Hawk–Dove: The Flagship Game

Every deep idea in this course gets a home in one game, and this is it. Hawk–Dove is the Drosophila of evolutionary game theory — the fruit fly everyone breeds because it is small enough to work out with a pencil and rich enough to show the whole machine running. Maynard Smith and Price built it in 1973 to answer a question that had nagged biologists for decades: if fighting wins resources, why don’t animals fight to the death over every scrap? The stag roars and backs down; the snake wrestles but rarely bites; the fish flares its gills and lets the intruder go. Restraint is everywhere, and natural selection is supposed to be ruthless. What gives?

The answer, worked out below to the last decimal, is that restraint is not a favour a population does itself — it is an equilibrium selection is forced into. By the end you will be able to take any resource value and any injury cost, and predict the exact mix of aggressors a population will hold, forever, without anyone choosing it.

The setup: two strategies, one prize

Picture two animals arriving at the same patch of food, the same nest site, the same mate — one resource worth V in fitness (extra offspring) to whoever ends up with it. Each animal is hard-wired to one of two strategies:

• A Hawk always escalates. It fights, and keeps fighting until it wins or is injured.
• A Dove always displays — puffs up, postures, makes noise — but the moment the opponent actually fights, it flees unhurt.

Now walk through the four possible meetings, because the entire model lives in these four sentences:

• Hawk meets Dove. The Dove flees at the first sign of a real fight. The Hawk takes the whole resource, uninjured: Hawk gets V, Dove gets 0.
• Dove meets Hawk. Same encounter from the other side: Dove gets 0, Hawk gets V.
• Dove meets Dove. Two posturers, neither willing to fight. They split the resource (or each wins it half the time, which averages the same): each gets V/2.
• Hawk meets Hawk. Two escalators — a real fight. One wins the resource (V), the other is injured (pays C). Each is equally likely to be the winner, so the average payoff to a Hawk in this meeting is (V − C)/2.

That last line is the hinge of the whole model, so read it slowly. A Hawk facing another Hawk sometimes wins V and sometimes pays C, fifty-fifty, so on average it collects $(V-C)/2$. When injury is cheap that number is positive and fighting pays; when injury is dear it goes negative — the average Hawk-vs-Hawk meeting is a net loss. Which regime you’re in decides everything.

We can pack all four meetings into a payoff matrix. Each cell is the payoff to the row player when it meets the column player:

Case 1 — when fighting is cheap (V > C)

Start with the easy world: the prize is worth more than an injury costs, so V > C. A stag fighting over a whole harem when a scuffle only bruises; a bird contesting the last territory before winter when losing is death anyway. Here $(V - C)/2$ is positive — even a Hawk-vs-Hawk brawl pays off on average.

Let’s test whether pure Hawk is an ESS using the invasion test. Imagine a population of all Hawks and drop in a rare Dove mutant. Compare how each does against the resident Hawk crowd:

The invasion test’s first condition asks: is $E(\text{Hawk},\text{Hawk}) > E(\text{Dove},\text{Hawk})$? That is, is $(V-C)/2 > 0$? When $V > C$, yes — so pure Hawk is a strict best reply to itself, and it is the ESS. The rare Dove earns 0 while everyone around it earns a positive $(V-C)/2$, so selection erases it.

Worked numbers. Let V = 8, C = 4. Then a Hawk meeting a Hawk earns $(8 - 4)/2 = 2$, while a Dove mutant earns 0. Two beats zero; the Dove starves out. Everyone fights, and — because winning is worth more than getting hurt — everyone should.

Case 2 — when fighting is costly (C > V), the interesting case

Now flip the world. Injuries cost more than any single prize is worth: C > V. A real fight over one meal can leave you crippled for a season; the antlers can gore, the fangs can kill. This is the realistic regime for most animal contests, and it is where the model earns its fame — because now $(V - C)/2$ is negative. A Hawk-vs-Hawk fight is, on average, a loss.

