Hypothetical absolute evaluation and convergence properties

Let's suppose that the absolute positional evaluation of moves in all chess positions is known. The evaluation of kth move in a chess game G with n plies will be denoted by ev(k,G). So, ev(k,G)=d if it's the dth best move in the position after k-1 plies etc. Consider also the average / mean value

M(G)=(ev(1,G)+...ev(n,G))/n

Conjecture 1 (well known). The best game with M=1 (that is, when only the best moves are played) is a draw.

Conjecture 2. max(M(G))<1.5 for all draw games where white plays always the best move. It means that, if black tries to make a draw playing (in average) only 2nd best moves, white will win.

Problem 3. Find better estimations for max(M(G)) for all draw games where white plays always the best move.

Catalyst: yeah, good criticism. Instead of absolute evaluations, we should take evaluations with high enough but finite horizon (let's say, 25 plies).

Here's the 1st test of my conjecture 2 with official Rybka evaluations.  I choose always the 1st best move for white and the 2nd best move for black unless the 1st best move is forced. In the latter case, I'll compensate it at the next move.

1. e4 e5 (+0.11)
2. Nf3 Nf6 (+0.18)
3. Nxe5 Qe7 (+0.41)
4. d4 Nc6 (+0.41)

The 2nd test with QP

1. d4 d5 (+0.15)
2. c4 dxc4 (+0.18)
3. Nf3 a6 (+0.26)
4. e3 b5 (+0.26)
5. a4 c6 (+0.56)

3rd test with 1.Nf3

1. Nf3 d5 (+0.11)
2. d4 e6 (+0.15)
3. g3 c5 (+0.15)
4. Bg2 Nc6 (+0.18)
5. O-O cxd4 (+0.22)
6. Nxd4 Nf6 (+0.22)
7. c4 Qb6 (+0.37)

M(G) = 1.5 in all this games and the score (numerical evaluation) is growing quite rapidly.

Catalyst: all right, let me reformulate it in the following way. Let's suppose that the Almighty knows the exact value (W=win, L=lose, D=draw) of each move in every position.

By definition, the best move for white in a position P is a move minimizing the total number of draw lines for black. Subtler variants of this definition can be considered as well. Let's suppose that white always makes best moves starting from the initial position. Denote T the total number of draw lines for black and N the average number of moves in all draw lines.

Conjecture 2'. log_2(T) / N < 1

It means that black has on average no more than 2 moves to achieve a draw. This conjecture might be too strong.

Problem 3'. Find estimations for the average number A of moves for black to achieve a draw (supposing that white plays always best moves).

By definition, A^N=T or log_A(T) / N = 1.

Despite your reluctance to admit it, it can be easily estimated using numerical evaluations of any good chess engine. On average, black shouldn't lose more than 0.05 per move, otherwise the score will be higher than +2 after 40 moves which is almost certainly a win for white.

My first test is to count the total number of lines evaluated by Rybka within +0.20 after 3 moves. We can suppose that 1.e4 is the best move for white.

There are 3 answers within +0.20:

Sicilian: 1...c5; KP 1...e5 and French 1...e6

Let's suppose that 2.Nf3 is the best 2nd move in Sicilian and KP, and 2.d4 is the best 2nd move in French.

... to be continued

When it reads 0.05 on move 13 for example that's not to be taken as an absolute expression of the value of the position.  It's just the computer's best guess at the time.

Of course it's a very strong guess, but it will deviate more than 0.05 quite often.

My 2 cents.

waffllemaster: I know but we can suppose that a good engine will give reasonably good approximations.

Let's continiue to count.

Sicilian 1...e4 c5 2.Nf3, there are 4 answers within +0.20:

2...d6 (modern Sicilian), 2...Nc6 (old Sicilian), 2...e6 (Franco-Sicilian) and 2...g6 (hyperaccelerated dragon)

Modern Sicilian 1...e4 c5 2.Nf3 d6 3.d4

2 variations within +0.20: 3...cxd4 and 3...Nf6

Old Sicilian, Franco-Sicilian and Hyperaccelerated Dragon

1 acceptable variation 3...cxd4

So, we have 5 admissible lines in Sicilian after 3 moves.

whoaa!! you guys sound like geniuses at work! Can some1 be more ...let's just say, "explaining an idiot"-ish??

How many formulas do U need to accept it as a scientific theory?

Thanx, any concrete questions?

Well, it's reasonable to suppose that if white and black both play perfectly, it will lead to a draw. Now let only white play perfectly. The question is: how good black should play in order to achieve a draw? For this purpose, I introduced the threshold average number A of moves at each ply for black still leading to a draw. The goal is to make estimations for A. I thought before that it should be between 1 and 2. Now, after tests, I'm rather inclined to think that it's between 2 and 3.

In particular, it means that while playing against a perfect opponent, U should virtually always stay in top 3 moves at each ply.

I don't think Rybka is the best metric for measuring these positions. While it's arguably the strongest engine on the market, the past few iterations have been designed with improving its playing strength against the current version and other computers. The way they accomplish this is by trying different algorithms and weights on positions to increase the accuracy and speed of calculation.

So, even though Rybka 4 can actually be said to play positionally, the entire data set is biased based on the specific principles that are programmed into the engine in order to increase play strength against other engines. This works fairly well in middlegame play but as stated above, it's nigh impossible to handle openings and it still tends to suck at complicated endgames.

chess is always depicted as a 2 dimensional game tree, have you tried labelling your canidate moves X, Y and Z etc ?

Really? It seems to me that if you consider the objective value every move allowing 1+ drawing line is equally strong. If instead you consider the practical value, the defeinition is not always true. Let's say, W is += (but B draws with best play) in a position where he can force a perpetual. According to this definition forcing the perpetual is the best move since B can draw only by accepting the perpetual(1 drawing line), while pushing the += advantage for dozen of moves allows B to draw with very precise play precise play in more variations. Clearly the definition doesn't work. 

Thanx for this pertinent remark! Still, from the absolute viewpoint (let's say of Almighty God) forcing perpetual, in a position where black can achieve draw anyway, is indeed the best move! Black should find that unique draw line. In all other cases, they have more possibilities.

Hmmm... giving black the draw immediately in a way that cannot be missed is better than pressing forward while having the better practical chances? From an absolute point of view, the two options are =, in practice the second is better. In no way the first. My point is that "best move" is a concept often including practical considerations. Attempts to describe it with such a simple definition are bounded to fail. 

Well, the definition can be modified in many ways. At this moment, I work on other things.