Here is an example of how an equilibrial population operates.
Imagine a population consisting of 95% strategy A and 5% strategy B. Remember, for a strategy to be equilibrial it must be consistent with:
ESS rule #2a E(A,A) = E(B,A) (payoffs in common interactions are equal)
but not consistent with ESS rule #2b: E(A,B) > E(B,B) (rare interactions favor the common strategy)
Now if rule 2 is not true, the only possibility that is still consistent with ESS rule #2a is:
E(A,B) = E(B,B).
So this gives a payoff matrix with values that meet the two conditions (2a true, 2b not true)):
| A | B | |
Strategy |
(common payoffs) | (rare payoffs) |
| A | 1 | 0.5 |
| B | 1 | 0.5 |
(note any numerical values can be used as long as equilibrial conditions of 2a true, 2b not true are met)
Fitness of Each Strategy
Now the fitness (W) of A = E(A,A)*freq(A) + E(A,B)*freq(B)
and the fitness (W) of B = E(B,A)*freq(A) + E(B,B)*freq(B)
Now since E(A,A) = E(B,A) = 1 and E(A,B) = E(B,B) = 0.5 and
since freq(A) = 0.95 and freq(B) = 0.05, then:
W(A) = 1 * 0.95 + 0.5 * 0.05 and W(B) = 1 * 0.95 + 0.5 * 0.05
which are the same. Thus neither strategy has any advantage over the other.
But what happens if some additional B strategists enter by migration, mutation, learning (we're dealing with behavior after all) or any other mechanism consistent with neutrally adaptive evolution? (See genetic drift).
Now, the freq. of B increases and A drops. Just for the heck of it, let's say that these neutrally adaptive processes result in new frequencies of freq(A) = 0.9 and freq(B) = 0.1. Our fitness equations are now:
W(A) = (1 * 0.90) + (0.50 * 0.10) and W(B) = (1 * 0.90) + (0.50 * 0.10)
They are still equal and they always will be equal as long as the payoffs for common interactions equal each other as do the payoffs for rare interactions!
This is what an equilibrial population is -- the only possible evolution with respect to these strategies is some neutrally adaptive mechanism.