Indifference Principle for Mixed Strategies¶

Proposition¶

Let $\sigma^*$ be a profile of mixed strategies in a finite normal-form game, and let $S_i^+=\{s_i\in S_i\colon\sigma_i^*(s_i)>0\}$ be the support of $\sigma_i^*$ . Then $\sigma^*$ is a Nash equilibrium in mixed strategies if and only if, for every player $i$ , there is a number $\bar v_i$ such that

\begin{aligned} v_i(s_i,\sigma_{-i}^*)&=\bar v_i &&\text{for all } s_i\in S_i^+,\\ v_i(s_i',\sigma_{-i}^*)&\le \bar v_i &&\text{for all } s_i'\notin S_i^+. \end{aligned}

In that case $\bar v_i=v_i(\sigma^*)$ is player $i$ 's equilibrium payoff.

Interpretation¶

The reason is that $\displaystyle v_i(\sigma_i^*,\sigma_{-i}^*)=\sum_{s_i}\sigma_i^*(s_i)\,v_i(s_i,\sigma_{-i}^*)$ is a weighted average. If one strategy in the support earned less than another, an expected payoff maximizer could move its probability to the better one and gain. So all strategies used must tie, and none unused may do better.

Player 1's expected payoffs from U (3q) and D (2-q) against q: the lines cross at q=1/2 with value 3/2, above a third row B paying 1 and below a row paying 2.

The figure draws player 1's side of the $2\times 2$ example. Against $q=\sigma_2(L)$ $σ_{2} (L)$ , $U$ earns $3q$ and $D$ earns $2-q$ $q$ . Player 1 is willing to mix only where the two lines cross, at $q=\tfrac{1}{2}$ with value $\bar v_1=\tfrac{3}{2}$ $\frac{3}{2}$ ; this is how player 1's indifference pins down player 2's probability. The flat lines are the extra row of the second example: $B$ paying $1$ stays below $\tfrac{3}{2}$ , so the equilibrium survives, while a $B$ paying $2$ lies above it and violates the second condition.

In a two-player game, these conditions constrain one player's mixing probabilities through the other player's indifference equations. The usual error is to solve a player's own indifference for her own probabilities. An equation may be an identity rather than determine a unique probability: if every payoff is zero, all mixtures make both players indifferent and every profile is an equilibrium. With more than two players, a player's indifference equations can involve several opponents' mixtures together.

Indifference is not the whole test. The second line (no profitable strategy outside the support) must also be checked, and a strategy outside the support may earn exactly $\bar v_i$ . Indifference also means the player has no strict reason to use the equilibrium weights; the Harsanyi purification theorem gives one justification for them.

Key Results¶

The two conditions are necessary and sufficient for $\sigma^*$ to be a Nash equilibrium.
The equilibrium payoff $v_i(\sigma^*)$ can be computed from any single pure strategy in the support.
A strictly dominated strategy never lies in the support of an equilibrium strategy.
In a $2\times 2$ game with no pure-strategy equilibrium, there is a unique equilibrium; it is fully mixed, and each player's probability solves the other player's indifference equation.
Finding equilibria by support enumeration: guess the supports, solve the indifference equations, then check that probabilities lie in $[0,1]$ and that no strategy outside the support does better.

Examples¶

A $2\times 2$ game. Payoffs (player 1, player 2):

	$L$	$R$
$U$	$(3,1)$	$(0,2)$
$D$	$(1,3)$	$(2,0)$

There is no pure equilibrium: from $(U,L)$ player 2 switches to $R$ ; from $(U,R)$ player 1 switches to $D$ ; from $(D,R)$ player 2 switches to $L$ ; from $(D,L)$ player 1 switches to $U$ . Let $q=\sigma_2(L)$ and $p=\sigma_1(U)$ $σ_{1} (U)$ .

Player 1 indifferent: $3q=q+2(1-q)$ $q)$ , so $q=\tfrac{1}{2}$ $\frac{1}{2}$ .
Player 2 indifferent: $p+3(1-p)=2p$ $2 p$ , so $p=\tfrac{3}{4}$ $\frac{3}{4}$ .

Equilibrium payoffs: $v_1=3q=\tfrac{3}{2}$ and $v_2=2p=\tfrac{3}{2}$ $\frac{3}{2}$ .

Checking outside the support. Add a third row $B$ that gives player 1 the payoff $1$ against both columns. Against $q=\tfrac{1}{2}$ it earns $1<\tfrac{3}{2}$ $\frac{3}{2}$ , so the profile above (with $\sigma_1(B)=0$ $0$ ) is still an equilibrium. If $B$ instead gave player 1 the payoff $2$ against both columns, it would earn $2>\tfrac{3}{2}$ and the profile would fail the second condition.

Rock-Paper-Scissors. With payoffs $1$ (win), $-1$ (loss), $0$ (tie), each pure strategy earns $\tfrac{1}{3}(0)+\tfrac{1}{3}(1)+\tfrac{1}{3}(-1)=0$ against the uniform mixture, so uniform play by both players is an equilibrium with value $0$ .

Prerequisites¶

Sources¶

Game Theory · Week 2

Aliases: indifference principle, indifference condition, equal-payoff condition Keywords: support, make the opponent indifferent, solving for mixed equilibrium, 2×2 game, Rock-Paper-Scissors Symbols: element of, sigma, sum

Last updated: 27 October 2026 11:52