# Why does strategy-stealing not work for Go?

In Go passing is allowed. When the starting position is symmetrical (empty board, neither player has any points), this means that the first player could steal the second player's winning strategy simply by giving up the first move. Since the 1930s, however, the second player is typically awarded some compensation points, which makes the starting position asymmetrical, and the strategy-stealing argument will no longer work.

Which only states why it does not work for the first (black) player, as a symmetrical board would end in a tie and the second player (white) gets some more additional Komi-points (compensation). However, by passing the first move, the black player technically switches position with the white one to take advantage of strategy-stealing. Therefore, given the black player does not pass the first turn, white could use strategy-stealing, force a tie and then win with the additional compensation points. So why does strategy-stealing not work for Go in general (even if it would be the second instead of the first player)?

Edit: The strategy-stealing argument states that in a game, in which an extra move is never a disadvantage, the first player can always use the second players strategy. In Go a placed stone could be a disadvantage, but passing is an option. Therefore the strategy-stealing would work (as the extra move can be left out, and is therefore no disadvantage), if compensation points would not exist. As they do exist for the second player, it does not work for the first player (he could use it but would loose due to the compensation points). My question does not concern the strategy-stealing argument itself, but a similar inverted concept; The first player can not use the second ones strategy, but the second one could still use the first ones, forcing a tie (without compensation points) and win, considering his compensation points. I am thinking here of mimicking the opponents moves, which is less general than strategy-stealing (the way the term is used in the Wikipedia article), but, in a completely symmetrical game, one perfectly legitimate way of doing so.

Of course there is a trivial answer, as all used boards have an odd number of rows and columns. Therefore, there is at least one unique point (the middle if central inversion is used). The turn which takes this point can not be applied to the other player. Due to the nature of Go, this small difference can change the game dramatically. For example an adjacent stone could easily be captured by the moving player, whereas the same turn would capture no stone for the copying player. Therefore I restate the question: Is there any reason why the strategy-stealing argument would not work for a game of Go on a board with an even number of rows and columns?

• What is the strategy stealing argument? It sounds like the idea is that in a game with symmetrical starting positions, where passing is allowed, it's impossible for player two to have a winning strategy. Because player one could just pass, then use the winning strategy for player two. – littleO Dec 27 '14 at 14:43
• @littleO: As stated in the Wikipedia quote, the second player is awarded compensation points. If the game stays symmetrical until it ends it would be a tie, but it's not, because of the compensation points. My question concerns the second player. As The first one can never win with strategy-stealing (because of the compensation), he won't do it. But the second one still could. – marongaz Dec 27 '14 at 14:50
• It seems like you might be thinking of mimicking the opponent's moves, but I think that is not what Wikipedia meant by "strategy stealing." – littleO Dec 27 '14 at 14:56
• Ok. Strictly speaking I do not mean the strategy-stealing argument, which states that the first player can always use the second players strategy, if an additional move is no disadvantage. This does not work for Go because of the compensation points. The rest of the question tries to use the same concept for the second player. It is not the strategy-stealing argument itself anymore. I will clarify this in the question, thank you. – marongaz Dec 27 '14 at 15:02

• @marongaz On a $2\times 2$ board, symmetry is erased after 2 moves of both players. If you adopt a rule of not allowing repeating positions (some go-rules have such rule) then there is only one possible history of the game in which, surprisingly, black wins. How long do you play go? Try to do it in a $18\times 18$ or $\infty\times\infty$ board, it's not hard once you create a bit of mess around the center :) – Peter Franek Dec 27 '14 at 16:29