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Subject: Re: Proving something is better

Author: Peter Fendrich

Date: 16:07:04 12/20/02

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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.

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