Games Strategies and Decision Making 2nd Edition Harrington Solutions Manual
Games Strategies and Decision Making 2nd Edition
Harrington Solutions Manual
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7
Keep ’Em Guessing:
Randomized Strategies
1. Reproduced below is the telephone game from Section 4.2. Find all Nash equilibria in
mixed strategies.
The Telephone Game
Winnie
Colleen
Call
Wait
Call
0,0
2,3
Wait
3,2
1,1
ANSWER: Let p denote the probability that Colleen chooses call and q the probability that Winnie chooses call. For Colleen to be indifferent, Winnie must randomize in such a way that Colleen’s expected payoff from call is the same as her
expected payoff from wait:
1
q ⫻ 0 ⫹ (1 ⫺ q) ⫻ 2 ⫽ q ⫻ 3 ⫹ (1 ⫺ q) ⫻ 1 1 q ⫽ .
4
1
Thus, if Winnie calls with probability 4 , then Colleen is indifferent between calling and waiting. Of course, for Winnie to find it optimal to randomize, she must
receive the same expected payoff from her two pure strategies, which requires that
Colleen choose p, so that
1
p ⫻ 0 ⫹ (1 ⫺ p) ⫻ 2 ⫽ p ⫻ 3 ⫹ (1 ⫺ p) ⫻ 1 1 p ⫽ .
4
Since this is a symmetric game, it is not surprising that the equilibrium is
symmetric. The telephone game then has three Nash equilibria: (1) Colleen
chooses call and Winnie chooses wait; (2) Colleen chooses wait and Winnie
1
chooses call; and (3) both Colleen and Winnie choose call with probability 4 and
3
wait with probability 4 .
2. The count is three balls and two strikes, and the bases are empty. The batter wants to
maximize the probability of getting a hit or a walk, while the pitcher wants to minimize
this probability. The pitcher has to decide whether to throw a fast ball or a curve ball,
while the batter has to decide whether to prepare for a fast ball or a curve ball. The
strategic form of this game is shown here. Find all Nash equilibria in mixed strategies.
Baseball
Pitcher
Batter
Fastball
Curveball
Fastball
.35,.65
.3,.7
Curveball
.2,.8
.5,.5
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CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
ANSWER: There is no Nash equilibrium in pure strategies. Let us conjecture that
an equilibrium has both the batter and pitcher randomizing. Let p denote the frequency with which a pitcher throws a fast ball and q denote the frequency with
which a batter sets up for a fast ball. For p to be optimal, the pitcher must be indifferent between his two pure strategies, given the value for q. Also, given p, the
batter is indifferent between his two pure strategies. q is then defined by
q .65 (1 q) .8 q .7 (1 q) .5.
Solving this equation, q 6/7. When q 6/7, the expected payoff for the pitcher
is 47/70 whether he throws a fast ball or a curveball. p is defined by
p .35 (1 p) .3 p .2 (1 p) .5
and solving it gives us p 4/7. The expected payoff for the batter is 23/70, whether
he sets up for a fast ball or a curveball. In equilibrium, the pitcher throws a fast
ball 4/7 of the time and the batter sets up for a fast ball 6/7 of the time.
3. It’s spring break and you’re traveling south on Interstate 95, heading toward Fort
Lauderdale. Do you travel the legal limit of 65 miles per hour, or do you crank it up to 80
and hope that there’s no speed trap? And what about the state police? Do they set a speed
trap or instead head into town and find out whether the “Hot and Fresh” neon sign is lit
up at the Krispy Kreme? (Ouch, that’s a cheap shot!) The police like to nab speeders, but
they don’t want to set a speed trap if there won’t be any speeders to nab. A strategic form
for this setting is shown in the accompanying figure. The driver can either go the legal
limit of 65 mph or speed at 80 mph. The police officer can set a speed trap or head into
town and grab some of those delicious high-carb doughnuts. The best outcome for the
driver is that she speeds and isn’t caught; the payoff for that case is 70. The worst outcome is that she speeds and is nailed by the police, for which the payoff is 10. If she
chooses to drive the legal limit, then her payoff is 40 and is the same regardless of what
the state police do. (In other words, the driver doesn’t care about the caloric intake of the
trooper.) As for the police officer, his best outcome is setting a speed trap and nailing a
speeder, giving him a payoff of 100. His worst outcome is sitting out there in a speed trap
and failing to write a ticket; this outcome delivers a payoff of only 20. His payoff is 50
when he chooses to go to the Krispy Kreme. Find all Nash equilibria in mixed strategies.
Speed Trap and Doughnuts
State police officer
Driver
Speed trap
Krispy Kreme
80 mph
10,100
70,50
65 mph
40,20
40,50
ANSWER: Note that this game has no Nash equilibrium in pure strategies. If the
state police sets a speed trap, then the best reply of the driver is to drive 65 mph,
but if the driver doesn’t speed, then the officer doesn’t want to waste time with a
speed trap. Alternatively, if the officer chooses not to set a speed trap, then the
driver will plan to speed, but then if the driver speeds the officer wants to set a
speed trap. To find a Nash equilibrium in mixed strategies, let p denote the probability that the driver goes 80 mph and q denote the probability that the state
police sets a speed trap. For it to be optimal for the driver to randomize between
going 65 and 80, she must be indifferent between those two pure strategies. Hence,
q must be set at a level to make the driver’s expected payoff the same from her two
pure strategies:
1
q 10 (1 q) 70 40 1 q .
