The mating game

Among males and females there is an asymmetry regarding the care for the child. For most animals it is the female that give birth and take the responsibility for the child. A major concern for her is to get the male to stay and help out raising a family. If we view this in a game theoretic way we say there is a cost for raising a child and a gain which is the child itself. Initially there does not seem to be a game going on, the female chooses a man and he helps out raising the child. The thing is that this setting is not stable in an evolutionary sense because soon there will appear males that just leaves after the child is born. In a game theoretic sense it makes sense for them not to pay the cost and at the same time gain the benefits. How should the females respond to this? They want to protect themselves from the philandering males by testing the presumptive male beforehand, they start insisting on a period of courtship before mating. This makes it harder for the philanderers, if they meet a coy female(females insisting on courtship) there will be no babies. Now we have four players in our mating game. The so called fast females who don't insist on courtship, the coy ones, the faithful men who court the females and finally the philandering males who leaves after the mating and don't spend time courting. If we study this in a game theoretic sense we assign some values to the gains and the cost involved.

Suppose the payoff to each parent of babies is +15, and the total cost of raising babies is −20. Suppose the cost of a long courtship is −3 to each player. What would this figures lead to? If a coy female meets a faithful male there would be an equal net gain of 15-3-(20/2) = 2 points for each player. If a coy female meets a philanderer nothing happens and both gets zero points. If a fast female meets a faithful man both get 15-(20/2)=5 points. If a fast female meets a philanderer the female will gain 15-20= -5 points while the philanderer gains 15 points.

Looking for equilibrium, what would the relation be between the number of faithful and philandering males? This fraction can be calculated from the fact that a mixed strategy would be optimal when the expected payoff from either coy fast females are equal. Suppose the males are faithful with probability x, we would then have a probability (1-x) for the philanderers. Setting up the expected gain for the females we would have

Coy: 2*x + 0*(1-x) = 2x
Fast: 5*x + (-5)*(1-x) = 10x-5

From this we get x = 5/8 and (1-x)=3/8. So for every 8 males there would be 5 faithful ones and 3 philanderers.
Doing the same calculation for the expected gain for the men with x being the probability to encounter a coy female and (1-x) the probability to encounter a fast one

Faithful: 2*x + 5*(1-x) = 5-3x
Philanderer: 0*x + 15*(1-x) = 15-15x

From this we get x = 5/6 and (1-x) = 1/6. So for every 6 females there would be 5 coy ones and 1 fast one.

The interesting thing with this result is that it is not Pareto-optimal with regard to males and females as a whole. Not Pareto-optimal means that both parties can change strategies and gain more. With the figures above the net gain for the females is 5/4 points and 5/2 points for the males. If all females were fast and all men were faithful the parties would have a net gain of 5 points. Both males and females would struggle to get out of this mess. It is tough to be a bird.

More in depth info about this can be found in 'The selfish gene' by Richard Dawkins.