SolvedExercises_Chapter7_Solutions_Updated

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 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).

Games of Strategy , Fifth Edition Copyright © 2021 W. W. Norton & Company (c) This is an example of an assurance-type game because there are two Nash equilibria in pure strategies where the players coordinate on the same strategy. (d) There are two pure-strategy Nash equilibria: p = 1, q = 1 (Yes, Yes) and p = 0, q = 0 (No, No). There is also a mixed-strategy Nash equilibrium at p = 1/3, q = 1/3. (e) There are two pure-strategy Nash equilibria: p = 1, q = 1 (Yes, Yes) and p = 0, q = 0 (No, No).

Games of Strategy , Fifth Edition Copyright © 2021 W. W. Norton & Company S10. (a) To find Bart ’s equilibrium mix, set Lisa ’s payoffs from each of her pure strategies, against Bart ’s p -mix, equal to one another. In his p -mix, Bart will play Rock with probability p 1 , Scissors with probability p 2 , and Paper with probability 1 – p 1 – p 2 . Lisa ’s payoff from Rock against this p -mix is 10 p 2 – 10(1 – p 1 – p 2 ); her payoff from Scissors against the p -mix is – 10 p 1 + 10(1 – p 1 – p 2 ); and her payoff from Paper against the p -mix is 10 p 1 – 10 p 2 . Equating the last two of these payoffs yields – 10 p 1 + 10(1 – p 1 – p 2 ) = 10 p 1 – 10 p 2 , which simplifies to – 20 p 1 – 10 p 2 + 10 = 10 p 1 – 10 p 2 , or 10 = 30 p 1 or p 1 = 1/3. Then equating the first two payoffs yields 10 p 2 – 10(1 – p 1 – p 2 ) = – 10 p 1 + 10(1 – p 1 – p 2 ) or – 10 + 20 p 2 + 10 p 1 = – 20 p 1 – 10 p 2 + 10. Rearranging and substituting 1/3 for p 1 yields 30 p 2 = 10, or p 2 = 1/3 also. Then we get 1 – p 1 – p 2 = 1/3 as well. Bart ’s equilibrium mix entails playing each strategy 1/3 of the time or with probability 33 1/3. The symmetry of the game guarantees that Lisa ’s equilibrium mix is the same. (b) Check Bart ’s payoffs from each of his pure strategies against Lisa ’s mix : Payoff from Rock: (0)(0.4) + (10)(0.3) + ( – 10)(0.3) = 0 Payoff from Scissors: ( – 10)(0.4) + (0)(0.3) + (10)(0.3) = – 1 Payoff from Paper: (10)(0.4) + ( – 10)(0.3) + (0)(0.3) = 1 Bart ’s expected payoff from using the pure -strategy Paper exceeds his expected payoffs from his other two strategies. He should play only Paper when Lisa uses the mix described. Presumably, Lisa has chosen a nonequilibrium mix, so Bart is not indifferent among his available pure strategies. S11. (a) Table is below. Vendor 2 A B C D E Vendor 1 A 85, 85 100, 170 125, 195 150, 200 160, 160 B 170, 100 110, 110 150, 170 175, 175 200, 150 C 195, 125 170, 150 120, 120 170, 150 195, 125 D 200, 150 175, 175 150, 170 110, 110 170, 100 E 160, 160 150, 200 125, 195 100, 170 85, 85

Games of Strategy , Fifth Edition Copyright © 2021 W. W. Norton & Company (b) For both vendors, locations A and E are dominated. Thus, for a fully mixed equilibrium, we need only consider each vendor’s choice among locations B, C, and D. Let Vendor 1’s mixture probabilities be p B , p C , and (1 – p B – p C ). Similarly, let Vendor 2’s mixture probabilities be q B , q C , and (1 – q B – q C ). After simplifying, the p -mix and q -mix payoffs are as shown below. Vendor 2 B C D q -mix Vendor 1 B 110, 110 150, 170 175, 175 175 – 65 q B – 25 q C , 175 – 65 q B – 5 q C C 170, 150 120, 120 170, 150 170 – 50 q C , 150 – 30 q C D 175 , 175 150, 170 110, 110 110 + 65 q B + 40 q C , 110 + 65 q B + 60 q C p -mix 175 – 65 p B – 5 p C , 175 – 65 p B – 25 p C 150 – 30 p C , 170 – 50 p C 110 + 65 p B + 60 p C , 110 + 65 p B + 40 p C (c) To find the equilibrium p, set V endor 2’s payoffs equal: 175 – 65 p B – 25 p C = 170 – 50 p C = 110 + 65 p B + 40 p C 65 p B – 25 p C = 5 and 60 – 90 p C = 65 p B 60 – 90 p C – 25 p C = 5 Then p C = 55/115 = 11/23, p B = (25p C + 5)/65 = (275/23 + 115/23)/65 = (390/23) / 65 = 6/23, and p D = (1 – p B – p C ) = 1 – 11/23 – 6/23 = 6/23. Similarly, q B = 6/23, q C = 11/23, and q D = 6/23. One way to explain why A and E are unused in the equilibrium is to point out that they are (as noted above) dominated. This also implies that A and E are unused because they result in a payoff against the opponent’s equilibrium mixture that is lower than p roduced by choices B, C, and D. Specifically, when Vendor 2 uses the equilibrium mixture probabilities of (6/23, 11/23, 6/23), Vendor 1’s payoff from choosing A is 100(6/23) + 125(11/23) + 150(6/23) = 2,875/23. B is 110(6/23) + 150(11/23) + 175(6/23) = 3,360/23. C is 170(6/23) + 120(11/23) + 170(6/23) = 3,360/23. D is 175(6/23) + 150(11/23) + 110(6/23) = 3,360/23. E is 150(6/23) + 125(11/23) + 100(6/23) = 2,875/23.

