Hawk-Dove game ($e(p,q)=V(1-q)/2+p(V-Dq)/2$)

Hawk fights for a resource of value $V$, while Dove retreats from Hawk and shares with another Dove. Hawk–Hawk competition incurs damage cost $D$. The expected payoff difference between Hawk probabilities $p,r$ against population probability $q$ is $(p-r)(V-Dq)/2$. For positive $V,D$, pure Hawk is an evolutionarily stable strategy when $V>D$. At $V=D$, it remains stable because a mutant $p<1$ loses $V(1-p{)}^2/2$ against its own population. When $0<V<D$, the stable mixture has Hawk probability $p^{\ast }=V/D$: all first-condition payoffs tie, and its second-condition advantage is $D(p-p^{\ast }{)}^2/2>0$. For nonpositive resource values, pure Dove is stable if $V<0$ or if $V=0<D$. At $V=D=0$ all payoffs vanish and there is no evolutionarily stable strategy.

The Chicken payoff ordering describes a symmetric two-player bimatrix game in which aggression earns temptation $T$ against yielding, mutual yielding earns $R$, yielding to aggression earns $S$, and mutual aggression earns $P$, with the displayed strict ordering. Aggression is a best response to yielding, and yielding is a best response to aggression. Thus the two asymmetric action pairs are pure strategy Nash equilibria. The Hawk-Dove game has this ordering in its positive-resource regime where the damage cost exceeds the resource value. In the prisoner's dilemma, the ordering of $S$ and $P$ is reversed, making defection a strictly dominant strategy instead.