SolvedExercises_Chapter7_Solutions_Updated

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Games of Strategy , Fifth Edition Copyright © 2021 W. W. Norton & Company Solutions to Chapter 7 Exercises SOLVED EXERCISES S1. (a) The game most resembles an assurance game because the two Nash equilibria occur when the players use symmetric moves. (Here they both use the same moves to arrive at the Nash equilibria; in other games, they might use exactly opposite moves to arrive at the Nash equilibria.) In an assurance-type game, both players prefer to make coordinated moves, but there is also a preferred Nash equilibrium with higher payoffs for both players. In this game, (Risky, Risky) is the preferred equilibrium because it has higher payoffs, but there is a chance that the players will play the worse Nash equilibrium with lower payoffs. Even worse, the players might not play an equilibrium at all. Without convergence of expectations, these results can occur, and this is characteristic of an assurance-type game. (b) The two pure-strategy Nash equilibria for this game are (Risky, Risky) and (Safe, Safe). There is also a mixed-strategy Nash equilibrium in which each player chooses Safe with probability 2/5 and Risky with probability 3/5. Rowena’s p -mix (probability p on Safe) must keep Colin indifferent and so must satisfy 4p + 4(1 – p) = p + 6(1 – p) ; this yields p = 2/5 . The symmetry of the game guarantees that Colin’s equilibrium mix is the same. S2. (a) There is no pure-strategy Nash equilibrium here, hence the search for an equilibrium in mixed strategies. Rowena ’s p -mix (probability p on Up) must keep Colin indifferent and so must satisfy 16p + 20(1 – p ) = 6 p + 40(1 – p ); this yields p = 2/3 = 0.67 and (1 – p ) = 0.33. Similarly, Coli n’s q -mix (probability q on Left) must keep Rowena indifferent and so must satisfy q + 4 (1 – q ) = 2 q + 3(1 – q ); the correct q here is 0.5. (b) Rowena ’s expected payoff is 2.5. Col i n’s expected payoff = 17.33. (c) Joint payoffs are larger when Rowena plays Down, but the highest possible payoff to Rowena occurs when she plays Up. Thus, in order to have a chance of getting 4, Rowena must play Up occasionally. If the players could reach an agreement always to play Down and Right, both would get higher expected payoffs than in the mixed-strategy equilibrium. This might be possible with repetition of the game or if guidelines for social conduct were such that players gravitated toward the outcome that maximized total payoff.

Games of Strategy , Fifth Edition Copyright © 2021 W. W. Norton & Company S3. The two pure- strategy Nash equilibria are (Don’t Help, Help) and (Help, Don’t Help). The mixed-strategy Nash equilibrium has the following equilibrium mixtures: 2 p + 2(1 – p ) = 3 p + 0 p = 2/3 2 q + 2(1 – q ) = 3 q + 0 q = 2/3 That is, each player helps two-thirds of the time and doesn’t help one -third of the time. S4. (a) On the one hand, Evert does worse when using DL than she did before. On the other hand, Navratilova does better against DL than she did before and Evert’s p -mix must keep Navratilova indifferent. Importantly, the difference between Navratilova’s DL versus her CC against Evert’s DL has remained the same. This suggests that Evert’s p -mix may not change. (b) 70 p + 10(1 – p ) = 40 p + 80(1 – p ) p = 7/10 30 q + 60(1 – q ) = 90 q + 20(1 – q ) q = 2/5 Thus, the mixed-strategy Nash equilibrium occurs when Evert plays 7/10(DL) + 3/10(CC) and Navratilova plays 2/5(DL) + 3/5(CC). Evert’s expected payoff is 30(2/5) + 60(1 – 2/5) = 48. (c) Compared with the previous game, Evert plays DL with the same proportion, whereas Navratilova plays DL less, going from 3/5 to 2/5. Navratilova’s q -mix changes because her mix is dependent on Evert’s payoffs and Evert’s relative success against each of Navratilova’s choices has changed . On the other hand, Evert’s p - mix doesn’t change because Navratilova’s relative success with her two strategies against each of Evert’s strategies has remained unchanged. S5. (a) .7 p + .85(1 – p ) = .8 p + .65(1 – p ) p = 2/3 .3 q + .2(1 – q ) = .15 q + .35(1 – q ) q = 1/2 The mixed-strategy Nash equilibrium is Batter plays 2/3(Anticipate fastball) + 1/3(Anticipate curveball) and Pitcher plays 1/2(Throw fastball) + 1/2(Throw curveball). (b) Batter’s expected payoff = .3(1/2) + .2(1 – 1/2) = 0.25. Pitcher’s expected payoff = .7(2/3) + .85(1 – 2/3) = 0.75. (c) The mixed-strategy Nash equilibrium is now .75 p + .85(1 – p ) = .8p + .65(1 – p ) p = 4/5 .25 q + .2(1 – q ) = .15 q + .35(1 – q ) q = 3/5.