Watch both pure strategies fail the invasion test in turn.

Pure Hawk is not an ESS. In an all-Hawk world, each resident earns $E(\text{Hawk},\text{Hawk}) = (V-C)/2 < 0$ — everyone’s bleeding fitness into pointless fights. A rare Dove mutant earns $E(\text{Dove},\text{Hawk}) = 0$. Since $0 > (V-C)/2$, the Dove does better than the crowd and invades. Pure aggression collapses.

Pure Dove is not an ESS either. In an all-Dove world, each resident earns $E(\text{Dove},\text{Dove}) = V/2$ — everyone politely splitting resources, a little utopia. But a rare Hawk mutant walks in and earns $E(\text{Hawk},\text{Dove}) = V$, because every Dove it meets simply flees and hands over the whole prize. Since $V > V/2$, the Hawk invades. The peaceful utopia is a sucker’s paradise — one aggressor cleans up.

So we’re cornered: neither pure strategy can hold. Push the population toward all-Hawk and Doves invade; push it toward all-Dove and Hawks invade. Whenever both pure strategies are invadable like this, the ESS — if one exists — has to be a mixed strategy: a specific blend of Hawk and Dove that no rare mutant can beat. Finding that exact blend is the next section.

Solving for the mixed ESS: p* = V/C

Here is the payoff of the whole lesson. Let p be the fraction of the population playing Hawk (so 1 − p play Dove). We want the mix at which Hawk and Dove earn exactly the same — because at any other mix, one type earns more, spreads, and shifts the proportion. The stable point is where the tug-of-war balances.

Write the fitness of each type against a population that is a fraction p Hawk. A random opponent is a Hawk with probability p and a Dove with probability 1 − p:

$W_H(p) = p \cdot \frac{V - C}{2} + (1 - p) \cdot V$

$W_D(p) = p \cdot 0 + (1 - p) \cdot \frac{V}{2}$

A Hawk sometimes meets Hawks (earning $(V-C)/2$) and sometimes Doves (earning $V$). A Dove earns nothing against Hawks and $V/2$ against other Doves. At the mixed ESS these are equal — set $W_H(p) = W_D(p)$ and turn the crank:

$p \cdot \frac{V - C}{2} + (1 - p) \cdot V = (1 - p) \cdot \frac{V}{2}$

Move the right side over. Notice $(1-p)V$ on the left minus $(1-p)\tfrac{V}{2}$ on the right leaves $(1-p)\tfrac{V}{2}$:

$p \cdot \frac{V - C}{2} + (1 - p) \cdot \frac{V}{2} = 0$

Multiply everything by 2 to clear the halves:

$p(V - C) + (1 - p)V = 0$

Expand the second term:

$pV - pC + V - pV = 0$

The $pV$ and $-pV$ cancel — the model practically solves itself:

$V - pC = 0 \quad\Longrightarrow\quad V = pC \quad\Longrightarrow\quad \boxed{p^* = \dfrac{V}{C}}$

That is the headline result of evolutionary game theory: a population settles on a Hawk fraction of exactly V/C. The share of aggressors equals the ratio of what’s worth winning to what a fight costs to lose. No individual computes this. Selection grinds down any deviation until the proportion lands here — and it’s an ESS, uninvadable, because at $p^*$ every type earns the same and nothing rare can do better.

Worked example. Let V = 6, C = 10. Then $p^* = 6/10 = 0.6$ — the stable population is 60% Hawks and 40% Doves. Drop the animals at any other mix and selection walks them back to 60/40.

A second one. Let V = 4, C = 12. Then $p^* = 4/12 = 1/3 \approx 0.33$ — about 33% Hawks. Fighting is three times as costly as the prize is valuable, so only a third of the population can afford to be aggressive before Hawk-vs-Hawk collisions drag the aggressors’ payoff below the Doves’.

