46 11/01/01
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On October 31, 2001 at 18:17:39, James Swafford wrote: >On October 31, 2001 at 18:14:41, James Swafford wrote: > >> >>I'm playing around with a hyperbolic tangent function in order >>to predict a reward [-1 ... 1] based on my raw evaluation score >>of the principal variation. >> >>I've come up with predicted_reward = tanh(pawn_adv/300), where >>a pawn advantage of 1 pawn ---> pawn_adv=100. >> >>The following table shows the relationship between pawn_adv and >>predicted_reward: >> > >Actually, that's pawn_adv/100 vs. predicted_reward... > > >>pawn_adv predicted_reward >>.1 .033 >>.25 .083 >>.33 .11 >>.4 .133 >>.5 .165 >>.75 .245 >>1 .32 >>2 .58 >>3 .76 >>4 .87 >>5 .93 >>6 .964 >>7 .981 >>8 .990 >>9 .995 >>10 .9975 >>12 .9993 >>15 .9999 >> >>So... a 1 pawn advantage yields a predicted reward of .32. >> >>Has anybody done research, or know of research, that can tell me >>how close those figures are? i.e. if your program obtains a >>1 pawn advantage, do you know how likely it is to win? >> >>Tridgell and Baxter's paper says they give a one pawn advantage >>a predicted reward of .25, but it doesn't say why they chose >>that number. Maybe they pulled it out of thin air, I don't know. >> >>Comments? Anybody know of a better function than tanh() to >>do this? How does predicted reward convert into the expected percentage? Is it 50 + 50 * predicted_reward? If so, the predicted reward for small advantages seems a bit small. White is generally expected to score roughly 55%. On the above formula, this corresponds to a predicted reward of 0.1, but I don't think you would find many takers for the proposition that White's initial advantage is more than 0.25 pawns. Simon >> >>-- >>James
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