2
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CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
The left-hand expression is the expected payoff from speeding, which delivers a payoff of 10 with probability q (there is a speed trap) and a payoff of 70
with probability 1 q (no speed trap). The right-hand payoff of 40 is what she
gets from driving 65 mph. Solving this equation for the probability of a speed
1
trap, we find that q 2 . Thus, at a Nash equilibrium, the state police set a
speed trap 50% of the time. For the state police officer to find it optimal to
randomize over his two pure strategies—in particular, choosing a speed trap
50% of the time—the probability that the driver is speeding must equal the
expected payoff to the officer from setting a speed trap and doing a doughnut
run. This condition is
3
p 100 (1 p) 20 50 1 p .
8
4. A mugger and a victim meet on a dark street. The mugger previously decided whether
to bring a gun and, if he did, whether to show it during the robbery. If the mugger
does not show a gun—either because he doesn’t have one or has one and hides it—
then the victim has to decide whether to resist. (Note that if the mugger does have a
gun and shows it, then the victim’s payoff is 5 regardless of the strategy chosen,
because the victim’s strategy is what to do if no gun is shown.) The strategic form of
this situation is shown below. Note that all payoffs have been specified, except for the
mugger’s payoff when he chooses to have a gun and show it. Find a condition on x,
whereby a Nash equilibrium exists in which the mugger randomizes over the two
pure strategies gun, hide and no gun and the victim randomizes over resist and do
not resist.
Victim
Mugger
Resist
Do not resist
No gun
2,6
6,3
Gun, hide
3,2
5,4
Gun, show
x,5
x,5
ANSWER: Let p denote the probability that the victim chooses resist and q denote
the probability that the mugger chooses gun, hide. We are presuming the mugger
assigns probability zero to gun, show. The mugger is indifferent between gun, hide
and no gun when
1
p 3 (1 p) 5 p 2 (1 p) 6 1 p .
2
The victim is indifferent between resist and do not resist when
3
(1 q) 6 q 2 (1 q) 3 q 4 1 q .
5
Before we can conclude that this is a Nash equilibrium, it must be the case that
2
the mugger’s expected payoff from choosing no gun with probability 5 and gun,
3
hide with probability 5 is at least as great as choosing the pure strategy gun, show,
for that is another option for the mugger. This is the case if
1
1
3 5 x 1 4 x.
2
2
If instead x 4, then there is no Nash equilibrium in which the mugger randomizes
over no gun and gun, hide and the victim randomizes over resist and do not resist.
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CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
5. For the game below, find all mixed-strategy Nash equilibria.
Player 2
Player 1
x
y
z
a
2,3
1,4
3,2
b
5,1
2,3
1,2
c
3,7
4,6
5,4
d
4,2
1,3
6,1
ANSWER: Note that c strictly dominates a and y strictly dominates z Thus, any
Nash equilibrium in mixed strategies must assign zero probability to those dominated strategies. We can then eliminate them, so the reduced game is as shown in
the figure below.
Player 2
x
y
c
5,1
3,7
2,3
4,6
d
4,2
1,3
b
Player 1
For this reduced game, b strictly dominates d, so the latter can be deleted. The
reduced game is as shown in the figure below.
Player 2
Player 1
b
c
x
y
5,1
3,7
2,3
4,6
This game has no pure-strategy Nash equilibria. To find the mixed-strategy
Nash equilibria, let p denote the probability that player 1 chooses b and q denote
the probability that player 2 chooses x. The equilibrium conditions ensuring that
players want to randomize are
1
q 5 (1 q) 2 q 3 (1 q) 4 1 q .
2
1
p 1 (1 p) 7 p 3 (1 p) 6 1 p .
3
6. Find all Nash equilibria in mixed strategies for the game shown here.
Player 2
Left
Middle
Right
2,2
0,0
1,3
Player 1 Middle
1,3
3,0
1,0
Bottom
3,1
2,3
2,2
Top
SOLUTIONS MANUAL
CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
ANSWER: There is no pure-strategy Nash equilibrium for the game.
For player 1, top is strictly dominated by bottom. Hence, a Nash equilibrium in
mixed strategies has player 1 assigning zero probability to top. Note that, given top
is eliminated, middle weakly dominates right for player 2. Thus, as long as player 1
assigns positive probability to bottom, then middle yields a strictly higher expected
payoff than right for player 2. Next, note that there is not a Nash equilibrium in
which player 1 assigns zero probability to bottom. If she did, then she would be
choosing pure strategy middle (as top is strictly dominated) and player 2’s best reply
to that is left, but player 1’s best reply to player 2 choosing left is bottom. We conclude that, at a Nash equilibrium, player 1 must assign positive probability to bottom, in which case player 2 must assign zero probability to right. In sum, Nash
equilibrium entails player 1 randomizing over middle and bottom and player 2 randomizing over left and middle. Let b denote the probability that player 1 assigns to
bottom; thus probability 1 b goes to middle. For player 2, let l denote the probability assigned to left, so 1 l goes to middle. The conditions equating the expected
payoffs are
1
l 1 (1 l) 3 l 3 (1 l) 2 1 l .