See it settle: drag the population to the ESS

Reading $p^* = V/C$ is one thing; watching a population crawl to it from a lopsided start is another. In the lab below, each line is a type’s fitness against the current mix. Where they cross, Hawk and Dove earn the same — that crossing is the ESS at V/C. Load it hot at 95% Hawks and let selection run.

Notice what the third experiment shows: the formula $p^* = V/C$ only works while C > V. The moment fighting gets cheap, the crossing slides off the edge of the graph and we’re back in Case 1, where pure Hawk is the ESS. The lab makes visible what the algebra warned about — a pitfall we’ll pin down formally in a moment.

The cost of stability: stable ≠ optimal

Now for the twist that makes Hawk–Dove more than a curiosity. We’ve found the population settles at $p^* = V/C$. But is that a good place to be? Let’s compute what everyone actually earns there — and compare it to what they could have had.

At the ESS every type earns the same, so we can just evaluate the Dove’s fitness at $p^*$ (it’s easier — Doves never touch the messy $(V-C)/2$ term). Recall $W_D(p) = (1-p)\cdot \tfrac{V}{2}$, and substitute $p^* = V/C$:

$W_D(p^*) = \left(1 - \frac{V}{C}\right)\frac{V}{2} = \frac{V(C - V)}{2C}$

Worked numbers, V = 6, C = 10. The Hawk fraction is $1 - V/C = 1 - 0.6 = 0.4$, so each individual earns $0.4 \cdot \tfrac{6}{2} = 0.4 \cdot 3 = \mathbf{1.2}$.

Now the gut-punch. Suppose the whole population could somehow be all Doves — everyone politely sharing, nobody fighting. Each would earn $V/2 = 6/2 = \mathbf{3}$. That’s more than double the 1.2 they actually get at the ESS. The stable mix is leaving fitness on the table, hand over fist, forever.

The all-Dove utopia earns 3 each and the population would love to live there — but it can’t, because it isn’t stable. A single Hawk mutant in that peaceful world earns the full V = 6 by mugging every Dove it meets, so aggression spreads until the population is dragged down to the 1.2-earning ESS and pinned there. Every individual would be better off in a world with less fighting, and not one of them can unilaterally get there — a Dove who keeps yielding just gets exploited. The population is trapped burning fitness on fights it cannot collectively stop.

Two ways to read p* = V/C

One subtlety before the recap, and it foreshadows the next lesson. The result “60% Hawks” can describe two physically different worlds that a biologist cannot tell apart from the payoffs alone:

1. A monomorphic mixed strategy. Every animal is identical and plays Hawk with probability V/C and Dove otherwise — each individual flips a biased coin at every contest. The population looks 60% Hawk at any instant because each animal is Hawkish 60% of the time.
2. A stable polymorphism. A fraction V/C of the animals are pure, lifelong Hawks and the remaining 1 − V/C are pure, lifelong Doves. No individual randomises; the population is 60% Hawk because 60% of the bodies are Hawks.

Both give the identical population state, the identical payoffs, the identical stability. In Hawk–Dove they are indistinguishable at the level of fitness — which is why the ESS is sometimes an individual’s dice and sometimes a census of two fixed types. Lesson 3 makes this distinction sharp and shows when the two readings come apart; for now, just hold that “V/C Hawks” can mean one flexible animal or two rigid castes.

Three pitfalls that trip everyone

Spaced recall: back to the invasion test

Recap

You now own the flagship game end to end: from four sentences about who fights whom, to a payoff matrix, to the invasion test, to the exact stable fraction $p^* = V/C$ and the fitness it earns — and the sobering fact that this stable point wastes fitness the population can’t reclaim.

Next up — ESS and Nash: The Extra Bar. You’ve seen an ESS emerge as a stable mix; next we place it precisely alongside the Nash equilibrium you know from game theory (every ESS is a Nash equilibrium, but not every Nash equilibrium is an ESS), and we finally split apart the two readings of $p^* = V/C$ — one animal rolling dice versus two fixed types living side by side.