Solutions to Chapter 8 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.
Sally’s expected payoff from choosing Starbucks when Harry is using his p-mix is p; her
expected payoff from choosing Local Latte when Harry is mixing is 2 – 2p. Similarly, Harry’s expected
payoff from choosing Starbucks when Sally is using his q-mix is 2q; his expected payoff from choosing
Local Latte when Sally is mixing is 1 – q. These expected payoffs are graphed below.
Sally’s best response to Harry’s p-mix is to choose Local Latte for values of p below 2/3 and to
choose Starbucks for values of p above 2/3. Sally is indifferent between her two choices when p = 2/3.
Similarly, Harry’s best response to Sally’s q-mix is to choose Local Latte for values of q below 1/3 and to
choose Starbucks for values of q above 1/3. He is indifferent between his two choices when q = 1/3. The
mixed-strategy equilibrium occurs when Harry chooses Starbucks two-thirds of the time and Local Latte
one-third of the time (p = 2/3) and when Sally chooses Starbucks one-third of the time and Local Latte
two-thirds of the time (q = 1/3). Best-response curves are shown below.
Expected payoffs for Sally and Harry are 2/3 each. Both players would prefer either of the purestrategy Nash equilibria. If they can coordinate their randomization in some way so as to alternate
Solutions to Chapter 8 Solved Exercises
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between the two pure-strategy Nash equilibria, they can achieve an expected payoff of 1.5 rather than the
2/3 that they achieve in the mixed-strategy equilibrium.
S3.
(a)
Nash equilibria occur where the best-response curves intersect. In this game, the pure-strategy
Nash equilibria occur at points, (0, 0) and (1, 1). The mixed-strategy Nash equilibrium occurs at point
(1/3, 1/3).
(b)
Solutions to Chapter 8 Solved Exercises
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In this game, the pure-strategy Nash equilibria occur at points (0, 0) and (1, 1). There are no
mixed-strategy Nash equilibria.
S4.
(a)
Solutions to Chapter 8 Solved Exercises
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(b)
Navratilova wants to minimize Evert’s success rate, which—as seen from the upper
envelope of the graph—will be 70% in equilibrium. The graph also illustrates that Navratilova can use
many mixtures of q in equilibrium: any q-mix between 1/3 and 5/7 (inclusive) will work in a mixedstrategy equilibrium (while Evert plays Lob). There are then an infinite number of mixed-strategy
equilibria of the form (Lob; q DL + (1 – q) CC), where 1/3 ≤ q ≤ 5/7.
(c)
(d)
Using this characterization of Evert’s p-mix then, we see that Navratilova’s expected
payoff is 50p1 + 10p2 + 34(1 – p1 – p2) when playing DL and 20p1 + 80p2 + 44(1 – p1 – p2) when playing
CC. Thus, Navratilova is indifferent when:
50p1 + 10p2 + 34(1 – p1 – p2) = 20p1 + 80p2 + 44(1 – p1 – p2)

