## Section2.5Equilibrium Points

In this section, we will try to gain a greater understanding of equilibrium strategies in a game. In general, we call the pair of equilibrium strategies an equilibrium pair, while we call the specific payoff vector associated with an equilibrium pair an equilibrium point.

Determine the equilibrium point(s) for the following games.

1. $\left[\begin{matrix}(2, -2)\amp (-1, 1)\\ (2, -2)\amp (-1, 1) \end{matrix} \right]$

2. $\left[\begin{matrix}(0, 0)\amp (-1, 1)\amp (0, 0)\\ (-1, 1)\amp (0, 0)\amp (-1, 1)\\ (0, 0)\amp (1, -1)\amp (0, 0) \end{matrix} \right]$

What do you notice about the values of the equilibrium points of the games in Exercise 2.5.1?

The big question we want to answer is “Can two equilibrium points for a two-player zero-sum game have different values?” By experimenting with some examples, try to create an example of a game with two equilibrium points where those points have different values for one of the players. If you can successfully create such an example, you will have answered the question. But just because you can't find an example, that doesn't mean one doesn't exist!

After trying several examples, you might be beginning to believe that the answer to the above question is “no.” Now you are ready to try to prove the following theorem:

##### Solution Theorem for Zero-Sum Games.

Every equilibrium point of a two-person zero-sum game has the same value.

Let's start with the $2 \times 2$ case. We will use a proof by contradiction. We will assume the theorem is false and show that we get a logical contradiction. Once we reach a logical contradiction (a statement that is both true and false), we can conclude we were wrong to assume the theorem was false; hence, the theorem must be true. Make sure you are comfortable with the logic of this before moving on.

Since we want to assume the theorem is false, we assume we have a two-player zero-sum game with two different equilibrium values. Since we don't have a specific example of such a game, we want to represent the game in a general form. In particular, we can represent the general game

\begin{equation*} \left[\begin{matrix}(a, -a)\amp (c, -c)\\ (d, -d)\amp (b, -b) \end{matrix} \right]. \end{equation*}

Note that if $a$ is negative, then $-a$ is positive; thus, every possible set of values is represented by this matrix. We want to look at the possible cases for equilibria.

Explain what goes wrong if $(a, -a)$ and $(d, -d)$ are equilibria with $a \neq d\text{.}$

Hint

Think about the different cases, such as $a\lt d\text{,}$ $a>d\text{.}$

Generalize you answer to Exercise 2.5.3 to explain what goes wrong if the two equilibria are in the same column. Similarly, explain what happens if the two equilibria are in the same row.

Does the same explanation hold if the two equilibria are diagonal from each other? (Explain your answer!)

From your last answer, you should see that we need to do more work to figure out what happens if the equilibria are diagonal. So let's assume that the two equilibria are $(a, -a)$ and $(b, -b)$ with $a \neq b\text{.}$ It might be helpful to draw the payoff matrix and circle the equilibria.

Construct a system of inequalities using the fact that a player prefers an equilibrium point to another choice. For example, Player 1 prefers $a$ to $d\text{.}$ Thus, $a > d\text{.}$ List all four inequalities you can get using this fact. You should get two for each player– remember that Player 1 can only compare values in the same column since he has no ability to switch columns. If necessary, convert all inequalities to ones without negatives. (Algebra review: $-5 \lt -2$ means $5 > 2\text{!}$)

Now string your inequalities together. For example, if $a \lt b$ and $b \lt c$ then we can write $a \lt b \lt c\text{.}$ (Be careful, the inequalities must face the same way; we cannot write $a> b \lt c\text{!}$)

Explain why you now have a contradiction (a statement that must be false). We can now conclude that our assumption that $a \neq b$ was wrong.

Explain why you can conclude that all equilibria in a $2 \times 2$ two-player zero-sum game have the same value.

We just worked through the proof, step-by-step, but now you need to put all the ideas together for yourself.

Write up the complete proof for the $2 \times 2$ case in your own words.

Can you see how you might generalize to a larger game matrix? You do not need to write up a proof of the general case, just explain how the key ideas from the $2 \times 2$ case would apply to a bigger game matrix.

Hint

Think about equilibria in (a) the same row, (b) in the same column, or (c) in a different row and column.

We've seen that any two equlibrium points must have the same value. However, it is important to note that just because an outcome has the same value as an equilibrium point, that does not mean it is also an equilibrium point.

Give a specific example of a game matrix with two outcomes that are $(0, 0)\text{,}$ where one is an equilibrium point and the other is not.

Working through the steps of a mathematical proof can be challenging. As you think about what we did in this section, first make sure you understand the argument for each step. Then work on understanding how the steps fit together to create the larger argument.

The next section summarizes all our work with finding equilibrium points for zero-sum games.