Why do fights between animals of the same species so rarely end in death, even among animals with horns, fangs, and stings? In 1973 John Maynard Smith and George Price answered with "The Logic of Animal Conflict" in Nature, the paper that introduced the Hawk strategy and the evolutionarily stable strategy. A population does not need to be noble for the good of the species; a mix of fighters and displayers is simply stable, because once fighters are common, fighting stops paying. (Their gentle strategy was called Mouse in that paper. Dove came later.) Price was a strange and brilliant man, a chemist who taught himself evolution, wrote down an equation describing natural selection in general, gave away nearly everything he owned to homeless people in London, and died in 1975. The rock-paper-scissors lizards were found in California by Barry Sinervo in the 1990s.
Hawks and doves
Two animals want the same worm. Fighting wins it but can get you hurt. Backing off is safe but earns nothing when the other one fights. Nobody in the herd knows any game theory, and the herd finds the equilibrium anyway. You are going to predict it with one division, then watch 300 animals discover it on their own.
Two animals arrive at the same worm at the same moment. The worm is worth V points to whoever eats it. Each animal has one of two habits. A Hawk fights for the worm. A Dove puffs itself up and displays for a while, but backs off the moment the other one actually attacks.
So there are three kinds of meeting:
- Hawk meets Dove. The dove backs off. The hawk gets V, the dove gets 0.
- Dove meets Dove. Both display, nobody attacks, and they end up sharing: V/2 each.
- Hawk meets Hawk. They fight. The winner eats the worm, worth V. The loser gets hurt, which costs C points. Each hawk has a half chance of winning, so on average a hawk in a fight gets
(V − C)/2.
That is the whole Hawk-Dove game. The grid below shows it with V = 10 and C = 20. Look at the top-left cell: Hawk meets Hawk is −5 each. A fight over a 10-point worm costs more than the worm is worth, so two hawks both come out worse than if they had never met. Fighting is expensive. Move the sliders and watch every cell change.
Two ways to fight over a worm
Both animals are the same kind of animal, so the grid is symmetric.Colors in this chapter: the grid uses the site's row and column colors. Everywhere else, red is a hawk and blue is a dove.
What fraction of hawks?
Now picture a whole population of these animals. Every day each one meets a random other animal over a worm. Nobody decides to be a hawk or a dove. Each animal is born one or the other, and the animals that collect more points have more babies of their own kind. That sentence is the whole engine: a habit that earns more spreads, and a habit that earns less shrinks.
Say a fraction p of the population are hawks. (If p = 0.3, then 30 percent are hawks and 70 percent are doves.) A hawk's random partner is a hawk with probability p and a dove with probability 1 − p, so a hawk's average payoff is:
A dove gets 0 against a hawk and V/2 against a dove, so:
The letter E stands for expected value: the average you get if you play many times. If EH > ED, hawks are doing better, so next year there are more hawks and p goes up. If EH < ED, p goes down. The population settles where the two are equal. So set EH = ED and solve for p, one honest step at a time:
- Start with the two averages set equal:
p(V − C)/2 + (1 − p)V = (1 − p)V/2 - Subtract (1 − p)V/2 from both sides. The right side becomes 0. On the left, a whole (1 − p)V minus half of it leaves half of it:
p(V − C)/2 + (1 − p)V/2 = 0 - Multiply everything by 2 to clear the halves:
p(V − C) + (1 − p)V = 0 - Multiply out both brackets:
pV − pC + V − pV = 0 - The pV and −pV cancel:
V − pC = 0 - Add pC to both sides and divide by C:
pC = V, so p = V/C
The population settles at a fraction of hawks equal to the prize divided by the cost:
With V = 10 and C = 20, p* = 10/20 = 1/2: half hawks, half doves. A juicier worm (bigger V) means more hawks. Worse injuries (bigger C) mean fewer. And if C ≤ V, then V/C ≥ 1, which is not a fraction of anything: a hawk earns at least as much as a dove no matter what p is, and hawks take over completely.
The chart draws EH in red and ED in blue for every p from 0 to 1. Where the lines cross is p*. Slide p around and see who is winning on each side of the crossing.
Payoff versus p
Hawks spread where the red line is higher. Doves spread where the blue line is higher.Let it evolve
Time to stop calculating and let the animals do it. Below are 300 animals, red hawks and blue doves. Each generation, every animal is paired with a random partner and the pair plays the worm game once. Then the next generation is born: each kind gets a share of the 300 new animals in proportion to the total points its members collected. (Points can be negative, so the program adds C/2 to every score first. The Math corner explains why that changes nothing.) Nobody in the simulation knows what V/C is. Nobody solves an equation. They meet, eat, get hurt, and have babies.
Here is the deal. Before you press Run, work out V/C for the current sliders and type it into the prediction box, as a percent like 50% or a fraction like 1/2. Then run. The red line is the hawk fraction, generation by generation; once you start, a dashed line shows what the equation predicts. Land within 5 percentage points and that is a star. Then start from 5 percent hawks, then from 95 percent, then make the prize bigger than the cost and predict what happens to the doves.
The line wobbles because of luck: who bumps into whom changes a little each generation. Tick "No luck" to use exact averages instead, and the wobble disappears.
Let it evolve
300 animals. Random pairs. Babies in proportion to points.Rock, paper, scissors, lizard
Hawks and doves settle down. Some games never do. On dry hillsides in California lives the side-blotched lizard, Uta stansburiana, and its males come in three throat colors. Orange males are big, aggressive, and hold large territories with many females. Blue males hold a small territory and guard one mate closely. Yellow males have throats striped like a female's; they hold no territory and sneak in to mate while nobody is looking.
