Bimatrix game

$𝐀=\left[\begin{array}{ccc}c& d& e\\ f& g& h\end{array}\right]$	$𝐁=\left[\begin{array}{ccc}p& q& r\\ s& t& u\end{array}\right]$
$\begin{array}{cccc}& X& Y& Z\\ V& c,p& d,q& e,r\\ W& f,s& g,t& h,u\end{array}$ A payoff matrix converted from A and B where player 1 has two possible actions V and W and player 2 has actions X, Y and Z

In game theory, a bimatrix game is a simultaneous game for two players in which each player has a finite number of possible actions. The name comes from the fact that the normal form of such a game can be described by two matrices - matrix $A$ describing the payoffs of player 1 and matrix $B$ describing the payoffs of player 2.

Player 1 is often called the "row player" and player 2 the "column player". If player 1 has $m$ possible actions and player 2 has $n$ possible actions, then each of the two matrices has $m$ rows by $n$ columns. When the row player selects the $i$-th action and the column player selects the $j$-th action, the payoff to the row player is $A[i,j]$ and the payoff to the column player is $B[i,j]$.

The players can also play mixed strategies. A mixed strategy for the row player is a non-negative vector $x$ of length $m$ such that: ${\sum }_{i=1}^m{x}_i=1$. Similarly, a mixed strategy for the column player is a non-negative vector $y$ of length $n$ such that: ${\sum }_{j=1}^n{y}_j=1$. When the players play mixed strategies with vectors $x$ and $y$, the expected payoff of the row player is: $x^{𝖳}Ay$ and of the column player: $x^{𝖳}By$.

Nash equilibrium in bimatrix games

Every bimatrix game has a Nash equilibrium in (possibly) mixed strategies. Finding such a Nash equilibrium is a special case of the Linear complementarity problem and can be done in finite time by the Lemke–Howson algorithm.

There is a reduction from the problem of finding a Nash equilibrium in a bimatrix game to the problem of finding a competitive equilibrium in an economy with Leontief utilities.

Related terms

A zero-sum game is a special case of a bimatrix game in which $A+B=0$.