Mixed Strategies and the Indifference Condition

Watch the second player mix two strategies by probability. Slide the mixing probability back and forth.

As the rival's mix changes, my expected payoff from each strategy moves along a straight line. At the mix where the two lines cross, neither strategy is better. Only that indifferent mix can appear in a mixed equilibrium. Making your rival indifferent is what pins down your mix.

For example, say a 🧑 friend takes a penalty kick with a ⚽ ball against a goalkeeper friend. Always kicking left lets the keeper guard only left. So the kicker mixes left and right kicks. The mix makes the keeper equally unhappy guarding either side.

u_1(\text{Top}, q) = q\,u_{TL} + (1-q)\,u_{TR}

q^* : u_1(\text{Top}, q^*) = u_1(\text{Bottom}, q^*)

Part of these learning paths

The Game Theory Core

Examples & Exercises

In Matching Pennies, find the mix $q^*$ of player 2 that makes player 1 indifferent, and explain why.

u_1(H, q) = q \cdot 1 + (1-q) \cdot (-1) = 2q - 1

u_1(T, q) = q \cdot (-1) + (1-q) \cdot 1 = 1 - 2q

Matching Pennies — P1 indifferent at

q = 1/2

Show proof

Let player 2 play $H$ with probability $q$ .
Player 1's expected payoff from $H$ is $2q - 1$ , from $T$ is $1 - 2q$ .
The indifference condition $2q - 1 = 1 - 2q$ gives $q^* = 1/2$ .
The key insight: in a mixed equilibrium each player chooses the mix that "makes the opponent indifferent" — not the mix maximizing their own payoff, but the one giving the opponent the incentive to mix.
If $q < 1/2$ , $T$ is better ( $1-2q > 2q-1$ ); if $q > 1/2$ , $H$ is better.
Only at $q^*$ is every mix of player 1 (weakly) optimal.

In the Battle of the Sexes, find $q^*$ making player 1 indifferent and $p^*$ making player 2 indifferent.

u_1(O, q) = 3q, \qquad u_1(F, q) = 2(1-q)

u_2(p, O) = 2p, \qquad u_2(p, F) = 3(1-p)

Battle of the Sexes — P1 indifferent at

q^* = 0.4

Show proof

Player 1: $u_1(O,q) = 3q$ , $u_1(F,q) = 2(1-q)$ . From $3q = 2 - 2q$ , $q^* = 2/5 = 0.4$ .
Player 2: $u_2(p,O) = 2p$ , $u_2(p,F) = 3(1-p)$ . From $2p = 3 - 3p$ , $p^* = 3/5 = 0.6$ .
The fully mixed Nash equilibrium is $(p^*, q^*) = (0.6, 0.4)$ .
Expected payoffs are $3 \times 0.4 = 1.2$ for player 1 and $2 \times 0.6 = 1.2$ for player 2 — worse for both than either pure equilibrium $(3,2)$ or $(2,3)$ .
Mixed equilibria can be Pareto inferior.

8. Game Theory