Games where one player has only two courses of action while the other
has more than two, are called **2 X n or n X 2 games**.

If these games do not have a saddle point or are reducible by the dominance method, then before solving these games we write all 2 X 2 sub-games and determine the value of each 2 X 2 sub-game.

This method is illustrated by the following example.

Determine the solution of game for the pay-off matrix given below:

Player B | ||||
---|---|---|---|---|

Player A | I | II | III | |

I | -3 | -1 | 7 | |

II | 4 | 1 | -2 |

Solution.

Obviously, there is no saddle point and also no course of action dominates the other. Therefore, we consider each 2 X 2 sub-game and obtain their values.

(a)

Player B | |||
---|---|---|---|

Player A | I | II | |

I | -3 | -1 | |

II | 4 | 1 |

The saddle point is 1. So the value of game, V1 is 1.

(b)

Player B | |||
---|---|---|---|

Player A | I | II | |

I | -3 | 7 | |

II | 4 | -2 |

This game has no saddle point, so we use the algebraic method.

Value of game, V2 = | (-3) X (-2) - (7 X 4) ------------------------- (-3 - 2) - (7 + 4) |
= | 11 --- 8 |

(c)

Player B | |||
---|---|---|---|

Player A | II | III | |

I | -1 | 7 | |

II | 1 | -2 |

This game has no saddle point, so we use the algebraic method.

Value of game, V3 = | (-1) X (-2) - (7 X 1) ----------------------- (-1 - 2) - (7 + 1) |
= | 5 --- 11 |

The 2 X 2 sub-game with the lowest value is (c) and hence the solution to this game provides the solution to the larger game.

Using **algebraic method**:

A plays ( 3/11, 8/11)

B plays (0, 9/11, 2/11)

Value of game is 5/11.