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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
4.
(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
(c) If all the other males are Hawks, then a male who adopts the Hawk
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
2
4
6
Percentage of hawks
Payoff
8
Blue
Line
Red Line
e
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
4
I
D. (Assume for the time being that 5
4
I
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
1
2
5
4
I
D=5
8
I
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
I
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
(c) The equation you have just written should contain the expression I
D.
Solve for the numerical value of I
Dthat just equates the expected number
of successes for diggers and invaders. The answer is 3
5.
(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
e
Blue line
8-5(I/D)
5
Black line
I
_
D
(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
I
_
D
e
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
.
