What is the (finite) maximum number of Nash equilibria in a 2x2 game?

Question

I think the answer should be 3: 2 pure and 1 mixed. But that is just because I fail to construct more for at least 3 hours. I don't know how to prove this, there are just too many variables. Am I right? If yes, how would I go about proving it?

Answer 1

Consider the $2 \times 2$ game \begin{array}{c|cc|} & L & R \\ \hline T & a_1,b_1 & a_2,b_3 \\ B & a_3,b_2 & a_4,b_4 \\ \hline \end{array}

Assume no ties: $a_1 \ne a_3$ and $a_2 \ne a_4$. (For simplicity, I only discuss the Row player. Replace $a$ with $b$ to cover the Column player.)

There are four possible cases:

1.1) if $a_1>a_3$ and $a_2>a_4$, then the strategy $T$ is strictly dominant and thus is the only best reply for Row;

1.2) if $a_1 < a_3$ and $a_2<a_4$, then the strategy $B$ is strictly dominant and thus is the only best reply for Row;

1.3) if $a_1>a_3$ and $a_2<a_4$, then both T and B may be best replies for Row;

1.4) if $a_1<a_3$ and $a_2>a_4$, then both T and B may be best replies for Row.

Make the analogous list for Column (replace $a$ with $b$ and number items 2.x instead of 1.x), assuming she has no ties either. Combining the two lists, you get a total of 16 cases that can be grouped in three categories:

a) a unique equilibrium in pure strategies; f.i., 1.1 and 2.1 yield (T,L) as the unique equilibrium in pure strategies;

b) a unique equilibrium in mixed strategies; f.i., 1.3 and 2.4 yield (aunique equilibrium in mixed strategies;

c) two equilibria in pure strategies and one in mixed strategies; f.i., 1.3 and 2.3 yield (T,L) and (B,R) as equilibria in pure strategies and there is also an equilibrium in mixed strategies.

The above may be summarised as follows: generically (=assuming no ties), the maximum number of Nash equilibria in a $2 \times 2$ game is three.

If you allow for ties (as per your comment), the picture gets messy. You can have games with only two equilibria in pure strategies, such as: \begin{array}{c|cc|} & L & R \\ \hline T & 1,1 & 0,0 \\ B & 0,0 & 0,0 \\ \hline \end{array} where only $(T,L)$ and $(B,R)$ are equilibria.

More importantly for your question, you can have games with an infinite number equilibria such as \begin{array}{c|cc|} & L & R \\ \hline T & 0,1 & 0,1 \\ B & 0,1 & 0,1 \\ \hline \end{array} where any strategic profile (pure or mixed) is a Nash equilibrium. If you feel "cheated" by this example because every player is indifferent over every outcome, you might appreciate how particular (=non-generic) is the assumption that ties are possible.