Computer Chess Club Archives




Subject: Re: Proving something is better

Author: Peter Fendrich

Date: 16:56:58 12/20/02

Go up one level in this thread

On December 20, 2002 at 19:30:38, Uri Blass wrote:

>On December 20, 2002 at 19:07:04, Peter Fendrich wrote:
>>On December 20, 2002 at 12:16:25, Uri Blass 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.
>>>>>Hi Peter,
>>>>>If you had not noticed it, you can take a look at a similar program I have
>>>>>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.
>>>>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.
>>>>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
>>>If the target is only to find which programs is better we can throw draws.
>>>You can imagine the following game chessa:
>>>One game of chessa includes at least one game of chess.
>>>chessa is finished only when a chess game is finished in a win.
>>>if a chess game that is played as part of chessa is finished in a draw then
>>>chessa continues and the sides play chess with opposite colors.
>>>By these rules in both cases the winner won 10 games of chessa with no draws
>>>(draw in chessa cannot happen).
>>In that case you don't need anything more than the result.
>>What I'm doing is producing a statment like:
>>A is better than B with the probability of x%.
>>The 10-0 result will raise x very high but the 55-45 result will lower the
>>probability even if A is still regarded as the best.
>if the 55-45 is result of 90 draws then 55-45 give the same probability that the
>winner is better as 10-0.
>The draws are only relevant for estimate of the difference in rating but not for
>deciding about the better player.

That is essentially the same thing. Different estimates of rating gives
different probabilities of A beating B. The both are closely related.
If the ratings are changed the probabilties should be changed.

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