The following material is flaggedSuppose that I offer you two possible bargains: either I will give you $1, or I will flip a coin and give you $1 if it lands heads, nothing otherwise. It would make sense for you to pick the first offer, right?Green Level. It is intended to reflect material that the author believes to be a matter of consensus among experts in the field. This belief may be incorrect, however; and as the author is not an expert and does not have an expert fact-checking the article, errors may creep in.

Right. The first offer a) is never worse than the second offer, and b) is sometimes better than the second offer. Similar situations can happen in game theory. Let us look at the following:

A | B | |

A | $1 | $2 |

B | $-2 | $0 |

That's right. Since I know you will pick A, I will minimize my losses and pick A as well. If you were to change your choice in response to my choosing A, I would only be pleasantly surprised, and the same would still be true if I were to pick B. Thus, you are always better off picking A.

And this leads to the main subject of this post: dominating strategies. A

*strategy*(the game-theory term for a particular move or action) is considered

*dominating*over another one if it is a) never worse for the player choosing it, and b) better at least once for the player choosing it.

So, here's where the core principle of game theory gets into it. Because your opponent is acting intelligently, (or at least non-randomly), your opponent is allowed to choose strategies. Barring unusual circumstances (such as the iterated prisoners' dilemma) that will be mentioned later, a

*rational opponent*(that is, one acting in its own best interest) will never choose a dominated strategy. In a zero-sum game, it is generally best to assume that your opponent will never choose a strategy if one that always hurts you at least as much is available.

So, given the following game, what do you do?

A | B | C | |

A | $0 | $0 | -$1 |

B | $1 | $1 | $1 |

C | $1 | $0 | $2 |

B | C | ||

A | $0 | -$1 | |

B | $1 | $1 | |

C | $0 | $2 |

B | C | ||

B | $1 | $1 | |

C | $0 | $2 |

B | |||

B | $1 | ||

C | $0 |

B | |||

B | $1 | ||

And so we have the final outcome: both players will play B, resulting in you winning $1. Thus, for you, the

*value*of this game is $1, and the

*value*for your opponent is -$1.

<<Notation and Basic Concepts | Game Theory | Utility Functions>>

## No comments:

## Post a Comment

pleasebecivil

civilitymakesthingseasier

pleesebecivil

civiltymakesthingseasier

pleasebecivilized

civilitymakesthingseesier