Chicken Game¶

Definition¶

The chicken game is a symmetric two-player normal-form game in which each player chooses $S$ (swerve, yield) or $T$ (go straight, stand firm), and each player's payoffs satisfy

v_i(T,S)>v_i(S,S) >v_i(S,T)>v_i(T,T).

Standing firm is best if the other yields, yielding is best if the other stands firm, and mutual standing firm is the worst outcome. A standard numerical version:

	$S$	$T$
$S$	$0,0$	$-1,1$
$T$	$1,-1$	$-3,-3$

Interpretation¶

Chicken is an anti-coordination game: each player wants to do the opposite of the other. It models standoffs such as two drivers heading for each other, price wars, or a confrontation in which backing down is costly but collision is disastrous. In biology the same structure is called the hawk–dove game.

Best-response diagram for the chicken table, with p and q the probabilities of T for players 1 and 2; the two step-shaped curves meet at (S,T), (T,S) and the mixed point (1/3, 1/3).

In the figure $p$ and $q$ are the probabilities with which players 1 and 2 choose $T$ in the table above. Player 1's best response $BR_1$ is $T$ ( $p =$ $p=1$ $1$ ) when $q<\tfrac{1}{3}$ $\frac{1}{3}$ , $S$ ( $p =$ $p=0$ $0$ ) when $q>\tfrac{1}{3}$ $\frac{1}{3}$ , and any $p$ when $q=\tfrac{1}{3}$ $\frac{1}{3}$ ; player 2's $BR_2$ is the mirror image. Because each player wants the opposite of the other, both curves step downwards and cross three times: at the two pure equilibria and at the mixed equilibrium $(\tfrac{1}{3},\tfrac{1}{3})$ .

Raise the crash cost $c$ and watch the mixed equilibrium $q=w/(w+c-\ell)$ fall towards $0$ ; with $c<\ell$ $ℓ$ , standing firm becomes dominant and only $(T,T)$ remains.

Compare it with the Prisoner's Dilemma. There, defecting is best whatever the other does, so there is a single equilibrium. In chicken, no action is dominant, $(T,T)$ is worse than being the one who yields, and the game has three equilibria. The theory alone does not say which one is played; the two pure equilibria require the players to know who yields.

Chicken is a standard example for purification: if each player's toughness, that is the payoff from standing firm, is slightly private, the mixed equilibrium can be reinterpreted as pure behaviour by different types, and the game becomes a Bayesian game.

Key Results¶

The pure Nash equilibria are $(S,T)$ and $(T,S)$ ; $(S,S)$ and $(T,T)$ are not equilibria.
There is exactly one further Nash equilibrium in mixed strategies, symmetric, in which each player chooses $T$ with the probability that makes the other indifferent.
No action is strictly dominated, so iterated elimination of strictly dominated strategies removes nothing.
The mixed equilibrium is Pareto-dominated by $(S,S)$ , which is not an equilibrium: it is not Pareto efficient.

Examples¶

Mixed equilibrium in the table above. Let the opponent play $T$ with probability $q$ . By the indifference principle, $S$ yields $-q$ and $T$ yields $(1-q)-3q=1-4q$ $4 q$ ; equating gives $q=\tfrac{1}{3}$ $\frac{1}{3}$ . By symmetry each player plays $T$ with probability $\tfrac{1}{3}$ , and each earns $-\tfrac{1}{3}$ .

The crash $(T,T)$ occurs with probability $\tfrac{1}{9}$ .
Both players get $0>-\tfrac{1}{3}$ at $(S,S)$ , which confirms the Pareto comparison.
In the pure equilibria the payoffs are $(-1,1)$ and $(1,-1)$ .

General parameters. With $v_i(S,S)=0$ $0$ , $v_i(T,S)=w>0$ $0$ , $v_i(S,T)=-\ell<0$ and $v_i(T,T)=-c$ with $c>\ell$ $ℓ$ , indifference gives $-\ell q=w(1-q)-cq$ $c q$ , so $q=w/(w+c-\ell)$ $ℓ)$ . A more costly crash (larger $c$ ) lowers the probability of standing firm. The table above has $w=1$ $1$ , $\ell=1$ $1$ , $c=3$ $3$ , so $q=1/3$ $1/3$ .

Prerequisites¶

Sources¶

Game Theory · Week 5

Aliases: game of chicken, chicken, hawk–dove game, hawk-dove game, snowdrift game Keywords: anti-coordination game, swerve or go straight, two pure equilibria and one mixed equilibrium

Last updated: 22 October 2026 19:04