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Solutions to Chapter 3 Exercises
SOLVED EXERCISES
S1. (a) There is one initial node (I) for Hansel making the first move; three decision nodes (D)
(b) There is one initial node (I) for Hansel making the first move; four decision nodes (D)
(c) There is one initial node (I) for Hansel making the first move; five decision nodes (D)
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
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 on the trees are labeled 1, 2, and so
(a) Scarecrow has two strategies: (1) N or (2) S. Tinman has two strategies: (1) t if
(b) Scarecrow has two actions at three different nodes, so Scarecrow has eight strategies: 2 • 2 • 2
= 8. To describe the strategies accurately, we must specify a player’s action at each decision node.
Tinman has two actions at three different nodes, so Tinman also has eight strategies: 2 • 2 • 2 = 8.
(c) Scarecrow has two actions at three decision nodes, so Scarecrow has eight strategies: 2 •
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
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 S, so Scarecrow
(b) The graph below indicates which action Scarecrow and Tinman choose at each node.
(c) The graph below indicates which action Scarecrow, Tinman, and Lion choose at each
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
S4. The game tree is shown below:
Boeing prefers $300 million to losing $100 million, so Boeing will peacefully accommodate Airbus’s
S5. (a) For Barney to win the game, 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
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
(b) From part (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,
Another way 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
(c) The full game has 21 matchsticks, and Fred begins. As described in part (b), Fred wants
(d) Each player wants to leave the other player a multiple of 5 matchsticks. So on each turn,
S6. (a) The game tree is shown below:
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
(b) The graph in part (a) indicates the four rollback equilibria, which can be described as
(c) With 5 matchsticks at the beginning of the game, there is a second-mover advantage,
(d) There is more than one rollback equilibrium because, so long as Barney plays optimally,
S7. (a) Elroy will win if the running total is at least 91 on his turn, because he can bring the total
to exactly 100 from any number between 91 and 99. However, if the running total is 90, then Elroy cannot
win because he cannot get the total to 100 and after his turn the total will be at least 91. So, any player
(b) The player who goes first cannot win, so all choices between 1 and 9 are equally good (or
S8. (a) The game tree is shown below:
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
(b) The rollback equilibrium is Lion 1 eats the slave, Lion 2 does not eat Lion 1, and Lion 3
S9. (a) The game tree is shown below:
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
(b) Rollback pruning is illustrated by arrows on branches of the tree. The equilibrium entails
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
