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Subject: Re: Proving something is better

Author: Uri Blass

Date: 17:01:17 12/20/02

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On December 20, 2002 at 19:56:58, Peter Fendrich wrote:

>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.


It is not the same suppose player A beat B 1000-0 with 999999000 draws

you are going to have no doubt that A is better but if the result is
500000500-499999500 with no draw then it is clearly possible that the results
are random.

If you throw a fair coin 1000000000 times then in most of the cases the
difference between the number of heads and the number of tails is going to be
more than 1000.


if you

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