# Definition of a drawing strategy

Given the following definition:

Let $$M$$ be a move set and $$A,B \subset M^\omega$$ are disjoint, we define the winning conditions for the game $$G(A,B)$$ as follows: if the play of the game is $$x \in M^\omega$$, then player I wins if $$x \in A$$, player II wins if $$x \in B$$, and otherwise the game is a draw. A strategy $$\sigma$$ is winning or drawing in $$G(A,B)$$ for player I if for all strategies $$\tau$$ , we have that $$\sigma * \tau \in A$$ or $$\sigma * \tau \notin B$$, respectively. A strategy $$\tau$$ is winning or drawing in $$G(A,B)$$ for player II if for all strategies $$\sigma$$, we have that $$\sigma * \tau \in B$$ or $$\sigma * \tau \notin A$$, respectively.

what would a "drawing strategy in $$G(A, B)$$" mean? A strategy such that a player can at least play a draw (so either draw or win), or a strategy such that a player can force a draw (so the game always ends in a draw). I find the formulation of the definition a bit unclear in this regard (but maybe that's just me?), and I think these two options are not equivalent.

• Note that it may be possible that a player has no winning strategy but has a win-or-draw strategy and yet has no strategy that guarantees exactly a draw! There is almost no reason one would be interested in an exactly-drawing strategy. Sep 6, 2020 at 17:03

A drawing strategy is one which either draws or wins against every strategy - or perhaps more snappily, one which never loses. The point (looking at drawing strategies for player $$I$$ for simplicity) is that $$\sigma*\tau\not\in B$$ means that player $$II$$ does not win the play of $$\sigma$$ against $$\tau$$ - that is, that $$\sigma$$ either draws against $$\tau$$ or wins against $$\tau$$.
(In particular, $$\sigma*\tau\in A$$ does imply, but is not implied by, $$\sigma*\tau\not\in B$$.)