Coordination and the selection problem

When a game has one equilibrium the theory makes a prediction, and when it has three the theory says only that any of the three is consistent with rational play, which is a much weaker thing to say.

Wanting the same thing, and wanting different versions of it

The games so far have had players pulling against each other. A large and important class does the opposite: what each player most wants is to match the other, and the only question is on what.

The purest case has no conflict at all. Two people are cut off mid-call and each must decide whether to ring back or wait. Both ringing back gives an engaged tone, both waiting gives silence, and either mismatch reconnects them. There are two equilibria, "you ring, I wait" and "I ring, you wait", both players are perfectly happy with either, and yet pairs of people fail at this constantly. Nothing in the payoffs distinguishes the two equilibria, and nothing in the theory so far tells either person which to expect.

Conflict returns as soon as the players rank the matches differently. In the game traditionally called battle of the sexes, two people want to spend the evening together but disagree about where. Payoffs, Alex first:

Both cells on the diagonal are equilibria: given that Bo is at the opera, Alex does better at the opera than alone at the football, and symmetrically. Alex prefers one equilibrium and Bo the other, and the theory as it stands has nothing to say about which occurs, or about what happens if both stand firm.

The mixed equilibrium is worse than both

Wilson's odd-number result from the previous lesson says to look for a third equilibrium, and it is there.

Let $p$ be the probability Alex goes to the opera. Bo's expected payoff from the opera is $1\cdot p+0\cdot (1-p)=p$, and from the match it is $0\cdot p+3(1-p)=3-3p$. Setting these equal gives $4p=3$, so $p=0.75$. Now let $q$ be the probability Bo goes to the opera. Alex's payoff from the opera is $3q$ and from the match is $1-q$, equal when $4q=1$, so $q=0.25$.

Each player goes to their own preferred venue with probability 0.75. Alex's expected payoff is $3(0.25)=0.75$, and by symmetry so is Bo's. Compare that with the pure equilibria, which pay 3 and 1. The mixed equilibrium is worse for both players than either pure one, including the one each likes least.

The reason is visible in the probabilities. They meet at the opera with probability $0.75\times 0.25=0.1875$ and at the match with the same 0.1875, so they end up together only 37.5 per cent of the time and spend 62.5 per cent of their evenings apart. This is not a pathological case. In every coordination game the mixed equilibrium sits between the pure ones and delivers less than either, which is exactly what makes it an unattractive prediction and exactly what makes it a good description of two people who genuinely do not know what the other will do.

Example. Change the numbers so that each player gets 5 at their preferred venue and 2 at the other, with 0 for missing each other. What are the mixed equilibrium probabilities and how often do they meet?

Bo's indifference: opera pays $2p$, the match pays $5(1-p)$, so $7p=5$ and $p=5/7\approx 0.714$. By symmetry Bo goes to the opera with probability $2/7\approx 0.286$. They meet with probability $(5/7)(2/7)+(2/7)(5/7)=20/49\approx 0.408$, and each earns $5(2/7)=10/7\approx 1.43$, again below the 2 that even the worse pure equilibrium pays.

Now you. Suppose Alex's payoffs stay 3 and 1 but Bo becomes indifferent between venues, earning 2 at either when they are together and 0 apart. What is Alex's equilibrium mixing probability now?

Coordination with risk: the stag hunt

Rousseau, in the 1755 Discourse on the Origin of Inequality, described a hunting party in which each member must stay at their post to bring down a stag, and any one of them will abandon it to chase a passing hare. The game that carries his name is the most consequential of the coordination games, because the two equilibria differ in risk rather than in who gets more.

Both diagonal cells are equilibria. Mutual stag hunting pays 4 each and mutual hare chasing pays 3 each, so one equilibrium is better for everybody. But the hare is safe: it pays 3 whatever the other hunter does, while the stag pays 4 or nothing.

The mixed equilibrium quantifies the risk. Hunter 2 is indifferent when $4p=3p+3(1-p)$, that is when $4p=3$, so $p=0.75$. Read that as a threshold on belief: hunting the stag is a best response only if you assign at least a 0.75 probability to your partner hunting the stag too. Being 70 per cent confident in your partner is not enough.

Two named criteria pull in opposite directions here. Payoff dominance says pick the equilibrium that is better for everyone, which is stag. Risk dominance, in the form Harsanyi and Selten gave it in 1988, says pick the equilibrium that is the best response to complete ignorance about the other player, meaning a fifty-fifty belief. Against a coin flip, stag pays 2 and hare pays 3, so hare is risk dominant. The formal version compares the products of what each player loses by deviating: at the stag equilibrium the product is $(4-3)(4-3)=1$, at the hare equilibrium it is $(3-0)(3-0)=9$, and the larger product wins. Hare again.

The experiments side with risk. In John Van Huyck, Raymond Battalio and Richard Beil's 1990 study, groups of fourteen to sixteen subjects each chose an effort level from 1 to 7, with the payoff determined by the minimum effort anyone chose and by their own cost of effort, a game with seven equilibria at which everyone matches. Almost a third started at the highest effort. By the tenth round nearly three-quarters of subjects were choosing effort 1, the worst equilibrium for everybody, and essentially nobody was choosing 7. One low player is enough to punish the high ones, so a single unlucky draw sends the whole group down and it never recovers. In two-player versions the efficient equilibrium was reached often, because with one partner rather than fifteen the risk is far smaller.

