Author: Dieter Buerssner
Date: 16:04:39 08/04/03
Go up one level in this thread
On August 04, 2003 at 18:36:27, Dann Corbit wrote:
>If a player rated 500 points below a machine in real strength plays enough
>games, he will win one of them.
If we can trust the statistics behind the ELO system. I am not sure, if it is
true for bigger rating differences (in the tails of the distributions). Also, to
my knowledge, the statistics does not say anything about the distribution of
wins/draws/losses. If you are expected to get 1% of the points, you could do it
by just drawing every 50th game and don't win any game.
>It may take quite a lot of games. But this is not an opinion, it is a fact.
I think, it is not a "pure" fact. It (the percentage of points, you reach, not
necessarily the number of wins) might be a fact, when we take the statistics
assumed behind the Elo system as a fact. But still my above point about
distribution of wins and draws would be valid.
>This assumes simply playing games based on strength. If the human is clever at
>learning computer weakesses or playing anticomputer chess, it may tip the
>balance somewhat.
>
>Win expectency for a difference of 0 points is 0.5
>Win expectency for a difference of 100 points is 0.359935
>Win expectency for a difference of 200 points is 0.240253
>Win expectency for a difference of 300 points is 0.15098
>Win expectency for a difference of 400 points is 0.0909091
>Win expectency for a difference of 500 points is 0.0532402
>Win expectency for a difference of 600 points is 0.0306534
>Win expectency for a difference of 700 points is 0.0174721
>Win expectency for a difference of 800 points is 0.00990099
>Win expectency for a difference of 900 points is 0.00559197
>Win expectency for a difference of 1000 points is 0.00315231
I am no native English speaker. Win expectency sounds a bit misleading to my
ears, because it does not really mean, that you win that many games.
BTW. I calculate slightly different numbers. The correct formula is:
p = 0.5 - 0.5*erf(rating_difference/400.)
With:
x
-
2 | | 2
erf(x) = -------- | exp( - t ) dt.
sqrt(pi) | |
-
0
Or for the numerically interested people, for rating_difference big, the form
p = 0.5*erfc((rating_difference)/400.)
would be better, when a "good" erfc-function is available (erfc(x) =: 1-erf(x)).
It would avoid to calculate the difference of numbers of equal size.
Regards,
Dieter
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