3
3
b 1 (1 b) 3 b 3 (1 b) 0 1 b .
5
2
The unique Nash equilibrium has player 1 choose middle with probability 5 and
3
1
bottom with probability 5 , and player 2 choose left with probability 3 and middle
2
with probability 3 .
7. It is the closing seconds of a football game, and the losing team has just scored a
touchdown. Now down by only one point, the team decides to go for a two-point
conversion that, if successful, will win the game. The offense chooses among three
possible running plays: run wide left, run wide right, and run up the middle. The
defense decides between defending against a wide run and a run up the middle. The
payoff to the defense is the probability that the offense does not score, and the payoff to the offense is the probability that it does score. Find all mixed- strategy Nash
equilibria.
Two-Point Conversion
Defense
Defend against
wide run
Defend against
run up middle
Offense
Run wide left
Run up middle
Run wide right
.6,.4
.4,.6
.6,.4
.3,.7
.5,.5
.3,.7
ANSWER: Let l denote the probability that the offense chooses to run wide left, r
the probability it runs wide right, and 1 l r the probability it runs up the
middle. The defense is indifferent between its two strategies when
l .6 (1 l r) .4 r .6 l .3 (1 l r) .5 r .3 1 l r .25.
Letting d denote the probability that the defense defends against an outside run,
the offense’s expected payoff from its three pure strategies is
Wide left: d .4 (1 d) .7 .7 .3d
Middle: d .6 (1 d) .5 .5 .1d
Wide right: d .4 (1 d) .7 .7 .3d.
7-5
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CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
For the offense to be indifferent among them, it must be true that
.7 .3d .5 .1d 1 d .5.
A Nash equilibrium is then any strategy pair in which the defense defends
against the outside run with probability .5 and the offense runs up the middle
with probability .75. The offense can divide up .25 between left and right outside
in any way.
8. The childhood game of Rock–Paper–Scissors is shown in the accompanying figure.
(If you’re unfamiliar with this game, see Section 4.2.) Show that each player’s
assigning equal probability to his or her three pure strategies is a symmetric Nash
equilibrium.
Rock–Paper–Scissors
Lisa
Bart
Rock
Paper
Scissors
Rock
0,0
1,1
1,1
Paper
1,1
0,0
1,1
Scissors
1,1
1,1
0,0
ANSWER: As the game is symmetric, it is sufficient to consider one player; let it
be Lisa. We just need to show that if Bart randomizes by choosing each pure
1
strategy with probability 3 , Lisa’s expected payoff from each of her pure strategies
is the same. In the following expressions, the first term refers to Bart’s choosing
rock, the second term to his choosing paper, and the third term to his choosing
scissors.
1
1
1
Lisa’s expected payoff from rock: a b 0 a b (1) a b 1 0
3
3
3
1
1
1
Lisa’s expected payoff from paper: a b 1 a b 0 a b (1) 0
3
3
3
1
1
1
Lisa’s expected payoff from scissors: a b (1) a b 1 a b 0 0
3
3
3
As Lisa’s expected payoff is zero for all of her three pure strategies, her expected
1
payoff is zero for any mixed strategy. Thus, assigning probability 3 to each pure strategy is optimal. By symmetry, the same is true for Bart.
9. Each of three players is deciding between the pure strategies go and stop. The payoff to
120
go is m , where m is the number of players that choose go, and the payoff to stop is 55
(which is received regardless of what the other players do). Find all Nash equilibria in
mixed strategies.
ANSWER: There are at least seven Nash equilibria in mixed strategies. First, there
are three asymmetric pure-strategy Nash equilibria in which two players choose
go and the other one chooses stop. Each player who chooses go earns a payoff of
60, which exceeds the payoff of 55 from choosing stop. The player who chooses
stop earns 55, which exceeds the payoff from choosing go, which is 40.
Now consider a strategy profile in which one player chooses the pure strategy
go and the other two players symmetrically randomize, choosing go with probability p:
(1 p) 60 p 40 55 1 p .25.
SOLUTIONS MANUAL
CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
The left-hand expression is the expected payoff from choosing go and is equated
to the payoff from choosing stop. The solution for p is the mixed strategy for the
other player that makes this player indifferent between stop and go. It is a Nash
equilibrium for one player to use the pure strategy go and each of the other two
players to choose go with probability .25; this gives us another three Nash equilibria.
Finally, there is a symmetric mixed-strategy Nash equilibrium in which each
player chooses go with probability q that is defined by
(1 q) 2 120 2 q (1 q) 60 q2 40 55
40q2 120q 65 0
where (1 q)2 is the probability that the other two players both choose stop,
2q(1 q) is the probability that one of the other two players chooses go and q2 is
the probability that the other two players both choose go. Using the quadratic
formula, one finds that q is approximately .71. There is then a symmetric Nash
equilibrium in which each player chooses go with probability .71. It is possible that
there are asymmetric Nash equilibria in which two or more players randomize but
with different probabilities.
