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Solve the following game:

written 5.3 years ago by | modified 10 months ago by |

I | II | III | IV | |
---|---|---|---|---|

I | 6 | 2 | 4 | 8 |

II | 2 | -1 | 1 | 12 |

III | 2 | 3 | 3 | 9 |

IV | 5 | 2 | 6 | 10 |

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Solve the following game:

written 5.3 years ago by | modified 10 months ago by |

I | II | III | IV | |
---|---|---|---|---|

I | 6 | 2 | 4 | 8 |

II | 2 | -1 | 1 | 12 |

III | 2 | 3 | 3 | 9 |

IV | 5 | 2 | 6 | 10 |

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written 5.3 years ago by |

By rules of dominance:

- Column II dominates column III. The matrix reduces to:

I | II | IV | |
---|---|---|---|

I | 6 | 2 | 8 |

II | 2 | -1 | 12 |

III | 2 | 3 | 9 |

IV | 5 | 2 | 10 |

- Column II dominates column IV. The matrix reduces to:

I | II | |
---|---|---|

I | 6 | 2 |

II | 2 | -1 |

III | 2 | 3 |

IV | 5 | 2 |

- Row III dominates row II. Matrix reduces to:

I | II | |
---|---|---|

I | 6 | 2 |

III | 2 | 3 |

IV | 5 | 2 |

- Row I dominates row IV. Matrix reduces to:

I | II | |
---|---|---|

I | 6 | 2 |

III | 2 | 3 |

By method of oddments:

$ \text{Player 1’s strategy} = [\dfrac15 , 0, \dfrac45 , 0] \\ \text{Player 2’s strategy} = [\dfrac15 , \dfrac45 , 0, 0] \\ \text{Value of the game} = 6× \dfrac15 + 2× \dfrac45 = \dfrac{14}{5}$

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