Consider the Prisoners’ Dilemma that is played for two rounds. Is there a NE of finitely-repeated...

Consider the Prisoners’ Dilemma that is played for two rounds.

Is there a NE of finitely-repeated Prisoner’s Dilemma which is different from the SPE?

Is there a NE outcome which is different from the SPE outcome?

Solutions

Expert Solution

If the game is repeated for any finite number of times, the SPE outcome will always be the Nash equilibrium outcome where both the players defect.

This can be explained through the idea of backward induction.

If the players know that the game is played for a finite number of times, then in the final round, they will surely defect (i.e. play the Nash equilibrium outcome). Given this, it's optimal for the players to defect in the period before that,...and hence optimal for them to defect in the period even before that....and so on.

Ultimately, we can say that it's optimal for the players to defect in the first period only. Hence, if the prisoners dilemma game is played for any finite number of times, the SPE outcome will always be the Nash equilibrium outcome.

Given this, if the game is played for two periods (which is a finite number), the SPE outcome will be the Nash equilibrium outcome where both of them defect.

Rahul Sunny answered 1 year ago

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