Review of Rules for Finding an ESS

Using The Payoff Matrix to Predict a Pure ESS
in Two Strategy Games

In two strategy games it is a relatively simple matter to determine if one of the strategies is a pure ESS, provided certain very reasonable assumptions are met. In this section, we will review the procedure for making this determination and the logic behind this procedure.

Recall that a pure ESS is a strategy that is unbeatable by other known strategies. This means that:

a pure ESS is immune to invasion by any other known strategy
and, a strategy that is a pure ESS is also capable of invading and displacing other known strategies.

If we can show that either of the statements above is true, then we have shown that the strategy is a pure ESS (either one is fine, they are essentially equivalent as far as the mathematics of the game are concerned).

We can use the payoff matrix and a simplifying assumption to make the determination. Let's get away from the caller / satellite model and instead define two abstract strategies: A and B (the purpose of this switch is simply to get you more familiar with manipulating and using the payoff matrix).

As always, the matrix lists the relative payoffs to each strategy for each type of encounter. In this example we will assign a value to each payoff. Thus:

Opponent's Strategy

Focal Strategy
A

B

A
E(A,A) = 0 E(A,B) = 1

B
E(B,A) = - 0.5 E(B,B) = 0.5

Let's assume that:

The population initially consists entirely of individuals who use strategy, A.
A very small number of strategy B invaders appear. In the simplest case, this could be a single individual invading a large group of A strategists. This invasion could be the result of mutation, learning, or immigration.
Meetings that lead to contests between different individuals occur at random. Thus, there is no tendency for individuals to collect by strategy in certain places and increase certain types of interactions over what would happen by chance encounter.

What types of interactions occur and how frequent are they?

The most common contests will involve A strategists. Why is this the case? The answer is that nearly everyone is an A strategist and meetings and conflicts with an alternative strategy are directly related to the freq. of that strategy. Thus:

A vs. A conflicts will be the most common for A strategists and will therefore largely determine the fitness of A strategists. From the point of view of A, these occur at the frequency of A.
B vs. A conflicts are the most common for B strategists and therefore will largely determine the fitness of B strategists. In the most extreme case where there is a single B invader, B vs. A will be the only type of encounter that matters with respect to its fitness.

! Any B vs. A conflict can also be viewed as an A vs. B conflict! Put another way, such a conflict involves payoffs to both strategies (E(B,A) to strat. B and E(A,B) to strat. A). Notice however, that from the point of view of A, interactions with B are extremely rare as compared to those with A. Thus, we will assume that we can ignore the fitness contribution of A vs. B interactions to the overall fitness of strat. A. You will have a chance to look at this assumption in more detail further down on this page.

! A Note of Warning

Remember that we are attempting to calculate strategy fitnesses. Thus, we are interested in the frequency of certain types of interactions from the point of view of the strategy. Since we are considering pairwise contests, the frequencies of each contest from the point of view of one contestant (strategist) will be equal to the frequency of the opponent in the contest.

Sometimes students who are familiar with basic probability and population biology assume that the frequency of a particular payoff equals a term in a binomial expansion of the strategy frequencies. For example, if a = freq Strat A and b = freq Strat B, then (a + b)^2 is expanded to predict that:

AA conflicts would occur at the frequency a * a,
A vs. B conflicts at 2 * a * b and
B vs. B conflicts at b * b.

This sort of formulation is true if one wants to estimate the rate of occurrence of these interactions in the whole population. But it is not correct when we are only interested in the frequency of interactions from the point of view of a particular strategy!

Now, if we consult the payoff matrix, we can see how this invasion turns out. In our example:

The payoff to strategy A when it is an A vs. A contest is 0. (Our convention is that this means that the interaction has no fitness consequences -- it neither increases nor decrease reproduction).
What about strategy B? All (if a single invader) of the contests it faces are B vs. A. From the matrix, we see that the payoff to strategy B in contests vs. strat. A is -0.5. That is, B loses some fitness as a result of this interaction.
It should now be rather obvious that B cannot invade A.

From the situation we just considered,we can construct a general rule to determine whether or not a two strategy game contains a pure ESS:

IF: E(A,A)> E(B,A) (the most common encounter for each strategy)

THEN: A IS STABLE vs. B (it is a pure ESS vs. B)

You may be wondering what would happen if the fitness consequences of the most common types of interactions are equal, i.e.,

E(A,A) = E(B, A)

Does that mean that neither is stable against the other? Not necessarily. In this one case, there is an additional test that must be performed before concluding whether or not there is a pure ESS.

If there is more than one B invader, there also may be some rare interactions with payoff E(B,B). Also, in this particular situation, the payoff E(A,B) starts to matter, even though it is still extremely rare.

Why now but not before? In the previous example, the A vs. B interaction was very rare in comparison with the common A vs. A contests. Thus, any effects on the overall fitness of A due to interactions with B were so small as to probably not matter. However in the case we are now considering, E(A,A) = E(B,A). Thus, the common A vs. A conflict confers no relative advantage or disadvantage. (The same logic applies to the most common contest B experiences, B vs. A). So the remaining interactions will decide whether or not there is a pure ESS.

Thus, if E(A,B) > E(B,B) A must still have an advantage over B and therefore it will be stable!

To review this, consider the following scenario. A population of A strategists is invaded by a small number of B strategists. In the most common types of contests for each strategy the payoffs E(A,A) and E(B, A) are equal. Thus, neither strategy is competitively aided or hindered by these contests. However, in the rare contests, A is doing better than B since E(A,B) > E(B,B) and so A will eventually out-compete B.

Summary of rules for Finding a Pure ESS
in a Two Strategy Game:

Assumption -- one strategy is very rare compared to the other. In this example, let A be the common strategy and we will determine whether or not it is immune from invasion by B.

Rule #1: E(A,A)> E(B,A)

or if:

Rule #2: E(A,A) = E(B, A) and E(A,B) > E(B,B)

There is a method that uses rule #1 to find an ESS by inspection of a payoff matrix. Click here to see how to use this method.

!! You may be uneasy about rule #1. Mathematically you can vaguely imagine cases where E(A,A) is greater than E(B,A) yet A is not stable against B! These situations require more than one B strategy invader so that all the payoffs might matter. Since more than one invader is not an unreasonable scenario, you become suspicious that game theorists are either intellectually shallow or are trying to sweep things under a rug.

OK, if you are that interested in the potential limitations of rule# 1, then PRESS HERE for a more complete discussion including a general discussion of appropriate simplification in modeling.

	Opponent's Strategy
*Focal Strategy*	A	B
A	E(A,A) = 0	E(A,B) = 1
B	E(B,A) = - 0.5	E(B,B) = 0.5