L567 Lecture 7
During the previous lecture we began thinking about Game
Theory. We were thinking in terms of two strategies, A and
B.
One way to organize the information is to put it into a
“payoff matrix
“Payoff” to
A
B
When interacting with
A
B
E(A,A)
E(A,B)
E(B,A)
E(B,B)
Where, for example, E(A,A) gives the expected payoff to A
when interacting with A.
We could rewrite the matrix to read
“Payoff” to
A
B
When interacting with
A
B
a
b
c
d
Note that
1) if a>c, then A is and ESS
2) if d>b, then B is and ESS
3) Either A or B could be an ESS. The outcome of selection
would then depend on initial conditions. Hence Natural
selection does not necessarily result in the “best” strategy
winning. Show this graphically. In addition, no claim is
made that either A or B represent an optimal strategy.
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L567 Lecture 7
4) if neither A nor B is an ESS, then selection is frequencydependent. Both strategies will increase when rare under the
action of natural selection, leading to a stable polymorphism.
Since by definition, there is no Evol. Stable Strategy, we call
this an Evolutionarily Stable State. It is a state of the
population that is stable. We would like to know the
frequency of A at the stable equilibrium point. How do we
solve for that?
Let q be the frequency of strategy A. Then 1-q is the
frequency of strategy B. In the space below, show how you
would solve for q at equilibrium, q̂ . Hint at equilibrium, the
fitness of strategy A must be equal to the fitness of strategy
B.
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L567 Lecture 7
Now consider a “mixed” strategy such that a single
individual could play either A or B. Let the name of the
mixed strategy be I.
Strategy I: play A with probability q
and play B with probability (1-q)
The Bishop-Cannings theorem: If I is an ESS, then at the
ESS, the following is true
E(A, I) = E(B, I) = E(I, I).
Why is that true?*
In addition, if I is an ESS, then
E(I, A) > E(A, A), and E(I, B) > E(B, B). (i.e., Maynard
Smith’s condition in equation 2.4b is met. Why is this true?
What does it mean??
*Hint: write out E(I, A) > E(A, A). Compare to previous
page
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L567 Lecture 7
Some definitions.
Strategy: specification of what an organism will do in a given
situation.
Pure strategy: conditional (e.g. Bourgeois) or unconditional
(e.g. Hawk) fixed response.
Mixed strategy: stochastic strategy (e.g., play Hawk with
probability q and Dove with probability 1-q.
THE HAWK-DOVE GAME
We assume that two individuals pair up at random in a large
population. Hawks will always fight for the contested
resource. Doves will retreat from Hawks, and split the
resource with other doves.
Let V = the Value of the resource. We assume V>0.
Let C = the Cost of fighting
Here is the payoff matrix
“Payoff” to
Hawk
Dove
When interacting with
Hawk
Dove
E(H,H)
E(H, D)
E(D,H)
E(D, D)
“Payoff” to
Hawk
Dove
When interacting with
Hawk
Dove
(V-C)/2
V
0
V/2
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L567 Lecture 7
Okay, now answer the following questions:
1. When is Hawk and ESS?
2. When is Dove an ESS?
3. When would the population be polymorphic at
equilibrium? What would be the frequency of hawks at
equilibrium?
4. When would a mixed strategy be favored by selection.
What would be the probability that an individual plays Hawk
in this mixed strategy at equilibrium?
Now consider a new strategy, Bourgeois. Bourgeois behaves
like a hawk if it is the owner of the territory (e.g. sunspot),
and it behaves like a dove, if it is the intruder on the territory.
To get the simplest payoff matrix, we assume that Bourgeois
is a territory owner one half of the time.
“Payoff” to
Hawk
Dove
Bourgeois
Hawk
(V-C)/2
0
(V-C)/4
When interacting with
Dove
Bourgeois
0.5[((V  C) / 2)  V ]
V
V/2
V/4
3
V/2
V
4
Convince yourself that the new gray entries are correct. Then
show that Bourgeois is the only ESS if C>V.
Finally, How does the Bourgeois strategy fit in with the
findings in the Davies paper on the speckled wood
butterflies. What was the definitive test of “play hawk if
owner”?
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