Author: Peter Fendrich
Date: 05:53:58 01/09/04
Go up one level in this thread
On January 07, 2004 at 12:23:00, Dieter Buerssner wrote: >Perhaps I can myself clearer by another example. Assume again the 1000 games, >W=510, D=0, L=490. Player a wins as white 255 games and as black 255 games. >With your forumula we get for the result 0.51 +/- 0.01586 (67% margin). Why 67% ??? >If I simualtate this match (1 million such matches), and plot p(m) vs. m, and I >fit a curve > >1/(sqrt(2*pi)*sigma*n)*exp(-0.5*(x-x0)^2/sigma^2) > >(I kept n constant at 1000, because the integral must be 1/1000), I get >sigma=0.01581. As we can expect, practically the same number as calculated by >your formula. But now assume Player A won all his games with white, and >additionally 20 games with black. No draws. Again m=0.51. W/D/L the same as >above. Just to check if I got it right: You mean that A plays 490 white games and 510 black games? >But we can already assume now (we know more about the match than just >W/D/L), that here player a is most probably better (very similar to the W/D/L >20/980/0 case). The game might be a game, where typically white wins, and only >when he does a sublte fault will lose. Player A showed, that he was clearly >better, because he allowed B not a single win with black. The distribution of m >will be much narrower. Actually, the fit yields sigma=3.1e-3 (although now - >again not too surprisingly - the distribution is slightly tilted from the normal >distribution; I did not try to do the mathematics, but I guess, it is a >binominal distribution). This also shows, that player A ist stronger with big >confidence. >Your formula would not give an adequate error margin in this case, because it >ignores the more detailled knowledge about performance as white and black. The formulas I've given is general for a stochastic variable X with 3 outcomes. You could redefine the X to [wW, wD, wL, bW, bD, bL] all from player A's perspective. Apply the same formulas but with the 6 outcomes instead. I'm not sure if this is the trick you are looking for... If not I associate to 2 different tecniques: ANOVA: Split up you sample in groups and analyse the Variance and Mean differences between groups. I think the 2-way ANOVA is applicable here. Bayesian statistics: Gives you some good techniques to include preknown knowledge in the model in a more general fasion. Use Google! There a numerous descriptions. Finally you can always try the method I used to evaluate a result. That was before Rémi found a more elegant solution ignoring the draws. I think I've sent you the text describing it otherwise give me a hint. I suppose this method can be used to estiamte all variables. It's really a Bayesian approach computerised. >>Maybe we can continue by email - I will send you one when things are calmed down >>here... > >No problem with me, to disuss it in email. I just thought, I post an followup >here, and try to explain it a bit clearer. Perhaps, we are not the only 2 >persons interested. Maybe so, I doubt there are more ... >>I did put your examples into my formulas and got slightly different Rating-Dif's >>than you have. Maybe it's just decimal errors. > >Indeed, I had a few rounding error. I wrote down some intermediate results, >rounded them (including one error I made in the last digit). And even more >embarrassing, I did some circular calculation (from m to Elo and back, and did >not recognize, that what came back was not exactly the same to what was put in) > >>I also used another formula than >>you did. > >I guess, you are referring to the Elo calculations. Yes >I think, the Elo just complicates things, without giving more insight. :-) I wouldn't put it that way. It's the other way around. The -400LG(x) formula is a simplification. But is it exact? I haven't tried to derivate it from Elo formulas but it seems to me as an approximation. /Peter
This page took 0.03 seconds to execute
Last modified: Thu, 07 Jul 11 08:48:38 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.