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Subject: Re: Pairing question about Ikarus

Author: Uri Blass

Date: 06:36:09 02/22/03

Go up one level in this thread


On February 22, 2003 at 09:16:07, Uri Blass wrote:

>On February 22, 2003 at 08:54:30, Rolf Tueschen wrote:
>
>>I read 6: Ikarus     3.5 /  6    2b= 13w+  1w=  4b=  3w- 14b+.
>>
>>So that means that Ikarus although playing the placed 1, 2, 3, 4 progs, it could
>>get full points against the last and pre-last. Placed om 14 and 13.
>>Is this ok? Something seems to be wrong or biased. Point is that a game against
>>14 is a SURE win. That is as if a top program after a loss or two draws got a
>>point for free. Note Ikarus had 2,5 pts before playing Matador with 0.5 pts.
>>
>>Could some expert explain why such things are still possible?
>>
>>Rolf Tueschen
>
>Can you suggest another pairing without repeating the sam game twice?
>
>Matador already played against all the programs that scored less than 50%
>It had to play against a program that scored at least 50%.
>
>Uri

I can add that Matador drew with a program that drew with a program
that drew with a program that drew with a program
that drew with a program that won the leader Fritz.

I have the following results based on the table.

Patzer-Matador    (round 1)      1/2-1/2
Patzer-Comet      (round 6)      1/2-1/2
Diep-Comet        (round 3)      1/2-1/2
Anaconda-Diep     (round 4)      1/2-1/2
Anaconda-Shredder (round 6)      1/2-1/2
Shredder-Fritz    (round 5)      1-0

Note that based on this record I have a better result against Kasparov
so it does not prove that Matador has practical chances against Fritz but
I think that the number of games are too small even to be sure that Fritz has a
sure win against Matador so I see no basis to the claim that a game against 14
is a sure win and all the beginning of the post has nothing to do with it.

Even in case that a program lose all the first 5 games in a tournament there is
no evidence for the claim that it has no chances.

Uri



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