Author: Uri Blass
Date: 10:29:01 02/02/01
Go up one level in this thread
On February 02, 2001 at 10:47:38, Andrew Dados wrote: >On February 02, 2001 at 06:59:06, Uri Blass wrote: > >>On February 02, 2001 at 02:55:44, Andrew Dados wrote: >> >>>On February 01, 2001 at 10:21:43, Uri Blass wrote: >>> >>>>On February 01, 2001 at 07:39:11, Leen Ammeraal wrote: >>>> >>>>>On January 31, 2001 at 20:17:17, Bruce Moreland wrote: >>>>> >>>>>>I expressed very forcefully that a 10-0 result was more valid than a 60-40 >>>>>>result. >>>>>> >>>>>>I've done some experimental tests and it appears that I'm wrong. >>>>>> >>>>>>I have no idea why. >>>>>> >>>>>>bruce >>>>> >>>>>According to the little Windows app Whoisbetter by Steve Maugham, >>>>>the certainty of being better for the winner is >>>>>97% with the score 60 - 40, and >>>>>99% with the score 7 - 0 >>>> >>>>It is clearly wrong. >>>>The probablities that you talk about are not the probability that the winner is >>>>better. >>>> >>>>We cannot know the probability that the winner is better unless we have more >>>>knowledge. >>> >>>We can calculate this and if we disregard draws the chance that winner of 60-40 >>>is better is indeed 97.4%. >> >> >>Even without draws we cannot calculate it because we need information about the >>apriori distribution of the probability of the better player to win. >> >>If we assume that it is 0.51 then we get 0.51^20/(0.51^20+0.49^20) >> >>If we assume that it is 0.6 we get 0.6^20/(0.6^20+0.4^20) >> >>If we assume that the probability of the better player to win has 50% >>probability to be 51% and 50% probability to be 60% then we get something else. >> >>Uri > >Hell(o) Uri, > >With all respect I think you're wrong. > >You can enumerate through all possible rating differences and scores. >(like, for score 60-40 chance of it for rating_diff 0 is d0, for rating_diff 1 >is d1 etc...) I agree that we can calculate it but it is not the probability that I mean to. I am interested to know p(A is better than B after knowing that A won 60-40) You calculate p(the result is 60-40 when you know that A is d0 points better than B). practically the data that I have is the 60-40 and I do not have the data which program is better. I need to do some guess about the distribution of the probability of A to win B that I call p or in other words the distribution of d that is the rating difference between the players. This guess is called the aprior distibution of p and you can translate is to an aprior distribution of d. You can get 97.4% for some d0 but if you change d0 the 97.4% will be changed. >You get the function P(score,rating_diff). Factoring that through all >rating_diff in some range gives you answer what is the chance that rating_diff >lies in that range (you need to normalize P first). I agree that you can use the aprior disribution of rating diff to calculate the probability that the difference in rating is at some range when you get the result when a private case is the case when you calculate the probability that the winner is better but you need an aprior distribution of d. I did not read your assumption about the aprior distribution of d when you claimed that the probability that the winner is better given the result 60-40 is 0.974. Practically your assumptions may be different in different cases. if you do a small change in a chess program(changing the value of pawn by 1%) your aprior distribution will be that d is small and you can be sure that it is not more than 10 elo. If you do a big change and add evaluation code or add search rules you can have assumtions that d can be big when it is possible that the new program is weaker by 200 elo because of some logical bug. The distribution will not be symmetric in this case and it also mean that 60-40 for A does not give the same confidence as 60-40 for B. Uri
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