978-0393123982 Chapter 30 Solution Manual Part 2

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NAME 373

(d) If the kicker kicks left with probability πK, then if the goalie jumps

left, the probability that the kicker will not score is πK.

(e) If the kicker kicks left with probability πK, then if the goalie jumps

(f) Find the probability πKthat makes the payoff to the goalie equal from

jumping left or jumping right. 1

1+p.

(g) The variable ptells us how good the kicker is at kicking the ball

into the left side of the goal when it is undefended. As pincreases, does

the equilibrium probability that the kicker kicks to the left increase or

decrease? Decreases. Explain why this happens in a way that

even a TV sports announcer might understand. The better the

30.8 (1) This problem is an illustration of the Hawk-Dove game de-

scribed in the text. The game was first used by biologist John Maynard

Smith to illustrate the uses of game theory in the theory of evolution.

Males of a certain species frequently come into conflict with other males

over the opportunity to mate with females. If a male runs into a situation

of conflict, he has two alternative strategies. If he plays “Hawk,” he will

fight the other male until he either wins or is badly hurt. If he plays

“Dove,” he makes a bold display but retreats if his opponent starts to

fight. If two Hawk players meet, they are both seriously injured in battle.

If a Hawk meets a Dove, the Hawk gets to mate with the female and the

Dove slinks off to celibate contemplation. If a Dove meets another Dove,

they both strut their stuff but neither chases the other away. Eventually

the female may select one of them at random or may get bored and wander

off. The expected payoffs to each male are shown in the box below.

374 GAME APPLICATIONS (Ch. 30)

The Hawk-Dove Game

Animal B

Hawk Dove

(a) Now while wandering through the forest, a male will encounter many

conflict situations of this type. Suppose that he cannot tell in advance

whether another animal that he meets is a Hawk or a Dove. The payoff to

adopting either strategy for oneself depends on the proportions of Hawks

and Doves in the population at large. For example, that there is one

Hawk in the forest and all of the other males are Doves. The Hawk would

find that his rival always retreated and would therefore enjoy a payoff of

10 on every encounter. Given that all other males are Doves, if

the remaining male is a Dove, his payoff on each encounter would be

(b) If strategies that are more profitable tend to be chosen over strategies

that are less profitable, explain why there cannot be an equilibrium in

strategy is sure to encounter another Hawk and would get a payoff of

−5.If instead, this male adoptled the Dove strategy, he would again

be sure to encounter a Hawk, but his payoff would be 0.

(d) Explain why there could not be an equilibrium where all of the an-

NAME 375

(e) Since there is not an equilibrium in which everybody chooses the

same strategy, we look for an equilibrium in which some fraction of the

males are Hawks and the rest are Doves. Suppose that there is a large

male population and the fraction pare Hawks. Then the fraction of any

player’s encounters that are with Hawks is about pand the fraction that

are with Doves is about 1 −p. Therefore with probability paHawk

meets another Hawk and gets a payoff of −5 and with probability 1 −p

he meets a Dove and gets 10. It follows that the payoff to a Hawk when

the fraction of Hawks in the population is p,isp×(−5) + (1 −p)×10 =

10 −15p. Similar calculations show that the average payoff to being

a Dove when the proportion of Hawks in the population is pwill be

(f) Write an equation that states that when the proportion of Hawks in

the population is p, the payoff to Hawks is the same as the payoffs to

(g) Solve this equation for the value of psuch that at this value Hawks

(h) On the axes below, use blue ink to graph the average payoff to the

strategy Dove when the proportion of Hawks in the male population who

is p. Use red ink to graph the average payoff to the strategy, Hawk, when

the proportion of the male population who are Hawks is p.Labelthe

equilibrium proportion in your diagram by E.

0255075

100

Percentage of hawks

Payoff

Blue

Line

Red Line

376 GAME APPLICATIONS (Ch. 30)

(i) If the proportion of Hawks is slightly greater than E,whichstrat–

egy does better? Dove. If the proportion of Hawks is slightly less

than E, which strategy does better? Hawk. If the more profitable

strategy tends to be adopted more frequently in future plays, then if the

strategy proportions are out of equilibrium, will changes tend to move the

proportions back toward equilibrium or further away from equilibrium?

