The hawk-dove game

Game theory is used both in biology and in economics to understand how individuals act when faced with tradeoffs between certain costs and benefits. Here we look at a competition between two behavioural strategies, one we call ’co-operate’ the other ’defect’. In general co-operators try to share resources and defectors try to take resources for themselves. To illustrate the problem, we construct a payoff table based on the interactions of co-operators (C) and defectors (D). We assume that when

• C meets C they both get payoff \(1\) (reward).

• D meets D they both get payoff \(0\) (punishment).

• C meets D then D gets payoff \(T\) (temptation), and C gets payoff \(S\) (sucker).

In table form this can be written

individual/opponent	C	D
C	1	\(S\)
D	\(T\)	0

giving the payoffs for the individual playing the game.

In this section, we will look at the following example known as the hawk-dove game,

individual/opponent	Dove	Hawk
Dove	1	\(S=1/4\)
Hawk	\(T=5/4\)	0

In a world full of doves (and no hawks) everyone gets a payoff of 1, while in a world of hawks everyone gets zero. But… the best payoff is for hawks who live in a world full of doves, they will get payoff 5/4 every time they meet a dove.

Evolutionary game theory considers how a population of individuals following these rules will evolve over time. We make the following assumptions:

• Pairwise contests occur between two individuals.

• Population is infinite. So every time an individual plays the game against a different person.

• Those with higher payoffs (fitness) increase in the population. (Darwin’ law of natural selection)

Let \(x\) be the proportion of the population who co-operate. Let the fitness of an individual be its expected payoff given that there are \(x\) co-operators in the population. We assume that the rate of increase of co-operators is proportional to their fitness minus average fitness of both co-operators and defectors.

\[\frac{dx}{dt} = x (\mbox{fitness of C - average fitness}) \label{repeq}\]

This is known as the replicator equation.

The fitness of C is

\[x.1+(1-x)\frac{1}{4} = \frac{1}{4} + \frac{3x}{4}\]

That is, if a co-operator chooses a person to play the game with at random then they will interact with another co-operator with probability \(x\) and get payoff 1 and they will interact with a defector with probability \(1-x\) and get payoff \(1/4\).

Likewise, the fitness of D is

\[x.\frac{5}{4} +(1-x).0 = \frac{5}{4} x\]

The average fitness is then

\[x(\frac{1}{4} + \frac{3x}{4}) + (1-x) \frac{5}{4} x\]

i.e. the

Thus

(1)\[\frac{dx}{dt} = f(x) = x (1-x) (\frac{1}{4} + \frac{3x}{4} - \frac{5}{4} x ) = \frac{1}{4} x(1-x)(1-2x)\]

is the replicator equation.

The steady states of equation (1) are \(x_*=0\), \(x_*=1\) or \(x_* = 1/2\).

We can determine the stability of these three steady states by differentiating equation [repeqf] with respect to \(x\) to get:

\[f'(x) = \frac{1}{4} \left((1-2x)^2 - 2x(1-x) \right)\]

Evaluating at the steady states we get

\[f'(0) = f'(1) = 1/4\]

so the 0 steady state (all defect) and the 1 steady state (all co-operate) are unstable. The steady state where there is a balance between co-operation and defect has

\[f'(1/2) = \frac{1}{4} \left(0 - 2\frac{1}{4} \right) = - \frac{1}{8}\]

so is stable.

The analysis thus reveals that the hawks and doves coexist at a proportion \(1/2\). Note that this is not optimal for the population as a whole. The mean fitness in this situation is

\[1/2(1/2+1/2 \times 1/4)+5/4 \times 1/2\times1/2=5/16+5/16=10/16\]

If all played Dove the mean fitness would be \(1\). In situations like this natural selection acts to maximise individual fitness and not group fitness.