Single Comment

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.