Final Exam: Evolutionarily Stable Strategies

An evolutionarily stable strategy is the move that, once common, no rare mutant can profitably invade — the equilibrium a population reaches by breeding, not by reasoning. Across this course you tested it with a single inequality, watched it settle Hawk–Dove into a mixed $p^* = V/C$ , tuned $1{:}1$ sex ratios, and explained why stubborn defection is as stable as any handshake. This final exam sweeps all five lessons at once. It is one-way: answer, and you have committed.

How this exam works

Questions appear one at a time. Once you submit an answer it is locked — there is no going back, no retry, and no restart. Some questions ask you to select every option that applies. Your score stays hidden until the very end, and you need 70% to pass. Read each stem carefully before you commit.

Question 1 of 24

A strategy I is an ESS. A rare mutant J appears and, played against the resident I, earns exactly what I earns against I — that is, E(I,I) = E(J,I). What additional condition must hold for I to remain an ESS?

E(J,J) > E(I,J): the mutant must do better against copies of itselfE(I,J) > E(J,J): when the tie is on the first line, I must do strictly better against the mutant than the mutant does against itselfNothing more is needed; equal payoffs already guarantee stabilityE(I,I) > E(J,J): the resident must beat the mutant in self-play

Select an answer to continue.

Course Recap

Big picture

Evolutionarily Stable Strategies — the whole arc

Evolutionarily Stable Strategies
- The invasion test
  - E(I,I) > E(J,I), OR tie then E(I,J) > E(J,J)
  - Derived from a fraction ε of mutants
  - AllD is an ESS; AllC is not
- Hawk–Dove
  - Hawk vs Hawk = (V − C)/2
  - V > C: pure Hawk is the ESS
  - C > V: mixed ESS at p* = V/C
  - ESS pays V(C − V)/(2C) < V/2 — wasteful
- ESS versus Nash
  - Every ESS is Nash; not every Nash is an ESS
  - Every strict Nash is an ESS
  - Mixed ESS are never strict Nash
  - Randomiser = stable polymorphism
- Frequency-dependent selection
  - Negative FD (rare-type advantage) → stable mix
  - Positive FD → lock-in
  - Fisher sex ratio → 1:1
  - RPS lizards → cycles, no ESS
- Cooperation and limits
  - TFT only neutrally stable; AllD is an ESS
  - Repeated dilemma is bistable — history picks the basin
  - Replicator dynamics: fitness − mean; ESS is an attractor
  - Limits: not optimal, may not exist, finite size, kin rB > C, space, naturalistic fallacy

Key Takeaways

What an ESS really claims

An evolutionarily stable strategy is the move that, once common, no rare mutant can profitably invade — a claim about resistance, not virtue. The whole test lives in one inequality: $E(I,I) > E(J,I)$ , or a tie there settled by $E(I,J) > E(J,J)$ . In Hawk–Dove that logic gives pure Hawk when $V > C$ and the mixed $p^* = V/C$ when $C > V$ — a mix everyone would gladly trade away, since it pays $V(C-V)/(2C)$ against all-Dove’s $V/2$ . ESS refines Nash: every ESS is Nash, every strict Nash is an ESS, but weak equilibria and mixed strategies fall between. Frequency dependence can stabilise a blend (sex ratios, scale-eating fish) or spin forever without a rest point (rock–paper–scissors lizards). And the honest fine print matters most: stability is not optimality, some games have no ESS while others have several, the model assumes an infinite well-mixed population of strangers, and “it evolved” never means “it is good.” Hold those together and you can read any strategic ecology — biological or social — for what actually resists invasion.