Games of Strategy , Fifth Edition Copyright © 2021 W. W. Norton & Company (c) For the goalie playing against the kicker using LL 37.8%, HC 24.4%, and LR 37.8% of the time, payoffs from each of his possible strategies are L .378(0.40) + .244(0.85) +.378(0.95) = .7177 C .378(0.95) + .244(0) +.378(0.95) = .7182 R .378(0.95) + .244(0.85) +.378(0.40) = .7177 (d) All three strategies give almost the same expected payoff, so the goalie could play all of them effectively in his mix. He will use all of L, C, and R, in his equilibrium mixture. (e) These mixed strategies are Nash equilibria, because each player’s strategy is a best response given the other player’s strategy. (f) The equilibrium payoff to the kicker is .717822 . S13. (a) Suppose Rowena plays Up with probability p and Down with probability (1 – p). Her p -mix must keep Colin indifferent between his two pure strategies, Left and Right. Thus it must be true that p A + (1 – p )C = p B + (1 – p )D, or p = (D – C)/[(A – B) + (D – C)]. (b) Colin must keep Rowena indifferent between her two pure strategies, Up and Down. Therefore: qa + (1 – q ) b = qc + (1 – q ) d , or q = ( d – b )/[( a – c ) + ( d – b )]. (c) In each case, the player’s equilibrium mixture probabilities are totally independent of his or her own payoffs. Rowena’s optimal p depends only on Colin’s payoffs and Colin’s equilibrium q depends only on Rowena’s payoffs. (d) To guarantee that a mixed-strategy Nash equilibrium exists, the values for p and q cannot be 0 or 1. Thus, we need C  D, (A – B)  (C – D), and A  B. Similarly, b  d , ( a – c )  ( b – d ), and a  c .

Games of Strategy , Fifth Edition Copyright © 2021 W. W. Norton & Company S14. (a) Consider any one of the young men. If he chooses to go after a Brunette, he gets a guaranteed payoff of 5. If he chooses to go after the Blonde, he gets 10 if none of the other ( n – 1) men choose Blonde, and 0 otherwise. Since each of the other ( n – 1) chooses Blonde with probability p , and they are choosing independently, the probability that none of them choose Blonde is (1 – p ) n – 1 . Therefore, the expected payoff to any one man from choosing Blonde is 10(1 – p ) n – 1 + 0[1 – (1 – p ) – 1 ] = 10(1 – p ) n – 1 . In the mixed-strategy equilibrium, this man must be indifferent between his two pure strategies. Therefore, it must be true that 10(1 – p ) n – 1 = 5, or (1 – p ) n – 1 = 1/2 = 2 – 1 , so p = 1 – 2 – 1/( n – 1) . (b) The same logic can be applied to the m young men who are choosing Blonde, so q = 1 – 2 1/( m – 1) . Consider any one of the ( n – m ) choosing Brunette. If he switched to Blonde, his payoff will be 10 if all the m -mixers happen to choose Brunette, and 0 otherwise. Therefore, his expected payoff is 10(1 – q ) m . So this pure Brunette chooser does not want to switch to a pure Blonde strategy if 10(1 – q ) m < 5. For this Brunette chooser, the payoff from any mixture will be an average between the two sides of this inequality, so the same inequality also rules out his wanting to switch to any mixed strategy. But we already have the condition of the mixed-strategy equilibrium 10(1 – q ) m – 1 = 5. Therefore, 10(1 – q ) m = 10(1 – q) m – 1 (1 – q) = 5(1 – q) < 5. So, the condition holds.