978-0393919684 Chapter 7 Lecture Note Part 3

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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.

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

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).

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.

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

2 1

–

2 1

–

2 1

–

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

12 2 2 1 3

2 1 1

2 1 2 1 1

2 1 1 3 2 1

12 1

12 1 3

2 1 2 1

2 1 2 2 2 1

2 1 2 1

2 1

pp p

pp p p

ppp

pp p p p

p p

é ù

æ ö

–+ – +

é ù

–

ê ú

ç ÷

+ê ú

è ø +

ê ú

––ê ú

ê ú

æ ö

–– + +

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.

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

p





bdca

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

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:

)(

)()(

bdca

bcaddbcad





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.)

Subcase 1: b < c < d

DeLne

bc 

and

cd 

. Then



and



, 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



and



as in subcase 1. (The only diMerence is that





, so

0)(  bccbdd

Subcase 3: c < b < d

DeLne

cb 

and

cd 

. Then



and



, 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

No matter the relationship of b and c, for Case 1 it is always true that

0)(  bccbdd

, so

0



, 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



is always positive is very similar.

The same methods are used to demonstrate that

0



0



, and

0



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.