10. A total of n 2 companies are considering entry into a new market. The cost of entry is
30. If only one company enters, then its gross profit is 200. If more than one company
enters, then each entrant earns a gross profit of 40. The payoff to a company that enters
is its gross profit minus its entry cost, while the payoff to a company that does not enter
is 60. Find a symmetric Nash equilibrium in mixed strategies.
ANSWER: Let p denote the symmetric probability that a company enters. Given
all other companies use this strategy, the expected payoff to a company for
entering is
(1 p) n1 (200 30) c 1 (1 p) n1 d (40 30).
(1 p)n1 is the probability that all other companies do not enter, and
1 (1 p)n1 is the probability that at least one other company enters. This must
be equated to that company’s payoff from not entering, which is 60:
(1 p) n1 (200 30) c 1 (1 p) n1 d (40 30) 60
1
5
1 p 1 a b n1 .
16
11. Sadaam Hussein is deciding where to hide his weapons of mass destruction (WMD),
while the United Nations is deciding where to look for them. The payoff to Hussein from
successfully hiding WMD is 5 and from having them found is 2. For the UN, the payoff
to finding WMD is 9 and from not finding them is 4. Hussein can hide them in facility
X, Y, or Z. The UN inspection team has to decide which facilities to check. Because the
inspectors are limited in terms of time and personnel, they cannot check all facilities.
a. Suppose the UN has two pure strategies: It can either inspect facilities X and Y (both
of which are geographically close to each other) or inspect facility Z. Find a Nash
equilibrium in mixed strategies.
ANSWER: Let x denote the probability that Saddam Hussein hides WMD in facility X, y is the probability for facility Y, and z 1 x y is the probability for facility Z. For the United Nations, a is the probability that it inspects X and Y, and
b 1 a is the probability that it inspects Z. If we conjecture that both players
randomize over all of their pure strategies, the expected payoffs from Saddam’s
three pure strategies are
Expected payoff from X: a 2 (1 a) 5 5 3a.
Expected payoff from Y: a 2 (1 a) 5 5 3a.
Expected payoff from Z: a 5 (1 a) 2 2 3a.
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7-8
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CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
a and b must equate these three expected payoffs:
1
5 3a 2 3a 1 a .
2
Turning to the UN, it is optimal for it to randomize over its two pure strategies
1
(in particular, assigning probability 2 to each) if and only if x and y equate the UN’s
two expected payoffs:
x 9 y 9 (1 x y) 4 x 4 y 4 (1 x y) 9 1 x y 1
2
where the left-hand expression is the expected payoff from inspecting facilities X
and Y. To be part of a Nash equilibrium, it is only necessary that Saddam hide
1
WMD at either facility X or Y with probability 2 . There is then an infinite number
1
of Nash equilibria, which differ only in terms of how this probability 2 is allocated
between facility X and facility Y. For example, one Nash equilibrium has Saddam
1
1
hide the WMD at X with probability 6 , Y with probability 3 , and Z with probability
1
1
1
, while the UN inspects X and Y with probability 2 and Z with probability 2 .
2
3
Another Nash equilibrium is for Saddam to hide the WMD at X with probability 8 , Y
1
1
with probability 8 , and Z with probability 2 , while the UN inspects X and Y with
1
1
probability 2 and Z with probability 2 .
b. Suppose the UN can inspect any two facilities, so that it has three pure strategies. The
UN can inspect X and Y, X and Z, or Y and Z. Find a Nash equilibrium in mixed
strategies.
ANSWER: Let x denote the probability that Saddam hides WMD in facility X, y the
probability for facility Y, and z 1 x y the probability for facility Z. For the
UN, a is the probability that it inspects X and Y, b is the probability that it inspects
X and Z, and c 1 a b is the probability that it inspects Y and Z. Conjecture
that both players randomize over all of their pure strategies. The expected payoffs
from Saddam’s three pure strategies are
Expected payoff from X: a 2 b 2 (1 a b) 5 5 3a 3b.
Expected payoff from Y: a 2 b 5 (1 a b) 2 2 3b.
Expected payoff from Z: a 5 b 2 (1 a b) 2 2 3a.
a and b must equate these three expected payoffs:
5 3a 3b 2 3b.
5 3a 3b 2 3a.
2 3a 2 3b.
The last condition implies a b; let the common value be denoted d. Then the
first (and second) condition become
1
5 3d 3d 2 3d 1 d .
3
Therefore, Nash equilibrium has the UN uniformly randomize over its three
1
pure strategies, assigning 3 to each of them. The UN’s expected payoffs are
Expected payoff from X and Y: x 9 y 9 (1 x y) 4 4 5x 5y.
Expected payoff from X and Z: x 9 y 4 (1 x y) 9 9 5y.
Expected payoff from Y and Z: x 4 y 9 (1 x y) 9 9 5x.
SOLUTIONS MANUAL
CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
For the UN to be content to randomize over its three pure strategies, it must be
the case that
4 5x 5y 9 5y.
4 5x 5y 9 5x.
9 5x 9 5y.
The last condition implies x y. Letting this common value be denoted w, then
the first condition becomes
1
4 5w 5w 9 5w 1 w .
3
Therefore, Nash equilibrium has Saddam uniformly randomize over its three pure
1
strategies; assigning 3 to each of them.
