Question

Suppose two players play the following prisoner's dilemma for 10 periods (periods 1 through 10).

	C	D
C	3,3	-1,4
D	4,-1	0,0

Suppose that players simultaneously choose their strategy before the repeated game, and can't change it once the repeated game has started. Players can choose one of the three following strategies as defined in class:

• Always Defect

• Grim-Trigger

• Tit-For-Tat

(a) Draw the 3×3 matrix game with the payoffs for each strategy pair calculated over the 10 periods. (Each row and each column represent one of the three strategies)

(b) What are the pure-strategy Nash equilibria of game in part (a)?

(c) Now suppose now that player 1 make a mistake in period 2. That is if his strategy is supposed to play C then he accidentally plays D, and if his strategy is supposed to play D then he accidentally plays C. Draw the 3×3 matrix game with the payoffs for each strategy pair calculated over the 10 periods.

(d) What are the pure-strategy Nash equilibria of game in part (c)?

(e) Now consider the new strategy called Win-Stay, Lose Shift (WSLS):

• Start with C • If the other person plays C, then I repeat MY last action

• If the other person plays D, then I switch MY last action

What sequence of actions is played if both players play WSLS against each other, and player 1 makes a mistake in period 2. What are the payoffs for each player?

(f) Is this better than when both players play TFT against each other, and player 1 makes a mistake in period 2?

Answer 1

There are three strategies possible:

(i) always defect: you always chose not to cooperate. i.e. both players always play D.

(ii) grim trigger: you start by cooperating and continue to cooperate as long as everyone has cooperated in the past. If someone has defected, then you defect forever.

(iii) tit fot tat: you follow a course of action consistent with your opponent's previous turn.

(a)

		Player 2
		always defect	grim trigger	tit for tat
Player 1	Always defect	0,0	3,-1	3/2 , -1/2[1]
	grim trigger	-1,3	20,20[2]	11,11[3]
	tit for tat	-1/2, 3/2	11,11	10, 10[4]

Pure strategy nash equilibria are (always defect, always defect), (grim trigger, grim trigger).

[1] There are two possible situations:

Player 2 starts with C or Player 2 starts with D

In the first case, the outcome is D₁C₁, D₂D₂, D₃,D₃.....D₁₀,D₁₀

Payoff of player 1 = 3 + 0*9 = 3, Payoff of player 2 = -1 + 0*9 = -1

In the second case, the outcome is D₁D₁, D₂D₂, D₃,D₃.....D₁₀,D₁₀

Payoff of player 1 = 0*10 = 0, Payoff of player 2 = 0*10 = 0

Both cases can occur with equal probability 1/2.

Expected outcome of player 1 =

Expected outcome of player 2 =

[2] Each player starts with C. Since no one defects, outcome is C₁C₁, C₂C₂, .....C₁₀C₁₀

Payoff of each player = 2*10 = 20

[3] There are two possible situations:

Player 2 starts with C or Player 2 starts with D

In the first case, the outcome is C₁C₁, C₂C₂,.....C₁₀C₁₀

Payoff of each player = 2*10 = 20

In the second case, the outcome is C₁D₁, D₂C₂, D₃D₃.....D₁₀D₁₀

Payoff of player 1 = -1 + 3 + 0*8 = 2, Payoff of player 2 = 3+ (-1) + 0*8 = 2

Both cases can occur again with equal probability 1/2.

[4] There are four possible cases:

C₁C₁, C₂C₂,.....C₁₀C₁₀

C₁D₁, D₂C₂, C₃D₃.....C₁₀D₁₀

D₁C₁, C₂D₂, D₃C₃.....D₁₀C₁₀

D₁D₁, D₂D₂, D₃,D₃.....D₁₀,D₁₀

(c)

		Player 2
		always defect	grim trigger	tit for tat
Player 1	Always defect	-10,30	20,20	37/2, 41/2
	grim trigger	-9,27	-5, 23	16,18
	tit for tat	-19/2, 57/2	-13/2, 45/2	10,10

Nash equilibrium is (grim trigger, always defect)