Evolutionary Game Theory

When an individual’s fitness depends on what everyone else is doing, evolution becomes a game. Instead of climbing a fixed fitness landscape, strategies compete against each other, and the “best” move depends on which moves are currently common.

Replicator dynamics of the Hawk–Dove game converge to the mixed ESS from any starting frequency.

Payoffs and strategies

A game is specified by a payoff matrix that gives the reward to each strategy when it meets each other strategy. Write $A_{ij}$ for the payoff to a player using strategy $i$ against an opponent using strategy $j$ . The canonical biological example is the Hawk–Dove game, in which two animals contest a resource of value $V$ , and escalated fights cost $C$ .

Hawks always escalate; Doves display but retreat if the opponent escalates. Two Hawks fight, split the value, and share the injury cost, for an expected payoff $(V-C)/2$ . A Hawk meeting a Dove takes the whole resource, $V$ ; the Dove gets nothing. Two Doves share peacefully, each getting $V/2$ .

$A = \begin{bmatrix} (V-C)/2 & V \\ 0 & V/2 \end{bmatrix}, \qquad \text{rows and columns ordered } (\text{Hawk}, \text{Dove}).$

Evolutionarily stable strategies

An evolutionarily stable strategy (ESS), introduced by Maynard Smith, is a strategy that, once adopted by (almost) the whole population, cannot be invaded by any rare mutant. Formally, strategy $s$ is an ESS if for every alternative $s'$ either its self-payoff is strictly larger, $A(s,s) > A(s',s)$ , or the payoffs tie against $s$ and $s$ does better against the mutant, $A(s,s) = A(s',s)$ and $A(s,s') > A(s',s')$ . The first condition says $s$ is a best reply to itself (a Nash equilibrium); the second breaks ties in favor of the resident.

An ESS is the game-theoretic version of an uninvadable phenotype, and it links directly to the optimum concepts of optimization: it is a fitness peak defined relative to the current population rather than in isolation.

The replicator equation

To see how strategy frequencies actually change, let $x_i$ be the frequency of strategy $i$ , with $\sum_i x_i = 1$ . Its expected payoff is $f_i(\mathbf{x}) = \sum_j A_{ij} x_j$ , and the population mean payoff is $\bar f(\mathbf{x}) = \sum_j x_j f_j(\mathbf{x})$ . The replicator equation says a strategy grows in proportion to how much it beats the average:

$\dot{x}_i = x_i\big(f_i(\mathbf{x}) - \bar f(\mathbf{x})\big).$

Strategies that earn more than the mean increase in frequency; those below the mean decline; and the simplex $\sum_i x_i = 1$ is preserved because $\sum_i \dot x_i = \bar f - \bar f = 0$ .

ESS and fixed points

The fixed points of the replicator dynamics are exactly the states where every surviving strategy earns the mean payoff, $f_i = \bar f$ for all $i$ with $x_i > 0$ . This connects the two viewpoints: an ESS is an asymptotically stable fixed point of the replicator dynamics, an equilibrium that neighbouring frequency mixtures flow back toward. Checking that stability is the same linear-stability calculation used for any dynamical system in equilibria and stability; like the coupled feedbacks of predator–prey models, the game can also produce cycles or coexistence rather than a single winner.

Worked example: the Hawk–Dove mixed ESS

Assume war is expensive, $C > V$ , so neither pure strategy is stable: a population of all Doves is invaded by Hawks (who take $V$ from every Dove), while a population of all Hawks is invaded by Doves (who avoid the ruinous $(V-C)/2$ ). Look instead for a mixed ESS: a fraction $\hat p$ of Hawks at which Hawk and Dove earn equal payoffs.

The expected payoffs, given a Hawk frequency $p$ , are

$f_H = p\,\frac{V-C}{2} + (1-p)\,V, \qquad f_D = p\cdot 0 + (1-p)\,\frac{V}{2}.$

Setting $f_H = f_D$ and simplifying, the $V$ and $pV$ terms cancel to leave $-pC + V = 0$ , so

$\hat p = \frac{V}{C}.$

The population settles where the fraction of Hawks equals the value-to-cost ratio. Cheap resources or costly fights (small $V/C$ ) yield a mostly peaceful population; valuable resources with mild costs (large $V/C$ ) yield mostly aggression. For $V = 2$ and $C = 3$ , $\hat p = 2/3$ : two-thirds Hawks is the uninvadable mix.

In code

We integrate the replicator dynamics for the Hawk–Dove game with $V = 2$ , $C = 3$ and confirm convergence to $\hat p = V/C = 2/3$ .

R

library(deSolve)

A <- matrix(c(-0.5, 2,      # Hawk-Dove payoffs, V = 2, C = 3
               0.0, 1),
            nrow = 2, byrow = TRUE)

repl <- function(t, x, p) {
  f    <- as.vector(A %*% x)   # payoff to each strategy
  fbar <- sum(x * f)           # mean payoff
  list(x * (f - fbar))
}

out <- ode(c(H = 0.9, D = 0.1), seq(0, 50, 0.1), repl, parms = NULL)
tail(out, 1)                   # H -> 0.6667 = V/C, D -> 0.3333

Python

import numpy as np
from scipy.integrate import solve_ivp

A = np.array([[-0.5, 2.0],     # Hawk-Dove payoffs, V = 2, C = 3
              [ 0.0, 1.0]])

def repl(t, x):
    f = A @ x
    return x * (f - x @ f)     # x_i (f_i - mean payoff)

sol = solve_ivp(repl, (0, 50), [0.9, 0.1], t_eval=[50])
print(sol.y[:, -1])            # -> [0.667, 0.333], Hawk freq = V/C = 2/3

[0.66673614 0.33326386]

Julia

A = [-0.5 2.0;                 # Hawk-Dove payoffs, V = 2, C = 3
      0.0 1.0]

x = [0.9, 0.1]; dt = 0.01
for _ in 1:5000
    f = A * x
    global x += dt .* (x .* (f .- x'f))   # x'f is the mean payoff
end
println(x)                     # -> [0.667, 0.333], Hawk freq = V/C = 2/3

Why it matters

Evolutionary game theory reframes natural selection for the common situation where fitness is frequency-dependent: mating systems, foraging and contest behaviour, sex ratios, and microbial public goods. It explains why populations often settle at mixtures rather than a single optimal type, and it supplies the machinery — payoff matrices, the ESS criterion, and the replicator equation — used to analyse the evolution of cooperation and the evolution of virulence.