Author: Albert Silver
Date: 06:12:06 01/26/02
Go up one level in this thread
On January 26, 2002 at 09:07:51, Albert Silver wrote:
>On January 26, 2002 at 08:54:10, Sune Fischer wrote:
>
>>On January 26, 2002 at 08:41:09, Albert Silver wrote:
>>
>>>On January 26, 2002 at 06:50:45, Sune Fischer wrote:
>>>
>>>>On January 25, 2002 at 17:25:39, Dann Corbit wrote:
>>>>
>>>>>On January 25, 2002 at 17:07:46, Albert Silver wrote:
>>>>>
>>>>>>
>>>>>>>>3.If the rating of perfect player is say x ;what would be the rating of
>>>>>>>>the stongest computer player ever(that is the best chessprogram that can be
>>>>>>>>ever contructed useing computer technology) .It would be x-?.Or would it be x?
>>>>>>>
>>>>>>>It would be zero, unless it was perfect also. The perfect player would win
>>>>>>>every game and get all the ELO points. The imperfect player would lose all the
>>>>>>>games and get an ELO of zero.
>>>>>>
>>>>>>Maybe. The imperfect player may not find all the best moves, but that doesn't
>>>>>>mean that all the moves it plays are losing.
>>>>>
>>>>>Being able to see 5900 plies ahead means that any microscopic slip along the way
>>>>>by the opponent will bring a loss if it can bring a loss. I hypothesize that a
>>>>>2800 player will score zero points against a perfect player. Playing around the
>>>>>clock, perhaps once in a trillion centuries, the imperfect player might gain 1/2
>>>>>of a point. Once in a trillion millenia maybe a full point. But it won't be
>>>>>enough to pull his ELO above zero.
>>>>
>>>>Realisticly a 2800 player probably has a branchfactor of no more than 2, ie. he
>>>>is able to always choose the best or second best move (on average).
>>>>If the average game lasts 100 moves, then that is still 10^30 plausible games of
>>>>which only a handfull will be good enough against *perfect* play.
>>>>Poor odds I agree with you :)
>>>
>>>You're presuming that anything other than one move, the best move, will lose
>>>forcibly to best play. I believe that more than one move is available to a
>>>non-loss thus perfect play would be often a flip of the coin between a few
>>>(perhaps three as I hypothesized in another post in the thread) moves. I have
>>>seen no evidence to suggest there is only one path to a non-loss and that a
>>>single path of perfect play is needed to avoid it. Everything we know whether
>>>from personal research or from the current tablebases suggests there are several
>>>paths. If this were accepted to be true, the question would be whether the 2800
>>>player is incapable of hitting on _one_ of these non-losing moves (according to
>>>perfect play).
>>>
>>> Albert
>>
>>You could interpet in an similar way; there is a 50% chance of the 2800 chooses
>>a move that is *good enough*.
>>It was just an estimate, probably way off :)
>>
>>Suppose that a *correct* move is done with 95% certainty (on average) and that
>>the average game length is only 60 moves, then he has a 0.95^60 = 4.6% chance of
>>a draw!
>>
>>This is perhaps more realistic?
>>
>>-S.
>
>Well, a few things come to mind. One is that there would be more than one
>correct move to hit on. Second that I wasn't aware that his chances changed with
>each move, so I don't think that the longer the game the worse his chances. Give
>a 2800 player a dead equal dry game and I don't think he will suddenly be in
>danger of losing just because it can take 40 moves to trade off the pieces and
>pawns and play the endgame to the end. There is more to chess than probability.
>
Just as an example, let's take the following classic chess composition:
[D]7K/8/k1P5/7p/8/8/8/8 w - -
You can throw as many stats you wish, but I will tell you flat out that the
chances of Kasparov not drawing this are 0.
Albert
>
>>
>>
>>>>
>>>>Maybe in the coming years we will see exactly how close the 2800 is to perfect
>>>>play, if it is possible for computers to crush 2800 guys the same way 2800
>>>>players crush 2400 players then it seems there is still some way to go.
>>>>
>>>>-S.
>>>>
>>>>>I further hypothesize that every chess game ever played to date has a mistake in
>>>>>it (and by both parties if at least 2 ply are completed). Does not mean that we
>>>>>can find it.
>>>>>
>>>>>Of course, if it turns out that 1. d4 always wins, and there are games that go:
>>>>>1. d4 {black resigns} 1-0
>>>>>Then I'm wrong.
>>>>>;-)
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