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1、 （b） There is one initial node （I） for Hansel making the first move; four decision nodes （D） including the initial node, which represent the nodes where Hansel or Gretel make a decision; and nine terminal nodes （T）. （c） There is one initial node （I） for Hansel making the first move; five decision no

2、des （D） including the initial node, which represent the nodes where Hansel or Gretel make a decision; and eight terminal nodes （T）.S2. For this question, remember that actions with the same label, if taken at different nodes, are different components of a strategy. To clarify the answers, the nodes

3、on the trees are labeled 1, 2, etc. （in addition to showing the name of the player acting there）. Actions in a strategy are designated as N1 （meaning N at node 1）, etc. The trees are below in the solutions to Exercise S3. Numbering of nodes begins at the far left and proceeds to the right, with node

4、s equidistant to the right of the initial node and numbered from top to bottom. （a） Scarecrow has two strategies: （1） N or （2） S. Tinman has two strategies: （1） t if Scarecrow plays N, or （2） b if Scarecrow plays N. （b） Scarecrow has two actions at three different nodes, so Scarecrow has eight strat

5、egies, 222=8. To describe the strategies accurately, we must specify a players action at each decision node. Scarecrow decides at nodes 1, 3, and 5, so we will label a strategy by listing the action and the node number. For example, to describe Scarecrow choosing N at each node, we write （N1, N3, N5

6、）. Accordingly, the eight strategies for Scarecrow are: （N1, N3, N5）, （N1, N3, S5）, （N1, S3, N5）, （S1, N3, N5）, （N1, S3, S5）, （S1, N3, S5）, （S1, S3, N5）, and （S1, S3, S5）. Tinman has two actions at three different nodes, so Tinman also has eight strategies, 222=8. Tinmans strategies are: （n2, n4, n6

7、）, （n2, n4, s6）, （n2, s4, n6）, （s2, n4, n6）, （n2, s4, s6）, （s2, n4, s6）, （s2, s4, n6）, and （s2, s4, s6）. （c） Scarecrow has two actions at three decision nodes, so Scarecrow has eight strategies: 222=8. Scarecrows strategies are: （N1, N4, N5）, （N1, N4, S5）, （N1, S4, N5）, （S1, N4, N5）, （N1, S4, S5）, （

8、S1, N4, S5）, （S1, S4, N5）, and （S1, S4, S5）. Tinman has two strategies: （t2） and （b2）. Lion has two strategies: （u2） and （d2）. S3. （a） Beginning with Tinman, we see that Tinman prefers a payoff of 2 over 1, so Tinman chooses t. With Tinman choosing t, Scarecrow receives a payoff of 0 for N and 1 for

9、 S, so Scarecrow chooses S. Thus, the rollback equilibrium is Scarecrows choosing S and Tinmans choosing t （even though he wont have a chance to play it）. Tinmans action does not affect the rollback equilibrium, because Scarecrow expects Tinman to choose t, so Scarecrow best responds by choosing S.

10、（b） The graph below indicates which action Scarecrow and Tinman choose at each node. Scarecrows equilibrium strategy is S1, S3, N5, and Tinmans is n2, n4, s6. yielding the equilibrium payoff （4,5）. （c） The graph below indicates which action Scarecrow, Tinman, and Lion choose at each node. Scarecrows

11、 equilibrium strategy is N1, N4, N5; Tinmans is b; and Lions is d, yielding the payoff （2, 3, 2）.S4. The game tree is shown below. Boeing prefers $300 million to losing $100 million, so Boeing will peacefully accommodate Airbuss entry into the market. Airbus expects Boeing to accommodate its entry p

12、eacefully, so it can make $300 million by entering, or nothing by not entering, so Airbus will enter the market. Thus, the rollback equilibrium is Airbuss entering the market and Boeings peacefully accommodating, with a payoff to each firm of $300 million in profit.S5. （a） For Barney to win the game

13、, he must remove the last matchstick, which means that if he leaves Fred 1 to 4 matches, Fred can remove all of them and Barney would lose, so Barney must leave more than 4 matchsticks. Because there are 6 matchsticks and Barney must take at least 1, Barney should remove only 1 matchstick, which wil

14、l leave Fred with 5 matchsticks. No matter what Fred does, Barney, on his next turn, will be able to remove all the remaining matchsticks to win the game. More precisely, Barney should take 1 matchstick on his first turn. If Fred takes f matchsticks, Barney should take （5f） matchsticks. （b） From par

15、t （a）, we know that whomever is left with 5 matchsticks will lose the game, so Barney should remove enough matchsticks to leave Fred only 5. If Barney leaves Fred 6 to 9 matches, then Fred will leave Barney with 5, and Barney will lose, so Barney must leave Fred with more than 9 matches. Also, if Ba

