Punishment, Detection, and Forgiveness in Repeated Games

The Stage Game Prisoners’ dilemma structure applies in many situations – Lovers or roommates – Colluding oligopolists – Arms control agreements – Common-pool resources – Cooperating vampire bats – …many more

Working Example: Prisoners’ Dilemma R, R -1, T T, -1 0, 0 CooperateDefect Cooperate Defect PLAyER 1 PLAyER 1 Player 2 Assume T>R>0

Stage Game In the stage game, Defect is a dominant strategy for both players. So the only Nash equilibrium has them both playing defect and each getting a payoff of 0 Both would be better off if they both cooperated, but how to enforce that?

Repeated Play Suppose that after each round of play, players are told their payoff on the previous round and with probability d>0, they go on to play another round. Can we get cooperative play by having each player threaten to punish a defection.

Punishment and forgiveness Grim trigger: (No forgiveness) I will cooperate until you defect, but If you ever defect, I will defect in all future rounds. Conditional N-period punishment. If you defect, I will start to defect and I will keep defecting until I have seen you cooperate N times in a row. Then I will cooperate so long as you do not defect.

Symmetric SPNE with Grim Trigger Suppose that the other player is playing Grim Trigger. If you play Grim Trigger as well, then you will both cooperate as long as the game continues and and you will each receive an expected payoff of R×(1+d +d 2 + d 3 + d 4 + ….+ )=R/(1-d)

What if you defect against Grim Trigger If you defect and the other guy is playing Grim Trigger, you will get a payoff of T>R the first time that you defect. But after this, the other guy will always play defect. The best you can do, then is to always defect as well. You both get zero when you both defect, so expected payoff from defecting is just T+0=T

Cooperate vs Defect If other guy is playing Grim trigger and nobody has yet defected, your expected payoff from playing cooperate is R/(1-d) If other guy is playing Grim trigger and nobody has yet defected, your expected payoff from playing defect is T+Pd/(1-d) Cooperate is R/(1-d) better for you if R/(1-d)>T+Pd/(1-d) which implies d>(T-R)/(T-P) Example If T=10, R=5, P=2, then condition is d>5/8. If d is too small, it pays to “take the money and run”

Other equilbria? Grim trigger is a SPNE if d is large enough. Are there other SPNEs? Yes, for example both play Always Defect is an equilibrium. If other guy is playing Always Defect, what is your best response in any subgame? Another is Play Defect the first 10 rounds, then play Grim Trigger.

The “Folk Theorem”: A general result The “good news”: In a repeated game with complete information, where the probability d that it will be continued to the next round is sufficiently close to 1, an efficient outcome can always be sustained as a subgame perfect Nash equilibrium.

More about the Folk Theorem “Not-so-good-news” In a repeated game of incomplete information with d close to one, not only can efficient outcome can be sustained as a Nash equilibrium, so can almost anything else. Possible explanation for why men wear neckties or women wear absurdly painful high heels.

Details of a Folk Theorem Consider a repeated game with an inefficient Nash equilibrium. Consider a strategy called Strategy A: “Do some quite arbitrary sequence of plays” so long as everybody else does their specified drill. If anyone fails to do so, revert to your inefficient Nash equilibrium action. If everybody prefers the result when all follow the arbitrary sequence to the inefficient Nash equilibrium, then for d close to 1, the strategy profile. Everybody uses Strategy A is a subgame perfect Nash equilibrium.

Forgiveness Does the grim trigger strategy have to be so unrelenting? In the real world, why might it not be a good idea to have an unforgiving punishment? What if you get a noisy signal about other player’s action? What if other player made a one-time mistake? This question is much wrestled with in religion and in politics.

Tit for Tat: a more forgiving strategy What is both players play the following strategy in infinitely repeated P.D? Cooperate on the first round. Then on any round do what the other guy did on the previous round. Suppose other guy plays tit for tat. If I play tit for tat too, what will happen?

