指定习题和参考答案Chapter 2: 1, 21. (a) D. There is so much amount and kinds of yogurt that each shopper need not consider whatothers do.(b) G. A prom is likely to be small enough that any participant can have some conceivable influence on others. For example, these two should prevent from dressing the same. However, in a BIG CITY, this may not be a case.(c) D. If he is not restricted to any specific university.(d) G. There are so few competitors for this specific product.(e) G. He must consider the interaction between him and his running mate. Even more, he must consider the response of the constituency when he chooses the mate.2. (a) Z; R/NR; I; F; NC.(b) NZ; NR; A; F; C.(c) NZ;NR; I; F; CChapter 3: 1, 2, 6, 71. (a) D=3, T=6(b) D=4, T=9(c) D=5, T=82. (a) #S(A)=2, S(A)={N,S}; #S(B)=2, S(B)={t, b}(b) #S(A)=2*2*2=8, S(A)={NNN, NNS, NSN, NSS, SNN, SNS, SSN, SSS}(left, right-up,right-down)#S(B)=2, S(B)={t, b}; #S(C)=2, S(C)={u,d}(c) #S(A)=2*2*2=8, S(A)= {NNN, NNS, NSN, NSS, SNN, SNS, SSN, SSS}(left, middle,right)#S(B)=8, S(B)={nnn, nns, nsn, nss, snn, sns, ssn, sss}6. (a) At the final run, B must take dimes. At the run next to final, if A choose to pass, given Btake dime at the final run, he can only get the reward accumulating at a speed of 5 cents per turn, initially 0 cents. If he takes dimes, he gets the dimes accumulating at a speed of 20 cents per turn, initially 10 cents. He surely takes dimes. Given this, B must take dimes at the run second next to final (B’s payoffs are the same as in our initial game) . Then A has the same reasoning as at the next to final run. And so on. So the rollback equilibrium strategy is always taking dime, for both players.(b) Using rollback method, let’s consider from the second round.No matter what payoffs both have got in the first round, they have sunk and cannot affect both players’ decisions in the second round. So two players play the second round as if it is a single-round centipede. So each plays is always choosing to take dimes, resulting payoffs of10 to A and 0 to B.Keeping this in both players’ mind, the first round is played almost as a single-roundcentipede, except that A always gets a reward of 10, whether she passes the dime or take it.But the fact that you always get something will neither affect your decision. So finally the first round game is exactly as a single-round centipede.Use rollback to this reduced single-round centipede. Notice the game is essential the same as before in the first 6 runs when the total amount is not accumulated to 50 when it’s player A’ s turn to play. If we can decide that the game after the 6th run will have an equilibrium of both players always taking dimes, we can be sure that B will get at most 20 when the 7th run comes and A (at most) takes 50.Now consider this ‘small’ equilibrium with only the last 4 runs. B will take dimes at in the final run (with a payoff of 90, otherwise 0). In any round before for A to play, taking dimes will leave her at least 50, but passing it will leave her at most 100-90=10 if B takes dimes then. In any round before for B, taking dime will leave him at least 60, but passing it will leave her at most 100-50=50 if A takes dimes then. Given the final run’s action of B (taking dimes), we can derive that both will always take dime in any round before, but after 6th run.Now let’s go back to our 6th run. B will take dimes at the 6th run (with a payoff of 60, otherwise 20). Given B’s choice, the game before the 6th run goes like the original one.The equilibrium is that both players always take dimes.