What is the biggest odds that engines will be able to beat top humans?

[Uri Blass]

at classical time control(assume for the discussion 90+30)

People from the leala team(including larry kaufman) believe that in the future leela will be able to beat top humans at classical time control with rook odds(of couse not today and today it is going to lose against them even with knight odds).
I believe most chess players believe that even with knight odds it is impossible.

I guess it is something between 2 random black pawns and a white knight and I think that if with some odds top humans get 50% at classical time control then they are going to learn from their experience and get more than 50% in a rematch.

[Ajedrecista]

Hello Uri:

at classical time control(assume for the discussion 90+30)

People from the leala team(including larry kaufman) believe that in the future leela will be able to beat top humans at classical time control with rook odds(of couse not today and today it is going to lose against them even with knight odds).
I believe most chess players believe that even with knight odds it is impossible.

I guess it is something between 2 random black pawns and a white knight and I think that if with some odds top humans get 50% at classical time control then they are going to learn from their experience and get more than 50% in a rematch.

Do you refer to win one single game in infinite trials; or to win a N-game match? If it is a single game, the negative binomial distribution comes to mind, where there are N-1 failures (draws and loses) after the only success (the win) at the end of the experiment.

Code: Select all

Negative binomial distribution.
X is counting n trials, given w=1 success.
W := probability of a win by the handicapped side (the engine).

Notation of the binomial coefficient: C(a , b) = a! / [ b! · (a - b)! ]

Probability mass function:
f(n;w,W) := Prob.(X=n) = C(n-1 , w-1) · W^w · (1-W)^(n-w)
f(n;1,W) = C(n-1 , 0) · W · (1-W)^(n-1)
f(n;1,W) = W · (1-W)^(n-1)

------------

I know that you are refering to the biggest odd to make a win possible in practice, and not how much long to wait for the win.

Regarding odds or handicaps, please take a look on this old thread where I copied Aquarium Demo's estimated Elo differences for a variety of material handicaps.

Then, you can estimate a probability of win based on Elo difference:

Code: Select all

W := win ratio ; D := draw ratio.
Elo_diff = 400·log10[(W + D/2)/(1 - W - D/2)]
W + D/2 = 1/[1 + 10^(-Elo_diff/400)]

Assuming D > 0:
W < 1/[1 + 10^(-Elo_diff/400)]

You wrote about 'something between 2 random black pawns and a white knight'. Looking to the table of the old thread, it says -500 Elo for two pawns (it is true that are white pawns and not black pawns) and -800 Elo for a white knight. The full rook odd goes up to -1250 Elo (a white rook odd).

Let us take something intermediate like -600 Elo to get W + D/2 ~ 3.07% and W = 2% because I want without entering in draw models, just to get a simple number for W such as 0.02 = 1/50. Then:

Code: Select all

Probability of win the last game and no one more after n games, assuming a probability of 2% of win a single game:
f(n;1,0.02) = 0.02 · (0.98)^(n-1)
F(n;1,0.02) := cumulative distributive function [sum of f(n;1,0.02) from n=1 onwards]

Probabilities rounded up to the nearest 0.01%:

 n     f(n;1,0.02)     F(n;1,0.02)
==================================
  1       2.00 %          2.00 %
 10       1.67 %         18.29 %
100       0.27 %         86.74 %

It looks more like a matter of 'when' rather than 'yes/no', but it is also true that Elo rating system was not designed for such large differences. Do you feel that you can win 2% of the games (1 out of 50) and draw 2.13% of games (circa 1 out of 47) against someone 600 Elo stronger than you? Me not.

Fatigue of the human player ought to play a role, so it is better to play no more than one or two games per day with that time control; or play against a pool of human players of the same strength (like going from single CPU to multiple CPU in computer chess to play more games in less time). The same with power outages: imagine that the computer is finally winning and a blackout happens... I guess that the game could be resumed later in this case and count as a legitimate win if the engine converts the advantage.

The negative binomial distribution assumes independent trials, so not learning involved. This exercise was just a simple model to obtain a rough estimate.

------------

If you thought about a N-game match, I guess that some calculations with the trinomial distribution can be made, then restricting to wins > loses for the engine playing the weak side.

Regards from Spain.

Ajedrecista.

[Dann Corbit]

At some point, it may be possible for the computer to win with 8 pawns and a king. We would all cringe, but I think Philodor would like it.

[Ajedrecista]

Hello Dann:

At some point, it may be possible for the computer to win with 8 pawns and a king. We would all cringe, but I think Philodor would like it.

I think it is too much against a top human or even a normal human: just sac/trade a piece/pawn for a pawn before the human flags and the computer will not have enough material to win any game. Other than that, it would be a good experiment with a random mover playing the normal side and top-notch computer playing the pieceless side with the TC proposed by Uri. How many games until the top-notch computer wins? Probably very few, but the top-notch computer does not know that the other side is a random mover and might start offering easy pawn trades, thus diminishing its winning chances. It is like when you need points from faults in snooker: is easier and common practice to get them keeping the table crowded (more balls on).

Regarding Philidor's famous quote, it began here (c'eſt celle de bien jouer les Pions: Ils ſont l'ame des Echècs). No less than in 1749!

Regards from Spain.

Ajedrecista.

[Werewolf]

Hello Dann:

At some point, it may be possible for the computer to win with 8 pawns and a king. We would all cringe, but I think Philodor would like it.

I think it is too much against a top human or even a normal human: just sac/trade a piece/pawn for a pawn before the human flags and the computer will not have enough material to win any game. Other than that, it would be a good experiment with a random mover playing the normal side and top-notch computer playing the pieceless side with the TC proposed by Uri. How many games until the top-notch computer wins? Probably very few, but the top-notch computer does not know that the other side is a random mover and might start offering easy pawn trades, thus diminishing its winning chances. It is like when you need points from faults in snooker: is easier and common practice to get them keeping the table crowded (more balls on).

Regarding Philidor's famous quote, it began here (c'eſt celle de bien jouer les Pions: Ils ſont l'ame des Echècs). No less than in 1749!

Regards from Spain.

Ajedrecista.

Surely he was joking.
Anything more than a knight and the computer is toast no matter how strong it is.

[Ajedrecista]

Hello:

Surely he was joking.
Anything more than a knight and the computer is toast no matter how strong it is.

That was a joke, for sure; but far of useless, let me search, find, learn and share about the origin of one of the most famous and more repeated quotes about chess. It was a positive approach.

The idea of the random mover versus a top-notch computer still holds if anyone is interested. How fast the random mover would waste its initial, huge advantage? Are we speaking of some games? Are we speaking of some moves within the very first game?

Regards from Spain.

Ajedrecista.

[acase]

Has anyone ever tried an old engine against like perhaps Fritz 5.32 against Stockfish at Knight odds and long time controls?