Question about the $* = \{0\mid 0\}$ Game in “On Numbers and Games”

combinatorial-game-theorysurreal-numbers

Why is the game $* = \{0\mid 0\}$, where $0=\{\;\mid\;\}$ is the empty game, mentioned on p.72 of Conway's "On Numbers and Games", a win for the 1st player?

It seems to me that if the 1st player to move is Left, then Left must play $0$, Right must play $0$, and Left loses. Analogously if the 1st player to move is Right.

The game is also mentioned in Hamkins' new book "Lectures on the Philosophy of Mathematics".

What am I not understanding?

Best Answer

Once Left plays $0$, the position is now $\{\ \mid\ \}$, from which neither player has any valid moves. Since it is now Right's turn, Right loses.

It seems like you might be misinterpreting $\{0\mid0\}$ to mean that Left and Right both have a "supply" and take turns choosing from that supply, but that is not the case. Instead, the positions listed are those to which the game will be reduced.

Here's another example. Consider the game $${\uparrow}=\{0\mid\star\}$$ From this position, Left has one move, namely to $0$, from which Right will lose. On the other hand, if Right moves first, they have one move, which is to $\star$, which is a win for the next player (i.e., Left). Thus, Left will win either way.