2

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

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.

3
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.