978-0393919684 Chapter 3 Solution Manual Part 2

Solutions to Chapter 3 Exercises

S10. (a) The game tree is shown below:

(b) The Proposer has one node with 11 actions; thus the Proposer has 11 complete strategies.

The Responder has 11 nodes with actions accept or reject at each node; thus the Responder has

(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 rejecting when

(e) There are many possible utilities that may represent Rachel’s 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%

(f) One possible explanation is that the average Responder values not only the money, but

UNSOLVED EXERCISES

U1. A first-mover advantage is not a necessary property of sequential games. There are sequential

games with a first-mover advantage, a second-mover advantage, or no advantage. Consider the matchstick

U2. For this question, remember that actions with the same label, if taken at different nodes, are

(a) Albus has three actions at only one node, so he has only three complete strategies, which

(b) Albus has the same two actions at two different nodes, so he has 22 = 4 complete

(c) Albus has the same two actions at two different nodes, so he has 22 = 4 complete

U3. (a) The equilibrium strategies are (N1) for Albus and (b2, a3, b4) for Minerva. The rollback

payoff is (2, 1).

U4. (a) The tree is shown below. Rollback equilibrium entails Congress passing a bill containing

both Proposals A and B and the president signing the bill. Payoffs are (3, 3).

(b) The tree

(c) The president’s payoff is worse when he has the line-item veto (LIV) than when he does

not. This outcome arises because Congress knows that without the LIV, they can get their preferred

U5. (a) Amy will win if there are 1 to 10 pennies on her turn, because she can remove all of

them. However, if there are 11 pennies, then Amy cannot win, because after her turn, there will be at most

This process may be continued all the way back to 99 pennies, revealing that whoever is left with

99 pennies will lose the game. Thus, if Amy is moving first, she can simply remove 1 penny first, and

If both players play optimally, then Amy will win, because there is a first-mover advantage to the

game. By going first, Amy can always ensure that Beth is left with a multiple of 11 pennies.

(b) Amy should begin by removing 1 penny and then remove (11 – b) pennies thereafter. If

U6. (a) For this new game, if Amy is left with 1 penny, she loses, but if there are 2 to 11, she

wins, because she can remove enough to leave only 1 penny for Beth, forcing her to lose. But if there are

(b) As explained in part (a), each player is trying to leave the other with a multiple of 11 plus

U7. (a) The game has a second-mover advantage. Fozzie can guarantee victory by exactly

(b) When the difference in the number of pennies in each jar is between 1 and 10 inclusive,

each player’s optimal move is to remove the difference from the jar with more pennies, leaving the same

U8. (a) The game tree is shown below:

(b) In the rollback equilibrium, Lion 4 would eat Lion 3 if given the opportunity. Knowing

(c) The additional lion is good for the slave, who is not eaten in the new equilibrium. In this

game, the final lion always prefers to eat, so the second-to-last lion must not eat in order to defend itself.

U9. (a) The game tree is shown below:

(b) The rollback equilibrium for the game is for Bart to choose the amusement park (A), for

(c) Bart has two actions at one node, so he has two complete strategies, which are A and S.

U10. (a) If Rich chooses to continue at the initial node, then there are two ways he can win: (1) he

(b) If Rich gives up at the initial node, he can only win if Kelly wins immunity (probability

(c) Rich’s probability of winning if he were to continue is xp + yq. His probability of

(d) Using the given values and the inequality from part (c), we see that quitting is better

when zq < xp + yq, or z(0.6) < (0.45)(0.4) + (0.5)(0.6), which simplifies to z < 0.8. This means that when

(e) Rich should choose Give Up when zq > xp + yq. Using the values from Figure 3.11, this

inequality becomes (0.9)q > (0.45)p + (0.5)q  (0.4)/(0.45) > p/q. That is, Rich should choose Give Up