The general case

In this section we analyse the general form of 2-player games.

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

gives the payoffs for the individuals playing the game. The replicator equations

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

can be found by first noting that the fitness of C is

\[x.1+(1-x)S\]

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

Likewise, the fitness of D is

\[xT+(1-x).0 = xT\]

The average fitness is then

\[x(x+(1-x)S) + (1-x)xT\]

Thus

(1)\[\begin{split}\begin{aligned} \frac{dx}{dt} = f(x) & = & x (x+(1-x)S - x(x+(1-x)(S+T))) \nonumber \\ & = & x (1-x) (x+(1-x)S - Tx) \end{aligned}\end{split}\]

is the replicator equation for the general case.

The steady states of equation (1) are \(x_*=0\), \(x_*=1\) and the solution to

\[x_*+(1-x_*)S = Tx_*\]

or

\[x_* = \frac{S}{S+T-1}\]

In order for this steady state to lie between 0 and 1 we need either

• \(T>1\) and \(S>0\)
• or \(T<1\) and \(S<0\)

We can determine the stability of the three steady states by differentiating equation (1) with respect to \(x\) we get:

\[f'(x) = x(1-x)(1-S-T) + (1-2x)(x+(1-x)S-Tx)\]

Evaluating at the steady states we get

\[f'(0) = S\]

so the 0 steady state is stable if \(S<0\).

\[f'(1) = T - 1\]

So the 1 steady state is stable if \(T<1\).

\[\begin{split}\begin{aligned} f'\left(\frac{S}{S+T-1}\right) & = & -S (1 - \frac{S}{S+T-1}) \\ & = & \frac{S(T-1)}{1-S-T} \end{aligned}\end{split}\]

Thus the co-existence steady state is stable if both \(T>1\) and \(S>0\), but unstable if \(T<1\) and \(S<0\).

Below we illustrate how the stability is determined in the \(ST\) plane.

FIGURE HERE

We now consider some further examples. The prisoners dilemma is

individual/opponent	Keep quiet	Blame other
Keep quiet	1	\(S=-1/4\)
Blame other	\(T=5/4\)	0

Here \(S<0\) and \(T>1\) and the stable strategy is to keep quiet. Here the situation is very bad for the population as a whole. The group could do best by all co-operating but by acting rationally or following natural selection all will defect.

The stag hunt is

individual/opponent	Group	Self
Group	1	\(S=-1/4\)
Self	\(T=1/2\)	0

thus \(S<0\) and \(T<1\), so all defect and all co-operate are both stable. The steady state

\[\frac{S}{S+T-1} = \frac{1}{3}\]

thus if \(x(0)>1/3\) then all defect, if \(x(0)<2/3\) then all co-operate. Stag hunts are hard to establish but when established are stable to defection.