What rating gap we need for 1920-0 FRC result?

[Uri Blass]

In rating list I see only games when the rating difference is small and I wonder what is the minimal difference that we need to get 1920-0 results from all FRC chess position(960 positions multiplied by 2 games).

I think that as long as we do not get this result it may be interesting to allow FRC matches between engines and I wonder if it is going to change the order of the engines because some engines are relatively better against weak engines or not.

[Ajedrecista]

Hello Uri:

The Elo rating system was not design for such huge differences. Anyway, if we take the following example without considering error bars:

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GAP > 400*log10(1919.5/0.5) ~ 1433.7 Elo
GAP > 1433 Elo

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If we consider error bars, I would bound µ + z·s < 1 (100%) and µ - z·s > 1919.5/1920 (µ being the expected score, s being the sample standard deviation and z being the z-score of the normal distribution). Then:

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µ + z·s < 1
µ - z·s > 1919.5/1920

Since both values are symmetric respect µ, then:
µ = (1 + 1919.5/1920)/2 = 1919.75/1920

GAP > 400*log10(1919.75/0.25) ~ 1554.1 Elo
GAP > 1554 Elo

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Lastly, if we remember the trinomial distribution:

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W := probability of a win.
D := probability of a draw.
L := probability of a lose.
Prob.(wins, draws, loses) = { [ (wins + draws + loses)! ] / [ (wins)! · (draws)! · (loses)! ] } · (W^wins) · (D^draws) · (L^loses)

Bounding again:

a)
W^1920 > 1919.5/1920
W > 10^{ [ log10( 1919.5 / 1920 ) ] / 1920 }
W > 1 - 1.36·10^(-7)

b)
L = 0 ; D = 1 - W
W^1919 · (1 - W) < 0.5/1920
The former bound of W is enough to satisfy this condition.

GAP > 400*log10[ (W + D/2) / (L + D/2) ] = 400*log{ [ ( W + (1 - W)/2 ) ] / [ (1 - W)/2 ] }
GAP > 400*log10[ (1 + W) / (1 - W) ] ~ 2867.4 Elo
GAP > 2867 Elo

As you see, the required gap with either method is in the order of the thousands of Elo, as expected, where the Elo rating system was not designed for. The tails of the distribution ought to be different from real, probably with gap < 1433 Elo still enough to clinch the 1920-0 perfect score.

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I only tried 1920-0 result. This could be applied to other extreme results such as 648-0 (Chess 324) or so. Please do not apply to not extreme results such as 2-0.

Regards from Spain.

Ajedrecista.

[Uri Blass]

The question is practical.

There are computer rating lists like
https://computerchess.org.uk/ccrl/404FRC/

What is the minimal gap that you need to get 1920-0 result.

I may do some similiar test with significantly faster games of fixed nodes per move but I need a simple tool to allow me to play 1920 games between opponents like Stockfish 16.1 4096nodes per move and Clover6.1 512 nodes per move when the idea is both players play all games white and black.