# The multiplication game on the free group

Fix $W\subseteq\mathbb F_2$ and consider the following two-person game: Player 1 and Player 2 simultaneously choose $x$ and $y$ in $\mathbb F_2$. The first player wins, say one dollar, iff $xy\in W$. Does this game admit Nash equilibria?

Of course, for very particular choices of $W$ it does, for instance if $W$ is either finite or cofinite. But I am not interested in these cases, I am interested in solvability of the multiplication game for all $W$.

One possible mathematical reformulation (game theorists would say that the mixed extension of this game is not uniquely defined and then there are several different mathematical reformulations) is the following:

Question. Do there exist, for all $W\subseteq\mathbb F_2$, finitely additive probability measures $\overline\mu,\overline\nu$ (of course depending on $W$) on the power set of $\mathbb F_2$ such that for all other finitely additive probability measures $\mu,\nu$ one has $$\int\int\chi_W(xy)d\mu(x)d\overline\nu(y)\leq\int\int\chi_W(xy)d\overline\mu(x)d\overline\nu(y)\leq\int\int\chi_W(xy)d\overline\mu(x)d\nu(y)$$ ?

Motivation. If we play the same game on a (countable) amenable group, one can show that a Nash equilibrium exists and is given by a right-invariant mean $\overline\mu$ and a bi-invariant mean $\overline\nu$. I am wondering what can happen for non amenable groups... I suspect (motivated by some possibly wrong speculative consideration) that the multiplication game is actually solvable for all $W$ only on amenable groups, giving an hopefully nice new characterization of amenability. However, any approach to proving that solvability implies amenability so far got stuck.

I'm confused. If $W$ is the whole group, player 1 wins, otherwise player 2 wins. Right? –  Lee Mosher May 31 '12 at 17:07
Why should P2 win? Notice that the choices are done independently (simultaneously, if you want), so that P2 cannot choose a suitable $y$ such that $xy\in W$, because (s)he does not know $x$. –  Valerio Capraro May 31 '12 at 18:01
I've edited, saying explicitly that the choices of $x$ and $y$ are simultaneous. –  Valerio Capraro May 31 '12 at 18:09