Example. Suppose the stag is worth 6 to each hunter rather than 4, with everything else unchanged. What is the new belief threshold, and which equilibrium is risk dominant?

Indifference now requires $6p=3$, so $p=0.5$: a hunter needs only even odds on their partner to be worth joining. The deviation products are $(6-3)(6-3)=9$ for stag and $(3-0)(3-0)=9$ for hare, so they are exactly tied and the criterion gives no answer. Sweetening the stag by 50 per cent moved the game from clearly risk dominated to marginal.

Now you. With the original payoffs, suppose the hare is worth only 2 rather than 3, so the grid reads 4, 4 on the stag diagonal, 0, 2 and 2, 0 off it, and 2, 2 on the hare diagonal. What is the belief threshold, and which equilibrium is risk dominant?

Anti-coordination: chicken

The mirror image is a game where each player wants to do the opposite of the other. Two drivers approach head on, and each may swerve or hold their line. Holding while the other swerves wins the contest, worth 1; swerving while the other holds loses it, worth -1; both swerving is a draw at 0; both holding is a collision, worth -10.

The two pure equilibria are the asymmetric ones, in which exactly one driver swerves. Each driver prefers the equilibrium where the other yields, which puts this game close to bargaining. The mixed equilibrium is more interesting. Let $p$ be the probability the other driver holds. Swerving pays $-p$; holding pays $(1-p)(1)+p(-10)=1-11p$. Equal when $10p=1$, so $p=0.1$.

So in the symmetric mixed equilibrium each driver holds one time in ten, and the two of them collide with probability $0.1\times 0.1=0.01$. Collisions are rare and they are not zero, which is the structural point: in equilibrium the disaster has positive probability, and it must, since a driver who was certain the other would swerve would never swerve themselves.

Now raise the cost of the collision from 10 to 20. Indifference becomes $-p=1-21p$, giving $p=0.05$, and the collision probability falls to 0.0025. Doubling the severity of the crash halves the aggression and quarters the accident rate, which is the same comparative static as the audit game and this time in the direction common sense expects, because here the payoff being changed belongs to both players at once.

The biological version, introduced by John Maynard Smith and George Price in 1973, is hawk-dove, and it reads the mixture as a population: 10 per cent hawks and 90 per cent doves, a stable proportion maintained by nothing more than the fact that hawks do badly in a population of hawks.

Example. In the original chicken game, what is each driver's expected payoff in the mixed equilibrium?

At $p=0.1$ each driver is indifferent, so the expected payoff equals the payoff of swerving, which is $-p=-0.1$. Both drivers do worse than the 0 they would get by both swerving every time, and they do it while colliding one time in a hundred.

Now you. With the collision cost at 20, what is the expected payoff, and how does the pair's total fare compared with the original game?

What actually selects an equilibrium

Nothing inside the model does. Every argument that picks one equilibrium out of several is an import, and it is more honest to name the import than to pretend the equilibrium concept did the work.

Focal points. Thomas Schelling, in The Strategy of Conflict in 1960, observed that people coordinate on whatever is conspicuous, and that conspicuousness has nothing to do with payoffs. Asked where they would meet a stranger in New York with no way to communicate, most of his subjects named the information booth at Grand Central Station, and almost all named noon. Later laboratory work by Judith Mehta, Chris Starmer and Robert Sugden in 1994 confirmed the effect systematically: asked to match a partner by naming heads or tails, the overwhelming majority named heads; asked for a positive number, most named 1. Salience is cultural, arbitrary and completely effective, and it is invisible to a payoff table.

Risk dominance and history. Where salience is absent, groups tend to the safe equilibrium, as the minimum-effort experiments showed, and once a group has settled somewhere, the history itself becomes the reason to stay. Which side of the road a country drives on is a pure coordination game solved permanently by precedent, and the cost of Sweden's changeover on the morning of 3 September 1967, planned for years and executed with the roads closed, is what overriding a precedent costs.

Communication. Talk that binds nobody, called cheap talk, is powerless in the prisoner's dilemma, where a player who intends to defect is happy to promise cooperation. In a coordination game it is close to decisive, because a player who announces "opera" has no incentive to lie: they want to be believed and then matched. Laboratory coordination rates rise sharply with one-way announcements, which is why the same institution, a pre-play meeting, is useless for a cartel enforcing prices and effective for two firms agreeing a technical standard.

Where this leaves us

Multiplicity is the standing weakness of equilibrium analysis, and it gets worse rather than better as the course goes on: the lesson on repeated games produces infinitely many equilibria in a game as simple as the prisoner's dilemma.

There is exactly one large class of games where the problem does not arise at all. If the players' interests are perfectly opposed, so that one player's gain is precisely the other's loss, then every equilibrium of the game gives the same payoffs, and mixing an equilibrium strategy of one with an equilibrium strategy of the other still gives an equilibrium. In those games the prediction is unique and the theory is at its strongest, which is why it was solved twenty years before Nash. That is the next lesson.