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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.

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  • $\begingroup$ 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. $\endgroup$
    – user21820
    Sep 6, 2020 at 17:03

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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, $\sigma*\tau\not\in B$.)

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  • $\begingroup$ OK, this is what I guessed. But a strategy that forces draws exclusively is strictly stronger than a drawing strategy? Is there any game theoretic interest in this kind of strategy? $\endgroup$
    – Jori
    Sep 5, 2020 at 22:48
  • $\begingroup$ @Jori I wouldn't say that. The condition of forcing a draw is strictly stronger than forcing a non-loss, but a strategy which forces a draw is less strong, in the sense of quality of strategy, than a strategy which forces a draw-or-win but sometimes allows a win. If we have the option we want to win more. So I don't see immediately what the interest in that stronger condition would be, but of course I could be missing something. $\endgroup$ Sep 5, 2020 at 23:16

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