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H.G.Muller wrote on Sun, Apr 27, 2008 07:27 PM UTC:

I missed this one:

Hans Aberg:
| You say that a pawn ahead is always a win with only some 
| exceptions. So is a rook. So set values of these the same - 
| does not work to predict generic rook against pawn end-games.

Why would you want to set the values the same? Because both a Pawn and a
Rook advantage in the end-game is 100% won?

In the first place they aren't. An extra Pawn will win 90%, an extra Rook
99.9-100%. But, for the sake of argument, change the example to an extra
Rook or an extra Queen, and say that both now win 100%. Does that prove
the piece value of a Rook and a Queen are the same?

Of course NOT. 100% is far outside the linear regime, and obviously the
score becomes meaningless there. I never take into account the results of
thest that are more than 70% biased. I don't have to, as a full Pawn
advantage in the opening in the linear range only adds 12%, (so I get a
62% score for pure Pawn odds), and in principle I can always balance the
material such that I get scores below that by adding a Pawn.

So what would NOT work, and what I thus do NOT do, is first giving full
Archbishop odds, and then giving Chancellor odds. Because both these odds
would lead to ~95% scores, and trying to peel a piece value out of the
fact if they are 95% or 96% falls far below the resolution (determined by
statistical noise), and would be extremely sensitive to the exact initial
position (as the 5% non-wins would be due to tactical freak accidents like
quick mates or perpetuals).

What I do is play C vs A or CC vs AA (nearly equal) to see who wins. Or Q
vs A+P and Q vs C+P (to see who does better). Or A vs R+R or A vs R+B or A
vs R+N+P, to see if A typically beats those combinations (which it does,
but never by more than 60% or so, or I would give stronger material to the
weak side).

But if Q beats A by a larger score (say 59%) than A+P beats Q (say 53%), I
consider it evidence that Q-A > 0.5 Pawn. Because I know (from observation)
that adding another 50cP advantage to the imbalance in favor of the weak
side (like replacing an unpaired Bishop by a Knight) will in general
simply adjust the existing score by 6% at the expense of the side who you
give the weakness. And that duplicating a weakness (like giving one side
two unpaired Bishops in stead of the other's two Knights) will simply
double the score excess to 12%, the same as for Pawn ods, and can be
exactly compensated by giving the Knight-side an extra Pawn. So Q+N will
still beat A+B(unpaired), consistent with Q-A in the point system being
more poinst than the B-N difference of 0.5 Pawn, and A+P-Q being less.

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Comments

Aberg variation of Capablanca's Chess. Different setup and castling rules. (10x8, Cells: 80)

💡📝Hans Aberg wrote on Sat, May 3, 2008 04:43 PM UTC:

H.G.Muller:
| But the point is that this does not alter the piece values.

Right, though that might be just a preferred way to structure theory because it suits human thinking. Essentially, define contexts, and attach values to them. First define piece values in nutralk settings. Then observe that the bishop pair gets an added value. Then try to figure out values for good and poor bishops. And so on. By contrast, computers tend to be very poor at handling such contexts, so other methods might be suitable for programs.

Derek Nalls wrote on Sat, May 3, 2008 03:26 PM UTC:

Muller:

You have my best regards toward your worthwhile effort to publish your
empirical, statistical method for obtaining the material values of pieces
in the ICGA Journal.  My assessment is that it will surely be a much
better paper than the junk [name removed] published in the same journal
regarding piece values.

[The above has been edited to remove a name and/or site reference. It is
the policy of cv.org to avoid mention of that particular name and site to
remove any threat of lawsuits. Sorry to have to do that, but we must
protect ourselves. -D. Howe]

H. G. Muller wrote on Sat, May 3, 2008 03:22 PM UTC:

Sure, this is what people do and have done for ages. It is well known that the advantage of having the move is worth 1/6 of a Pawn, (corresponding in normal Chess to a white score of 53-54%) and that, by inference, wasting a full move is equivalent to 1/3 of a Pawn.

But the point is that this does not alter the piece values. It just adds to them, like every positional advantage adds to them. In my test the advantage of having the lead move is neutralized by playing every position both with white to move and black to move.

💡📝Hans Aberg wrote on Sat, May 3, 2008 11:50 AM UTC:

H.G.Muller:
| As piece values are only useful as strategic guidelines for quiet
| positions, they cannot be sensitive to who has the move.

At the beginning of the game, white is thought to have a slight advantage, and the first task of black will be attempting to neutralize that. And it might be possible set a picee value to that positional advantage, just as when reasoning in terms of getting positional compensation for a sacrifice. Somewhat less han a pawn, perhaps. If one knows about the black/white winning statistics, one might be able to set a value on it that way. It may not be usable for a computer program as it does not change sides, but only computes the relative values of moves.

H. G. Muller wrote on Sat, May 3, 2008 10:34 AM UTC:

As piece values are only useful as strategic guidelines for quiet positions, they cannot be sensitive to who has the move. A position where it matters who has the move is by definition nont quiet, as one ply later that characteristic will have essentially changed. So at the level of piece-value strategies, Chess is a perfectly symmetric game.

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