In a population consisting entirely of Doves, a single Hawk mutant appears. What happens, and what does this reveal about the stability of the all-Dove strategy?
AThe Hawk is outcompeted because cooperative Dove populations are more productive overall
BThe Hawk invades successfully — it wins every contest, getting V instead of V/2, so pure Dove is not an ESS
CThe Hawk is neutral — it gets the same fitness as Doves when it is rare because it rarely meets other Hawks
DThe Hawk invades only if V > C; otherwise it does no better than Doves
In an all-Dove population, every contest ends in display and resource sharing — each Dove gets V/2. A rare Hawk wins every contest because Doves retreat immediately, so the Hawk always gets V > V/2. This positive fitness advantage means the Hawk can invade and spread. The all-Dove state is therefore not an evolutionarily stable strategy (ESS) regardless of the values of V and C. Option D conflates the condition for a mixed ESS with the condition for Hawk to invade an all-Dove population — those are different questions.
Question 2 Multiple Choice
Under what condition is a pure Hawk strategy itself evolutionarily stable (resistant to invasion by Dove mutants)?
AWhen C > V — injury cost exceeds resource value, making fighting too risky for pure Hawks
BWhen V > C — resource value exceeds injury cost, so Hawks in an all-Hawk population still earn positive expected payoffs that exceed what a rare Dove earns
CNever — Hawk is always invaded by Doves regardless of V and C
DWhen V = C — the costs and benefits exactly balance, making Hawk neutral
In an all-Hawk population, each individual earns (V − C)/2 on average. A rare Dove earns 0 (it always retreats against Hawks). If V > C, then (V − C)/2 > 0 > 0... wait, the Dove earns exactly 0 regardless, and Hawks earn (V−C)/2. When V > C, (V−C)/2 > 0 > Dove's payoff? No — Dove earns 0, which equals Dove's payoff. Actually when V > C, Hawks get (V-C)/2 which is positive, and a rare Dove gets 0. Since 0 < (V-C)/2, Doves cannot invade — pure Hawk is an ESS. When C > V, (V-C)/2 is negative, so rare Doves get 0 which is better, and Dove can invade. This is why C > V is the condition for a mixed equilibrium.
Question 3 True / False
At the mixed ESS in the Hawk-Dove game (with C > V), the average fitness of Hawks exactly equals the average fitness of Doves, so neither strategy has a selective advantage.
TTrue
FFalse
Answer: True
The mixed equilibrium frequency V/C is defined by the condition that fitness is equalized across strategies. At this frequency, the payoff for playing Hawk (averaged over encounters with Hawks and Doves in proportion V/C and 1−V/C) equals the payoff for playing Dove. This equality is precisely what makes the equilibrium stable: if Hawks become too common, their average fitness drops below Doves' (because they encounter costly Hawk-Hawk fights more often), reducing their frequency back toward V/C. Conversely, if Doves become too common, Hawks do better, restoring the ratio.
Question 4 True / False
Natural selection consistently favors the most aggressive strategy in any population, because aggression provides direct access to resources.
TTrue
FFalse
Answer: False
The Hawk-Dove game demonstrates the opposite: fitness is frequency-dependent, not absolute. When Hawks are rare, aggression is highly profitable (they mostly meet Doves and win). As Hawks become common, they increasingly encounter each other, paying injury costs that erode fitness. At frequency V/C, Hawk fitness falls to equal Dove fitness and further spread stops. Natural selection does not maximize aggression — it produces a stable balance point where the costs and benefits of fighting exactly counterbalance each other. Species with dangerous weapons (snakes, deer) often show ritualized rather than lethal conflict, consistent with high C relative to V.
Question 5 Short Answer
Explain why neither pure Hawk nor pure Dove is an evolutionarily stable strategy when the cost of injury (C) exceeds the resource value (V), and describe the equilibrium that evolves instead.
Think about your answer, then reveal below.
Model answer: Pure Dove is unstable because a rare Hawk mutant always outcompetes Doves — it wins every contest (getting V vs. Dove's V/2) and can invade. Pure Hawk is unstable when C > V because in an all-Hawk population each individual earns (V−C)/2, which is negative; a rare Dove mutant earns 0 by retreating, which is better than a negative payoff, so Doves can invade. Since each pure strategy can be invaded by the other, neither is an ESS. The stable outcome is a mixed equilibrium: the proportion of Hawks in the population (or the probability any individual plays Hawk) settles at V/C, the frequency at which the average fitness of Hawks equals the average fitness of Doves, so neither strategy has a net selective advantage.
This illustrates frequency-dependent selection: the fitness of a strategy depends on how common it is. The equilibrium is self-correcting — any deviation from V/C creates a selective pressure that returns the population to that ratio.