The web browser you are using does not have features required by the tutorials and game simulators.
Please use the most recent version of one of the following:
This tutorial shows how to find stable equilibria in asymmetric games. It assumes that you have already completed the Stable Strategies tutorial for symmetric games and have a basic understanding of asymmetric games, from starting either the Conflict II or Parental Care tutorial. If you work through all the example problems in detail, this tutorial should take about 30 minutes.
In a symmetric game, you can calculate the evolutionarily stable strategy (ESS). In an asymmetric game, since there are two roles with different strategy sets, stability consists of a pair of strategies, one for each role. A stable state in an asymmetric game is called a Nash equilibrium, and its calculation is a bit different from that for an ESS. Mathematically, the ESS is actually a special case of the Nash equlibrium. Since the two concepts are related, let’s start with a recap of the ESS.
An evolutionarily stable strategy is a strategy that cannot be invaded by another strategy.
That is, if the entire population plays the ESS strategy, a mutation that made some members play another strategy would be eliminated. The idea behind the Nash equilibrium is the similar, although it is stated in terms of players switching strategies, rather than invasion of a population by a rare mutant1. For stability in evolutionary games, we want to find strict Nash equilibria, defined as follows:
A pair of strategies is a strict Nash equilibrium if neither player can unilaterally switch to another strategy without reducing its payoff.
Here is a simple example (remember, the payoff to the left of the comma is for the strategy to the left against the strategy above; the payoff to the right of the comma is for the strategy above against the strategy to the left):| Male | ||||
|---|---|---|---|---|
| A | B | C | ||
| Female | X | 2 , 3 | 1 , 2 | 1 , 1 | Y | 1 , 1 | 2 , 1 | 3 , 2 |
| Z | 1 , 2 | 2 , 2 | 2 , 1 | |
Male A Female X (the upper-left cell) is a strict Nash equilibrium. Looking at the male’s payoffs (to the right of the comma) in the X row, we see that his payoff would drop from 3 to 2 or 1 if he switched from A to B or C. Looking at the female’s payoffs (to the left of the comma) in the A column, we see that her payoff would drop from 2 to 1 if she switched from X to Y or Z. Neither player can unilaterally switch to another strategy without reducing its payoff.
20
You can quickly pick out strict Nash equilibria by eye: (1) Find the single greatest payoff to the left of the comma in each column and circle it. If there is no single highest payoff (as in column B below), then there are no Nash equilibria in that column. (2) For each cell with a circled payoff, look across its row; if the payoff to the right of the comma in this cell is the single highest payoff in its row, then this cell is a strict Nash equilibrium.
| Male | ||||
|---|---|---|---|---|
| A | B | C | ||
| Female | X | 2 , 3 | 1 , 2 | 1 , 1 | Y | 1 , 1 | 2 , 1 | 3 , 2 |
| Z | 1 , 2 | 2 , 2 | 2 , 1 | |
| Male | ||||
|---|---|---|---|---|
| A | B | C | ||
| Female | X | 2 , 3 | 1 , 2 | 1 , 1 | Y | 1 , 1 | 2 , 1 | 3 , 2 |
| Z | 1 , 2 | 2 , 2 | 2 , 1 | |
For practice, find the strict Nash equilibria in each of the following payoff matrices. Check the box in any cell that is a strict Nash equilibrium and then click the ✓ button to check your answer.
40
Some of the practice games had two or more strict Nash equilibria. What does it mean to have two equilibria? Explore the original example, in which A-X and C-Y are equilibria, in the simulator below. Change the initial proportions of the six strategies and see how it affects the outcome.
50
This game has complex behaviors. It’s no surprise that the outcome depends on the initial proportions of strategies involved in the strict Nash equilibria (X and A, C and Y). However, even changing the intial proportions of B and Z can flip the outcome from one equilibrium to the other.
Games modeling real behaviors seldom have simple numeric payoffs. How can we find Nash equilibria in terms of variables? It’s not clear when looking at a matrix like this one whether a payoff would increase or decrease when strategies are switched.
| Male | |||
|---|---|---|---|
| A | B | ||
| Female | X | r , s/2 | s/2 , r |
| Y | r/2 , s | s , r | |
We must look at each cell in the matrix and ask under what conditions it would be a strict Nash equilibrium. Going back to the definition, a strategy pair is a strict Nash equilibrium if neither player can switch to another strategy without reducing its payoff. We assume (as in most games) that all variables are greater than 0.
