An $\infty$-Borel set is a set $X\subseteq\mathbb{R}$ which has an $\infty$-Borel code - a set $r$ coding the construction of $X$ via open sets, complementation, and well-ordered unions (see http://en.wikipedia.org/wiki/Infinity-Borel_set). Under $ZFC$, every set of reals is $\infty$-Borel; however, this is not provable in $ZF+AD$ (it follows from a stronger axiom, $AD_+$).
Consider the following statement:
$$\text{$(*)$ Every $\infty$-Borel set (viewed as a game in the usual way) is determined.}$$
Clearly $(*)$ is independent (assuming consistency of large cardinals) of $ZF$, and is implied by $AD$. My first question is whether this is sharp:
(1) Is $(*)$ strictly weaker than $AD$?
(I'm mostly asking this for actual implication - that is, whether $ZF+(*)\vdash AD$ - but I'm also interested in the question of their relative consistency strengths.)
EDIT: Andres' comments below completely answer this question. However, I'll leave the following paragraph from the pre-comments question, for the record:
I suspect the answer is "yes"; however, the only approach I see is to start with a model of $AD$, and add an undetermined set of reals without adding any new $\infty$-Borel sets of reals. For example, if the $\infty$-Borel codes in a model of $ZF+AD$ could be reduced to individual reals, then starting with $V\models ZF+AD$ and adding an undetermined game via countably closed forcing would do the trick; but I don't see how to justify the hypothesis, and without it I don't know what to do.
Assuming the answer to (1) is "yes," I also have the following question:
(2) Is $(*)$ an interesting principle on its own?
Obviously this is somewhat vague; I am asking both about the principle $(*)$ itself, and the theory $ZF+(*)$ (the difference being, maybe $ZF+(*)$ happens to be some well-known theory but the specific sentence $(*)$ is not particularly interesting).
