Computer Chess Club Archives




Subject: Re: Proving something is better

Author: Uri Blass

Date: 16:30:38 12/20/02

Go up one level in this thread

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.


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