Author: Rémi Coulom
Date: 11:20:22 12/20/02
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On December 20, 2002 at 11:03:14, Peter Fendrich wrote: >On December 20, 2002 at 04:10:35, Rémi Coulom wrote: > >>On December 19, 2002 at 19:28:01, Peter Fendrich wrote: >>> >>>I did, some 15-20 years ago, in the Swedish "PLY" a couple of articles that >>>later became the basics for the SSDF testing. >>>A year or so ago you posted a question about how to interpret results with very >>>few games. In a another thread I posted a new theory for this as an answer >>>"Match results - a complete(!) theory (long)". >>>I also made a program to use for this that can be found at Dann's ftp site. >>>/Peter >> >>Hi Peter, >> >>If you had not noticed it, you can take a look at a similar program I have >>implemented: >>http://remi.coulom.free.fr/WhoIsBest.zip >>Basically, I started with the same theory as you did, but I went a bit farther >>in the calculations. In particular, I proved that the result does not depend on >>the number of draws, which is intuitively obvious once you really think about >>it. I also found a more efficient way to estimate the result. I checked the >>results of my program against yours and found that they agree. >> >>Rémi > >Hi, >For me it's not so obvious that you can through the draws out. >I just took a short look at your paper and maybe I misunderstood some of it. > >Take this example: A wins to B by 10-0 >Compared with: A wins to B by 10-0 and with additional 90 draws. >Not counting the draws will get erronous results. No, results are the same. You can try with your own program to test this: 10 wins and 0 losses should produce the same probability whatever the number of draws. > >The results between our programs shouldn't agree, I think, because I heavily >relies on the trinomial distribution (win/draw/lose). One can use the binomial >function (win/lose) and add 0.5 to both n1 and n0 for draws. That will probably >give a fairly good approximate value but the only correct distribution is the >trinomial. If you add 0.5 to both n0 and n1 for draws, then, yes, the result will be inaccurate. In order to obtain the exact value, you have to simply ignore the number of draws, not add them to n0 and n1. That is Trinomial(n1,n0,n0.5) == Binomial(n1,n0) > >/Peter
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