12. Consider the two-player game below. Find all of the mixed-strategy Nash equilibria.
Player 2
Player 1
Slow
Fast
Small
2,0
3,8
Medium
3,7
2,1
Large
3,4
5,6
ANSWER: Since Large strictly dominates Small then we know that all Nash
equilibria assign probability zero to Small. Thus, the Nash Equilibria of this
game is equivalent to the Nash equilibria of:
2
1
Slow
Fast
Medium
3,7
2,1
Large
3,4
5,6
Let m denote the probability that 1 assigns to Medium and s denote the probability that 2 assigns to Slow. Let us derive each player’s best reply. Note that Large
weakly dominates Medium. Hence, if 2 assigns any probability to Fast then 1
strictly prefers Large. If s denotes the probability that 2 assigns to Slow then 1’s
expected payoff from Medium is
s 3 (1 s) 2 2 s,
and her expected payoff from Large is
s 3 (1 s) 5 5 2s.
Large is strictly preferred when
5 2s 2 s or 1 s.
Hence, if s 1 then 1’s best reply is m 0; that is, the pure strategy Large.
If s 1 then either pure strategy gives a payoff of 3 and, in addition, any mixed
strategy gives a payoff of 3. Thus, if s 1 then m is a best reply for all values of
m, 0 m 1. Now consider player 2. Given m, player 2’s expected payoff from
pure strategy Slow is
m 7 (1 m) 4 4 3m,
7-9
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CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
player 2’s expected payoff from pure strategy Fast is
m 1 (1 m) 6 6 5m.
Slow is strictly preferred to Fast when
4 3m 6 5m or m 1
,
4
and Fast is strictly preferred to Slow when
4 3m 6 5m or m 1
.
4
1
Player 2 is indifferent between Fast and Slow when m 4 and, furthermore, all
mixed strategies give the same payoff of 4.75. Thus, player 2’s best reply is s 0
1
1
1
when m 4 , all values of s when m 4 , and s 1 when m 4 . The best reply
functions are plotted in the figure below, and we can see that (m, s) is a Nash
1
equilibrium if (m, s) (0, 0) or (m, s) (m, 1) and m 4 . Alternatively stated, if
s 1 then there is a unique best reply for player 1 of m 0 and the best reply of
player 2 is s 0. Hence, (m, s) (0, 0) is a Nash equilibrium. If s 1 then any
mixed strategy for player 1 is a best reply; however, s 1 is a best reply for player 1
1
1
if and only if m 4 . Thus, (m, s) (m, 1) is a Nash equilibrium and m 4 .
s
1
Best reply of player 1
Best reply of player 2
0
0
1
1
4
m
13. Consider the two-player game below. Find all of the mixed-strategy Nash equilibria.
Player 2
Player 1
Left
Right
Top
1,2
0,2
Bottom
1,0
3,4
ANSWER: There are two pure-strategy Nash equilibrium: (T,L) and (B,R), where
T refers to Top, B to Bottom, L to Left, and R to Right. In fact, these are the only
mixed-strategy Nash equilibria.
SOLUTIONS MANUAL
CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
Notice that B weakly dominates T for player 1 and R weakly dominates L for
player 2. Thus, if 2 assigns any probability to R then 1 strictly prefers pure strategy
B; and if 1 assigns any probability to B then 2 strictly prefers R.
Suppose 2 assigns probability one to L then 1’s best reply is any mixed strategy
but only if 1 chooses T with probability one is it a best reply for 2 to assign probability one to L. This gives us the Nash equilibrium (T,L). Now suppose 2 assigns
probability between 0 and 1 to L. Now 1’s best reply is B. Given that 2’s best reply
to 1 choosing B is to choose R for sure, and this contradicts 2 choosing R with
some probability between 0 and 1, we conclude there is not a Nash equilibrium in
which 2 randomizes.
Finally, suppose 2 choose R for sure. Then 1’s best reply is B and, given that R
is the best reply for 2 in that case, this gives us the Nash equilibrium (B,R).
Remember that “almost all” games have a finite and odd number of Nash
equilibria. This is one of those peculiar games with a finite and even number of
Nash equilibria.
14. Consider the two-player game below. Assume players are allowed to randomize.
a. Derive players’ best-reply functions.
Player 2
Player 1
x
y
a
3,3
4,2
b
6,3
2,6
c
5,3
3,2
ANSWER: Notationally, player 1 plays A with probability a, plays B with probability b and plays C with probability (1 a b). Similarly, player 2 plays X probability x and plays Y with probability (1 x)
Player 2 strictly prefers X over Y if and only if (iff ):
a 3 b 3 (1 a b) 3 a 2 b 6 (1 a b) 2
3 2a 6b 2 2a 2b 1 1 4b 1
1
b.
4
1
Thus, player 2’s best reply is to choose pure strategy X when 0 b 4 , any mixed
1
1
strategy when b 4 , and pure strategy Y when 4 b 1.