30.9 (2) Economic ideas and equilibrium analysis have many fascinating

applications in biology. Popular discussions of natural selection and bio-

logical fitness often take it for granted that animal traits are selected for

the benefit of the species. Modern thinking in biology emphasizes that

individuals (or strictly speaking, genes) are the unit of selection. A mu-

tant gene that induces an animal to behave in such a way as to help the

species at the expense of the individuals that carry that gene will soon

be eliminated, no matter how beneficial that behavior is to the species.

A good illustration is a paper in the Journal of Theoretical Biology,

1979, by H. J. Brockmann, A. Grafen, and R. Dawkins, called “Evo-

lutionarily Stable Nesting Strategy in a Digger Wasp.” They maintain

that natural selection results in behavioral strategies that maximize an

individual animal’s expected rate of reproduction over the course of its

lifetime. According to the authors, “Time is the currency which an animal

spends.”

Females of the digger wasp Sphex ichneumoneus nest in underground

burrows. Some of these wasps dig their own burrows. After she has dug

her burrow, a wasp goes out to the fields and hunts katydids. These

she stores in her burrow to be used as food for her offspring when they

hatch. When she has accumulated several katydids, she lays a single egg

in the burrow, closes off the food chamber, and starts the process over

again. But digging burrows and catching katydids is time-consuming. An

alternative strategy for a female wasp is to sneak into somebody else’s

burrow while she is out hunting katydids. This happens frequently in

digger wasp colonies. A wasp will enter a burrow that has been dug by

another wasp and partially stocked with katydids. The invader will start

catching katydids, herself, to add to the stock. When the founder and

the invader finally meet, they fight. The loser of the fight goes away and

never comes back. The winner gets to lay her egg in the nest.

Since some wasps dig their own burrows and some invade burrows

begun by others, it is likely that we are observing a biological equilibrium

in which each strategy is as effective a way for a wasp to use its time for

producing offspring as the other. If one strategy were more effective than

the other, then we would expect that a gene that led wasps to behave

in the more effective way would prosper at the expense of genes that led

them to behave in a less effective way.

Suppose the average nesting episode takes 5 days for a wasp that

digs its own burrow and tries to stock it with katydids. Suppose that the

NAME 377

average nesting episode takes only 4 days for invaders. Suppose that when

they meet, half the time the founder of the nest wins the fight and half

the time the invader wins. Let Dbe the number of wasps that dig their

own burrows and let Ibe the number of wasps that invade the burrows

of others. The fraction of the digging wasps that are invaded will be

about 5

D. (Assume for the time being that 5

D<1.) Half of the diggers

who are invaded will win their fight and get to keep their burrows. The

fraction of digging wasps who lose their burrows to other wasps is then

D=5

D. Assume also that all the wasps who are not invaded by other

wasps will successfully stock their burrows and lay their eggs.

(a) Then the fraction of the digging wasps who do not lose their burrows

Therefore over a period of 40 days, a wasp who dug her own bur-

row every time would have 8 nesting episodes. Her expected number of

(b) In 40 days, a wasp who chose to invade every time she had a chance

would have time for 10 invasions. Assuming that she is successful half the

Write an equation that expresses the condition that wasps who always dig

their own burrows do exactly as well as wasps who always invade burrows

Solve for the numerical value of I

Dthat just equates the expected number

of successes for diggers and invaders. The answer is 3

(d) But there is a problem here: the equilibrium we found doesn’t appear

to be stable. On the axes below, use blue ink to graph the expected num-

ber of successes in a 40-day period for wasps that dig their own burrows

every time where the number of successes is a function of I

D. Use black

ink to graph the expected number of successes in a 40-day period for in-

vaders. Notice that this number is the same for all values of I

D. Label the

point where these two lines cross and notice that this is equilibrium. Just

to the right of the crossing, where I

Dis just a little bit bigger than the

equilibrium value, which line is higher, the blue or the black? Black.

