What do they tell us about our species?

Curiously, I think the answer is that human males are generally monogamous.

Before getting on to that we should ask: what is the excess adipose tissue carried by females of the human species actually for?  There are two naive answers which I've heard expressed in the past:

1. To provide milk for the human infant
2. To attract a mate

Neither of these explanations hold water.  Every other mammal that I'm aware of only has swollen glands whilst the offspring are young enough to breastfeed. At all other times the females look more or less similar to the males.  The second explanation leaves open an obvious question too.    Why should a male be attracted to these things in the first place?

What does seem to be true, is that - all other things being equal - possession of the aforesaid is an encumbrance, as any well-endowed cave-woman that has ever tried to flee a sabre-toothed tiger might attest.

There are cases in nature, however, where encumbrances are carried purely to improve the likelihood of selection.  An example of this is the peacocks tail.  However, in almost all examples these compromises - which improve the chance of being selected for mating, but reduce the chance of individual survival - are carried by the male, not the female.  The human species appears to be an outlier.

Let's try to understand this using game theory.  Imagine a world with just one male and one female.  There are two female strategies, namely $B$ and $¬B$, for "has excess adipose tissue" or "does not have excess adipose tissue" $^\dagger$.  Likewise the male can play $LB$ or $¬LB$, for "likes such things" and "does not like them".  This is essentially a coordination game, like the driving game where two players have to co-ordinate on which side of the road to use.  If they co-ordinate on $B/LB$ or $¬B/¬LB$ then they can mate, which is coded by an award of $+10$ utils.  Conversely, if they do not co-ordinate, they do not mate, the species dies out, and they lose $10$ utils.  What makes this slightly different from the driving game is that one player - the female - loses an additional point for playing one of her strategies, because it makes her less able to run away from sabre-toothed tigers.  Here's the payoff table

	¬LB	LB
B	-11\-10	9\ 10
¬B	10\ 10	-10\-10

The first thing to note is that there are two pure Nash Equilibria.  These are strategy pairs where neither player can improve their payoff by changing strategy unless the other one does as well.  The pure equilibria are represented by the bottom left and top right outcomes in the table.

There is also a mixed Nash Equilibrium.  This is where the female plays $B$ with probability $p$ and the male plays $¬LB$ with probability $q$. The expected payoff for the male if he plays $LB$ is
$$
\begin{align}
\langle M \lvert LB \rangle &= +10p - 10(1-p) \\
&= +20p - 10
\end{align}
$$
Similarly, the other expected payoffs are
$$
\begin{align}
\langle M \lvert ¬LB \rangle &=  -20p + 10 \\
\langle F \lvert B \rangle &= -20q + 9 \\
\langle F \lvert ¬B \rangle &= +20q - 10 \\
\end{align}
$$
To find $p$ and $q$ that give a mixed Nash Equilibrium we just need to set the first two expected payoffs equal to eachother and solve for $p$ and set the 2nd two equal and solve for $q$.  This tells us that the equilibrium is where
$$
p = \frac{1}{2} \\
q = \frac{19}{40}
$$
We can create a phase diagram for the $p,q$ space and imagine that each point represents a possible state for the genomes of the population at large.   For example, $(0.1, 0.9)$ means that one tenth of the females are endowed and nine tenths of the males prefer them not to be.  At each point in the phase space we can attach an arrow because, for each value of $p$ it is either advantageous to males to increase or to decrease $q$ and vice versa.  The result are a bunch of trajectories, as shown below

All the trajectories tend towards one or other of the two pure Nash equilibria, and the mixed Nash equilibrium has a basin of attraction of size zero.  What is surprising though, is that despite the fact that one of the pure Nash equilibria is clearly worse for one of the parties (and no better for the other) the basin of attraction for each is almost the same size!

Right at the beginning I claimed that the existence of the $B$s demonstrated that males are in general monogamous in our species?  How is that so?

When we built the payoff table we implicitly assumed pair bonding.  In non-monogamous species like peacocks and peahens, the males make very little commitment when they mate.  Since they make little commitment the peacocks are not very choosy - in fact the only thing that determines whether a peacock mates with a peahen is whether or not she lets him.  Likewise, if we were non-monogamous a female would fairly easily find a mate whatever strategy she played.  The result is that the arrows in the evolutionary diagram above would all point downwards towards $¬B$, whether they were in the left or the right hand side of the phase space.  This would make $¬B$ the only Evolutionary Stable Strategy.

So, the fact that $B$ is, evidently, an ESS for humans, proves that human males must be monogamous, in general!

FOOTNOTES

• $\dagger$ : I'm trying to avoid Google's algorithms labelling this a dodgy site.