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U11. (a) Yes. The pure-strategy Nash equilibrium is (Land, Sea).
(b) Let p1 be the probability that Patton plays Air and p2 be the probability that Patton
plays Sea, so that 1 – p1 – p2 is the probability that he plays Land. Similarly, let q1, q2 , and 1 – q1
– q2 be the probabilities that MacArthur plays Air, Sea, and Land, respectively.
MacArthur’s indifference conditions are
3p1 + 4p2 + 3(1 – p1 – p2) = 0p1 + 6p2 + 4(1 – p1 – p2)
and 3p1 + 4p2 + 3(1 – p1 – p2) = 7p1 + 0p2 + 3(1 – p1 – p2).
A little algebra shows that the second condition impies that p1 = p2. Substituting in the first
condition, we have 7p1 + 3(1 – 2p1) = 6p1 + 4(1 – 2p1) p1 = 1 – 2p1. Thus p1 = 1/3, p2 = 1/3, and
1 – p1 – p2 = 1/3.
The indifference conditions for Patton are
0q1 + 2q2 + 1(1 – q1 – q2) = 2q1 + 0q2 + 2(1 – q1 – q2)
and 0q1 + 2q2 + 1(1 – q1 – q2) = 1q1 + 2q2 + 0(1 – q1 – q2).
The first condition implies that q1 = 3q2 – 1. Substituting this into the second condition yields 2q2
+ 1(1 – (3q2 – 1) – q2) = 1(3q2 – 1) + 2q2 2 – 4q2 = 3q2 – 1. Thus q2 = 3/7, q1 = 2/7, and 1 – q1 –
q2 = 2/7.
The mixed-strategy equilibrium of this game occurs where Patton plays 1/3(Air),
1/3(Sea), and 1/3(Land), while MacArthur plays 2/7(Air), 3/7(Sea), and 2/7(Land). The expected
payoffs are 8/7 and 10/3 to Patton and MacArthur, respectively.
(c) For one player to play a pure strategy while the other plays a mixed strategy, it
must be true that the second player receives exactly the same payoff against the first player’s pure
strategy when playing either (or any) of the strategies in his equilibrium mixture. This game
offers two possibilities for this situation: first, Patton plays pure Land while MacArthur plays a
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
mixture of Air and Land, and second, MacArthur plays pure Sea while Patton mixes over Air and
Land. In the first case, if Patton is playing pure Land, MacArthur’s best response is to play pure
Sea, so he will not mix over Air and Land. That leaves the second case.
When MacArthur plays pure Sea, Patton is indifferent between playing Air and Land and will not
play Sea. If Patton plays Air with probability p and Land with probability 1 – p, MacArthur will
play pure Sea if
0p + 4(1 – p) 3p + 3(1 – p) = 3 and 0p + 4(1 – p) 7p + 3(1 – p)
p ≤ 1/4 and p ≤ 1/8.
There are thus an infinite number of mixed-strategy equilibria where Patton plays p(Air) and (1 –
p)(Land)—when 0 < p ≤ 1/8—and MacArthur plays pure Sea. The expected payoffs are 2 and 4 –
4p to Patton and MacArthur, respectively.
U12. (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.
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
U13. (a) From the table given in the solution to Exercise S11 of Chapter 4, it is clear that
buying a $30 ticket is a (weakly) dominated strategy. Thus, any equilibrium mixture will only
involve the contestants buying either no ticket or a $15 ticket. For each player the only possible
payoff from playing $0 (that is, not buying a ticket) is $0, so to be indifferent between playing $0
and $15 in equilibrium it must be true that the expected payoff for each player is $0. If one player
were to buy a $30 ticket with positive probability, the expected value of buying a $15 ticket
would fall below $0 for the other two players; that is, they would no longer be indifferent
between playing $0 and playing $15. Thus, the weakly dominated strategy, $30, is not played by
any of the players in equilibrium.
(b) Let Larry’s probability of not buying a ticket be p (so that his probability of
buying a $15 ticket is 1 – p). Similarly, let q and r be the probabilities that Moe and Curly do not
buy tickets respectively.
Larry’s indifference condition, in terms of Moe’s q and Curly’s r, is
10qr + 5q + 5r – 5 = 0, or 2qr + q + r – 1 = 0.
Moe’s indifference equation, in terms of Larry’s p and Curly’s r, is
10pr + 5p + 5r – 5 = 0, or 2pr + p + r – 1 = 0.
Curly’s indifference equation, in terms of Larry’s p and Moe’s q, is
10pq + 5p + 5q – 5 = 0, or 2pq + p + q – 1 = 0.
