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Solutions to Chapter 12 Exercises
UNSOLVED EXERCISES
U1. (a) Payoff table is shown below. When two fighters meet, each has a 50% probability of
Let x represent the population proportion of sharers (S). Then the fitness of a fighter (F) is F(F) =
(b) Payoff table is shown below. With x defined as above, now F(F) = 200x – 50(1 – x) =
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
Column
Fight Share
Column
Fight Share
(c) In part (a), V = 200 and C = 100; note that C represents the total cost of the fight, which is
50 calories for each fighter. Here, V > C. Fighters are similar to hawks and sharers are similar to doves.
U2. (a) Let D represent a perpetual defector, and T a tit-for-tat player.
Player 2
T D
(b) F(D) = p(2n) + (1 – p)(2n + 2) = 2(n + 1 – p)
(d) F(D) > F(T) when 2n + 2 – 2p > 3n – np – p, or 2 – p > n – np. This yields np – p > n – 2,
(e) See the following diagram. There are three possible equilibria. In one, p = 0; the entire
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
(f) The population will move toward the all-tit-for-tat equilibrium (in which cooperation
Intuitively, the possible advantage of being hardwired to play tit-for-tat is that it gives a player the
chance to establish long-lasting (and beneficial) cooperation if she meets another tit-for-tat player. The
U3. (a)
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
Payoff table is shown below:
(b) Remember that payoffs are years in jail here, so smaller numbers represent better fitness.
Then, no matter what mixture makes up the overall population, S is (strictly) fitter than N, so N will die
(c) Suppose N has died out. Then S strictly dominates N in a population consisting of A, T,
The (only) ESS in this game is A. In Exercise S3 (which had no S strategy), T could be an ESS
(provided that there were enough Ts in the population at the beginning of the game). In this problem, T
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
Column
A T N S
Note that if a fully rational player were matched against a player she knew to be a T type, the
U4. In the Tennis-Point game there are two species: servers and receivers. Let x be the proportion of
servers that always serve down the line (DL), so that 1 – x servers that always serve crosscourt (CC).
Similarly, let y be the proportion of receivers that always receive down the line (DL), so that 1 – y
receivers always receive crosscourt (CC). The payoff table of this game follows:
Fitness of a DL server = 50y + 80(1 – y) = 80 – 30y
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
Receiver
DL[y] CC[1 – y]
The diagram below shows the dynamics of the game. There is one equilibrium, at the point x = 0.7,
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
U5. (a) Recall that the payoff table is as follows. If this were a rationally played game,
the mixed-strategy equilibrium would entail Row playing Fight with probability p where p solved –50p +
(b) The payoff table is shown below:
Column
Fight Share Mix
–50(2/3) + 200(1/3) = 100/3,
–50(2/3) + 200(1/3) = 100/3
0(2/3) + 100(1/3) = 100/3
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
Column
Fight Share
(c) If the population is mostly mixers with a small proportion (x) of mutant sharers, F(M) =
(500/3)x + (100/3)(1 – x) and F(S) = 100x + (100/3)(1 – x). The final terms in these fitness expressions
U6. The payoff table for the Baker-Cutler game follows:
(a) Fitness of type F = 50s + 50(1 – s) = 50
Fitness of type G = 90s + 9(1 – s) = 9 + 81s
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
Cutler
F[f] G[1 – f]
(f) The unique ESS occurs at f = 0, s = 1. Interestingly, along the line f = 1, 0 ≤ s ≤ 41/81
there are an infinite number of points that are nearly stable. Slight perturbations from a point on this line
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
