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.

1 Answer 1


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.

  • Yeah, 0.0 is about 1000 moves behind the horizon :-) (I still got a position which is +99 on LiChess, so you are not even close :P) Oct 20, 2022 at 18:31
  • 1
    There will shortly be many queens on the board, and it's ""dead as it can be"?? Wait, is the diagram upside down???
    – bof
    Oct 20, 2022 at 22:50
  • @bof yes, it's from Black's point of view
    – Hauptideal
    Oct 20, 2022 at 23:21

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