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```					Solutions to Chapter 7 Exercises

SOLVED EXERCISES

S1.      False. A player’s equilibrium mixture is devised in order to keep her opponent indifferent
among all of her (the opponent’s) possible mixed strategies; thus, a player’s equilibrium mixture
yields the opponent the same expected payoff against each of the player’s pure strategies. Note
that the statement will be true for zero-sum games, because when your opponent is indifferent in
such a game, it must also be true that you are indifferent as well.

S2       (a)     The game most resembles an assurance game because the two Nash equiibria
occur when both players play the same move. In an assurance game, both players prefer to make
the same move, but there is also a preferred Nash equilibrium with higher payoffs for both
players. In this game, (Risky, Risky) is the preferred equilibrium because it has higher payoffs,
but there is a chance that the players will play the worse Nash equilibrium with lower payoffs.
Even worse, the players might not play an equilibrium at all. Without convergence of
expectations, these results can occur, and this is characteristic of an assurance game.
(b)     The two pure-strategy Nash equilibria for this game are (Risky, Risky) and (Safe,
Safe).

S3.      (a)     There is no pure-strategy Nash equilibrium here, hence the search for an
equilibrium in mixed strategies. Row’s p-mix (probability p on Up) must keep Column indifferent
and so must satisfy 16p + 20(1 – p) = 6p + 40(1 – p); this yields p = 2/3 = 0.67 and (1 – p) = 0.33.
Similarly, Column’s q-mix (probability q on Left) must keep Row indifferent and so must satisfy
q + 4 (1 – q) = 2q + 3(1 – q); the correct q here is 0.5.
(b)     Row’s expected payoff is 2.5. Column’s expected payoff = 17.33.
(c)     Joint payoffs are larger when Row plays Down, but the highest possible payoff to
Row occurs when Row plays Up. Thus, in order to have a chance of getting 4, Row must play Up
occasionally. If the players could reach an agreement always to play Down and Right, both would
get higher expected payoffs than in the mixed-strategy equilibrium. This might be possible with
repetition of the game or if guidelines for social conduct were such that players gravitated toward

Solutions to Chapter 7 Solved Exercises                                                      1 of 6
the outcome that maximized total payoff.

S4.       The two pure-strategy Nash equilibria are (Don’t Help, Help) and (Help, Don’t Help).
The mixed-strategy Nash equilibrium has the following equilibrium mixtures:
2p + 2(1 – p) = 3p + 0          p = 2/3
2q + 2(1 – q) = 3q + 0          q = 2/3
That is, each player helps two-thirds of the time and doesn’t help one-third of the time.

S5.       (a)     Evert will play DL the same as before, because her p-mix depends upon
Navratilova’s payoffs.
(b)     70p + 10(1 – p) = 40p + 80(1 – p)        p = 7/10
70q + 40(1 – q)= 10q + 80(1 – q)         q = 2/5
So the mixed-strategy Nash equilibrium occurs when Evert plays 7/10(DL) + 3/10(CC) and
Navratilova plays 2/5(DL) + 3/5(CC).
Evert’s expected payoff is 70(2/5) + 40(1 – 2/5) = 52.
(c)     Compared with the previous game, Evert plays DL with the same proportion,
whereas Navratilova plays DL less, going from 3/5 to 2/5. Navratilova’s q-mix changes because
her mix is dependent on Evert’s payoffs. The changes in Evert’s payoffs lead to a change in
Navratilova’s q-mix. On the other hand, Evert’s p-mix doesn’t change because Navratilova’s
payoffs have remained unchanged.

S6.       (a)     0p – 1(1 – p) = 1p – 10(1 – p)    p = 9/10
0q – 1(1 – q) = 1q – 10(1 – q)    q = 9/10
In the mixed-strategy Nash equilibrium James plays 9/10(Swerve) + 1/10(Straight), and Dean
plays 9/10 (Swerve) + 1/10(Straight). James and Dean play Straight less often than in the
previous game.
(b)     James’ expected payoff = 9/10 – 10(1 – 9/10) = – 1/10
Dean’s expected payoff = 9/10 – 10(1 – 9/10) = – 1/10
(c)     If James and Dean collude and play an even number of games where they
alternate between (Swerve, Straight) and (Straight, Swerve), their expected payoffs would be 0.
This is better than the mixed-strategy equilibrium, because their expected payoffs are
– 1/10.

