Evolutionary Dynamics of Biological Games

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We’ll start with looking at a simple game, the game of Hawk and Dove. It’s a model
of how any kind of animal can use aggression to its own advantage. After analyzing
that, we’ll turn to a more interesting game: sex. Why are there about as many
males as females born? Why are eggs so much larger than sperm? Why are there
only two sexes?
In the game of Hawk and Dove, there’s a species that competes for some resource,
and getting this resource helps the winner produce more offspring – technically, it
increases the evolutionary fitness of the winner. The resource could be anything:
access to food, access to nesting space, access to potential mates. What’s important
is that an animal in this species is one of two kinds: a Hawk will always fight when
challenged, and a Dove will always back down.
When a Hawk meets a Hawk, one will win the resource, and gain a fitness G; the
other will be badly injured, and lose fitness, -C. Amongst all Hawks, the average
change in fitness is 12 (G − C). We are going to assume, from now on, that this is
negative, that is, G < C.
When a Dove meets a Dove, one will win the resource, and gain a fitness G. The
other will back down before injury, and gain nothing. The average change in fitness
this time is 12 (G).
Finally, when a Hawk meets a Dove, the Hawk will win the resource, and gain a
fitness G. The Dove will back down before injury, and gain nothing. The average
change in fitness this time is zippo for the Dove, G for the Hawk.
This information is usually presented in a matrix, called the payoff matrix:
W (H, H) =
1
(G − C)
2
W (D, H) = 0
W (H, D) = G
W (D, D) =
1
G
2
If the population were mostly hawks, there would be trouble: lots if fights, lots
of casualties. If the population were mostly doves, peace reigneth. There a point,
somewhere in between, where the two can co-exist.
1) Assume a population has a proportion of x Hawks and (1 − x) Doves. we’ll
represent this as a population strategy, S, defined as S = xH + (1 − x)D. Show
that in such a population,
1
W (H, S) =x (G − C) + (1 − x)G
2
1
W (D, S) =(1 − x) G
2
2) Show further that if x < G/C, the population will evolve towards more Hawks,
while if x > G/C, the population will evolve towards more Doves.
There’s another concept of stability we can examine: stability of strategies. For
example, if our population was pure Dove, and a mutation arose that was a Hawk,
the Hawks would quickly spread. The pure Dove population would be said to have
an evolutionarily unstable strategy.
Here’s how we’ll analyze this. Say we have a population with strategy I, and then
a few individuals with strategy J intrude. Say a percentage . Then the new
population mix will be R = J + (1 − )I. For I to ward off this invasion, it would
have to have a fitness that is superior to that of J, that is,
W (J, R) < W (I, R)
If this holds, we say I is an evolutionarily stable strategy, an ESS. If we let  → 0,
we get a condition for stability against invasion (SAI):
W (J, I) ≤ W (I, I)
3) This condition is necessary: if it doesn’t hold, then J will invade I. Show that
H is not SAI by I, and I is not SAI against H.
4) Let p = G/C. Assume we have a mutant R which employes strategy H with
probabilty p and strategy D with probability (1 − p). Show that R cannot be
invaded by H or by D:
W (H, R) ≤W (R, R)
W (D, R) ≤W (R, R)
Now for sex: we want to show that reproduction with an equal number of males
and females is an ESS. To do this, we’ll have ton compute fitnesses, this time the
number of grandchildren that a strategist produces.
We’ll start with a population that has a strategy M of producing a proportion
m males and (1 − m) females. That is, if the first generation of children has N1
members, then nN1 of them are male and (1 − m)N1 are female.
Now for grandkids. Say there are N2 of these. Then a typical male parent will have
N2 /(mN1 ) kids, and a typical female N2 /((1 − m)N1 ) kids.
5) Now we want to see the fitness of a different strategy, P . Say R has a male child
with probability p and a female with probability (1 − p). Show that W (P, M ) is
proportional to
1−p
p
+
m 1−m
6) Maximize this with respect to p. What does this mean?
An old problem in biology is anisogeny – that is, not-equal-gametes. Many species,
including humans but most obviously including chickens – have eggs that are enormously larger than the sperm that fertilize them. The egg contains materials needed
to nourish the developing animal, and the sperm contains little more than some
DNA and a delivery system. Why? Why don’t the sperm also contain food for the
young? Why are female gametes larger than male gametes?
Now this is a bad way of asking the question: if we define ’female’ to be the one
with the big gamete, the question goes away. But then another question arises: why
are there only two sexes? That is, why are there only two gamete sizes?
Here’s a possible answer: let’s say all gamete sizes started out equal, with some
random fluctuation in size. Suppose it came about that gamete size was heritable
and dominant. By pure chance, there would arise a gamete a fifth the size of the
average, and the individual producing this ultra-small gamete could then afford to
produce five times as many as average. It would have five times as many offspring,
and its offspring would have five times as many offspring and ... and you can see
where this is heading.
6) Can you show, using game theory, that equal-size gametes is not a strategy that
is SAI against small-sized gametes?

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