# (DT) Segment 3: Monty Hall

## To Calculate

We can approach this problem by enumerating all possible scenarios which can take place in the game. These are shown in the three tables below. The entries of the table indicate what is behind each Door, the contestant's initial choice, the Door opened by Monty, and the final outcome being a success or a failure based upon the contestant's decision to switch or stay.

Case 1: Car is behind Door 1.

Door 1 | Door 2 | Door 3 | Initial pick | Door opened | Switch | Success |
---|---|---|---|---|---|---|

Car | Goat | Goat | Door 1 | Door 2 | Yes | No |

Car | Goat | Goat | Door 1 | Door 2 | No | Yes |

Car | Goat | Goat | Door 1 | Door 3 | Yes | No |

Car | Goat | Goat | Door 1 | Door 3 | No | Yes |

Car | Goat | Goat | Door 2 | Door 3 | Yes | Yes |

Car | Goat | Goat | Door 2 | Door 3 | No | No |

Car | Goat | Goat | Door 3 | Door 2 | Yes | Yes |

Car | Goat | Goat | Door 3 | Door 2 | No | No |

Case 2: Car is behind Door 2.

Door 1 | Door 2 | Door 3 | Initial pick | Door opened | Switch | Success |
---|---|---|---|---|---|---|

Goat | Car | Goat | Door 1 | Door 3 | Yes | Yes |

Goat | Car | Goat | Door 1 | Door 3 | No | No |

Goat | Car | Goat | Door 2 | Door 1 | Yes | No |

Goat | Car | Goat | Door 2 | Door 1 | No | Yes |

Goat | Car | Goat | Door 2 | Door 3 | Yes | No |

Goat | Car | Goat | Door 2 | Door 3 | No | Yes |

Goat | Car | Goat | Door 3 | Door 1 | Yes | Yes |

Goat | Car | Goat | Door 3 | Door 1 | No | No |

Case 3: Car is behind Door 3.

Door 1 | Door 2 | Door 3 | Initial pick | Door opened | Switch | Success |
---|---|---|---|---|---|---|

Goat | Goat | Car | Door 1 | Door 2 | Yes | Yes |

Goat | Goat | Car | Door 1 | Door 2 | No | No |

Goat | Goat | Car | Door 2 | Door 1 | Yes | Yes |

Goat | Goat | Car | Door 2 | Door 1 | No | No |

Goat | Goat | Car | Door 3 | Door 1 | Yes | No |

Goat | Goat | Car | Door 3 | Door 1 | No | Yes |

Goat | Goat | Car | Door 3 | Door 2 | Yes | No |

Goat | Goat | Car | Door 3 | Door 2 | No | Yes |

Now, we are interested in the probability of winning a car () given that the contestant switches every time (). Denote by the car being behind the Door, by Monty opening the Door, and by the contestant's initial pick being the Door.

where

So, basically, in the numerator we have all the cases where the contestant picks a Door, and after Monty opens a Door, the contestant decides to switch and actually wins the car. Similarly, in the denominator we have, in addition to the cases where the contestant wins the car, the cases where the contestant initially picks the Door with the car behind it, and as a result loses by switching.

It is quite clear from the tables given above that, assuming that Monty doesn't care about which door he chooses to open whenever he has more than 1 door at his disposal,

and that,

so we will calculate only for one of the cases. Moreover, we also have, assuming that the car is behind Door 1,

, and similarly for cases with the Car behind other doors, thereby simplifying our calculations.

For the other terms in the denominator, we have,

Then, assimilating the above results we have,

Thus, the probability of winning a car given that the contestant decides to switch is .