Lecture 10
So far, we’ve mostly modeled agents as making decisions against an anonymous market: a consumer takes prices as given, a competitive firm takes the market price as given. Nobody’s individual action affects the environment that others face.
Game theory is the toolkit for situations where that assumption fails, where the payoff to your action depends on what others do, and they know the same about you.
Every game has three ingredients:
We’ll also need to specify the information structure (who knows what when) and the timing (who moves when).
A game is fully specified once you know all of these.
Games come in two standard forms:
Both representations can describe the same game. The choice is about which aspect, strategies or timing, you want to emphasize.
Two prisoners are separated into individual rooms and cannot communicate with each other. Each has two options: Cooperate (stay silent) or Defect (testify). The payoffs in years they will have to serve are:
| B Cooperates | B Defects | |
|---|---|---|
| A Cooperates | (-1, -1) | (-3, 0) |
| A Defects | (0, -3) | (-2, -2) |
Each player does better by defecting regardless of what the other does:
A strategy is strictly dominant for a player if it yields a strictly higher payoff than any alternative, regardless of what opponents do.
In the Prisoner’s Dilemma, Defect is strictly dominant for both players. So the prediction is (Defect, Defect), giving both players 2 years each, even though (Cooperate, Cooperate) would give them 1 year each.
This is the central puzzle: individually rational behavior leads to a collectively bad outcome.
The Prisoner’s Dilemma shows up everywhere: arms races, price wars, overfishing, tax evasion, doping in sports.
Not all games have dominant strategies. We need a more general solution concept.
Nash equilibrium (Nash, 1950): a profile of strategies \((s_1^*, \dots, s_n^*)\) such that each player’s strategy is a best response to the others’. No player has a profitable unilateral deviation.
\[ u_i(s_i^*, s_{-i}^*) \geq u_i(s_i, s_{-i}^*) \quad \text{for all } s_i, \text{ for all } i\]
Interpretation: a Nash equilibrium is a “no regrets” profile. Given what everyone else is doing, no one wishes they had chosen differently.
To find Nash equilibria in a normal-form game:
In the Prisoner’s Dilemma, (Defect, Defect) is the unique Nash equilibrium: both cells where the other defects have Defect underlined, and both cells where the other cooperates also have Defect underlined.
Two friends want to meet but can’t communicate. Each can go to a Cafe or a Bar.
| B: Cafe | B: Bar | |
|---|---|---|
| A: Cafe | (2, 2) | (0, 0) |
| A: Bar | (0, 0) | (1, 1) |
Two pure-strategy Nash equilibria: (Cafe, Cafe) and (Bar, Bar).
The game has multiple equilibria, raising the question of how players coordinate on one, and possibly whether some equilibria are more “focal” than others.
Same structure, but the players prefer different outcomes. A prefers Opera, B prefers Football, but both prefer being together.
| B: Opera | B: Football | |
|---|---|---|
| A: Opera | (2, 1) | (0, 0) |
| A: Football | (0, 0) | (1, 2) |
Two pure-strategy Nash equilibria: (Opera, Opera) and (Football, Football). Coordination is good, but there’s conflict about which equilibrium to land on.
A zero-sum game. Player A wants the pennies to match, Player B wants them to differ.
| B: Heads | B: Tails | |
|---|---|---|
| A: Heads | (1, −1) | (−1, 1) |
| A: Tails | (−1, 1) | (1, −1) |
No pure-strategy Nash equilibrium. For any pure profile, one player wants to switch.
We need mixed strategies.
A mixed strategy is a probability distribution over pure strategies.
A mixed-strategy Nash equilibrium is a profile of randomizations such that each player is indifferent between the pure strategies in the support of their mix, given the opponents’ mix.
Why indifference? If a player were strictly better off playing one pure strategy, they would play it with probability 1, not mix. Mixing is only rational when you are indifferent.
Let \(p\) = probability A plays Heads, \(q\) = probability B plays Heads.
A is indifferent between Heads and Tails when B makes both equally attractive:
\[ q \cdot 1 + (1 - q)(-1) = q(-1) + (1 - q)(1) \]
\[ 2q - 1 = 1 - 2q \implies q = 1/2 \]
By symmetry, \(p = 1/2\). The unique Nash equilibrium is \((p, q) = (1/2, 1/2)\): each player randomizes 50-50.
Expected payoff: 0 for each player. Exactly what you’d guess by symmetry.
Nash’s theorem (1950): Every finite game (finitely many players, finitely many pure strategies) has at least one Nash equilibrium, possibly in mixed strategies.
This is reassuring: we can always find some equilibrium. But the theorem doesn’t tell us which one is selected in games with multiple equilibria, or whether players will actually play it.
So far, our games have been simultaneous. But many real situations are sequential: one player moves first, and the second responds after observing the first move.
The extensive form uses a game tree:
An entrant (E) decides whether to Enter or Stay Out of a market. If E enters, the incumbent (I) decides to Fight or Accommodate.
| Path | Payoffs (E, I) |
|---|---|
| Stay Out | (0, 10) |
| Enter, Accommodate | (2, 5) |
| Enter, Fight | (−3, 0) |
What happens? It depends on what E believes I will do if entry occurs.
Solve from the end of the tree back to the beginning.
Step 1: If E enters, what does I do? Accommodate (5) beats Fight (0). So I accommodates.
Step 2: Given that I will accommodate, what does E do? Enter (2) beats Stay Out (0). So E enters.
Prediction: (Enter, Accommodate), payoffs (2, 5).
In the entry game, the normal form actually has two Nash equilibria: (Enter, Accommodate) and (Stay Out, Fight).
At (Stay Out, Fight), neither player wants to deviate: given that E stays out, I’s strategy of “Fight if entry occurs” is never tested; and given I’s threat to fight, E prefers to stay out.
But the Fight threat is not credible: if entry actually occurred, I would not carry it out. Backward induction rules out such non-credible threats.
Subgame perfect equilibrium (SPE): a strategy profile that induces a Nash equilibrium in every subgame, not just the whole game.
In finite games of perfect information, the SPE is exactly what backward induction gives you. It rules out Nash equilibria that rely on non-credible threats off the equilibrium path.
In the entry game, (Enter, Accommodate) is the unique SPE; (Stay Out, Fight) is Nash but not subgame perfect.
Schelling (1960): sometimes you are better off limiting your own options. A credible commitment device can change the equilibrium.
Commitment is valuable not despite restricting your future choices but because it does.
In the one-shot Prisoner’s Dilemma, the only Nash equilibrium is (Defect, Defect). What happens if the same game is played repeatedly?
Finitely repeated: backward induction still unravels cooperation. In the last round, both defect. Knowing that, they defect in the second-to-last round. And so on. Defection in every period is the unique SPE.
Infinitely repeated (or unknown horizon): cooperation can be sustained as an equilibrium if players are patient enough, using strategies that punish defection (e.g., tit-for-tat or grim trigger).
In an infinitely repeated game with a sufficiently patient discount factor, essentially any individually rational payoff can be supported as a subgame perfect equilibrium.
Intuition: if defection today triggers enough future punishment, players prefer to cooperate.
The flip side: repeated interaction gives us cooperation, but at the cost of a huge multiplicity of equilibria. Predicting what will actually happen is hard.