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Crime, Punishment, and Forgiveness

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Crime, Punishment, and Forgiveness

Econ 171

- Suppose 2 players play repeated prisoners dilemma, where the probability is d<1 that you will play another round after the end of each round.
- The grim trigger strategy is to play cooperate on the first round and play cooperate on every round so long as the other doesn’t defect.
- If the other defects, the grim trigger strategy plays defect on all future rounds.

- Suppose that the other player is playing Grim Trigger.
- If you play Grim Trigger as well, then you will cooperate as long as the game continues and and you will receive a payoff of R.
Your expected payoff from playing Grim Trigger if the other guy is playing Grim Trigger is therefore

R(1+d +d2 + d3 + d4 + ….+ )=R/(1-d)

- 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.
- Your expected payoff from defecting is therefore T+ P(d +d2 + d3 + d4 + ….+ )
=T+Pd/1-d

- 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 isR/(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”

- 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.

Number of Number of Own

Own Boats Other People’sPAYOFF

Boats

1 2 25

1 3 20

1 4 15

22 45

2 3 35

2 4 20

- All send one boat.
- All send two boats.
- There is more than one Nash equilibrium for the stage game.
- There are no pure strategy Nash equilibria, but there is a mixed strategy Nash equilibrium for the stage game.
- There are no pure or mixed strategy Nash equilibria for the stage game.

- 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 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+d2+d3 +…)=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 45+ 20(d+d2+d3 +…)
- Grim trigger is bigger if
- 20<5 (d+d2+d3 +…)
- This means 20<5d/(1-d) which implies d>4/5

The stage game:

- Payoff to player 1 is V1(x1,x2)=5+x1-2x2
- Payoff to player 2 is V2(x1,x2)=5+x2-2x1
- Strategy set for each player is the interval [1,4]
What is a Nash equilibrium for the stage game?

- Both players choose 4
- Both players choose 3
- Both players choose 2
- Both players choose 1
- There is no pure strategy Nash equilibrium.

- 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.

- 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.

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+1d2+…=9-y+d/1-d)

Works out that V(y,y)>V(4,y) if d(8-y)>4

- 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?
- This question is much wrestled with in religion and in politics.

- 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?

- If you play tit for tat when other guy is playing tit for tat, you get expected payoff of
R(1+d +d2 + d3 + d4 + ….+ )=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 +d2 + d3 + d4 + ….+ )
=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)

- 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+d2R(1+d +d2 + d3 + d4 + ….+ )=T+ Sd+d2R/(1-d).

- 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.