Payoffs If you play tit for tat when other guy is playing tit for tat, you get expected payoff of R(1+d +d 2 + d 3 + d 4 + ….+ )=R/(1-d) Suppose instead that you choose to play “Always defect” when other guy is tit for tat. You will get T+ P(d +d 2 + d 3 + d 4 + ….+ ) =T+Pd/1-d Same comparison as with Grim Trigger. Tit for tat is a better response to tit for tat than always defect if d>(T-R)/(T-P)

Another try Sucker punch him and then get him to forgive you. If other guy is playing tit for tat and you play D on first round, then C ever after, you will get payoff of T on first round, S on second round, and then R for ever. Expected payoff is T+ Sd+d 2 R(1+d +d 2 + d 3 + d 4 + ….+ )=T+ Sd+d 2 R/(1-d).

Which is better? Tit for tat and Cheat and ask forgiveness give same payoff from round 3 on. Cheat and ask for forgiveness gives T in round 1 and S in round 2. Tit for tat give R in all rounds. So tit for tat is better if R+dR>T+dS, which means d(R-S)>T-R or d>(T-R)(R-S) If T=10, R=6, and S=1, this would mean if d>4/5. But if T=10, R=5, and S=1, this would be the case only if d>5/4, which can’t happen. In this case, tit for tat could not be a Nash equilibrium.

Some Problems

Problem 1: Chapter 13 A FISHERMAN’S CATCH AND PAYOFF Number of Number of Own Own Boats Other People’sPAYOFF Boats

What is the Nash equilibrium for the stage game for the three fishermen? A)All send one boat. B)All send two boats. C)There is more than one Nash equilibrium for the stage game. D)There are no pure strategy Nash equilibria, but there is a mixed strategy Nash equilibrium for the stage game. E)There are no pure or mixed strategy Nash equilibria for the stage game.

Can efficiency be sustained by the Grim Trigger? Suppose that the other two fishermen are playing the grim trigger strategy of sending one boat until somebody sends two boats and if anybody ever sends two boats, you send two boats ever after. If you and the others play the grim trigger strategy, you will always send 1 boat and so will they.

If others are playing grim trigger strategy, would you want to? If you play grim trigger, you will always send 1 boat. Your payoff will be 25 in every period. Assume that a fisherman discounts later profits at rate d. Value of this stream is then 25(1+d+d 2 +d 3 +…)=25(1/1-d) If instead you send 2 boats, you will get payoff of 45 the first time, but only 20 thereafter. Value of this stream is (d+d 2 +d 3 +…) Grim trigger is bigger if 20<5 (d+d 2 +d 3 +…) This means 20 4/5

Problem 7 The stage game: Payoff to player 1 is V 1 (x 1,x 2 )=5+x 1 -2x 2 Payoff to player 2 is V 2 (x 1,x 2 )=5+x 2 -2x 1 Strategy set for each player is the interval [1,4] What is a Nash equilibrium for the stage game?

A)Both players choose 4 B)Both players choose 3 C)Both players choose 2 D)Both players choose 1 E)There is no pure strategy Nash equilibrium.

Part b (i) If the strategy set is X={2,3}, when is there a subgame perfect Nash equilibrium in which both players always play 2 so long as nobody has ever played anything else. Compare payoff v(2,2) forever with payoff v(3,2) in first period, then v(3,3) ever after. That is, compare 3 forever with 4 in the first period and then 2 forever.

Part b(ii) X=[1,4] When is there a subgame perfect equilibrium where everybody does y so long as nobody has ever done anything differently and everybody does z>y if anyone ever does anything other than y? First of all, it must be that z=4. Because actions after a violation must be Nash for stage game. When is it true that getting V(y,y) forever is better than getting V(4,y) in the first period and then V(4,4) forever.

Comparison V(y,y) forever is worth V(y,y)/(1-d)=(5-y)/(1-d) V(4,y) and then V(4,4) forever is worth 9-y+1d+1d 2 +…=9-y+d/1-d) Works out that V(y,y)>V(4,y) if d(8-y)>4