Author: Peter Fendrich
Date: 16:31:37 12/20/02
Go up one level in this thread
On December 20, 2002 at 14:20:22, Rémi Coulom wrote: >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. No my program will not give the same result otherwise it would be seriously wrong. Even draws contains information. Look at the extreme with a match result 10-0 and x draws where x->Inifinite. Limes such a function give the probability of 50%. In practice the differences are small but they are still there. I will read your PDF document sometime during the weekend. Maybe we are talking about different things. Peter >>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
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.