Now, find x that makes player 1 indifferent to her three pure strategies
Payoff from pure strategy A equals the payoff from pure strategy B when
x 3 (1 x) 4 x 6 (1 x) 2 1 4 x 2 4x 1 x 2
5
Payoff from pure strategy A equals the payoff from pure strategy C when
x 3 (1 x) 4 x 5 (1 x) 3 1 4 x 3 2x 1 x 1
3
Payoff from pure strategy B equals the payoff from pure strategy C when
x 6 (1 x) 2 x 5 (1 x) 3 1 2 4x 3 2x 1 x 2
1
2
It follows that A is strictly preferred to B iff x 5 ; A is strictly preferred to C iff
1
1
x 3 ; and B is strictly preferred to C iff x 2 .
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1
Thus, if x 3 then A is strictly preferred to B and C and is the unique best
1
reply; if x 3 then 1 is indifferent between A and C and both are strictly preferred
1
2
to B so 1’s best reply is any mixed strategy over A and C; if 3 x 5 then C is
strictly preferred to A and A is strictly preferred to B so the unique best reply is C;
2
if x 5 then 1 strictly prefers C to A and is indifferent between A and B so 1’s
2
1
unique best reply is C; if 5 x 2 then C is strictly preferred to B and B is strictly
1
preferred to A so 1’s unique best reply is C; if x 2 then 1 is indifferent between
B and C and strictly prefers both to A so 1’s best reply is any mixed strategy over
1
B and C; and if 2 x 1 then B is strictly preferred to C and C is strictly preferred
to A so 1’s unique best reply is B. Summarizing, the best reply probabilities
attached to pure strategies A, B, and C for player 1 are:
(a, b, c) 1
x 3
(1, 0, 0)
if 0
(z, 0, 1 z) for any z
if x 3
(0, 0, 1)
if 3 x 2
(0, z, 1 z) for any z
if x 2
(0, 1, 0)
if 2 x
1
1
1
1
1
1
b. Find all of the mixed-strategy Nash equilibria.
ANSWER: Let us first show that there is no Nash equilibrium for which player 2
uses a pure strategy. If player 2 plays X (so x 1) then, by the answer in part (a),
player 1 responds by playing pure strategy B. But, if that is the case, then player
2’s best reply is pure strategy Y, not X. Thus, there is no Nash equilibrium in which
player 1 uses pure strategy X. Next consider player 2 using pure strategy Y. Player
1’s optimal strategy is pure strategy A but then player 2’s optimal response to that
is pure strategy X, not Y.
Given that there is no Nash equilibrium in which player 2 uses a pure strategy,
the only possibility left is for player 2 to use a fully mixed strategy. For player 2 to
play a fully mixed strategy, by the preceding analysis, player 1 must assign
1
probability 4 to B. This means that 1 must randomize with respect to B and either
A and/or C (though we know from 1’s best reply function that it is never optimal to randomize over all three pure strategies). For player 1 to choose B with
1
probability between 0 and 1, x must equal 2 in which case 1 randomizes between
B and C. In order to make player 2 content to randomize, we must have B chosen
1
with probability 4 and, for that to be consistent with 1’s optimal behavior, player 2
must give equal probability to each of his strategies. Therefore, there is a unique
1 1
1 3
Nash equilibrium: (x,1 x;a,b,1 a b) ( 2 ,2 ;0,4 ,4 )
15. Phil, Stu, and Doug are deciding which fraternity to pledge. They all assign a payoff
of 5 to pledging Phi Gamma and a payoff of 4 to Delta Chi. The payoff from not
pledging either house is 1. Phi Gamma and Delta Chi each have two slots. If all three
of them happen to choose the same house, then the house will randomly choose
which two are admitted. In that case, each has probability 2/3 of getting in and probability 1/3 of not pledging any house. If they do not all choose the same house, then
all are admitted to the house they chose. Find a symmetric Nash equilibrium in
mixed strategies.
ANSWER: Since we are looking for a symmetric Nash equilibrium, let p denote
the probability of pledging Phi Gamma for each of the three. The expected payoff
from choosing Phi Gamma is given by :
p2 a
2
1
11
4
5 1b (1 p2 ) 5 p2 a b 5 5p2 5 a b p2
3
3
3
3
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CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
Note that p2 is the probability with which all three choose Phi Gamma conditional
on one of them having chosen Phi Gamma. Similarly, the expected payoff from
choosing Delta Chi is
(1 p) 2 a
2
1
4 1b (1 (1 p) 2 ) 4 3(1 p) 2 4 4(1 p) 2
3
3
4 (1 p) 2
If they are to randomize in equilibrium then the expected payoff from the two pure
strategies over which they are randomizing must be the same, which requires:
4
4
5 a b p2 4 (1 p) 2 1 1 a b p2 (1 p) 2 0 1 p .873
3
3
Thus, there is a unique symmetric Nash equilibrium that has a student pledging
Phi Gamma with probability .873 and Delta Chi with probability .127.
16. Three retail chains are each deciding whether to locate a store in town A or town B. The
profit or payoff that a chain receives depends on the town selected and the number of
other chains that put stores in that town; see accompanying table.
Chain’s Own Location
Number of Other Chains with
Stores in That Town
Chain’s Profit
A
0
10
A
1
3
A
2
1
B
0
8
B
1
4
B
2
2
a. Find a symmetric mixed-strategy Nash equilibrium in which chains randomize.