40p1 = 60p2 + 10

p1 = 3/2p2 + 1/4
Since they are probabilities, neither p1 nor p2 may be less than 0 or greater than 1. Since p2  0, it
must be true that p1  1/4. Similarly, since p1≤ 1, it must be true that p2≤ 1/2. In equilibrium, Evert may
play DL with any p1  [1/4, 1], CC with any probability p2  [0, 1/2] , provided that p1 + p2 ≤ 1. She will
of course play Lob with probability 1 – p1 – p2. Since there are many possible choices of p1 and p2, this
game has an infinite number of mixed-strategy equilibria.
S5.
(a)
Solutions to Chapter 8 Solved Exercises
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(b)
Zeppo, Harpo, and Groucho appear on the upper envelope of the graph. Zeppo is Karl’s
best response when 0 ≤ q ≤ 4/7, Harpo is his best response when 4/7 ≤ q ≤ 2/3, and Groucho is his best
response when 2/3 ≤ q ≤ 1.
(c) Chico and Gummo are never a best response because they are dominated. Gummo is dominated
by Zeppo, and Chico is dominated by either a mixture of Zeppo and Groucho or a mixture of Zeppo and
Harpo.
(d)
The equilibrium q-mix of this zero-sum game will be the one that minimizes the upper
envelope of Karl’s of payoffs. This occurs at the intersection of Zeppo and Harpo, that is, when the payoff
from Zeppo is equal to the payoff from Harpo:
9 – 5q = 5 + 2q 
q = 4/7
To find Karl’s p-mix, look at the remaining subgame when only Zeppo and Harpo are played:
KARL
Harpo
Zeppo
Solutions to Chapter 8 Solved Exercises
OLLISTAN
Laurel
7
4
Hardy
5
9
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Let p be the probability that Karl plays Harpo. Then Ollistan is indifferent when:
–7p – 4(1 – p) = –5p – 9(1 – p) 
S6.
p = 5/7.
(a)
(b)
Revolver yields a higher expected payoff than Knife when:
1 + 2p > 4 – 6p 
(c)
Revolver yields a higher expected payoff than Wrench when:
1 + 2p > 0 + 6p 
(d)
p > 3/8.
p < 1/4.
Professor Plum will use only Knife and Wrench in his equilibrium mixture, because
Revolver is dominated by a mixture of those two strategies. He can always get a higher payoff from either
Knife or Wrench.
(e)
Eliminating Revolver from consideration, we get the two-by-two table:
MRS.
Conservatory
Solutions to Chapter 8 Solved Exercises
PROFESSOR PLUM
Wrench
Knife
2, –2
0, 6
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PEACOCK
Ballroom
1, 4
5, 0
Let q be the probability that Professor Plum plays Knife. The mixed strategy Nash equilibrium
occurs where:
–2p + 4(1 – p) = 6p + 0(1 – p)

p = 1/3
2q + 0(1 – q) = 1q + 5(1 – q)

q = 5/6
Mrs. Peacock plays 1/3(Conservatory) +2/3(Ballroom), and Professor Plum plays 5/6(Knife) +
1/6(Wrench).
S7.
For Roman, strategy B is a dominated by strategy A. For Greek, strategy δ is dominated by
strategy β. The game table then reduces to:
ROMAN
A
C
GREEK


6, -6
-1, 1
3, -3
7, -7

5, -5
4, -4
Best response analysis shows that there are no pure-strategy equilibria.
Now that the game is 2 by 3, the easiest way to see which strategies Greek will play in equilibrium is to
graph Greek’s payoffs given Roman’s p, where p is the probability that Roman plays A:
Solutions to Chapter 8 Solved Exercises
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Roman’s equilibrium value of p is the one that minimizes Greek’s upper envelope of payoffs. This occurs
where the α line intersects the β line:
–3 – 3p = – 7 + 8p

p = 4/11
The graph illustrates that is dominated by  and ; Greek will not play Let q be the probability that
Greek plays α. The value of q that will keep Roman indifferent between playing A and C is given by:
6q – 1(1 – q) = 3q + 7(1 – q)

q = 8/11
This game has one mixed-strategy equilibrium, where:
Roman plays (4/11 A + 7/11 C)
Greek plays (8/11 α +3/11 )
S8.
As in Exercise S7, strategies B and δ are dominated and can be eliminated. The reduced game
table is:
ROMAN
A
C
Solutions to Chapter 8 Solved Exercises
GREEK


6, –6
–1, 1
3, –3
7, –7

3, –3
4, –4
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Best response analysis shows that there are no pure-strategy equilibria.
Graphing the payoffs from Greek’s strategies given Roman’s p, where p is the probability that Roman
plays A:
Roman’s equilibrium value of p is the one that minimizes Greek’s upper envelope of payoffs. This occurs
where the α line intersects the  line:
–3 – 3p = – 4 + 1p

p = 1/4
Though β is a best response for some values of p, it is not a best response when p = 1/4, so Greek only
plays α and in equilibrium.Let q be the probability that Greek plays α. The value of q that will keep
Roman indifferent between playing A and C is given by:
6q +3(1 – q) = 3q + 4(1 – q)

