In this section, we will talk about the **algebraic method** used to solve **mixed strategy games**. Here we have provided formulas and examples of algebraic method.

Consider the zero sum two person game given below:

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

Player A | I | II | |

I | a | b | |

II | c | d |

The solution of the game is:

A play’s (p, 1 - p)

where:

p = | d - c -------------------- (a + d) - (b + c) |

B play’s (q, 1 - q)

where:

q = | d - b ------------------- (a + d) - (b + c) |

Value of the game, V = | ad - bc -------------------- (a + d) - (b + c) |

Consider the game of matching coins. Two players, A & B, put down a coin. If coins match (i.e., both are heads or both are tails) A gets rewarded, otherwise B. However, matching on heads gives a double premium. Obtain the best strategies for both players and the value of the game.

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

Player A | I | II | |

I | 2 | -1 | |

II | -1 | 1 |

Solution.

This game has no **saddle point**.

p = | 1 - (-1) ----------------------- (2 + 1) - (-1 - 1) |
= | 2 ---- 5 |

1 – p = 3/5

q = | 1 - (-1) ----------------------- (2 + 1) - (-1 - 1) |
= | 2 ---- 5 |

1 – q = 3/5

V = | 2 X 1 - (-1) X (-1) -------------------------- (2 + 1) - (-1 - 1) |
= | 1 ---- 5 |

Example 2: Algebraic Method in Game Theory

Solve the game whose payoff matrix is given below:

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

Player A | I | II | |

I | 1 | 7 | |

II | 6 | 2 |

Solution.

This game has no **saddle point**.

p = | 2 - 6 ----------------------- (1 + 2) - (7 + 6) |
= | 2 ---- 5 |

1 – p = 3/5

q = | 2 - 7 ----------------------- (1 + 2) - (7 + 6) |
= | 1 ---- 2 |

1 – q = 1/2

V = | 1 X 2 - (7 X 6) -------------------------- (1 + 2) - (7 + 6) |
= | 4 |