Solutions to Chapter 7 Solved Exercises                                                       2 of 6
(d)     James’ expected payoff = 1/2[(0 –1)/2)] + 1/2[(1 – 10)/2)] = – 5/2
Dean’s expected payoff = 1/2[(0 –1)/2)] + 1/2[(1 – 10)/2)] = – 5/2
These expected payoffs are much worse than the collusion example or the mixed-strategy
equilibrium. In this case both players are mixing with the wrong (that is, not the equilibrium)
mixture. Neither is player is best responding to the other’s strategy, and in this situation—with
the very real possibility of reaching the – 10 payoff—the expected consequences are dire.

S7.     (a)     See the table below. Best responses are underlined. Note that there is no cell
where both players are mutually best responding:
COLUMN

Left                      Right

Up               2, 1                      – 1, 4
ROW
Down             0, 3                      3, 2

There are thus no pure-strategy Nash equilibria.
(b)     p + 3(1 – p) = 4p + 2(1 – p)       p = 1/4
2q – 1(1 – q) = 3(1 – q)           q = 2/3
The mixed-strategy Nash equilibrium occurs where Row plays 1/4(Up) + 3/4(Down) and Column
plays 2/3(Left) + 1/3(Right).
(c)     Row’s p-mixture would remain the same, because Column’s payoffs have
remained the same. Changes to Row’s payoffs affect only Column’s equilibrium q-mix.
(d)     –q –1(1 – q) = 3(1 – q)            q = 3/4
So the new mixed-strategy Nash equilibrium would be:
Row plays 1/4(Up) + 3/4(Down)
Column plays 3/4(Left) + 1/4(Right)

S8.     (a)     .7p + .85(1 – p) = .8p + .65(1 – p)          p = 2/3
.3q + .2(1 – q) = .15q + .35(1 – q)          q = 1/2
The mixed-strategy Nash equilibrium is:
Batter plays 2/3(Anticipate fastball) + 1/3(Anticipate curveball)

Solutions to Chapter 7 Solved Exercises                                                       3 of 6
Pitcher plays 1/2(Throw fastball) + 1/2(Throw curveball)
(b)      Batter’s expected payoff = .3(1/2) + .2(1 – 1/2) = 0.25
Pitcher’s expected payoff = .7(2/3) + .85(1 – 2/3) = 0.75

S9.     The mixed-strategy Nash equilibrium is now:
.75p + .85(1 – p) = .8p + .65(1 – p)         p = 4/5
.25q + .2(1 – q) = .15q + .35(1 – q)         q = 3/5
The pitcher’s new expected payoff is .75(4/5) + .85(1 – 4/5) = 0.77, which is indeed
greater than his previous expected payoff. The pitcher can increase his expected payoff because
the batter is forced to adjust his equilibrium strategy in a way that favors the pitcher.

S10.    (a)      The home team must play the 20-yard play if they have 20 yards to go for fourth
down. The rival’s team best-response would be to anticipate the 20-yard play.
(b)      The home team’s expected payoff when there are 20 yards to go on fourth down
is (0.5)(1) + (0.5)(0) = 1/2.
(c)      0.2p + 0(1 – p) = 0p + 0.5(1 – p)             p = 5/7
0.8q + 1(1 – q) = q + 0.5(1 – q)              q = 5/7
That is, the mixed-strategy Nash equilibrium is:
Home Team plays 5/7(10-yard play) + 2/7(20-yard play)
Rival Team plays 5/7(Anticipate 10-yard play) + 2/7(Anticipate 20-yard play)
(d) The home team’s expected payoff is: (5/7)(.8) + (2/7)(1) = 6/7.
(e) With probability 0.5, the home team’s 20-yard play will succeed when a 20-yard play
is anticipated, resulting in payoffs of (1, 0). However, with probability 0.5, the 20-yard play will
fail, leaving the home team in the situation described in part (a): fourth down with 20 yards to go.
In part (b) we saw that the expected payoff to the home team in this situation was 1/2. The
expected payoff to the home team from using the 20-yard play on third down is the weighted
average of the outcomes when it succeeds and when it fails:
0.5(1) + 0.5(1/2) = 3/4.
(f) With probability 0.8, a 10-yard play succeeds when a 10-yard play is anticipated, and
with probability 0.2 it fails. When it succeeds the home team is left with 10 yards to go on fourth
down, which, as seen in part (d), yields an expected payoff of 6/7 to the home team. When it fails,
the home team is left with 20 yards to go on fourth down, which, as seen in part (b), yields an