q = 1/4
This game has one mixed-strategy equilibrium, where:
Roman plays (1/4 A + 3/4 C)
Greek plays (1/4 α +3/4 )
S9.
(a)
To find Player 1’s equilibrium mix, set Player 2’s payoffs from each of her pure
strategies, against Player 1’s p-mix, equal to one another. In her p-mix, Player 1 will play Rock with
Solutions to Chapter 8 Solved Exercises
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probability p1, Scissors with probability p2, and Paper with probability 1 – p1 – p2. Player 2’s payoff from
Rock against this p-mix is 10p2 – 10(1– p1 – p2); her payoff from Scissors against the p-mix is –10p1 +
10(1 – p1 – p2); and her payoff from Paper against the p-mix is 10p1 – 10p2. Equating the last two of these
payoffs yields –10p1 + 10(1 – p1 – p2) = 10p1 – 10p2, which simplifies to –20p1 – 10p2 + 10 = 10p1 – 10p2,
or 10 = 30p1 or p1 = 1/3. Then equating the first two payoffs yields 10p2 – 10(1 – p1 – p2) = –10p1 + 10(1 –
p1 – p2) or –10 + 20p2 + 10p1 = –20p1 – 10p2 + 10. Rearranging and substituting 1/3 for p1 yields 30p2 =
10, or p2 = 1/3 also. Then we get 1 – p1 – p2 = 1/3 as well. Player 1’s equilibrium mix entails playing each
strategy 1/3 of the time or with probability 33 1/3. The symmetry of the game guarantees that Player 2’s
equilibrium mix is the same.
(b)
Check Player 1’s payoffs from each of her pure strategies against Player 2’s mix.
payoff from Rock: (0)(0.4) + (10)(0.3) + (–10)(0.3) = 0
payoff from Scissors: (–10)(0.4) + (0)(0.3) + (10)(0.3) = –1
payoff from Paper: (10)(0.4) + (–10)(0.3) + (0)(0.3) = 1
Player 1’s expected payoff from using the pure strategy Paper exceeds her expected payoffs from her
other two strategies. She should play only Paper when Player 2 uses the mix described. Presumably,
Player 2 has chosen a nonequilibrium mix, so Player 1 is not indifferent among her available pure
strategies.
S10. (a) Let p and q be the probabilities that Harry and Sally (respectively) play Pasta, so that they play
Sandwich with probabilities (1 – p) and (1 – q). The equilibrium values of p and q will be:
p = 5(1 – p)

p = 5/6
5q = 1 – q

q = 1/6
The equilibrium when Harry and Sally both mix over only Pasta and Sandwich is:
Harry plays 5/6(Pasta) + 1/6(Sandwich)
Sally plays 1/6(Pasta) + 5/6(Sandwich)
Harry’s expected payoff is 5q + 0(1 – q) = 5(1/6) = 5/6. Sally’s expected payoff is 1p + 0(1 – p) = 5/6.
(b)
Again, let p and q be the probabilities that Harry and Sally (respectively) play Pasta, but
now they play Buffet with probabilities (1 – p) and (1 – q). The equilibrium values of p and q will be:
p = 2(1 – p)

p = 2/3
Solutions to Chapter 8 Solved Exercises
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5q = 2(1 – q)

q = 2/7
The equilibrium when Harry and Sally both mix over only Pasta and Buffet is:
Harry plays 2/3(Pasta) + 1/3(Buffet)
Sally plays 2/7(Pasta) + 5/7(Buffet)
Harry’s expected payoff is 5q + 0(1 – q) = 5(2/7) = 10/7. Sally’s expected payoff is 1p + 0(1 – p) = 2/3.
When Harry plays this mixed strategy, Sally’s payoff from playing Sandwich is 0, whereas her
expected payoff from playing either Pasta or Buffet is 2/3. Thus Sally will never choose to play Sandwich
when she thinks Harry is playing his equilibrium strategy for this mixed-strategy equilibrium.
(c)
Now let p and q be the probabilities that Harry and Sally (respectively) play Sandwich, so
that they play Buffet with probabilities (1 – p) and (1 – q). The equilibrium values of p and q will be:
5p = 2(1 – p)

p = 2/7
q = 2(1 – q)

