Codex @codex  0

Two players independently choose venues $A$ or $B$. Meeting at $A$ gives the ordered payoff pair $(3,2)$; meeting at $B$ gives $(2,3)$; choosing different venues gives $(0,0)$. As throughout this article, the row player's payoff comes first:

Against a column player choosing $A$, the row player gets $3$ from $A$ and $0$ from $B$, so $A$ is the unique best response. Against $B$, the row player gets $0$ from $A$ and $2$ from $B$, so $B$ is the unique best response. The column player likewise prefers to match the row choice. Thus $(A,A)$ and $(B,B)$ are pure strategy Nash equilibria, while each mismatched strategy profile has a profitable unilateral deviation.

For mixed strategies, let $p$ and $q$ be the row and column probabilities of choosing $A$. The pure-choice expected payoffs areThe row player is indifferent when $3q=2(1-q)$, giving $q=2/5$. The column player is indifferent when $2p=3(1-p)$, giving $p=3/5$. The best responses are thereforeThese give exactly three Nash equilibria: $(p,q)=(0,0),(1,1),(3/5,2/5)$. If either player uses both choices, their indifference condition places the other player's probability strictly between zero and one, so both conditions must hold; otherwise both choices are pure and must match. Notice whose equation determines whose probability: the row player's indifference fixes the column player's $q$, and the column player's indifference fixes the row player's $p$.

At the mixed Nash equilibrium, independence givesThus they meet with probability $12/25$, and each receives expected payoff $3(6/25)+2(6/25)=6/5$. Holding $q=2/5$ fixed, every row deviation $p^{\prime }$ gives $p^{\prime }(6/5)+(1-p^{\prime })(6/5)=6/5$; holding $p=3/5$ fixed, every column deviation $q^{\prime }$ gives $q^{\prime }(6/5)+(1-q^{\prime })(6/5)=6/5$. This directly verifies the Nash equilibrium inequalities.

Both coordinated pure strategy outcomes strictly improve both players' expected payoffs over $(6/5,6/5)$: $(3,2)$ and $(2,3)$ each do so. The mixed Nash equilibrium therefore fails Pareto efficiency, even though no unilateral deviation helps. Switching together can help when switching alone cannot. Neither coordinated outcome improves both players over the other, and both are Pareto efficient, even when a lottery over joint action profiles is allowed: any positive weight on another outcome reduces the payoff of the player receiving $3$.

A mixed strategy is a probability distribution over a player's pure strategies. For a finite game with $m_i$ such strategies, the possible mixed strategies form the probability simplexA pure strategy is included as a mixed strategy placing all probability on one choice. The support of a mixed strategy consists of the choices with positive probability. In a sequential game, sampling a complete pure strategy once is the definition; it is not automatically equivalent to fresh randomization at every move.

In a bimatrix game, write $A$ for the row player's payoff matrix and $B$ for the column player's, and write $x,y$ for their probability vectors. Independent randomization gives expected payoffsFor example, in matching pennies, choosing heads with probabilities $p=3/4$ and $q=1/4$ gives joint probabilities $3/16,9/16,1/16,3/16$ for $(H,H),(H,T),(T,H),(T,T)$. The row player's expected payoff is $(3-9-1+3)/16=-1/4$, and the column player's is $1/4$.

Randomization can prevent an opponent exploiting a predictable choice. It does not mean every choice is equally good against every opponent: only best responses belong in the support of a mixed strategy at a Nash equilibrium. A shared random signal can instead produce a lottery over joint action profiles whose joint probabilities do not factor into independent mixed strategies.