TITLE: Are Vitali-type nonmeasurable sets determinate?
QUESTION [6 upvotes]: Here, by a Vitali set, I mean the following.  Call $f_1,f_2:\omega\rightarrow 2$  tail-equivalent if  {$n| f_1(n)\not=f_2(n)$}$<\infty$.  Vitali sets (existence via AC) contain one such $f$ from each tail-equivalence class.  
Since any Vitali set $V$, as a subset of $2^\omega$, has inner measure 0 and positive outer measure, $V$ gives rise to an indeterminate game roughly of the following sort: Alice must name successive binary digits of a number $v$ in $V$ while Bob names some increasingly very thin open sets $U_i$ hoping that in the end, for some $i$, $v\in U_i$ ("$U_i$ captures $v$").
My question: does any $V$, in itself, define the payoff set of an indeterminate game?  (So now, a turn for either Alice or Bob consists of naming the next digit of one common number that Alice, but not Bob, would like to steer into $V$.)
(Easy observation: each tail-equivalence class contains many functions $f$ with $f(1)=0$, and a $V$ containing only these would give Bob a win as early has his first turn, thus making the game trivially determinate.  So one can't expect all such $V$ to define indeterminate games.  But how about all dense Vitali sets?)

REPLY [4 votes]: To answer the last (parenthetical) piece of the question, which Joel seems to have ignored: Density doesn't help; a dense Vitali set $V$ can still have an easy winning strategy for Bob.  The point is that, if $C$ is a tail-equivalence class and $B$ is any basic open set (the set of all $f\in 2^\omega$ that extend a particular finite sequence $b$), then there is an element $f\in C\cap B$ for which the odd-numbered terms $f(2n+1)$ ("Bob's moves") are not all zero --- just define $f$ to agree with $b$ where $b$ is defined, to have a 1 in some later odd-numbered position, and thereafter to behave in a way that puts it into $C$.  Now associate, arbitrarily, to each basic open set $B$, one tail-equivalence class $C_B$.  For each of these, pick one $f\in C_B\cap B$ as above; for all the other tail-equivalence classes (those not associated to any $B$) pick any $f\in C$ whose odd-numbered terms are not all zero.  Then the chosen $f$'s constitute a dense Vitali set, for which Bob has the winning strategy "always play 0".