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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.

A resident strategy is evolutionarily stable if every distinct mutant earns lower expected fitness when introduced at sufficiently small positive frequency. The upper bound on the frequency may depend on the mutant. In a finite symmetric matrix game, $e(x,y)=x^{T}Ay$ is the expected payoff, with the second argument representing the population strategy. Affineness in this argument yields the two-condition criterion for evolutionary stability. Every evolutionarily stable strategy is a symmetric Nash equilibrium, but a tied best response must also be strictly outperformed against its own mutant population. The Hawk-Dove game supplies an interior stable mixture despite equality of all payoffs against the resident mixture.

In a two-player symmetric finite game, represent mixed strategies by probability vectors and let $e(x,y)=x^{T}Ay$ be the payoff to a player using $x$ against one using $y$. A resident strategy $x^{\ast }$ is an evolutionarily stable strategy if every distinct mutant strategy $y$ has an invasion barrier ${\epsilon }_0(y)>0$ such that, for $0<\epsilon <{\epsilon }_0(y)$,Thus when mutants are sufficiently rare, the resident has strictly larger fitness against the population mixture. The barrier may depend on the mutant; a common barrier is not part of this definition.

For fixed $y\ne x^{\ast }$ putBy bilinearity the resident-minus-mutant fitness difference isIf this is positive for all sufficiently small positive $\epsilon$, taking $\epsilon \downarrow 0$ gives $a_y\ge 0$. If $a_y=0$, strict positivity requires $b_y>0$. Conversely, if $a_y>0$, choose $0<{\epsilon }_0\le a_y/[2(a_y+|b_y|)]$; then the difference is at least $a_y-\epsilon (a_y+|b_y|)>a_y/2>0$. If $a_y=0$ and $b_y>0$, it is $\epsilon b_y>0$ for every $0<\epsilon <1$. This proves the two-condition criterion for evolutionary stability: for every $y\ne x^{\ast }$, eitherorThe weak first comparison is the symmetric Nash equilibrium condition; the second excludes neutrally competitive mutants when that comparison is an equality.

For the Hawk-Dove game, let $x^{\ast }=(1/2,1/2)$ and $y=(p,1-p)$, $0\le p\le 1$. Against the resident, both pure strategies give $1/2$, soDirect multiplication of the payoff matrix givesConsequentlyThe equality case of the two-condition criterion for evolutionary stability therefore applies to every distinct mutant. More explicitly, the fitness difference in a mixed population is $2\epsilon (p-1/2{)}^2$, positive for every $0<\epsilon <1$. Hence $\boxed{(1/2,1/2)\text{ is an evolutionarily stable strategy}}$, with the common invasion barrier one.

In the usual positive-parameter case $0<V<D$, set $p^{\ast }=V/D\in (0,1)$. Every mutant ties the resident against the resident population, because $V-D{p}^{\ast }=0$. Against a mutant population, however,The second condition of the two-condition criterion for evolutionary stability holds, soIndeed its advantage at invasion frequency $ϵ$ is exactly $ϵD(p^{\ast }-p{)}^2/2$. The interior frequency balances the two pure-strategy payoffs in the Hawk-Dove game.

For completeness, if $V<D$ but $V<0$, pure Dove is an evolutionarily stable strategy by the strict inequality $e(0,0)-e(p,0)=-pV/2>0$ for every mutant $p>0$. If $V=0<D$, that difference is zero, but $e(0,p)-e(p,p)=D{p}^2/2>0$, so Dove again satisfies the second condition. Thus for the remaining allowed values $\boxed{x^{\ast }=(0,1)\text{ when }V\le 0<D\text{ or }V<D\le 0}$; equivalently, Dove works whenever $V<D$ and $V\le 0$.

For $V=D>0$, choose pure Hawk. Every mutant $p<1$ ties it against a resident Hawk population, so the second condition must be checked. The Hawk-Dove game formula givesConsequentlyThe actual invasion advantage is $ϵV(1-p{)}^2/2>0$, despite the tie at invasion frequency zero.

If the equal parameters are negative, pure Dove is instead an evolutionarily stable strategy: for a mutant $p>0$, $e(0,0)-e(p,0)=-pV/2>0$. If $V=D=0$, every payoff is zero and no resident has a strict advantage over any mutant. Thus the full equal-parameter answer is Hawk for $V=D>0$, Dove for $V=D<0$, and no ESS for $V=D=0$.

The payoff $e(x,z)$ is the expected fitness of an individual using strategy $x$ against a population using strategy $z$. The resident $x^{\ast }$ is initially common; the mutant $y$ is introduced at small frequency $ϵ$. Convexity of the strategy set makes their population mixture admissible. If residents have higher expected fitness in every sufficiently small invasion by a fixed mutant, selection works against that mutant: this is an evolutionarily stable strategy. The allowed upper bound on $ϵ$ may depend on $y$.

For expected payoffs of mixed strategies, $e(x,z)$ is affine in the population strategy $z$; in a finite matrix game it is $x^{T}Az$. This property is the basis of the two-condition criterion for evolutionary stability. For fixed $y\ne x^{\ast }$, putAffineness gives the exact invasion differenceIf this is positive for all sufficiently small $ϵ>0$, taking $ϵ\downarrow 0$ gives $A_y\ge 0$. If $A_y=0$, the difference is $ϵB_y$, so necessarily $B_y>0$. Conversely, if $A_y>0$, it remains positive for sufficiently small $ϵ$: for example take $ϵ<A_y/(A_y+|B_y|)$. If $A_y=0$ and $B_y>0$, it is positive for every $ϵ>0$. Applying this argument separately to every $y$ provesThus an evolutionarily stable strategy is a symmetric Nash equilibrium with an additional condition against tied mutants. For a general nonlinear interaction function on a convex set, affineness would be an additional necessary hypothesis for this particular equivalence; here it follows from the mixed-strategy expected-payoff interpretation. For example, on $X=[0,1]$ take $e(x,z)=xz(x/2-z)$ and resident $x^{\ast }=0$. For every $y>0$, $A_y=0$ and $B_y=y^3/2>0$, but at population $ϵy$ the mutant earns $ϵy^3(1/2-ϵ)>0$ while the resident earns zero. Thus the two conditions alone would not suffice for an arbitrary nonlinear $e$; the expected-payoff assumption is used explicitly.

For the Hawk-Dove game, let $p$ be an individual's Hawk probability and $q$ the population's Hawk probability. The given matrix yieldsWe use this formula in each case. In the usual interpretation, resource value $V$ and damage cost $D$ are positive. The parts below also specify the boundary or negative-value alternatives because positivity is not written explicitly in the PDF.