The question asks whether, for any subset ( W \subseteq \mathbb{F}_2 ), there exist finitely additive probability measures ( \overline{\mu} ) and ( \overline{\nu} ) on ( \mathbb{F}_2 ) that form a Nash equilibrium in the multiplication game.

Key points:

1. Game Setup: The game is zero-sum, with Player 1 maximizing the probability that ( xy \in W ) and Player 2 minimizing this proba...

Short answer: Not always. In fact, the property “the multiplication game admits a Nash equilibrium for every W ⊂ F2 (with finitely additive mixed strategies)” holds for a group G if and only if G is amenable. Since F2 is non-amenable, there exists some W ⊂ F2 for which no Nash equilibrium exists.

Details:

1. Formulation via lower and upper values For any subset W ⊂ G, define the lower and upper ...

This is a fascinating question that connects game theory, measure theory, and geometric group theory. Let me analyze the key aspects of this problem.

The Game Setup

The multiplication game on (the free group on two generators) is defined as follows:

• Players 1 and 2 simultaneously choose elements
• Player 1 wins if for some fixed subset $W \s...

Yes – if the players are allowed to mix with finitely additive probability measures ( “charges’’ on the power set of F₂ ) the game always has a saddle point, hence a Nash equilibrium, no matter what subset
W ⊂ F₂ is chosen. In particular the phenomenon has nothing to do with amenability; it holds for every group and for every infinite two–person zero–sum matrix game once finitely additive mixed...

Nash Equilibria in the Multiplication Game on Free Groups

This question explores an intriguing intersection between group theory and game theory, focusing on a game played on the free group .

Understanding the Problem

In the "multiplication game":

• Players 1 and 2 simultaneously choose elements and from
• Player 1 wins if the product belongs to a ...

First, I recall that the free group F₂ is non-amenab...

An excellent and deep question connecting game theory, group theory, and the concept of amenability. The answer is no, this game does not always admit a Nash equilibrium on , and your suspicion is correct: this property is a characterization of amenability.

Let's break down the proof that for the free group , one can choose a winning set for which no Nash equilibri...

Yes – once one allows the players to mix with finitely–additive probability measures, the multiplication game has a Nash equilibrium for every group and for every subset
W. Nothing like amenability is needed.

I. How to interpret strategies and pay–offs

Let G be any discrete group (countable or not) and let

K := { μ ∈ ℓ∞(G)* : μ is positive, μ(1)=1 }

be the set of finitely–additive prob...

Here is a fairly standard “minimax‐in‐infinite‐dimensions’’ argument which shows that nothing special about amenability is needed – as soon as you allow arbitrary finitely‐additive probabilities on your countable group, the multiplication‐game always has a value and a saddle‐point (hence a Nash equilibrium). In particular one does not get a new characterization of amenability by asking “solvable ...