ESS and Nash: The Extra Bar

So far you have two tools. The invasion test tells you when a strategy $I$ is an ESS: for every mutant $J$, either $E(I,I) > E(J,I)$, or the payoffs tie against the resident ($E(I,I) = E(J,I)$) and $I$ wins the tie-breaker $E(I,J) > E(J,J)$. And the Hawk–Dove game handed you a mixed ESS at $p^* = V/C$. This lesson connects that machinery to the most famous idea in game theory — the Nash equilibrium — and shows exactly where the two concepts agree, where they part ways, and why the difference is not a technicality but the whole point.

The headline, in one sentence: an ESS is a Nash equilibrium that has passed one extra exam. Nash asks, “would a rational player want to switch?” ESS asks that too — and then asks the tougher question, “and can a lucky, drifting mutant creep in even when switching is a wash?”

First, what a symmetric Nash equilibrium actually says

Before we can say “ESS refines Nash,” pin down the Nash side in the exact language we’ve been using. In a symmetric one-population game, a strategy $I$ is a Nash equilibrium (NE) if it is a best reply to itself:

$E(I,I) \ge E(J,I) \quad \text{for all } J.$

Read it aloud: if everyone around you is playing $I$, no alternative strategy $J$ earns strictly more against that crowd than $I$ does. Nobody has a reason to unilaterally deviate. That is the entire content of Nash — a no-regret condition. It is about a lone rational agent staring at a fixed population and asking, “given what everyone else does, can I do better by changing my mind?” If the answer is no, you’re at a Nash equilibrium.

Notice the inequality is weak ($\ge$, not $>$). That single choice of symbol is where all the drama lives. A strict Nash equilibrium uses $>$: every deviation does strictly worse. A weak Nash equilibrium allows ties: some deviation does exactly as well. Hold that distinction — it returns with a vengeance.

ESS ⊆ Nash: every stable strategy is an equilibrium

Now the first big claim, and it’s cleaner than you’d expect: every ESS is a Nash equilibrium. Set-wise, $\text{ESS} \subseteq \text{NE}$.

Why must this be true? Look at the two ways a strategy $I$ can pass the invasion test against a mutant $J$:

1. Condition 1 fires: $E(I,I) > E(J,I)$. That already gives you $E(I,I) \ge E(J,I)$ — with room to spare.
2. Condition 1 ties, Condition 2 fires: here $E(I,I) = E(J,I)$, which also gives you $E(I,I) \ge E(J,I)$ — with equality.

Either branch delivers $E(I,I) \ge E(J,I)$. Since $I$ has to beat every mutant $J$ to be an ESS, it satisfies $E(I,I) \ge E(J,I)$ for all $J$ — and that is precisely the Nash condition. So an ESS is automatically a best reply to itself.

The intuition is almost too obvious once you see it: if $I$ were not a best reply to itself, some $J$ would earn strictly more against the resident $I$-crowd. That $J$ would then have a fitness edge the instant it appeared and would invade immediately. A strategy that a better reply can walk straight through cannot possibly be uninvadable. Uninvadability requires, at minimum, being a best reply to yourself. ESS demands that floor and then some.

A Nash equilibrium that is NOT an ESS

Here’s the payoff for all that setup: the inclusion goes only one way. Not every Nash equilibrium is an ESS. The ESS piles on Condition 2, the tie-breaker, and that extra bar knocks out equilibria that are stable to reasoning but not to drift and mutation.

Let’s build the smallest possible counterexample and grind through the arithmetic. Two strategies, $A$ and $B$. Payoffs to the row player:

Read the table the same way every time: the row is you, the column is your opponent, the cell is your payoff. So $E(A,A)=1$, $E(A,B)=1$, $E(B,A)=1$, $E(B,B)=2$.

Step 1 — Is $A$ a Nash equilibrium? Compare $A$ against the only alternative, $B$, when the resident is $A$:

• $E(A,A) = 1$
• $E(B,A) = 1$

They tie. No strategy earns strictly more than $A$ against an $A$-crowd, so $A$ is a best reply to itself. $A$ is a (weak) Nash equilibrium. A rational player in an all-$A$ world has no strict reason to switch — deviating to $B$ pays exactly the same 1.

Step 2 — Is $A$ an ESS? Condition 1 already failed (it’s a tie, not a strict win), so everything rides on Condition 2: we need $E(A,B) > E(B,B)$.

