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)

When does grim trigger sustain cooperation? 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 So both paying grim trigger and always cooperating is a SPNE if T<R/(1-d) – For example if d=.9, grim trigger sustains cooperation if T<10R.

What did we learn? Cooperation can be sustained if T<R/(1-d) Equivalently if (1-d)T<R This is the case if temptation is not too big and if the probability d of playing again is not too small.

Forgiveness Does a punishment strategy have to be as unrelenting as grim trigger? 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 or was subjected to unusual temptation This question is much wrestled with in religion and in politics.

Foregiveness and religion There is a tension in religious prohibitions. To make people act as the priests would like them to, it might seem useful to tell them that they will be eternally punished for actions the priests don’t like. But if they do that, people who have violated the rules might as well continue to do so, since they are damned anyway. Hence religions often claim a “forgiving” deity in hopes of bringing lost sheep back into the herd.

What if temptation varies over time? Suppose that in the play of this game, the

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.

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.