Let me refer to Jech's "Set Theory" Chap. 33 Determinacy:


"With each subset A of $\omega^\omega$ we associate the following game $G\_A$, played by two players I and II. First I chooses a natural number $a\_0$, then II chooses a natural number $b\_0$, then I chooses $a\_1$, then II chooses $b\_1$, and so on. The game ends after $\omega$ steps; if the resulting sequence $\langle a\_0, b\_0, a\_1, b\_1, ...\rangle$ is in A, then I wins, otherwise II wins.


A strategy (for I or II) is a rule that tells the player what move to make depending on the previous moves of both players. A strategy is a winning strategy if the player who follows it always wins. The game $G\_A$ is determined if one of the players has a winning strategy.


The Axiom of Determinacy (AD) states that for every subset A of $\omega^\omega$, the game $G\_A$ is determined."


Now, there is some apparent lack of symmetry in the definition of the $G\_A$ game: the player who plays first (I) attempts for a sequence in A. 


What happens if we interchange the roles of both players? I. e. if we let the player who plays first attempt for a sequence not in A? Let us call this game $G'\_A$


Is it the case that for every subset A, A is determined wrt $G\_A$ iff A is determined wrt $G'\_A$?



Determinacy interchanging the roles of both players
The answer is no. The game $G'(A)$ you describe is just the same as the complementary game $G(A^c)$ from I's point of view. So the question is: if one of the players has a winning strategy in $G(A)$, does one of the players have a winning strategy in $G(A^c)$? Using AC one can construct a set $A$ for which this is not the case.


We do a recursive construction, much as in the usual recursive construction of a nondetermined set. The main difference is set out in advance a strategy $\sigma$ for I to be a winning strategy for $A$. Fix any strategy $\sigma$ for I (so it takes a finite sequences of even length as input and gives natural numbers as output). For a given infinite sequence $y$ of moves for II we let $\sigma\*y$ denote the member of $\omega^\omega$ resulting when I follows $\sigma$ and II plays out $y$.


Set $Z=\{\sigma\*y:y\in\omega^\omega\}$. We will make sure $Z\subseteq A$, then certainly $\sigma$ will be a winning strategy for I in $G(A)$.


So we also define $\langle a\_\alpha\rangle\_{\alpha<2^\omega}$ and $\langle b\_\alpha\rangle\_{\alpha<2^\omega}$ so that no $a\_\alpha$ is equal to any $b\_\beta$ and both sets are disjoint from $Z$. Let $\langle\sigma\_\alpha\rangle\_{\alpha<2^\omega}$ enumerate all of I's strategies and let $\langle\tau\_\alpha\rangle\_{\alpha<2^\omega}$ all of II's strategies. Suppose $\langle a\_\beta\rangle\_{\beta<\alpha}$ and $\langle b\_\beta\rangle\_{\beta<\alpha}$ have been defined. First consider $\sigma\_\alpha$; if it equals $\sigma$ we don't do anything. Otherwise it disagrees with $\sigma$ on some $s$; there are $2^\omega$ many $x$ which are extensions of $s$, for each such $x$ we have $\sigma\_\alpha\*x\not\in Z$ so pick one which also isn't equal to an $a$ or a $b$ so far and make it $a\_\alpha$.


Next consider $\tau\_\alpha$. There are continuum many $x$ that I can play so that $\tau\_\alpha\*x$ does not land in $Z$ (just have $x$ give a first move different from that prescribed by $\sigma$). Using a cardinality argument we get some $x$ such that $\tau\_\alpha\*x$ is not in $Z$ and not equal to anything chosen so far, and set it equal to $b\_\alpha$.


At the end let $A$ equal $Z$ together with all of the $a\_\alpha$. Then $\sigma$ is a winning strategy for I in $G(A)$ because every play according to $\sigma$ lands in $Z$. But no strategy wins $A^c$. We've made sure that for every strategy for I there is a play which equals an $a\_\alpha$ and thus lands in $A$ and so that strategy cannot be winning in $G(A^c)$. And similarly we've made sure that for every strategy for II there is a play equalling one of the $b\_\beta$ and thus landing in $A^c$ and not winning for II in $G(A^c)$.