16、rney leaves 11 matches, Fred can ensure he is left with 6 to 9 matches by choosing only 1 match, leaving Barney with 10 matches and no way to keep Fred from having 6 to 9 matches. Thus, Barney must take 2 matches, leaving Fred with 10, and must choose （5f） matches on each subsequent turn. Another wa

17、y to view this problem is that a player will lose if his turn begins with 5 matches. Thus, each player always wants to remove matchsticks to leave his opponent with 5. Since we know what happens when 5 matchsticks remain, we can divide the number of remaining matchsticks into units of 5. For example

18、, 12 matchsticks can be divided into two units of 5 with 2 extra. Barney wants to force Fred to have some multiple of 5, so Barney removes 2 matches at first, and then （5f） matches in each of his subsequent turns. （c） The full game has 21 matchsticks, and Fred begins. As described in （b）, Fred wants

19、 to leave Barney with some multiple of 5, and 21 is 4 units of 5 with 1 extra. So Fred should remove 1 matchstick on his first turn, and then （5b） matchsticks on his subsequent turns, where b is the number of sticks that Barney has just removed. With optimal play, Fred will win every time. （d） Each

20、player wants to leave the other player a multiple of 5 matchsticks. So on each turn, the player should divide the remaining matchsticks by 5 and remove the remainder. If the remainder is 0 and more than 4 matchsticks remain, then the player is stuck with a multiple of 5. So that player should random

21、ly choose 1 to 4 matchsticks, hoping that the opponent will make a mistake on a subsequent turn. If 4 or fewer matchsticks remain, then the player should remove all of them to win the game.S6. （a） The game tree is: （b） The graph in part （a） indicates the four rollback equilibria, which can be descri

22、bed as Freds taking 1 to 4 matchsticks, and then Barneys removing all remaining matchsticks. Letting the first number represent the number of matchsticks removed by Fred and the second by Barney, the four rollback equilibria may be described as: （1, 4）, （2, 3）, （3, 2）, and （4, 1）. （c） With 5 matchst

23、icks at the beginning of the game, there is a second-mover advantage, because no matter what quantity the first mover removes, the second mover can remove all remaining matchsticks to win the game. （d） There is more than one rollback equilibrium, because so long as Barney plays optimally, any of Fre

24、ds four actions at the initial node leads to the same payoff. Thus in equilibrium, he is indifferent among those four actions at that node.S7. （a） The game tree is shown below. （b） The rollback equilibrium is Lion 1 eats the slave, Lion 2 does not eat Lion 1, and Lion 3 would eat Lion 2 if given the

25、 opportunity （which he is not in equilibrium）. （c） There is a first-mover advantage, because Lion 3 will always eat Lion 2 if able, so Lion 2 has an incentive to not eat Lion 1 in order to protect himself from Lion 3. （d） Each lion has two actions at a single node （eat, dont eat）, so each has two co

26、mplete strategies.S8. （a） The game tree is given below. （b） Rollback pruning is illustrated by arrows on branches of the tree. The equilibrium entails Friedas choosing Rural; Big Giants always choosing Urban （UR, or U if U and U if R）; and Titans choosing Urban unless both Friedas and Titan have cho

27、sen Rural （UUUR）. The equilibrium payoffs are （2, 5, 5） to the stores in order of their moves.S9. （a） The game tree is shown below. （b） The Proposer has one node with 11 actions; thus the Proposer has 11 complete strategies. We can list these as the split proposed, with the first number indicating t

28、he portion for the Proposer and the second for the Responder. The 11 complete strategies are: 0/10, 1/9, 2/8, 3/7, 4/6, 5/5, 6/4, 7/3, 8/2, 9/1, and 10/0. The Responder has 11 nodes with actions Accept or Reject at each node; thus the Responder has 211, or 2,048 complete strategies. Some examples of

29、 possible strategies include accepting only 5/5, accepting only 10, accepting only odd numbers, and rejecting all offers. （c） Assuming that the players care only about their cash payoffs means that the Responder will definitely accept any positive offer and will be indifferent between Accepting and

30、Rejecting when offered nothing. If the Proposer assumes that the Responder will accept an offer of $0 when indifferent, then the rollback equilibrium is to offer $0 and for it to be accepted. However, although the Proposer may be unsure of the Responders action in the face of indifference, he can ex

31、pect the Responder to accept an offer of $1. If there is uncertainty about the Responders action when shes indifferent, the rollback equilibrium occurs where the Proposer offers $1 and that offer is accepted. （d） Because Pete knows Rachel will accept any offer of $3 or more, Pete can maximize his payoff by offering only $3. （e） There are many possible utilities that may represent Rachels utility. One common utility is fairness, in which Rachel receives a utility equal to the dollar amount if the offer is within 40% to 60% of the total am