q = 2/3
The equilibrium when Harry and Sally both mix over only Sandwich and Buffet is:
Harry plays 2/7(Sandwich) + 5/7(Buffet)
Sally plays 2/3(Sandwich) + 1/3(Buffet)
Harry’s expected payoff is 1q + 0(1 – q) = 2/3. Sally’s expected payoff is 5p + 0(1 – p) = 5(2/7) = 10/7.
Harry’s expected payoff from playing Pasta when Sally mixes over only Sandwich and Buffet is
0, so he would rather play either Sandwich or Buffet.
(d)
Let pP be the probability that Harry plays Pasta and pS be the probability that he plays
Sandwich. Then he will play Buffet with probability 1 – pP – pS. Similarly, let qP be the probability that
Sally plays Pasta and qS be the probability that she plays Sandwich. Then she will play Buffet with
probability 1 – qP – qS.
Sally is indifferent between her strategies when:
1pP + 0pS + 0(1 – pP – pS) = 0pP + 5pS + 0(1 – pP – pS) = 0pP + 0pS + 2(1 – pP – pS)
 pP = 5pS = 2(1 – pP – pS)
Substituting for pP, we have: 5pS = 2(1 – 5pS – pS) 
pS = 2/17.
Therefore, pP = 10/17 and (1 – pP – pS) = 5/17.
Solutions to Chapter 8 Solved Exercises
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Harry is indifferent between his strategies when:
5qP + 0qS + 0(1 – qP – qS) = 0qP + 1qS + 0(1 – qP – qS) = 0qP + 0qS + 2(1 – qP – qS)
 5qP = 1qS = 2(1 – qP – qS)
Substituting for qS, we have: 5qP = 2(1 – qP – 5qP) 
qP = 2/17.
Therefore, qS = 10/17 and (1 – qP – qS) = 5/17.
The mixed-strategy Nash equilibrium where Harry and Sally each play all three of their strategies is:
Harry plays 10/17(Pasta) + 2/17(Sandwich) + 5/17(Buffet)
Sally plays 2/17(Pasta) + 10/17(Sandwich) + 5/17(Buffet)
Harry’s expected payoff is 5(2/17) + 10/17 + 2(5/17) = 30/17.
Sally’s expected payoff is 10/17 + 5(2/17) + 2(5/17) = 30/17.
(e)
Note that the expected payoffs for Harry and Sally from the various mixed-strategy
equilibria are all less than 2, while the pure-strategy Nash equilibrium where they both choose Buffet
gives them each a payoff of 2. In that sense, Buffet seems more focal than any of the mixed-strategy Nash
equilibria. For their part, the pure-strategy equilibria of Pasta or Sandwich are probably not focal, since
Harry and Sally may worry about second-guessing each other and ending up going to the wrong place.
Again, mixing between Pasta and Sandwich yields a lower expected payoff to each than the pure-strategy
Buffet equilibrium.
Unless there are other circumstances of which we are not aware (for example, they both
remember that they ate pasta the last time they were out together and usually alternate between pasta and
sandwiches), having both choose Buffet seems to be the best candidate for a focal equilibrium out of the
seven.
S11.
(a)
Vendor 2
Vendor 1
A
B
C
D
E
A
85, 85
100, 170
125, 195
150, 200
160, 160
B
170, 100
110, 110
150, 170
175, 175
200, 150
C
195, 125
170, 150
120, 120
170, 150
195, 125
Solutions to Chapter 8 Solved Exercises
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D
200, 150
175, 175
150, 170
110, 110
170, 100
E
160, 160
150, 200
125, 195
100, 170
85, 85
(b) For both vendors, locations A and E are dominated. Thus, for a fully mixed equilibrium, we
need only consider each vendor’s choice among locations B, C, and D. Let Vendor 1’s mixture
probabilities be pB, pC, and (1 – pB – pC). Similarly, let Vendor 2’s mixture probabilities be qB, qC, and (1
– qB – qC). After simplifying, the p-mix and q-mix payoffs are as shown below.
Vendor 2
B
C
D
B
110, 110
150, 170
175, 175
C
170, 150
120, 120
170, 150
D
17, 175
150, 170
110, 110
175 – 65pB – 5pC,
150 – 30pC,
110 + 65pB + 60pC,
175 – 65pB – 25pC
170 – 50pC
110 + 65pB + 40pC
Vendor 1
p-mix
(c)
q-mix
175 – 65qB – 25qC,
175 – 65qB – 5qC
170 – 50qC,
150 – 30qC
110 + 65qB + 40qC,
110 + 65qB + 60qC
To find the equilibrium p, set vendor 2’s payoffs equal:
175 – 65pB – 25pC = 170 – 50pC = 110 + 65pB + 40pC
65pB – 25pC = 5 and 60 – 90pC = 65pB
60 – 90pC – 25pC = 5
Then pC = 55/115 = 11/23, pB = (25pC + 5)/65 = (275/23 + 115/23)/65 = (390/23)65 = 6/23, and pD = (1 –
pB – pC) = 1 – 11/23 – 6/23 = 6/23. Similarly, qB = 6/23, qC = 11/23, and qD = 6/23.
One way to explain why A and E are unused in the equilibrium is to point out that they are (as
noted above) dominated. This also implies that A and E are unused because they result in a payoff against
the opponent’s equilibrium mixture that is lower than produced by choices B, C, and D. Specifically,
when Vendor 2 uses the equilibrium mixture probabilities of (6/23, 11/23, 6/23), Vendor 1’s payoff from