ANSWER: Given the other two players choose A with probability a, the expected
payoff to a player from choosing pure strategy A is
A: a2 1 2a(1 a) 3 (1 a) 2 10 a2 6a 6a2 10 20a 10a2
10 14a 5a2
and from choosing pure strategy B is
B: a2 8 2a(1 a) 4 (1 a) 2 2 8a2 8a 8a2 2 4a 2a2
2 4a 2a2
Equate the two expected payoffs and solve for a:
10 14a 5a2 2 4a 2a2 1 8 18a 3a2 0 1 a 0.48
b. Find all mixed-strategy Nash equilibria in which one of the chains puts a store in
town A for sure.
ANSWER: Suppose chain 1 puts a store in town A for sure. If chain 3 chooses A
with probability a then chain 2’s expected payoffs are:
A: a 1 (1 a) 3 3 2a
B: a 8 (1 a) 4 4 4a
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Note that 4 4a 3 2a for all positive values of a which implies that chain 2
strictly prefers B regardless of the value of a. There is then no Nash equilibrium
in which 1 chooses A and 2 and 3 both randomize. It is straightforward to verify
that 1 choosing A and both 2 and 3 choosing B is a Nash equilibrium. By this argument, the only Nash equilibria in which one of the chains puts a store in A for sure
has the other two chains putting stores in B for sure.
c. Find all mixed-strategy Nash equilibria in which one of the chains puts a store in
town B for sure.
ANSWER: Suppose chain 1 puts a store in town B for sure. If chain 3 chooses A
with probability a then chain 2’s expected payoffs are:
A: a 3 (1 a) 10 10 7a
B: a 4 (1 a) 2 2 2a
10 7a 2 2a 1 a 8
9
Finally, we need to check that chain 1 choosing B for sure is optimal.
A: a2 1 2a(1 a) 3 (1 a) 2 10 a2 6a 6a2 10 20a 10a2
10 14a 5a2
B: a2 8 2a(1 a) 4 (1 a) 2 2 8a2 8a 8a2 2 4a 2a2
2 4a 2a2 2(1 a) 2
We need:
2 4a 2a2 10 14a 5a2
8
8 2
18a 3a2 8 1 18a b 3a b 8
9
9
which is true. Hence, it is a Nash equilibrium for chain 1 to choose B and for
8
chain 2 and chain 3 to choose A with probability 9 .
17. A factory is suspected of hiring illegal immigrants as workers. The authority is deciding
whether to conduct an inspection. If the factory has illegal workers and an inspection
takes place, the workers will be discovered. The cost of an inspection to the government
is 100. The benefit from the inspection is 500 if illegal workers are found, but 0 if none
are found. The payoff to the authority from conducting an inspection is the benefit
minus the cost, while the payoff from not inspecting is 0. For the factory, the payoff from
having illegal workers and not getting caught is 200, from not using illegal workers is 0,
and from using illegal workers and getting caught is 300. A factory must decide
whether or not to use illegal workers, and the government must decide whether or not
to conduct an inspection. Find all mixed-strategy Nash equilibria.
ANSWER: The game can be represented in the following strategic form:
Authority
Factory
Inspect
Don’t inspect
Hire
⫺300,400
200,0
Not hire
0,⫺100
0,0
Let us first show that there are no pure-strategy Nash equilibria. If the factory uses
illegal workers then the best reply of the government is to inspect (as that delivers
a payoff of 400 which exceeds the zero payoff from not inspecting), but then the
SOLUTIONS MANUAL
CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
factory would prefer not to use illegal workers (as it yields a payoff of 0 which
exceeds 300 which is the payoff from using them and getting caught). Thus, there
is no Nash equilibrium in which the factory uses illegal workers for sure.
Next consider the factory not using illegal workers. In that case, the best reply
of the government is not to inspect (which yields a payoff of 0 rather than 100
from inspecting and not finding any illegal workers), but then the factory wants to
use illegal workers if the government is not going to inspect. Thus, there is no Nash
equilibrium in which the factory does not use illegal workers for sure.
Finally, let’s find a mixed-strategy Nash equilibrium, and let p denote the
probability that the factory uses illegal workers and q denote the probability that
the government inspects. The expected payoff from using illegal workers is
q (300) (1 q) 200 and from not using illegal workers is q 0 (1 q) 0. The factory is indifferent between its two pure strategies if and only if:
q (300) (1 q) 200 q 0 (1 q) 0 1 q 2
5
The expected payoff of the government from inspecting is p 400 (1 p) (100)
and from not inspecting is p 0 (1 p) 0. The government is indifferent
between its two pure strategies if and only if:
p 400 (1 p) (100) p 0 (1 p) 0 1 p 1
5
There is then a unique mixed-strategy Nash equilibrium in which there is a 20%
chance the factory uses illegal workers and a 40% chance the government/authority inspects.
18. For the game below, find all of the mixed-strategy Nash equilibria. The first payoff in a
cell is for player 1, the second payoff is for player 2, and the third payoff is for player 3.