Solutions to Chapter 7 Solved Exercises                                                        4 of 6
expected payoff of 1/2 to to the home team. The expected payoff of running the 10-yard play in
the third down is the weighted average of these two outcomes:
0.2(1/2) + 0.8(6/7) = 1/10 + 24/35 = 55/70 = 11/14.
(g)        If the home team runs a 10-yard play on third down when the rival team
anticipates a 20-yard play, the 10-yard play automatically succeeds. The home team is then faced
with 10 yards to go on fourth down, which, as seen in part (d), yields an expected payoff of 6/7 to
the home team.
(h)        See the table below.
Rival Team
Anticipate 10-yard play   Anticipate 20-yard play
Home Team                   10-yard play             11/14, 3/14               6/7, 1/7
20-yard play             1, 0                      3/4, 1/4

(i)        3/14p + 0(1 – p) = 1/7p + 1/4(1 – p)              p = 7/9
11/14q + 6/7(1 – q) = 1q + 3/4(1 – q)             q = 1/3
(j)        The home team’s expected payoff is: (11/14)(1/3) + (6/7)(2/3) = 5/6.
The rival team’s expected payoff is thus: 1 – 5/6 = 1/6.
Facing third down with 20 yards to go, the home team has an 83.3% chance of winning the game.

S11.    (a)        – (1 – p) = kp – 10(1 – p)       p = 9/(k + 9)
– (1 – q) = kq – 10(1 – q)       q = 9/(k + 9)
Both James and Dean play Swerve with probability 9/(k + 9)
Both play Swerve less as k increases.
(b)        The expected value of the game for both James and Dean is: – 1(1 – p) = – 1(1 –
q) = -k/(k + 9).
(c)        k must be equal to 9 for both James and Dean to mix 50-50 in their mixed-
strategy Nash equilibrium.
(d)        k would have to be greater than 1 if James and Dean decide upon playing the
alternating scheme discussed in part (c) of Exercise S6. In that game, k = 1, and their expected
payoffs were 0 for the alternating scheme. Any value of k greater than 1 would lead to positive
expected payoffs for both players.

Solutions to Chapter 7 Solved Exercises                                                        5 of 6
S12.    (a)     Row has a dominant strategy if C > A; Column has a dominant strategy if C > B.
(b)      If neither of the conditions in part (a) hold, then we have C < A and C < B. In
this case there can be no Nash equilibrium in pure strategies.
(c)      The conditions for part (b)—A > C and B > C—imply that there is no Nash
equilibrium in pure strategies, but there will be an equilibrium in mixed strategies.
(d)      The mixed-strategy equilibrium value of p (probability of choosing Up) for Row
must satisfy pA = (B – C)(1 – p), or p/(1 – p) = (B – C)/A. Then p = (B – C)/(A + B – C).

S13.    (a)     50p + 10(1 – p) = 20p + 80(1 – p)                  p = 0.7
5√2q + 4√5(1 – q) = 3√10q + 2√5(1 – q)             q = √2 / (3 + √2 - √5) ≈ 0.65
The mixed-strategy Nash equilibrium is:
Evert plays .7(DL) + .3(CC)
Navratilova plays .65(DL) + .35(CC)
(b)     Evert’s equilbrium mixture does not change, but Navratilova now plays DL with
greater frequency. The change in payoffs for Evert leads Navratilova to change her equilibrium
mixture so that Evert can be kept indifferent between playing either of her two strategies.

Solutions to Chapter 7 Solved Exercises                                                       6 of 6

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