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Since the question in this form is somewhat ill-defined, I elaborate on it so it can be solved by experiment:

What is the maximum centipawn value you can come up with so that the weaker side holds a draw?

Assume two very strong deterministic (i.e., you only need one game) computers play out a position which is very bad (with value c centipawns) for one side, but the weaker side still draws. Say, two Leela copies. (Fortresses would be the first thing I'd try. But they neither may be solvable by tablebases nor a draw by 50 move/3 repetitions inside the horizon, otherwise it's an immediate 0.0!)

P.S. For a "normal" position I expect a guaranteed win somewhere in the evaluation range +2 to +3.

Hauke Reddmann
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1 Answers1

5

This position looks familiar, doesn't it? ;)

As you can check on lichess, the evaluation is nearly +8, and yet the position is as dead as it can be...

[Title "Diagram from Black's perspective"]
[FEN "kR6/p1p1p1p1/P1P1P1P1/p1p1p1p1/8/8/PKP1P1P1/8 b - - 0 1"] 
[startlfipped ""]

There are countless such examples, check this and this.

If you want to know the practical winning centipawn threshold (arising in realistic games), my guess would be something near 1.5.

SecretAgentMan
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Hauptideal
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