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Animals of the same species fight over food and mates, but rarely to the death. Too much aggression and everyone gets injured; too little and the meek get pushed aside. Evolution settles on a mix, and game theory predicts exactly what fraction.

Evolutionarily stable strategies

Nash equilibrium assumes reasoning. Evolution needs none: strategies that do well reproduce, those that do badly die out. An evolutionarily stable strategy is one that, once common, resists invasion by any rare mutant. In Hawk-Dove it is a precise mix of aggression and restraint.

From earlier lessons

Two quick ones before we start.

A mixed strategy is a probability distribution over...

At a mixed equilibrium a player is...

Predict, then play

Hawks fight; Doves back down. Fighting over a resource worth $V$ risks an injury cost $C$ . Set the values and let the population evolve: the higher-fitness type reproduces faster.

Why did that happen?

Take a guess, then reveal the reason

The mix is self-correcting. When Hawks are scarce, they mostly meet Doves and grab the resource uncontested, so they thrive and their share rises. When Hawks are common, they mostly meet other Hawks and pay the injury cost, so they do worse than Doves and their share falls. The only fraction where neither type has an edge, so neither grows, is $V/C$ . No animal is reasoning; the arithmetic of reproduction does the work.

The math

Definition, evolutionarily stable strategy

A strategy is an ESS if, when almost everyone plays it, no rare mutant playing anything else can earn higher average fitness and invade. Formally $s$ is an ESS if for every alternative $s'$ , either $u(s,s) > u(s',s)$ , or $u(s,s) = u(s',s)$ and $u(s,s') > u(s',s')$ . Every ESS is a Nash equilibrium, but the invasion condition is strictly stronger.

Hawk-Dove payoffs: two Hawks split the resource and share the injury, $(V-C)/2$ ; a Hawk against a Dove takes it all, $V$ ; a Dove against a Hawk gets nothing; two Doves share peacefully, $V/2$ . With $V = 2$ , $C = 6$ , that is Hawk-Hawk $= -2$ , Hawk-Dove $= 2$ , Dove-Hawk $= 0$ , Dove-Dove $= 1$ .

When $C > V$ there is no pure ESS: all-Hawk is invadable by Doves (Hawks injure each other), and all-Dove is invadable by Hawks (a lone Hawk cleans up). The stable mix sets Hawk and Dove fitness equal. At Hawk fraction $p$ , Hawk fitness is $2 - 4p$ and Dove fitness is $1 - p$ ; equating gives $p^* = V/C = 1/3$ .

Work it out

Find the stable Hawk fraction

At a Hawk fraction $p$ , a Hawk meets a Hawk with probability $p$ (payoff $-2$ ) and a Dove with probability $1-p$ (payoff $2$ ), so its fitness is $-2p + 2(1-p) = 2 - 4p$ . A Dove earns $0$ against Hawks and $1$ against Doves, so its fitness is $1 - p$ .

Set them equal: $2 - 4p = 1 - p$ , giving $3p = 1$ , so $p^* = 1/3 = V/C$ . Below $1/3$ , Hawks earn more and their share grows; above $1/3$ , they earn less and it shrinks. So $1/3$ is stable against invasion from either side, exactly the ESS. Raising the injury cost $C$ lowers the stable Hawk share, since fighting becomes more dangerous.

Key idea

An evolutionarily stable strategy resists invasion by any rare mutant, a stronger test than Nash. In Hawk-Dove with costly injuries, the stable population is a mix, with a Hawk fraction of exactly V/C.

Check yourself

Check yourselfapply

Injury cost rises from C = 6 to C = 10 (value still 2). What happens to the stable Hawk fraction?

Check yourselfanalyse

Why is all-Hawk not evolutionarily stable when C > V?

Check yourselfbrings back: mixed strategies

How does the ESS relate to the mixed Nash equilibrium of Hawk-Dove?

Sources

Maynard Smith, J., and Price, G. R. (1973). The logic of animal conflict. Nature, 246, 15-18.