Mixed-Strategy Equilibrium

A mixed strategy is a probability distribution over a player’s pure strategies, and a mixed-strategy Nash equilibrium is a profile of such distributions in which each player’s mix makes the opponent indifferent among the pure strategies they use (the indifference principle). In brief: in matching pennies the only equilibrium has both players choosing heads and tails with probability 1/2 each. In the penalty kick game below, the kicker aims left with probability 1/2 and the goalkeeper dives left with probability 5/8, and the kicker scores with probability 0.65.

Many games have no equilibrium in pure strategies, because any predictable choice can be exploited. Randomizing solves this: Nash’s theorem guarantees an equilibrium in mixed strategies for every finite game. This guide explains expected payoffs of mixed strategies, the indifference principle for finding equilibria, the cases of zero-sum and non-zero-sum games, and how to interpret randomization, with worked examples and a table.

What is a mixed strategy?

A mixed strategy assigns a probability to each pure strategy, the probabilities being nonnegative and summing to 1. The player uses a random device and the opponent knows the probabilities but not the outcome. The expected payoff of a pure strategy against the opponent’s mix is the sum of its payoffs weighted by the opponent’s probabilities. The expected payoff of a mix is the weighted average of the expected payoffs of its pure strategies, so a mix can never be better than the best pure strategy it contains, and it is a best response only if every pure strategy used in it is itself a best response.

What is the indifference principle?

In a mixed equilibrium, a player who randomizes over several pure strategies must be indifferent among them (otherwise the player would prefer to use the best one only). So to find the probabilities of one player, solve for the values that make the other player indifferent. For a 2 × 2 game this is one linear equation for each player.

How do you solve a zero-sum 2 × 2 game?

Example 1: matching pennies

The first player wins 1 if the pennies match and loses 1 otherwise. Let the first player show heads with the probability p. Against a second player who shows heads, the first player’s expected payoff is p · 1 + (1 − p)(−1) = 2p − 1; against tails it is p · (−1) + (1 − p) · 1 = 1 − 2p. For the second player to be indifferent, these must be equal: 2p − 1 = 1 − 2p, so p = 1/2. By symmetry the second player’s mix is also 1/2. The value of the game is 0.

Example 2: penalty kicks

The kicker aims left or right, the goalkeeper dives left or right. The probability that the kick scores: aim left and dive left 0.5; aim left, dive right 0.9; aim right, dive left 0.8; aim right, dive right 0.4. The kicker wants to maximize and the goalkeeper to minimize this probability. Let the kicker aim left with the probability p. Against a left dive the scoring probability is 0.5p + 0.8(1 − p) = 0.8 − 0.3p; against a right dive, 0.9p + 0.4(1 − p) = 0.4 + 0.5p. Equating: 0.8 − 0.3p = 0.4 + 0.5p, so p = 0.5, and the scoring probability is 0.65. Let the goalkeeper dive left with the probability q. If the kicker aims left, the scoring probability is 0.5q + 0.9(1 − q) = 0.9 − 0.4q; if right, 0.8q + 0.4(1 − q) = 0.4 + 0.4q. Equating: 0.5 = 0.8q, so q = 0.625, again with the value 0.65 ✓. Both players use mixed strategies and neither can improve.

Example 3: rock-paper-scissors

With a win worth +1, a loss −1 and a tie 0, the mix (1/3, 1/3, 1/3) gives every pure strategy of the opponent the expected payoff (1 − 1 + 0)/3 = 0. So each is a best response to the other, and this is an equilibrium with the value 0. Any deviation from the uniform mix can be exploited: a player who shows rock too often loses to paper more often.

How do you solve a non-zero-sum game?

Example 4: chicken

Two drivers head for a collision. Each can swerve (S) or drive straight (D). Payoffs (A, B): (S, S) = (0, 0), (S, D) = (−1, 1), (D, S) = (1, −1), (D, D) = (−10, −10). There are two pure equilibria, (S, D) and (D, S), each favoring one driver. For the mixed equilibrium, let A swerve with the probability p. B’s expected payoff from S is 0 · p + (−1)(1 − p) = p − 1, from D it is 1 · p + (−10)(1 − p) = 11p − 10. Indifference: p − 1 = 11p − 10, so p = 0.9. By symmetry, B swerves with the probability 0.9. The expected payoff to each is p − 1 = −0.1, and the probability of a crash is 0.1 · 0.1 = 0.01.

Example 5: a surprising effect

In a non-zero-sum game, a player’s equilibrium mix depends on the opponent’s payoffs, not on their own. In chicken, if the crash penalty for B increases from −10 to −100, A’s probability of swerving changes: B’s indifference becomes p − 1 = 101p − 100, hence p = 0.99. A’s own payoffs have no influence on A’s mix, which is determined by B’s incentives.

What about larger games?

For larger games, the equilibrium requires choosing the supports (sets of pure strategies used) and solving a system of linear equations for the probabilities, then checking that no unused strategy gives a higher payoff. The count of supports grows quickly, and general algorithms (such as the Lemke–Howson algorithm) solve two-player games. For zero-sum games, the minimax theorem of von Neumann ensures that the equilibrium value exists and can be found by linear programming.

How should randomization be interpreted?

Randomization is literal in sports (serves in tennis, penalty kicks) and in inspections (random audits). In other settings, a mixed equilibrium can be read as the frequency in a population, with a fraction of players using each pure strategy, or as the opponent’s uncertainty about what a player will do (Harsanyi’s purification). In the chicken example, 90% swervers and 10% drivers in a population meeting at random produce the same average outcomes.

Table of mixed equilibria

Where does the theory come from?

Mixed strategies were introduced by Émile Borel in the early 1920s and John von Neumann proved the minimax theorem for zero-sum games in 1928. Nash’s 1950 existence theorem extended the idea to all finite games. The empirical study of randomization in professional sports, for example by Mark Walker and John Wooders for tennis serves (2001), found that behavior is close to the predictions of mixed equilibria, though with some serial correlation that a perfectly random player would not show.

Which mistakes are common?

Using a player’s own payoffs to find their mix

A player’s probabilities are determined by the opponent’s indifference, so solve the opponent’s equations.

Forgetting to check unused strategies

In larger games, verify that every strategy outside the support gives no higher payoff.

Equating mixing with “being unpredictable by whim”

The probabilities must be exactly the equilibrium ones. Human attempts to “act randomly” are usually not random enough.

Expecting a mixed equilibrium to maximize payoffs

It balances incentives and can give poor payoffs, such as the crash risk in chicken or the miscoordination in the battle of the sexes.