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In some lizard populations, orange males beat blue, blue beat yellow, and yellow beat orange, and the three morphs rise and fall in endless cycles. It is Rock-Paper-Scissors written in biology, and the replicator equation traces the loop.

Replicator dynamics

The ESS asks where a population rests. Replicator dynamics asks how it moves: the share of a strategy grows in proportion to how much its fitness beats the average. Sometimes it converges, and sometimes, as in Rock-Paper-Scissors, it circles forever without settling.

From earlier lessons

Two quick ones before we start.

A Rock-Paper-Scissors opponent always throws Rock. Which move exploits that pattern?

An evolutionarily stable strategy resists invasion by...

Predict, then play

Watch a population of Rock, Paper, and Scissors players evolve on the simplex. Each strategy grows when it beats the field. A tie payoff bends the path.

Why did that happen?

Take a guess, then reveal the reason

Whatever is common gets beaten: lots of Rock invites Paper, which invites Scissors, which invites Rock again. So no strategy can hold a lead, and the population chases its own tail around the even mix. With fair ties the chase is a perfect closed loop. Change the tie payoff and you tilt the balance: penalizing ties makes matching the crowd costly, nudging the population inward to the stable center; rewarding ties does the reverse and flings it toward the boundary.

The math

Definition, replicator dynamics

Let $x_i$ be the share of strategy $i$ and $f_i$ its expected fitness against the current population. The replicator equation is $\dot{x}_i = x_i\,(f_i - \bar{f})$ , where $\bar{f} = \sum_j x_j f_j$ is average fitness: strategies above average grow, those below shrink. Its rest points include every Nash equilibrium, and every evolutionarily stable strategy is an asymptotically stable rest point.

For fair Rock-Paper-Scissors (win $+1$ , lose $-1$ , tie $0$ ) the game is zero-sum, so average fitness is zero and the dynamics conserve a quantity: the trajectories are closed orbits circling the interior equilibrium $(\tfrac13, \tfrac13, \tfrac13)$ , which is neutrally stable, neither attracting nor repelling.

Give ties a payoff $e$ . A negative $e$ penalizes playing like the crowd, favours diversity, and turns the orbits into an inward spiral toward the center, now asymptotically stable. A positive $e$ rewards conformity and spirals the population outward toward the edges, cycling between near-pure phases. Adding the Hawk-Dove game instead, the replicator converges to the ESS Hawk fraction $V/C$ , linking this lesson to the last.

Work it out

Read off the direction of motion

Take the population $(0.5, 0.3, 0.2)$ for (Rock, Paper, Scissors) with fair ties. A strategy’s fitness is its expected score against this mix: Rock scores $x_S - x_P = 0.2 - 0.3 = -0.1$ , Paper scores $x_R - x_S = 0.5 - 0.2 = 0.3$ , Scissors scores $x_P - x_R = 0.3 - 0.5 = -0.2$ .

Paper is above the average, so its share grows; Rock and Scissors shrink. But as Paper rises, Scissors (which beats Paper) starts to pay, and the lead rotates. That rotation, with no strategy able to keep its advantage, is the cycle. With fair ties the loop is closed; a small tie penalty would instead pull each loop a little tighter until the population spirals into the even mix.

Key idea

Replicator dynamics grow strategies that beat the average. In fair Rock-Paper-Scissors the population cycles forever on closed orbits; a small tie penalty stabilizes the even mix, a reward destabilizes it.

Check yourself

Check yourselfapply

Under fair Rock-Paper-Scissors, what do the replicator trajectories do?

Check yourselfanalyse

Penalizing ties (a negative tie payoff) changes the RPS dynamics how?

Check yourselfbrings back: ESS

How do replicator dynamics connect to evolutionarily stable strategies?

Sources

Taylor, P. D., and Jonker, L. B. (1978). Evolutionarily stable strategies and game dynamics. Mathematical Biosciences, 40(1-2), 145-156.