# Computer Chess Club Archives

## Messages

### Subject: Re: Proving something is better

Author: Peter Fendrich

Date: 16:31:37 12/20/02

Go up one level in this thread

```On December 20, 2002 at 14:20:22, Rémi Coulom 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.
>>>>/Peter
>>>
>>>Hi Peter,
>>>
>>>If you had not noticed it, you can take a look at a similar program I have
>>>implemented:
>>>http://remi.coulom.free.fr/WhoIsBest.zip
>>>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.
>>>
>>>Rémi
>>
>>Hi,
>>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.
>
>No, results are the same. You can try with your own program to test this: 10
>wins and 0 losses should produce the same probability whatever the number of
>draws.

No my program will not give the same result otherwise it would be seriously
wrong. Even draws contains information.
Look at the extreme with a match result 10-0 and x draws where x->Inifinite.
Limes such a function give the probability of 50%.
In practice the differences are small but they are still there.

I will read your PDF document sometime during the weekend. Maybe we are talking
Peter

>>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
>>trinomial.
>
>If you add 0.5 to both n0 and n1 for draws, then, yes, the result will be
>inaccurate. In order to obtain the exact value, you have to simply ignore the
>number of draws, not add them to n0 and n1. That is
>Trinomial(n1,n0,n0.5) == Binomial(n1,n0)
>
>>
>>/Peter

```