PROBLEM 5 Consider a prisoners\' dilemma game of N prisoners, where: If all pris

Question

PROBLEM 5 Consider a prisoners' dilemma game of N prisoners, where: If all prisoners cooperate C, then each prisoner gets a payoff of N. If a single prisoner defects D, while all the other prisoners cooperate C, then that prisoner gets N2, while the others get-1 If more than one prisoner defects, all prisoners who defected get zero, while any prisoner who remains coop- erated gets-1 Now, answer the following questions a) What is the Nash Equilibrium of this game (b) If the above N-prisoners game in repeated infinitely, determine the minimum 6 such that cooperation is sustained c) Based on your answer to Part (b), determine the value of 8 as N-o What do you conclude? Explain Dr. M. W. Baidas

Explanation / Answer

(a).

All prisoner's defects is the Nash equilibrium of the game. It is a strictly dominant strategy . Suppose all other prisoner's are cooperate; by defecting, the prisoner(lets say X) get a payoff of N2,and from silent, prisoner(X) get a payoff of N , so defecting is better. Suppose at least one prisoner is defects; by remaining cooperate, the prisoner(X) get a payoff of 1, while by defecting the prisoner(X) get a payoff of zero, so defects is better. Therefore, no matter what the opponents are doing, the prisoner(X) get a higher payoff by defecting, so it is a strictly dominant strategy.


(b)
Let us consider the following stratergies,

i. after any history in which all prisoner's choose cooperate in all previous periods, play cooperate this period.

ii. After any other history, defects.

The discounted pay off of cooperating is

N+N+ * N+** N+.......= N/(1-)

Cooperating and defecting gives a payoff ,

N2+ *0+ ** 0+ ***0+....=N2

Cooperating is better than defecting if,
N/(1-)N2

Which gives, N/(1-)N
That is,

(1-1/N)

Therefore the minimum is (1-1/N)

(c)
As N increases, we get (2) = 1 1/2 = 1/1, (3) = 2/3, (4) = 3/4, ... , which tends to 1.

Since 1/N 0 as N , when the game is really big, the discount factor will be close
to 1, making defecting more difficult.

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