• $E(A,B) = 1$
• $E(B,B) = 2$
• Is $1 > 2$? No. Condition 2 fails.

So $A$ is NOT an ESS, even though it is a genuine Nash equilibrium.

Now the evolutionary story, which is the part rationality-based Nash simply cannot see. Because $B$ does equally well against $A$ (both score 1), a $B$-mutant is selectively neutral in an $A$-world — selection neither punishes nor rewards it. So $B$ can drift upward in frequency by pure chance, generation after generation, with no fitness penalty stopping it. And the moment $B$ becomes common enough that $B$‘s meet other $B$‘s, the trap springs: $B$-vs-$B$ pays $2$, beating the $1$ that $A$-vs-$A$ pays. $B$ now has the edge, and it sweeps to fixation. $A$ was Nash, and evolution walked right past it.

And to complete the picture, check the other strategy, $B$, as a resident:

• $E(B,B) = 2$ versus the mutant $E(A,B) = 1$. Since $2 > 1$, Condition 1 fires: $B$ is a strict best reply to itself. $B$ IS an ESS (and a strict Nash equilibrium to boot).

Strict Nash ⇒ ESS: the shortcut, and where the ties live

We’ve seen a weak Nash equilibrium fail to be an ESS. The mirror-image fact is a genuinely useful shortcut: every strict Nash equilibrium is an ESS.

Suppose $I$ is a strict Nash equilibrium — meaning $E(I,I) > E(J,I)$ for every $J \ne I$. But look: that inequality is ESS Condition 1. It holds for all mutants, so $I$ passes the invasion test outright — Condition 2 never even has to be consulted. Strict NE → Condition 1 for all $J$ → ESS. Done.

So where do the interesting ESSs live — the ones that need Condition 2 at all? Exactly where strategies tie against the resident. If $E(I,I) = E(J,I)$ for some mutant $J$, then $I$ is not a strict best reply to itself, so the strict-NE shortcut doesn’t apply, and stability has to be rescued by the tie-breaker $E(I,J) > E(J,J)$. In those cases Condition 2 is doing all the stability work.

And the headline example of “strategies that tie against the resident” is the mixed ESS. Here’s the key structural fact, worth carving into stone:

Bring back Hawk–Dove. Its mixed ESS $p^* = V/C$ — play Hawk with probability $V/C$ — is a non-strict, mixed Nash equilibrium. At that mix, pure Hawk and pure Dove earn exactly the same expected payoff against the equilibrium population (that’s why nobody is tempted to become all-Hawk or all-Dove). Because those payoffs tie, $p^*$ flunks the strict test; it qualifies as an ESS only through Condition 2, which checks that the resident mix does better against a rare mutant than the mutant does against itself. The tie-breaker isn’t decoration in Hawk–Dove — it’s the entire reason the mix is stable.

The two faces of a mixed strategy

There’s one more idea in this lesson, and it’s the kind of distinction that separates people who use ESS from people who merely quote it. When we say Hawk–Dove settles at ”$p^* = V/C$,” what does the population physically look like? There are two completely different biological pictures that produce the identical math — and confusing them causes real errors.

Face 1 — the randomising individual. Every animal is a single mixed player: an identical stochastic gambler who, in each contest, plays Hawk with probability $p^* = V/C$ and Dove otherwise. Everyone is the same; each individual carries a coin and flips it. Variation lives within each animal, across its many encounters.

Face 2 — the stable polymorphism. Nobody randomises at all. The population is split into pure types: a fraction $p^*$ are lifelong pure Hawks, the remaining $1 - p^*$ are lifelong pure Doves. Each individual is deterministic; the population is the mixture. Variation lives between animals, not within them.

Here’s the twist: in the standard model these two pictures are game-theoretically equivalent. They generate the same population-level frequency of Hawk-plays, the same expected payoffs, the same equilibrium. The math can’t tell them apart. But biologically they are different claims about the world — is the variation you observe generated inside each individual (Face 1) or across different individuals (Face 2)? That is an empirical question with a real answer.

And nature has cast its vote. Real cases lean heavily toward polymorphism. The side-blotched lizard has three discrete, inherited throat-colour morphs; the left/right scale-eating fish (you’ll meet them next lesson) come in two fixed mouth-asymmetry types. These are pure, heritable strategies carried by different individuals — not one animal privately flipping a coin before each fight. When you see a stable mix in the wild, your default guess should usually be a polymorphism of pure types, not a population of identical gamblers.

Recap