Orange beats blue: orange males simply take over blue territories. Blue beats yellow: a male guarding one mate closely spots the sneaker. Yellow beats orange: an orange male cannot watch a huge territory, so the sneakers get through. That is exactly rock, paper, scissors. Whenever one color becomes common, the color that beats it spreads, and then gets beaten in turn by the third. The biologists Barry Sinervo and Curt Lively watched a real population for six years, saw the three colors take turns being the most common, and published it in Nature in 1996.
The simulation below uses the same babies-in-proportion-to-points rule with a rock-paper-scissors grid: beating a color is worth +1, losing to one is −1, a tie is 0, plus 4 to keep every score positive. One dose of reality: 1 newborn in 200 gets a random color, so no color ever vanishes. Watch the three lines chase each other.
Three lizards
Orange beats blue, blue beats yellow, yellow beats orange.The algebra in one line. Subtract the two averages and simplify:
This is positive when p < V/C and negative when p > V/C. Hawks gain when they are rarer than V/C and lose when they are more common, so the population is pushed toward p* = V/C from both sides, whichever side it starts on.
Why adding C/2 is fair. The simulation's rule is pnext = p · FH / (p · FH + (1 − p) · FD), where each F is the average plus C/2. A hawk's average is never below (V − C)/2, so every F is at least V/2, which is positive. Adding the same number to both does not change which is bigger, so p still rises exactly when EH > ED and stays put when they are equal. Biologists call this rule the replicator equation, and the whole subject evolutionary game theory.
Evolutionarily stable strategy. (say: ee-vuh-LOO-shun-air-uh-lee) A strategy is an ESS if, once nearly everyone uses it, any small group doing something else earns less and dies out. When C > V the ESS is the mix "be a hawk with probability V/C": extra hawks push p above V/C and lose, and extra doves push it below and lose too. When C ≤ V, Hawk is the ESS: a lone dove among hawks gets 0 while the hawks get (V − C)/2, which is at least 0, and when a hawk meets that dove it takes the whole worm.
Back to Chapter 4. There you found a mixed strategy by choosing the odds that make the other player indifferent. EH = ED is that same trick: hawk and dove earn the same, so nobody has a reason to switch. And p = V/C is exactly the mixed Nash equilibrium (Chapter 3) of the one-shot Hawk-Dove game, in which each player plays Hawk with probability V/C. Same equation, same answer. The difference is who does the mixing. In Chapter 4 a player mixes on purpose. Here no animal mixes at all: each is a pure hawk or a pure dove, and the population as a whole sits at the Nash odds. Evolution did the arithmetic.
A population that keeps playing the same game and copies whatever earns the most drifts to an equilibrium with nobody calculating anything. You find that equilibrium the same way you find a mixed strategy: set the two average payoffs equal and solve. For hawks and doves the answer is p* = V/C. The fraction of fighters is the prize divided by the cost of losing.
1. V = 6 and C = 12. What fraction of the population ends up hawks?
Answer
p* = V/C = 6/12 = 1/2. Half hawks, the same as V = 10, C = 20, because only the ratio matters.
2. V = 8 and C = 6. What happens?
Answer
V/C = 8/6 is more than 1, so there is no fraction where hawks and doves tie. The gap (V − pC)/2 = (8 − 6p)/2 is at least (8 − 6)/2 = 1 for every p up to 1. Hawks always earn more and take over completely. A fight over an 8-point worm that costs only 6 to lose is worth having.
3. V = 10, C = 20, and right now p = 1/4. Who is doing better, hawks or doves, and by how much?
Answer
EH = (1/4)(−5) + (3/4)(10) = 6.25 and ED = (3/4)(5) = 3.75. Hawks are ahead by 2.5, so p is rising toward 1/2. The shortcut agrees: (V − pC)/2 = (10 − 5)/2 = 2.5.
Add a third kind of animal, Bourgeois (say: boor-ZHWAH): if it reached the worm first, it fights like a hawk; if it arrived second, it backs off like a dove. At every meeting each animal is equally likely to have been first. (a) Show that two Bourgeois animals never fight and each gets V/2 on average. (b) With V = 10, C = 20, and a population settled at half hawks, half doves, work out what one Bourgeois animal earns. Careful: it is a tie, not a win. Then argue why Bourgeois invades anyway. (c) Once everyone is Bourgeois, show that neither a lone hawk nor a lone dove can invade.
Answer
(a) One was first and one was second. The first fights, the second backs off: no fight. Each animal is first half the time, so it gets V half the time and 0 the other half, V/2 on average, with no injuries ever.
(b) At p = 1/2, hawks and doves both earn (1/2)(5) = 2.5. A Bourgeois animal fights half the time and backs off half the time, so to animals who cannot see who was first it is a coin flip between hawk and dove, and it earns half of 2.5 plus half of 2.5: a tie. So at first it does not spread. But two Bourgeois meeting each get 5, while a hawk or a dove meeting a Bourgeois gets only 2.5 (a hawk: half of −5 and half of 10; a dove: half of 0 and half of 5). Every extra Bourgeois helps the other Bourgeois and nobody else, so once there are a few of them they pull ahead.
(c) Among Bourgeois, a lone hawk meets an owner half the time (a fight, −5 on average) and an intruder half the time (a free worm, 10): average 2.5. A lone dove gets 0 from owners and shares 5 with intruders: 2.5. The residents earn 5, so both invaders lose: Bourgeois is an ESS whenever C > V. John Maynard Smith named it, and it is one reason animals so often respect "I was here first".
Stars in this chapter
Earn them by doing the clever thing, not by clicking around.
- Predict the hawk fraction before the simulation finds it
- Set a prize worth fighting for and call the hawk takeover
- Watch the lizards cycle for 200 generations