Bimatrix game

A = [ c d e f g h ] {\displaystyle \mathbf {A} ={\begin{bmatrix}c&d&e\\f&g&h\\\end{bmatrix}}} B = [ p q r s t u ] {\displaystyle \mathbf {B} ={\begin{bmatrix}p&q&r\\s&t&u\\\end{bmatrix}}}
X Y Z V c , p d , q e , r W f , s g , t h , u {\displaystyle {\begin{array}{c|c|c|c}&X&Y&Z\\\hline 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 {\displaystyle A} describing the payoffs of player 1 and matrix B {\displaystyle B} describing the payoffs of player 2.[1]

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

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

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.[2]

Related terms

A zero-sum game is a special case of a bimatrix game in which A + B = 0 {\displaystyle A+B=0} .

References

  1. ^ a b Chandrasekaran, R. "Bimatrix games" (PDF). Retrieved 17 December 2015.
  2. ^ Codenotti, Bruno; Saberi, Amin; Varadarajan, Kasturi; Ye, Yinyu (2006). "Leontief economies encode nonzero sum two-player games". Proceedings of the seventeenth annual ACM-SIAM symposium on Discrete algorithm - SODA '06. p. 659. doi:10.1145/1109557.1109629. ISBN 0898716055.