Mixed Strategy¶

Definition¶

Let player $i$ in a normal-form game have a finite set of pure strategies $S_i=\{s_{i1},\ldots,s_{ik}\}$ ${s_{i 1}, \dots, s_{ik}}$ . A mixed strategy for player $i$ is a probability distribution $\sigma_i$ on $S_i$ : a function $\sigma_i\colon S_i\to[0,1]$ with

\sum_{s_i\in S_i}\sigma_i(s_i)=1.

The set of all mixed strategies of player $i$ is written $\Delta S_i$ . The support of $\sigma_i$ is the set of pure strategies with $\sigma_i(s_i)>0$ $0$ . A pure strategy $s_i$ is the degenerate mixed strategy that puts probability $1$ on $s_i$ . When $S_i$ is an interval (for example, prices or bids), a mixed strategy is given by a cumulative distribution function on $S_i$ .

A mixed-strategy profile $\sigma=(\sigma_1,\ldots,\sigma_n)$ assumes that players randomize independently. Player $i$ 's payoff is the expectation

v_i(\sigma)=\sum_{s\in S}\Big(\prod_{j\in N}\sigma_j(s_j)\Big)\,v_i(s),

where $S=S_1\times\cdots\times S_n$ $S_{n}$ .

Interpretation¶

A mixed strategy can be read literally, as a player who uses a randomizing device, or as a description of the opponents' uncertainty about what player $i$ will do. Either reading leads to the same calculations; it relies on players being expected payoff maximizers, so that only expected payoffs matter.

Player 1's expected payoff in Matching Pennies against q = 0.4 as a straight line in p, falling from 0.2 at T (p = 0) to -0.2 at H (p = 1), with the mixture p = 0.7 marked at -0.08.

The figure fixes player 2's mixture at $q=0.4$ in the Matching Pennies example and plots player 1's expected payoff against her probability $p$ of $H$ . Because $v_1(\sigma_1,\sigma_2)=p(2q-1)+(1-p)(1-2q)$ is a weighted average of the two pure-strategy payoffs, the graph is a straight line between $v_1(T,\sigma_2)=0.2$ at $p=0$ and $v_1(H,\sigma_2)=-0.2$ at $p=1$ $1$ . The mixture $p=0.7$ earns $-0.08$ , in between; the maximum is reached at a pure strategy, here $T$ .

Move $p$ : player 1's expected payoff runs along a straight line between her two pure-strategy payoffs, so no mixture beats the better pure strategy. Set $q=\tfrac12$ and the line becomes flat.

Mixing never lets a player beat her own best pure strategy against fixed opponents. Because the expected payoff is a weighted average of the payoffs of the pure strategies in the support, it is at most the largest of them. Randomization matters for a different reason: it changes what the opponents face, which is why games such as Matching Pennies have an equilibrium only in mixed strategies.

The independence assumption matters: a mixed-strategy profile cannot describe correlated play such as "both players follow the same coin toss".

Key Results¶

With $k$ pure strategies, $\Delta S_i$ is the simplex $\displaystyle \{p\in\mathbb{R}^k\colon p_j\ge0,\ \sum_j p_j=1\}$ $1}$ , a compact convex set; for $k\ge2$ it contains infinitely many mixed strategies.
By linearity of expectation, $\displaystyle v_i(\sigma_i,\sigma_{-i})=\sum_{s_i\in S_i}\sigma_i(s_i)\,v_i(s_i,\sigma_{-i})$ $s_{i} \in S_{i} \sum σ_{i} (s_{i}) v_{i} (s_{i}, σ_{- i})$ , which is linear in $\sigma_i$ .
Hence $\max_{\sigma_i\in\Delta S_i}v_i(\sigma_i,\sigma_{-i})=\max_{s_i\in S_i}v_i(s_i,\sigma_{-i})$ $max_{s_{i} \in S_{i}} v_{i} (s_{i}, σ_{- i})$ , and $\sigma_i$ is a best response if and only if every pure strategy in its support is a best response.
A pure strategy can be strictly dominated by a mixed strategy even when no pure strategy dominates it.

Examples¶

Matching Pennies. Player 1 wins $1$ if the coins match and loses $1$ otherwise. Let player 1 play $H$ with probability $p$ and player 2 play $H$ with probability $q$ . Then $v_1(H,\sigma_2)=q-(1-q)=2q-1$ $1$ , $v_1(T,\sigma_2)=1-2q$ $2 q$ , and

v_1(\sigma_1,\sigma_2)=p(2q-1) +(1-p)(1-2q) =(2p-1)(2q-1).

For $p=0.7$ $0.7$ , $q=0.4$ $0.4$ : the coins match with probability $0.28+0.18=0.46$ $0.46$ , so $v_1=0.46-0.54=-0.08=(0.4)(-0.2)$ $(0.4) (- 0.2)$ .

Dominance by a mixture. Player 1 has payoffs $U\colon(3,0)$ , $M\colon(0,3)$ , $D\colon(1,1)$ against player 2's $L$ and $R$ . Neither $U$ nor $M$ dominates $D$ , but $\tfrac{1}{2}U+\tfrac{1}{2}M$ yields $(1.5,1.5)$ , strictly more than $D$ against both columns.

Prerequisites¶

Sources¶

Game Theory · Week 2

Aliases: randomized strategy, mixed strategies, randomised strategy Keywords: σ_i, ΔS_i, randomization, support of a mixed strategy, mixed-strategy profile, Matching Pennies Symbols: delta, element of, product, real numbers, sigma, sum

Last updated: 22 October 2026 19:04