We’ll start by looking at A-X (upper left cell). For this to be a strict Nash equilibrium, the female’s payoff for X must be greater than her payoff for Y when the male plays A. That is, r > r/2. That condition is clearly met when r > 0. In addition, the male’s payoff for A must be greater than his payoff for b when the female plays X. That is, s/2 > r. Whether this condition is met depends on the values of s and r. So we conclude that A-X is a strict Nash equilibrium when s/2 > r.
Now consider B-X. The female’s payoff for X must be greater than her payoff for Y when the male plays A, i.e. s/2 > s, which is never true, so B-X cannot be an equilibrium. Similarly with A-Y, the female’s payoff for Y must be greater than her payoff for X when the male plays A, i.e. r/2 > r, which is never true, so A-Y cannot be an equilibrium.
Finally, there’s B-Y. The female’s payoff for Y is greater than for X when the male plays B, since s > s/2. In addition, the male’s payoff for B must be greater than his payoff for A when the female plays Y. That is, r > s. Whether this condition is met depends on the values of s and r. B-Y is a strict Nash equilibrium when r > s.
This game has two potential strict Nash equilibria, A-X when s/2 > r and B-Y when r > s.
Considering every cell in a large matrix may seem daunting. In a 3×3 matrix, there are 9 pairs of strategies and two inequalities per strategy. A shortcut can make this more manageable. First, scan each column and circle every payoff to the left of the comma that could possibly be the greatest in that column. For example, column A below contains the payoffs s, s/2, and r; you’d circle s and r because either one could be the greatest (depending on variable values) but s/2 can never be greater than s. Next, scan each row and circle every payoff to the right of the comma that could possibly be greatest in that row. Only those cells in which both payoffs are circled could be strict Nash equilibria.
| Male | ||||
|---|---|---|---|---|
| A | B | C | ||
| Female | X | s , r/2 | s/2 , 0 | s/4 , r/4 |
| Y | s/2 , r | s−t , s/2 | s/2−t/2 , s | |
| Z | r , r/2 | s−t/2 , r | s/2 , r/4 | |
In the example above, our shortcut identified two possible equilibria, A-X and B-Z. What conditions would make each a strict Nash equilibrium? For the female in A-X, s must be greater than s/2 and r, while for the male, r/2 must be greater than 0 and r/4. Thus A-X is a strict Nash equilibrium when s > r. For the female in B-Z, s−t/2 must be greater than s−t and s/2, while for the male, r must be greater than r/2 and r/4. After algebraic simplification, this comes to s > t2.
Use the simulator below to test our conclusion about this game. Can you verify that A-X is a strict Nash equilibrium when s > r? That B-Z is a strict Nash equilibrium when s > t? What happens when neither of these sets of conditions is met (e.g. s < r and s < t)? What happens when both sets of conditions are met (s > r and s > t)? You may need to vary the initial proportions to see the full impact of changing variables.
Try the two examples below. Find any strict Nash equilibria in each matrix, enter them in the text area, and then click the ✓ button to check your answer.
Real evolutionary games are usually simpler than the hypothetical examples above. Their matrices are often sparse, with payoffs that are null or otherwise clearly smaller than the alternatives. In many cases, rather than being determined by a complex inequality, a strategy pair is clearly never or always a strict Nash equilibrium.
90
Just as not every symmetric game has a pure ESS, not every asymmetric game has a strict Nash equilibrium. Here’s a simple example:
| Male | |||
|---|---|---|---|
| A | B | ||
| Female | X | 1 , 0 | 0 , 1 |
| Y | 0 , 1 | 1 , 0 | |
Try it out in the simulator. How does changing the payoffs or initial proportions affect the game dynamics? Can you make the game stable by changing only the initial proportions?
Unstable asymmetric games are fairly common in nature, including oscillations like these. For example, populations of predators and prey often rise and fall in alternation like this. A good year for prey causes more predators to survive, thus reducing the prey population. Lack of food causes the predator population to crash, which makes life easier for prey, and so on. The same can be true of parasites and hosts.
What about mixed strategies? In symmetric games, an equilibrium may be either a single strategy or a mix of two strategies (as in the Hawk-Dove game, where playing Hawk with a probability of v/c is the ESS when v < c). It turns out that no mixed strategies can be strict Nash equilibria3. When you see an apparently stable mix of strategies in an asymmetric game in the simulator, those are actually simple, rather than strict, Nash equilibria4. They are unstable and can be upset by small changes in initial proportions or payoff values.