Player 3: Hug
Player 2
Player 1
Kiss
Slap
Cuddle
7,1,5
1,2,4
Poke
2,2,1
5,3,2
Player 3: Shove
Player 2
Player 1
Kiss
Slap
Cuddle
5,0,5
3,4,1
Poke
3,3,3
0,5,4
ANSWER: First note that Slap strictly dominates Kiss for player 2. Thus, any Nash
equilibrium must have player 2 use pure strategy Slap. There are no dominated
strategies for the other two players so the game looks like:
Player 3
Player 1
Hug
Shove
Cuddle
1,2,4
3,4,1
Poke
5,3,2
0,5,4
Note that there is no pure-strategy Nash equilibrium. Let us then look to derive
a Nash equilibrium in which players 1 and 3 randomize. Letting h denote the
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CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
probability that player 3 hugs then player 1 is indifferent between her two pure
strategies when
3
Cuddle: h 1 (1 h) 3 h 5 (1 h) 0: Poke 1 h .
7
Letting c denote the probability that player 1 cuddles then player 3 is indifferent
between her two pure strategies when
2
Hug: c 4 (1 c) 2 c 1 (1 c) 4: Shove 1 c .
5
Thus, the unique mixed-strategy Nash equilibrium has player 1 cuddle with
2
3
probability 5 , player 2 slap for sure, and player 3 hug with probability 7 .
19. For the Avranches Gap game in Figure 7.10, find the security strategies and security
payoffs for General Bradley and General von Kluge.
ANSWER: Letting r denote the probability that Bradley attaches to pure strategy
reinforce, his expected payoff when von Kluge chooses attack is
r 3 (1 r) 0 3r
and when von Kluge chooses withdraw is
r 2 (1 r) 4 4 2r.
Bradley’s payoff is minimized by von Kluge choosing attack when
4 2r 1 r
3r
4
,
5
4
and is minimized by von Kluge choosing withdraw when r 5 . Thus, if Bradley is
pessimistic then he wants to choose r to maximize
3r
if 0
4 2r
if 5
4
r
4
5
r
1
4
which occurs when 3r 4 2r or r 5 .
4
Thus, Bradley’s security strategy is r 5 and his security payoff is 2.4. Letting
a denote the probability that von Kluge attaches to pure strategy attack, his
expected payoff when Bradley chooses reinforce is
a 0 (1 a) 3 3 3a
and when Bradley chooses eastward is
a 5 (1 a) 2 2 3a
von Kluge’s payoff is minimized by Bradley choosing reinforce when
3 3a
1
2 3a 1 a ,
6
and is minimized by Bradley choosing eastward when a
pessimistic then he wants to choose a to maximize
2 3a
if 0
3 3a
if 6
1
a
1
6
a
1
1
. Thus, if von Kluge is
6
s Strategies and Decision Making 2nd Edition Harrington Solutions Manual
SOLUTIONS MANUAL
CHAPTER 7: KEEP ’EM GUESSING: RANDOMIZED STRATEGIES
1
which occurs when 2 3a 3 3a or a 6 . Thus, von Kluge’s security strategy
1
is a 6 and his security payoff is 2.5. Note that the security strategies are different
2
1
from the Nash equilibrium strategies of a 5 and r 2 . This is because the game
is not one of pure conflict.
20. Consider the modified Rock−Paper−Scissors below. Find a symmetric mixed-strategy
Nash equilibrium.
Player 2
Player 1
Rock
Paper
Scissors
Rock
0,0
–2,2
1,–1
Paper
2,–2
0,0
–1,1
Scissors
–1,1
1,–1
0,0
ANSWER: Note that this is a game of pure conflict and there is no pure strategy
Nash equilibrium. We know that every strategic form game has at least one mixed
strategy Nash equilibrium. Also, we know that in games of pure conflict, the
Nash equilibrium strategy and the maximin strategy of players coincide. Hence,
the Nash equilibrium strategies can be found by finding the maximin strategies
of the players.
Suppose Player 1 plays R with probability a, P with probability b, S with probability (1 a b). If player 2 plays R for sure then the payoff of Player 1 is
0 a 2 b (1 a b) a 3b 1. If player 2 plays P for sure then the
payoff of Player 1 is 2 a 0 b (1 a b) 1 3a b. If player 2 plays
S for sure then the payoff of Player 1 is a b 0 (1 a b) a b.
Given this is a game of pure conflict then player 2’s optimal strategy minimizes
the expected payoff of player 1. Thus, player 2 plays R when
a 3b 1
1 3a b and a 3b 1
ab 1 ab
1
and b
2
1
.
4
In this case, player 1 chooses a strategy to maximize a 3b 1 which means
1
1
a 4 and b 4 and the payoff is 0. Player 2 plays P when
1 3a b
a 3b 1 and 1 3a b
ab 1 ab
1
1
and a .
2
4
In this case, player 1 chooses a strategy to maximize 1 3a b which means
1
1
a 4 and b 4 and the payoff is 0. Player 2 plays L when
a 3b 1 a b and 1 3a b a b 1 b 1
and a
4
1
.
4
In this case, player 1 chooses a strategy to maximize 1 3a b which means
1
1
a 4 and b 4 and the payoff is 0. Hence, the best player 1 can achieve is 0 and
1
1
that occurs by playing a 4 and b 4 . Therefore, the maximin strategy for player
1
1
1 is to play R with probability 4 , play P with probability 4 , and play S with probability
1
which is also her Nash equilibrium strategy. By symmetry, player 2 has the same
2
Nash equilibrium strategy.
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