Nash Equilibrium in Mixed Strategies¶

Definition¶

In a finite normal-form game with players $i\in N$ $N$ , a profile of mixed strategies $\sigma^*=(\sigma_1^*,\ldots,\sigma_n^*)$ with $\sigma_i^*\in\Delta S_i$ is a Nash equilibrium if, for every player $i$ and every $\sigma_i\in\Delta S_i$ $Δ S_{i}$ ,

v_i(\sigma_i^*,\sigma_{-i}^*)\ \ge\ v_i(\sigma_i,\sigma_{-i}^*).

Equivalently, $\sigma_i^*\in BR_i(\sigma_{-i}^*)$ for every $i$ , where $BR_i$ is the best-response correspondence extended to mixed strategies of the opponents.

Interpretation¶

The definition is the Nash equilibrium condition applied to the mixed extension of the game: no player can raise her expected payoff by a unilateral change of mixture. In the belief interpretation, $\sigma_{-i}^*$ describes player $i$ 's correct conjecture about her opponents, and each player best responds to that conjecture.

Best-response diagram for Battle of the Sexes in the unit square of (p, q): the blue BR_1 steps at q = 1/3, the red dashed BR_2 steps at p = 2/3, and they cross at three green points.

The figure draws both best-response correspondences of the Battle of the Sexes example, with $p=\sigma_1(O)$ across and $q=\sigma_2(O)$ up. Player 1 plays $F$ ( $p =$ $p=0$ $0$ ) when $q<\tfrac{1}{3}$ $\frac{1}{3}$ , $O$ ( $p =$ $p=1$ $1$ ) when $q>\tfrac{1}{3}$ $\frac{1}{3}$ , and any $p$ at $q=\tfrac{1}{3}$ $\frac{1}{3}$ ; player 2 switches at $p=\tfrac{2}{3}$ in the same way. A Nash equilibrium is a point that lies on both graphs, so the three crossings are the three equilibria: $(F,F)$ , $(O,O)$ and the mixed profile $p=\tfrac{2}{3}$ $\frac{2}{3}$ , $q=\tfrac{1}{3}$ $\frac{1}{3}$ . The jump of $BR_1$ sits at a value set by player 2's mixture, and the jump of $BR_2$ at one set by player 1's, which is why each player's equilibrium probability is fixed by the opponent's indifference.

Every crossing of $BR_1$ and $BR_2$ is a Nash equilibrium. Edit player 2's payoffs: while a fully mixed equilibrium exists, its $p$ is fixed by player 2's indifference, while $q=\tfrac{1}{3}$ remains player 1's indifference threshold. If the fully mixed equilibrium disappears or its support changes, the equilibrium $q$ need not remain $\tfrac{1}{3}$ .

Each player chooses a mixture that maximizes her own expected payoff against the opponents' mixtures. To compute a fully mixed equilibrium of a $2\times2$ game, however, player 1's indifference equation constrains player 2's probability, and player 2's equation constrains player 1's. When these equations are nontrivial, they determine the probabilities. They need not have unique solutions in a degenerate game: if every payoff is zero, both equations are identities and every pair of mixtures is an equilibrium.

Key Results¶

Every pure-strategy Nash equilibrium is a mixed-strategy Nash equilibrium with degenerate mixtures.
It suffices to check pure deviations: $\sigma^*$ is a Nash equilibrium if and only if $v_i(\sigma^*)\ge v_i(s_i,\sigma_{-i}^*)$ for all $i$ and all $s_i\in S_i$ $S_{i}$ .
Every pure strategy in the support of $\sigma_i^*$ earns exactly $v_i(\sigma^*)$ against $\sigma_{-i}^*$ , and no pure strategy earns more (the indifference principle).
A pure strategy that is strictly dominated (by a pure or a mixed strategy) has probability $0$ in every mixed-strategy Nash equilibrium.
Every finite game has at least one Nash equilibrium in mixed strategies (Nash existence theorem).

Examples¶

Matching Pennies. Player 1 wins $1$ on a match and loses $1$ otherwise; player 2 gets the negative. There is no pure equilibrium: at every cell one player wants to switch. If player 2 plays $H$ with probability $q$ , player 1 is indifferent when $2q-1=1-2q$ $2 q$ , so $q=\tfrac{1}{2}$ $\frac{1}{2}$ ; symmetrically $p=\tfrac{1}{2}$ $\frac{1}{2}$ . The unique Nash equilibrium is $\big((\tfrac{1}{2},\tfrac{1}{2}),(\tfrac{1}{2},\tfrac{1}{2})\big)$ , with payoff $0$ to both.

Battle of the Sexes. Payoffs are $(2,1)$ at $(O,O)$ , $(1,2)$ at $(F,F)$ , and $(0,0)$ otherwise. Let $p=\sigma_1(O)$ and $q=\sigma_2(O)$ $σ_{2} (O)$ .

Player 1 indifferent: $2q=1\cdot(1-q)$ $q)$ , so $q=\tfrac{1}{3}$ $\frac{1}{3}$ .
Player 2 indifferent: $1\cdot p=2(1-p)$ $p)$ , so $p=\tfrac{2}{3}$ $\frac{2}{3}$ .

The game has three Nash equilibria: $(O,O)$ , $(F,F)$ and the mixed profile $p=\tfrac{2}{3}$ $\frac{2}{3}$ , $q=\tfrac{1}{3}$ $\frac{1}{3}$ . In the mixed one each player's expected payoff is $\tfrac{2}{3}$ (player 1: $2q=\tfrac{2}{3}$ $\frac{2}{3}$ ; player 2: $p=\tfrac{2}{3}$ $\frac{2}{3}$ ), which is lower than either player's payoff in both pure equilibria. The mixed equilibrium is therefore not Pareto efficient: miscoordination occurs with probability $\tfrac{2}{3}\cdot\tfrac{2}{3}+\tfrac{1}{3}\cdot\tfrac{1}{3}=\tfrac{5}{9}$ $\frac{5}{9}$ .

Prerequisites¶

Sources¶

Game Theory · Week 2

Aliases: mixed-strategy Nash equilibrium, mixed Nash equilibrium, MSNE, mixed-strategy equilibrium, equilibrium in mixed strategies Keywords: mixed equilibrium, Matching Pennies, Battle of the Sexes, σ*, BR_i(σ_-i) Symbols: delta, element of, sigma

Last updated: 27 October 2026 11:52