The system of three equations in three unknowns is
(1) 2qr + q + r – 1 = 0
(2) 2pr + p + r – 1 = 0
(3) 2pq + p + q – 1 = 0
Solving equation for q in equation (1), for r in equation (2), and for p in equation (3), we have
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
1
2 1
r
qr
-
=+
1
2 1
p
rp
-
=+
1
2 1
q
pq
-
=+
Plugging the value of r into the top expression and then plugging the resulting expression for q in
the bottom equation, we have
12 1 1
12 1 2 1
11
12 2 2 1 3
2 1 1
2 1 2 1 1
3
2 1 1 3 2 1
12 1
12 1 3
2 1 2 1
2 1 2 2 2 1
1
2 1 2 1
2 1
pp p
pp
pp p p
ppp
pp p p p
p
p
p
p p
pp
p
é ù
æ ö
-+ - +
é ù
-
ê ú
ç ÷
+ê ú
è ø +
ê ú
--ê ú
ê ú
æ ö
-- + +
ê ú
+
ê ú
ç ÷ -
+ê ú
+-
è ø
ë û ë û
= = = =
+ - + +
é ù é ù æ ö
æ ö
-+
-ç ÷
ê ú ê ú
ç ÷ +è ø
+
è ø +
ê ú ê ú
+- + +
ê ú ê ú
æ ö
-+
ê ú ê ú
ç ÷ +
ë û
+
è ø
ë û
.
We can see that p = q = r, confirming that the mixed-strategy equilibrium is symmetric. Finally,
we are left with the equation 2p2 + 2p – 1 = 0. Using the quadratic equation, we see that p = (–1 +
√3)/2 ≈ 0.366. The other root, (–1 – √3)/2, is negative, but p must be between 0 and 1 inclusive
because it represents a probability.
The symmetric mixed-strategy Nash equilibrium of this game occurs where Larry, Moe, and
Curly each refrain from buying a ticket with probability (√3 – 1)/2 ≈ 0.366. The expected payoff
for each of the three players is $0, as we reasoned previously.
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
U14. There is no Nash equilibrium in pure strategies, and a, b, c, and d are
all greater than or equal to zero. It is to be shown that an increase in any one
of a, b, c, or d cannot lead to a lower expected value of the game for Rowena.
The mixed-strategy equilibrium mixture p and q will be
dcab
dc
p
bdca
bd
q
The expected value, V, of the game for Rowena is
)1( qbaqV
[also,
)1( qdcqV
].
Substituting for q:
bdca
cab
bdca
bda
V
)()(
, [since
bdca
ca
bdca
bd
q
11
]
bdca
bcad
V
.
To show that an increase in a cannot lead to a decrease in V, it is suffcient to
show that the partial derivative V / a is always greater than or equal to 0.
Using the quotient rule:
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
2
)(
)()(
bdca
bcaddbcad
a
V
.
Note that the denominator is always positive. (It’s not zero; if it were, Rowena
would have a pure-strategy equilibrium.) Thus the partial derivative of V with
respect to a will be nonnegative if and only if the numerator is nonnegative.
The numerator can be rewritten as
bccbddbcbddcdbcaddbcad )()()( 2
.
But how do a, b, c, and d relate to each other? For there to be a
mixed-strategy equilibrium, it must be true that a – c and d – b have the
same sign, and that a – b and d – c have the same sign. There are thus four
possible cases to consider, as shown below:
a > b a < b
d > c d < c
a > c Case 1 Case 2
d > b
a < c Case 3 Case 4
d < b
In Case 2, we have a > c > d > b > a, which is a contradiction. It cannot be
that a > a.
Similarly, in Case 3, we have a > b > d > c > a, which is a contradiction.
Since Case 2 and Case 3 cannot happen, we are left with Case 1 and
Case 4, where there isn’t an obvious way to compare a and d or b and c.
Consider Case 1. Here we know that d > b and d > c, but we don’t
know if b is greater than, equal to, or less than c. (We’ll see that for our
present question it doesn’t matter.)
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
Subcase 1: b < c < d
DeLne
bc
1
and
cd
2
. Then
0
1
and
0
2
, and
)())(()(
112121
bbbbbbbccbdd
)())((
1221
bbbb
2 2 2 2
1 2 1 2 2 2 1 1 2 2
0.b b b b b b=DD - D +D - D + D - + D + =DD +D >
Thus
0)( bccbdd
.
Subcase 2: b = c < d
DeLne
1
and
2
as in subcase 1. (The only diMerence is that
0
1
.)
0
2
2
, so
0)( bccbdd
.
Subcase 3: c < b < d
DeLne
cb
1
and
cd
2
. Then
0
1
and
0
2
, and
ccccccbccbdd )())(()(
112121
cccc )())((
1221
2 2 2 2
1 2 1 2 2 2 1 1 2 2
0.c c c c c c
Thus
0)( bccbdd
.
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
No matter the relationship of b and c, for Case 1 it is always true that
0)( bccbdd
, so
0
a
V
, which means that an increase in a will always lead to a strict increase
in V, and never a decrease in V.
The demonstration for Case 4 that
a
V
is always positive is very similar.
The same methods are used to demonstrate that
0
b
V
,
0
c
V
, and
0
d
V
for
both Case 1 and Case 4.
Thus an increase in a, b, c, or d can never lead to a decrease in V, the expected value of
the game for Rowena.
Games of Strategy, Fourth Edition Copyright © 2015 W. W. Norton & Company
