go zero sum symmetric

[phillip1882]

is go a symmetric zero sum game? it would seem so at first. if black passes his first move, whatever strategy black plays that is optimal, would also be the same strategy for white. and any gain by one player is an equal loss by the other. if true, correct komi for white, then, would seem to be at most 1 point in area scoring, given that black has the first move and thus 1 extra stone. however, given that mirror go can be beaten even when the player isn't sloppy about it, suggests to me that go is not perfectly symmetric, there are points throughout the game where, all else being equal, a particular move for one player is not equivalent for the other. thoughts?

[tentano]

This question can't be answered properly without solving go.

If the solution is "black wins", then it's not symmetric.

If the solution is "white wins", then it's not symmetric.

Experience has taught us that black is favored because of the first move advantage, but that's only when go is played by imperfect players.

If perfectly solving go gives us a draw (without komi) then you'd be right.

[Bill Spight]

A zero sum game is not one where the value of the game is zero. It is a game in which the payoffs of the players add up to zero. Go is a zero sum game. If one player wins by X points, the other player loses by X points. X - X = 0. Or, since winning or losing are all that matters in most forms of go, if one player wins the other loses.

Go is not symmetric. The players do not have the same strategies.

[Pippen]

Let me try to prove already today and long before Go is solved that - in case of perfect play - Black cannot lose (and therefore White cannot win): Either Black can win the game or not. If not, it means White can win or force a tie, so Black just passes with his first move and therefore will be in White's "shoes" to either win or force a draw. So in the worst case Black can force a draw, Black cannot lose. From that it follows: Either Black can win or a draw is inevitable. Is there any way to prove one of those alternatives and what's your guess?

p.s. It's amazing that we can prove things with so little info, weirdly in Chess - a game inferior to Go (in terms of complexity) - we cannot prove sh*t^^.

[Krama]

Let me try to prove already today and long before Go is solved that - in case of perfect play - Black cannot lose (and therefore White cannot win): Either Black can win the game or not. If not, it means White can win or force a tie, so Black just passes with his first move and therefore will be in White's "shoes" to either win or force a draw. So in the worst case Black can force a draw, Black cannot lose. From that it follows: Either Black can win or a draw is inevitable. Is there any way to prove one of those alternatives and what's your guess?

p.s. It's amazing that we can prove things with so little info, weirdly in Chess - a game inferior to Go (in terms of complexity) - we cannot prove sh*t^^.

If black passes it's whites turn but white still has 6.5 komi so I don't think black could really pass.

[Pippen]

@krama: Yes, I think the proof doesn't work with komi involved.

[Krama]

@krama: Yes, I think the proof doesn't work with komi involved.

We would need to form a question like this.

Given that both players play the best moves and assuming that black has a slight advantage and always wins (no komi). What is the biggest margin that black can win by?

It would probably be around 5-7 points thus maybe the 6.5 is correct.

I believe that with 5.5 komi they saw the statistic and noticed black way maybe winning in 53% of the times so they decided to try to balance it by giving white 6.5 komi. Now if the statistics that I last saw were correct, white wins in 52% of cases (pro game statistics).

It is still hard to tell if these statistics mean anything. But if we could somehow solve the game and show that black can win by 6 points with perfect play from both sides then maybe 6.5 komi is good. (0.5 points in white favor since white goes second).