Games of Strategy , Fifth Edition Copyright © 2021 W. W. Norton & Company The pitcher’s new expected payoff is .75(4/5) + .85(1 – 4/5) = 0.77, which is indeed greater than his previous expected payoff. The pitcher can increase his expected payoff because the batter is forced to adjust his equilibrium strategy in a way that favors the pitcher. S6. (a) 0 p – 1(1 – p ) = 1 p – 10(1 – p ) p = 9/10 0 q – 1(1 – q ) = 1 q – 10(1 – q ) q = 9/10 In the mixed-strategy Nash equilibrium James plays 9/10(Swerve) + 1/10(Straight), and Dean plays 9/10 (Swerve) + 1/10(Straight). James and Dean play Straight less often than in the previous game. (b) James’ s expected payoff = 9/10 – 10(1 – 9/10) = – 1/10. Dean’s expected payoff = 9/10 – 10(1 – 9/10) = – 1/10. (c) If James and Dean collude and play an even number of games where they alternate between (Swerve, Straight) and (Straight, Swerve), their expected payoffs would be 0. This is better than the mixed-strategy equilibrium, because their expected payoffs are – 1/10. (d) James’ s expected payoff = 1/2[(0 – 1)/2] + 1/2[(1 – 10)/2] = – 5/2. Dean’s expected payoff = 1/2[(0 – 1)/2] + 1/2[(1 – 10)/2] = – 5/2. These expected payoffs are much worse than the collusion example or the mixed-strategy equilibrium. In this case both players are mixing with the wrong (that is, not the equilibrium) mixture. Neither player is best responding to the other’s strategy, and i n this situation — with the very real possibility of reaching the – 10 payoff — the expected consequences are dire. S7. Watson’s expected payoff from choosing St. Bart’s when Holmes is using his p -mix is p ; Watson’s expected payoff from choosing Simpson’s when Holmes is mixing is 2 – 2 p . Similarly, Holmes’s expected payoff from choosing St. Bart’s when Watson is using his q -mix is 2 q ; his expected payoff from choosing Simpson’s when Watson is mixing is 1 – q . Watson’s best response to Holmes’s p -mix is to choose Simpson’s ( q = 0) for values of p below 2/3 and to choose St. Bart’s ( q = 1) for values of p above 2/3. Watson is indifferent between his two choices when p = 2/3. Similarly, Holmes’s best response to Watson’s q -mix is to choose Simpson’s ( p = 0) for values of q below 1/3 and to choose St. Bart’s ( p = 1) for values of q above 1/3. He is indifferent between his two choices when q = 1/3. The mixed-strategy equilibrium occurs when Holmes chooses St. Bart’s two-thirds of the time and Simpson’s one-

Games of Strategy , Fifth Edition Copyright © 2021 W. W. Norton & Company third of the time ( p = 2/3) and when Watson chooses St. Bart’s one-third of the time and Simpson’s two-thirds of the time ( q = 1/3). Best-response curves are shown below: There are three Nash equilibria indicated by the three large black dots on the diagram. At the bottom left and top right corners are the two pure strategy Nash equilibria: (Simpson’s, Simpson’s) where p = q = 0, and (St. Bart’s, St. Bart’s) where p = q = 1. The mixed strategy Nash equilibrium occurs in the center at p = 2/3, q = 1/3. Expected payoffs for Holmes and Watson are 2/3 each in this last equilibrium. Both players would prefer either of the pure-strategy Nash equilibria. If they can coordinate their randomization in some way so as to alternate between the two pure-strategy Nash equilibria, they can achieve an expected payoff of 1.5 rather than the 2/3 that they achieve in the mixed-strategy equilibrium. S8. (a) When x < 1, No is a dominant strategy for both players, so (No, No) is the unique Nash equilibrium; there is no equilibrium in mixed strategies. (b) There is a mixed-strategy Nash equilibrium when x > 1. In that MSE, Yes will be played by both players with probability 1/ x . To solve for p , use px = p + 1(1 – p ) to find x = 1/ p . Similarly, for q : qx = q + 1(1 – q ) gives x = 1/ q . In the mixed-strategy Nash equilibrium, Rowena plays 1/ x (Yes) + (1 – 1/ x )(No) and Colin plays 1/ x (Yes) + (1 – 1/ x )(No).

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