Q8) Consider the following normal form of prisoner\'s dilemma. Suppose that ther
ID: 1124117 • Letter: Q
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Q8) Consider the following normal form of prisoner's dilemma. Suppose that there is no discounting, so total payoffs are just the sum of the stage payoffs 112 Defect Cooperate-2,7 DefectCooperate 0,0 7,-2 5,5 Suppose this stage game is played 5 times. Is cooperation possible in the repeated game? What will be the SPNE of the game. Explain your answer clearly but briefly. Now consider the following modified prisoner's dilemma stage game. 112 Defect Cooperate 1-27 Partial DefectCooperate Partial 0,0 7,-2 5,5 6,0 0,6 3,3 Suppose that this stage game is played twice i.e. T-2 a) Find pure strategy nash equilibrium(a) of the stage game b) There are many SPNE of this game. Write 3 possible SPNE. c) Is cooperation possible in this repeated game? If so, provide a possible strategy which would induce cooperation. Clearly state the equilibrium achieved in each round d) Prove using payoffs that the strategy provided above is indeed an equilibrium strategy. Clearly show your workingExplanation / Answer
No, cooperation is not possible in the repeated game. Since the players are playing for a finite number of times we can use backward induction to solve the problem. Since it is a finite process, the players treat each game as a one - shot game as if the next game would be their last game. The one - shot game is a Nash equilibrium and as is the case of Prisoner's dilemma, the Nash equilibrium or the strategy of the players would be ( Defect, Defect). Now if we go one step backwards and
consider the strategy previous to that. As the players always treat the game as if the next game would be their last and they could defect in that, hence in the previous step also they defect with each other. Hence, in Prisoner's dilemma in all rounds in a finite repeated process the outcome or Nash equilibrium is always ( Defect, Defect). Thus, (Defect, Defect) is the SPNE of the game.
Modified Prisoner's Dilemma Stage game:
0,0 7,-2 3,-1
-2,7 5,5 0,6
-1,3 6,0 3,3
The underlined are the best responses.
a) Pure Nash equilibrium is (0,0) and (3,3) which is (Defect,Defect) and (Partial, Partial)
b) By the previous calculation in step (a), first SPNE is (Defect,Defect), second SPNE is (Partial,Partial)
3rd SPNE : Play (Cooperate, Cooperate) for periods 1,....T-1 and at T play Partial if all players have been cooperating upto that point otherwise play (Defect,Defect) for the rest of the games.
c) Refer to explanation in previous point b.
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