Author: Albert Silver
Date: 07:11:02 01/16/02
Go up one level in this thread
On January 16, 2002 at 08:09:58, Graham Laight wrote:
>On January 16, 2002 at 07:49:25, Albert Silver wrote:
>
>>On January 16, 2002 at 07:41:28, Graham Laight wrote:
>>
>>>It has occurred to me that it is wrong to evaluate a position in terms of
>>>relative pawns (the "de facto" standard - whereby an evaluation of 2 means that
>>>you're approximately the equivalent of 2 pawns ahead).
>>>
>>>This means that many aspects of evaluation have to be squeezed into a dimension
>>>which is not appropriate at all.
>>>
>>>A better way would be to evaluate "winning probability". If a position was a
>>>draw, the value would be 0.50 (or 50%). If the player should win 3 out of 4
>>>times, the eval should be 75%. If the player must win from here, then the
>>>evaluation should be 100%.
>>>
>>>It seems strange when you think about it that all programmers have chosen to
>>>adopt the traditional "pawn equivalence" standard.
>>>
>>>-g
>>
>>Not so strange considering that chess is a game of absolutes, whether we know
>>them or not. A positions is either a win (with best play), a draw, or a loss.
>>How are you going to estimate how many times a player _should_ win? -->
>>
>>Hmmm... Normally, I'd say John is going to win this, but having seen him drink 5
>>beers during lunch shortly before the game, I'd say he only has a 60% chance....
>>:-)
>>
>
>I don't say that the evaluations would be more "accurate" - but I do say that
>the number produced, accurate or not, would be on a more sensible scale.
>
>-g
In backgammon, evaluations are done according to probability since there is no
way to control the outcome of the dice, so you can only calculate your chances.
Chess is a different animal altogether. In chess, there is a theoretical
absolute answer. In other words, if I were to calculate very single possibility,
I'd know what the result would be with best play. Either the position is a win,
a draw or a loss. If chess were to be solved, then the evals would indeed only
have 3 evals possible. There is no probability though, unless you consider the
human chance to err, as I joked about above. Objectively, there is never a 70%
chance to win since best play must _always_ be assumed in order to play the best
moves. Suppose I were choosing two moves. In move A, the theoretical absolute
outcome is a win though the road is a very thin line (a long series of "only"
moves - anything else in the entire road would in fact lose), so it must be
considered the best move. Move B however presents more opportunities for the
opponent to make a fatal mistake and lead to easier wins, however if the
opponent doesn't fall for any of these traps move B is theoretically an absolute
loss (with best play). Move A has less "probability" to win, and Move B has
more. Which should it be?
Albert
This page took 0 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.