Author: Stefan Meyer-Kahlen
Date: 04:55:59 05/23/99
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On May 22, 1999 at 14:22:26, Peter Fendrich wrote: >Sorry, I forgot to answear the 95% question... :) > >To estimate the standard deviation of the Performance Rating there are >lots of options, giving about the same result. You need the individual results >to know the the error margin and that's what this is all about. There is a >difference between a result of only wins and losses compared to a result with a >lot of draws. The SSDF method is best described as: > (1) s=SQRT(W(1-m)**2 + D(0.5-m)**2 + L(0-m)**2/(n-1)) > (2) A=1.96 * s/SQRT(n) > > where s is estimated standard deviation > n is number of games > m is score/n > W,D,L is number of Wins, Draws and Losts respectivelly > A is the margin of error (for score, not rating points) > 1.96 is fethed from the Normal Distribution table to get > 95% reliability > SQRT is the square root > ** is used as 'the power of' > > Now we have an 95% interval of *scores* from m-A to m+A > > Compute the ratings for each m-A and m+A and here we go! > These ratings are the end points in the interval. > >//Peter Yes, that's what I was looking for, thanks. I think (1) should be (1) s=SQRT((W(1-m)**2 + D(0.5-m)**2 + L(0-m)**2)/(n-1)) right? Let's assume A and B play 3 games, A wins 2, 1 is a draw. If I use your fromula I get S = 0.2886 and A = 0,3266 If a calculate m-A I get 0.5067, which is > 0.5! There must be somethin wrong, because I don't think you can say that A is better than B with 95% prob. just after 3 games. Stefan
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