Just need U3! Thank you so much!

83. Consider a population in which there are two phenotypes: natural-born cooperators (who do not confess under questioning) and natural-born defectors (who confess readily). If two members of this population are drawn at random, their payoffs in a single play are the same as those of the husbandwife prisoners' dilemma game of Chapter 4, reproduced below. In repeated interactions there are two strategies available in the population, as there were in the restaurant-dilemma game of Section 12.2. The two strategies are A (always confess) and T (play tit-for-tat, starting with not confessing). COLUMN - (a) Suppose that a pair of players plays this dilemma twice in succes- sion. Draw the payoff table for the twice-repeated dilemma. (b) Find all of the E555 in this game. (0) Now add a third possible strategy, N, which never confesses. Draw the payoff table for the twice-repeated dilemma with three possible strategies and nd all of the E335 of this new version of the game. U3. Suppose that in the twice-repeated prisoners' dilemma of Exercise 83, a fourth possible type (type S) also can exist in the population. This type does not confess on the rst play and confesses on the second play of each episode of two successive plays against the same opponent. (a) Draw the four-byfour tness table for the game. (b) Can the newly conceived type S be an ESS of this game? (c) In the three-types game of Exercise 83, A and T were both ESS, but T was only neutrally stable because a small proportion of N mutants could coexist. Show that in the four-types game here, T cannot be ESS