(c) All the logic of (b) applies here until we begin to consider the last 5 runs (not 4 runs at thistime) of the first round game. Use rollback.B takes dimes in the final run, but get only 40 and leave 50 (as we restricts) to A. In the9th(second to the last) run, A can choose either to pass or take dimes(getting 50 both). The game essential the same when it ends in 9th run or 10th run.However, the 8th run is crucial. B can choose either to pass or to take dimes (getting 40 both). If B chooses to take dimes, A will definitely chooses to take dimes in the 7th run, and so on. The cooperation totally breaks and results in taking dimes always (our old equilibrium outcome).In any round before, but after 5th run, if both expect complete cooperation throughout the future, then they can either choose to pass or take. Any single deviation from cooperation (taking dimes) will result in total breakdown. So the 6-8th runs are all crucial points, given cooperation afterwards.Consider both pass dimes. Given this, at the 5th run, A chooses to pass is at least as good as to take. Taking them leads to breakdown as well. Suppose she passes it. B will also be indifferent with passing or taking dimes. So the 4-5th runs are also crucial points.However the 3rd run and the earlier are not crucial. If cooperation always happens afterward, passing dimes are always strictly better than taking dimes. If breakdown has already happen afterwards, taking dimes are definitely better choices.Summary: this game has multiple equilibria. Always passing dimes is one of them. Any deviations in one of 4-8th runs, given cooperation afterwards, will lead to breakdown from the very beginning. ( The number of equilbria is (1+5)*2=12, taking into account in the 9th run, both passing or taking dimes are indifferent for A.)7. (a) Amy; Beth.The player who reaches exactly 89 first will win; thus the player who reaches 78 first will win;thus 67,56, 45, 34, 23, 12, 1. Obviously Amy will reach 1 first.The player who reaches exactly 99 first will win. Then 88, 77, 66, 55, 44, 33, 22, 11.Obviously Beth will reach 11 first.(b) For Amy: 1 in the first run, then 11 no matter what Beth chooses, and 22, and so on. ForBeth, she can choose any amount between 1 and10.For Amy: any amount between 1 and 10. For Beth, 11 in his first run, 22 in his second run, and so on.Chapter 4: 2(d), 3(d), 5, 6, 9, 122. (d) Up is dominated by Straight (for Row). Then Left and Middle are both dominated by Right(for Column). Given Column plays Right, Row chooses Straight. The only NE is (Straight, Right).3. (d) (North, East) (unique NE). No dominant or dominated for both. Best-response analysis.5. Dominance solvable. (Up, Left). (Level, Center) has the same payoffs but not NE. There arealso other strategy configurations having strictly higher payoffs but not NEs.6. (1)JapaneseNorthern Southern AmericanNorthern 2 2Southern 1 3 (2) “Southern” is weakly dominated by “Northern” for Japanese. (North, North) is the only NE.(Check if “Southern” for Japanese can be part of any NE before you eliminate it.)9. (a) 3 NEs: (1, 1), (2, 2), (3, 3). The first is less likely to be focal point. But the other two areequally likely to be.(b) 12.5; Yes; Mixed strategy (just like flipping coins) can be used.12. (a) Two NEs: (Brunette, Blonde), and (Blonde, Brunette), yielding payoffs (5, 10) and (10, 5)respectively.Player 2Blonde BrunettePlayer 15Blonde 0,0 10,Brunette 5, 10 5, 5(b)Player 3Blonde BrunettePlayer 2 Player 2Blonde Brunette Blonde Brunette Player 1Blonde 0, 0, 0 0, 5, 0 0, 0, 5 10, 5, 5Brunette 5, 0, 0 5, 5, 10 5, 10, 5 5, 5, 53 NEs: (Blonde, Brunette, Brunette), (Brunette, Blonde, Brunette), and (Brunette, Brunette, Blonde), with payoffs 10 for who wins the Blonde, 5 for who wins Brunette.(c) The only possible NEs are those with only one pursues and wins the only Blonde and all others pursues and wins somehow excessive Brunettes.Or, follow the assumption given in text. Given k>=1, for any player, his payoffs of pursuing the Blonde is 0 and definitely worse than pursuing the Brunette with payoff 5. However, if k=0, you’d better pursue the Blonde.Consider this for a while, you will get the NE solutions.Thus All players choose Brunette cannot be a NE.。