At this level of I

D, which is the more effective strategy for any individ-

ual wasp? Invade. Suppose that if one strategy is more effective

than the other, the proportion of wasps adopting the more effective one

increases. If, after being in equilibrium, the population got joggled just

a little to the right of equilibrium, would the proportions of diggers and

378 GAME APPLICATIONS (Ch. 30)

Success

Blue line

8-5(I/D)

Black line

(e) The authors noticed this likely instability and cast around for possible

changes in the model that would lead to stability. They observed that

an invading wasp does help to stock the burrow with katydids. This may

save the founder some time. If founders win their battles often enough

and get enough help with katydids from invaders, it might be that the

expected number of eggs that a founder gets to lay is an increasing rather

than a decreasing function of the number of invaders. On the axes below,

show an equilibrium in which digging one’s own burrow is an increasingly

effective strategy as I

Dincreases and in which the payoff to invading is

constant over all ratios of I

Success

30.10 (1) The Iron Chicken restaurant is located on a busy interstate

NAME 379

highway. Most of its customers are just passing through and will never

return to the Iron Chicken. But some are truck drivers whose routes take

them past the Iron Chicken on a regular basis. Sybil, the nearsighted

waitress at the Iron Chicken is unable to distinguish regular customers

from one-time customers. Sybil can either give a customer good service

or bad service. She knows that if she gives bad service to any customer,

then she will get a small tip. If she gives good service to a truck driver,

he will give her a large tip in the (futile) hope that she will recognize

him the next time he comes, but if she gives good service to a one-time

customer, he will still leave a small tip. Suppose that the cost to Sybil

of giving a customer good service rather than bad service is $1. The

tips given by dissatisfied customers and customers just passing through

average $0.50 per customer. A truck driver who has received good service

and plans to come back will leave a tip of $2. Sybil believes that the

fraction of her customers who are truck drivers who plan to come back

is x. In equilibrium we would expect Sybil to give good service if xis

greater than 2/3 and bad service if xis smaller.

30.11 (2) Mona is going to be out of town for two days and will not be

needing her car during this time. Lisa is visiting relatives and is interested

in renting her car. The value to Lisa of having the car during this time

is $50 per day. Mona figures that the total cost to her of letting Lisa

use the car is $20, regardless of how many days Lisa uses it. On the

evening before the first of these two days, Mona can send a message to

Lisa, offering to rent the car to her for two days for a specified price. Lisa

can either accept the offer or reject the offer and make a counteroffer.

The only problem is that it takes a full day for a counteroffer to be made

and accepted.

Let us consider the Rubinstein bargaining solution to this problem.

We start by working back from end. If Lisa rejects the original offer,

then the car can only be rented for one day and there will be no time

for Mona to make a counteroffer. So if Lisa rejects the original offer, she

can offer Mona slightly more than $20 to rent the car for the last day

and Mona will accept. In this case, Lisa will get a profit of slightly less

than $50 −$20 = $30. Mona understands that this is the case. Therefore

when Mona makes her original offer, she is aware that Lisa will reject the

offer unless it gives Lisa a profit of slightly more than $30 .Monais

aware that the value to Lisa of renting the car for two days is $100 .

Therefore the highest price for two days car rental that Lisa will accept

380 GAME APPLICATIONS (Ch. 30)

(a) Suppose that the story is as before except that Mona will be out of

town for three days. The value to Lisa of having the car is again $50

per day and the total cost to Mona of letting Lisa use the car is $20,

regardless of how many days Lisa uses it. This time, let us suppose that

Lisa makes the first offer. Mona can either accept the offer or refuse it and

make a counteroffer. Lisa, in turn, can either accept Mona’s counteroffer

or refuse it and make another counteroffer. Each time an offer is rejected

and a new offer is made, a day passes and so there is one less day in which

the car can be rented. On the evening before the first of these three days,

Lisa reasons as follows. “If Mona rejects the offer that I make tonight,

then there will be two days left and it will be Mona’s turn to make an

offer. If this happens, Mona will get a profit of slightly less than $50

(Hint: We found this answer above, for the two-day case.) Since her total

costs are $20, Mona will make a profit of $50 if I offer her a price of

$70 for the three days rental.” Since three days of car rental is worth

a profit of $50.

(b) Now suppose that the story is as before except that Mona will be out

of town for four days and suppose that Mona makes the first offer. Mona

knows that if Lisa rejects her first offer, there will be three days left, it

will be Lisa’s turn to make an offer and so Lisa can make a profit of

$200 to Lisa, so to the highest price that Mona can expect Lisa to accept