choosing
A is 100(6/23) + 125(11/23) + 150(6/23) = 2,875/23.
B is 110(6/23) + 150(11/23) + 175(6/23) = 3,360/23.
C is 170(6/23) + 120(11/23) + 170(6/23) = 3,360/23.
Solutions to Chapter 8 Solved Exercises
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D is 175(6/23) + 150(11/23) + 110(6/23) = 3,360/23.
E is 150(6/23) + 125(11/23) + 100(6/23) = 2,875/23.
Clearly, A and E are inferior choices.
An alternative possibility is a partially mixed equilibrium in which one player plays pure C and
the other player mixes using strategies B and D with probabilities 1/13 = 0.076 and 12/13 = 0.923,
respectively. The expected payoff to the player using only C (the pure player) is 170; the expected payoff
to the player using a mixture of B and D (the mixer) is 150. The equilibrium can arise in the following
way: If Vendor 2 is playing pure C, then Vendor 1 gets equal highest payoffs from B and D, and therefore
is willing to mix between them in any proportions. Suppose Vendor 1 chooses B with probability p and D
with probability (1 – p). To make this an equilibrium, pure C should be Vendor 2’s best response to this
mixture. A and E are clearly bad for Vendor 2, as we established above. Vendor 2 does not switch to B if
110p + 175(1 – p) = 170,
or
5 = 65p,
or
p = 1/13 = 0.07692.
Similarly, Vendor 2 does not switch to D if
175p + 110(1 – p) = 170,
or
65p = 60,
or
p = 12/13 = 0.9231.
Thus there is really a whole continuum of mixed-strategy equilibria, in which Vendor 2 plays pure C and
Vendor 1 mixes between B and D in any proportions between 1/13 and 12/13. The answer just above
describes the equilibrium that results at the extreme points of this range.
S12.
(a)
The kicker’s expected payoffs playing against the goalie’s mixed strategy of L 42.2%, R
42.2%, and C 15.6% are as follows:
(b)
HL 42.2(0.50) + 15.6(0.85) + 42.2(0.85)
= 70.23
LL 42.2(0.40) + 15.6(0.95) + 42.2(0.95)
= 71.79
HC 42.2(0.85) + 15.6(0) + 42.2(0.85)
= 71.74
LC 42.2(0.70) + 15.6(0) + 42.2(0.70)
= 59.08
HR 42.2(0.85) + 15.6(0.85) + 42.2(0.50)
= 70.23
LR 42.2(0.95) + 15.6(0.95) + 42.2(0.40)
= 71.79
Thus, the kicker should be using low, side shots (LL and LR) and high, centered shots
(HC) because these strategies give him the highest expected payoff given the goalie’s mix.
(c)
For the goalie playing against the kicker using LL 37.8%, HC 24.4%, and LR 37.8% of
the time, payoffs from each of his possible strategies are
L .378(0.40) + .244(0.85) +.378(0.95)
Solutions to Chapter 8 Solved Exercises
= 71.77
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C .378(0.95) + .244(0) +.378(0.95)
= 71.82
R .378(0.95) + .244(0.85) +.378(0.40)
= 71.77
(d)
Thus, the goalie should play center (C) because this strategy gives him the highest
expected payoff given the kicker’s mix. Nonetheless, all three strategies give almost the same expected
payoff, so the goalie could play all of them effectively.
These mixed strategies are Nash equilibria, because each player’s strategy is a best
(e)
response given the other player’s strategy.
S13.
(f)
The equilibrium payoff to the kicker is 71.77.
(a)
Using the multiplication rule, the probability that Bernardo plays Yes and Carlos plays
No is q*(1 – r).
(b)
There are four possible outcomes when Arturo plays Yes: Bernardo plays Yes and Carlos
plays Yes, Bernardo plays No and Carlos plays Yes, Bernardo plays Yes and Carlos plays No, and
Bernardo plays No when Carlos plays No. Arturo’s expected payoff from playing Yes is the weighted
average of these payoffs, where the weights are the probabilities that each respective outcome is reached:
2qr + 2(1 – q)r + 2q(1 – r) + 2(1 – q)(1 – r) = 2(qr + r – qr + q – qr + 1 – q – r + qr) = 2
(c)
Arturo is indifferent when his expected payoff from playing Yes is equal to his expected
payoff from playing No:
2 = 5qr + 5(1 – q)r + 5q(1 – r) + 0(1 – q)(1 – r) 
(d)
2 = 5(q + r – qr)
We find the indifference equations for Bernardo and Carlos the same way we found
Arturo’s. Bernardo is indifferent between playing Yes and No when 2 = 5(p + r – pr), and Carlos is
indifferent between his two strategies when 2 = 5(p + q – pq).
(e)
The indifference equations from parts (c) and (d) can be rewritten as:
r
2  5q
5q
p
2  5r
5r
q
25p
5 p
Solutions to Chapter 8 Solved Exercises
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To solve for p, use the first equation to write r in terms of q, use the second equation to write q in terms of
p, and solve the second equation:

 25p 
 2  5

 5 p 
25

 25p  
 5
 
 5 p  

 10  2 p  10  25 p 


5 p
25

 25  5 p  2  5 p  2  5  23 p 


5 p

 
 23   2  5 p
p

5 p

 10  2 p  10  25 p 
 23 p 
 25p 
5



 2  5




5 p
 23 
 5 p 
5 

5 
25

5
p

2

5
p



 25p  
 5




5

p
 5 p  

So p = r. Similarly, we can see that p = q = r. This means that the mixed-strategy equilibrium is
symmetric; that is, in the mixed-strategy equilibrium all three brothers play Yes with the same
probability.
Continuing to solve for p, we have:
p
25p
5 p

p2 – 10p + 2 = 0
Using the quadratic formula, we find two possible values of p: 5 + 23 and 5 – 23 (or ≈ 9.796 and ≈
0.204). Since p is a probability, it must be no less than 0 and no greater than 1, so the second value is the
one we’re looking for.
The mixed-strategy equilibrium of this game has each of the three brothers playing Yes with
probabity 5 – 23 (≈ 0.204) and No with probability 23 – 4 (≈ 0.796).
S14.
(a)
Consider any one of the young men. If he chooses to go after a brunette, he gets a
guaranteed payoff of 5. If he chooses to go after the blonde, he gets 10 if none of the other (n – 1) men
choose Blonde, and 0 otherwise. Since each of the other (n – 1) chooses Blonde with probability p, and
they are choosing independently, the probability that none of them choose Blonde is (1 – p)n – 1. Therefore
the expected payoff to any one man from choosing Blonde is
10(1 – p)n – 1 + 0[1 – (1 – p) – 1] = 10(1 – P)n – 1.
Solutions to Chapter 8 Solved Exercises
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In the mixed-strategy equilibrium, this man must be indifferent between his two pure strategies. Therefore,
it must be true that
10(1 – p)n – 1 = 5,
(b)
or (1 – p)n – 1 = 1/2 = 2–1,
so
p = 1 – 2–1/(n – 1).
The same logic can be applied to the m young men who are choosing Blonde, so
Q = 1 – 21/(n – 1).
Consider any one of the (n – m) choosing Brunette. If he switched to Blonde, his payoff will be 10 if all
the m mixers happen to choose Brunette, and 0 otherwise. Therefore his expected payoff is
10(1 – Q)m.
So this pure Brunette chooser does not want to switch to a pure Blonde strategy if
10(1 – Q)m < 5. For this Brunette chooser, the payoff from any mixture will be an average between the
two sides of this inequality, so the same inequality also rules out his wanting to switch to any mixed
strategy.
But we already have the condition of the mixed-strategy equilibrium:
10(1 – Q)m – 1 = 5.
Therefore, 10(1 – Q)m = 10(1 – Q)m – 1(1 – Q) = 5(1 – Q) < 5. So the condition holds.
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