## Monty Hall variations

A few months ago a couple of articles bubbled up on Marilyn vos Savant and her famous analysis of the “Monty Hall Problem”. Never being one to shy away from a famous puzzle, I sat down and tried to intuit my way though it. When I failed to do that, I started talking to my friends about it. Since most of my friends are all either game designers or people who married/date/hang out with game designers we ended up creating a bunch of alternate scenarios. I don’t know how helpful they were but I did find them entertaining so I thought I would share them.

**Variation 0: “The Original Monty Hall Problem”**

You are going on a game show hosted by Monty Hall. The grand prize, a Cadillac, is behind one of three closed doors. The other two of the doors both have a goat behind it. (In this scenario you would really like a car and don’t like goats.)

First you are asked to pick one of the doors. Then Monty, who knows what is behind each of the doors, opens up one of the two doors you didn’t pick revealing a goat. Finally you are given the opportunity to either “stay” and get what ever was behind the door you initially chose, or “switch” and get whatever was behind the other unopened door. *What is the best strategy to win the car?*

So the intuitive answer is “it doesn’t matter you have a 50/50 shot” and that is wrong. The correct and counter intuitive answer that earned vos Savant decades of abuse is “always switch”, which yields a 2/3’s chance of getting the car.

The permutation space is small enough that we can actually just use counting to prove vos Savant correct. In the chart below I enumerated all the possibilities for car placement and initial player choice. I also showed what the result of either switching or staying would be in each case.

Car | Choice | Switch | Stay |
---|---|---|---|

A | A | Goat | Car |

A | B | Car | Goat |

A | C | Car | Goat |

B | A | Car | Goat |

B | B | Goat | Car |

B | C | Car | Goat |

C | A | Car | Goat |

C | B | Car | Goat |

C | C | Goat | Car |

If we total up the Switch column we see that there are 3 goats and 6 cars. The stay column, predictably, has the inverse.

After reading though this explanation I saw that vos Savant was correct. It still didn’t make any sense to me. So I started coming up with other similar situations to see if any of them might help my understanding.

**Variation 1: “The Marvin Hall Problem”**

In this variation the game show industry is not what it used to be. The studio has hit some hard times and the quality of employees have gone down hill. Monty quit a couple of years ago, so now his slacker step brother Marvin Hall is hosting. You show up for the game and one of the stage hands left the middle door open and you can see there is a goat behind it.

Marvin:“God dammit Steve, you had ONE F****** JOB! … [takes a deep breath and turns to you]… Just pick one of the other two and I can still catch happy hour”

You:“ummm door A”

Marvin:“Now do you want to keep it or switch?”

Aside from the smell of cigarettes coming form Marvin’s polyester suit, this situation seems pretty similar to the original. You are still staring at a goat and choosing between two closed doors, one with a goat and one with a car. What is the ideal strategy?

Car | Choice | Switch | Stay | |
---|---|---|---|---|

A | A | Goat | Car | |

A | C | Car | Goat | |

C | A | Car | Goat | |

C | C | Goat | Switch |

We can see it is a 50/50 shot either way.

**Variation 2: “The Jay & Kay Problem”**

In this variation you are playing the original version and after you made your initial selection, Monty reveals a goat behind the middle door. Then an alien come running out from stage left and moments later explodes into a blue gooey mess. Once the smoke clears Tommy Lee Jones and Will Smith stand up from where they had been sitting in the audience and explain all about aliens and stuff. Once the goo is cleaned up they get the audience’s attention and neuralize everyone.

At this point Monty turns to you:

Monty:“I don’t remember which one you picked”

You:“Neither do I”

Monty:“Well, which one would you like?”

At this point we are at the Marvin Hall Problem yielding a 50/50 shot. But, if Will Smith leans in to you on his way out and says “You picked door A” then you would be back to the original Monty Hall Problem and your best bet is to switch to door C.

**Variation 3: “The Howie D Problem”**

After the success of “Lets Make a Deal” and “The Newlyweds Game” network execs announce a new gameshow called “Trust Me!” hosted by Howie “Howie D.” Dorough of the Backstreet Boys. In this show a team of two tries to win a Cadillac. You and your college roommate get a chance to compete.

The show starts with you in the “Booth of Silence” where you can’t see or hear anything else that is going on. While you are in the booth your roommate is presented with three doors, one has a car and the other two have goats. Your roommate chooses one of the doors and Howie D. opens one of the other doors revealing a goat. Then your roommate is sent to his own Booth of Silence and you are brought to the stage.

At this point Howie D. gives you two options: either trust your roommate and go with his choice or to ignore your roommates choice and pick between one of the remaining choices, without knowing what your roommate chose. *Which option will maximize your chances of getting the car?*

First lets look at what your odds are if you choose to trust your roommate:

Car | Choice | Trust |
---|---|---|

A | A | Car |

A | B | Goat |

A | C | Goat |

B | A | Goat |

B | B | Car |

B | C | Goat |

C | A | Goat |

C | B | Goat |

C | C | Car |

We see that you end up with a one in three chance of winning.

Now lets look at what our odds are if you choose not to trust you roommate and select one of the two closed doors at random. For the sake of space lets say Monty revealed a goat behind second door:

Car | Choice | Prize |
---|---|---|

A | A | Car |

A | C | Goat |

C | A | Goat |

C | C | Car |

We can see that the coin toss yields a 50/50 shot at the car. Which means you should never trust your roommate.

**Variation 4: “The Double Cadillac Problem”**

In this variation there are 4 doors, two with Goats and two with Cadillacs. You pick one, and Monty reveals a goat in one of the other doors. You can then choose to stay with your original choice or to switch to one of the other closed doors. Staying has 1/2 chance of getting a car while switching has 2/3 chance. Always switch.

If however Monty revealed a Cadillac instead of a goat, that would give switching 1/3 chance. So if Monty shows you a car, stick with your initial choice.

**Variation 5: “The Chauffeur Problem”**

You don’t have a driver’s license, and you don’t own a car. You are presented with three sets, each with two doors and you get to choose one set and keep what’s behind both doors of the set.

- One set has a Cadillac behind each door. (This does you no good since you can’t drive)
- Behind another there are two drivers. (This does you no good since you don’t have a car)
- Behind the third there is a Cadillac and a driver. (This is perfect)

You pick Set A. Monty flips a coin and opens one of the two doors revealing a Cadillac. *What are the odds you are still taking the bus home after the show?*

The intuitive answer to this would be that since you know you didn’t get the set with both of the drivers, there is a 50/50 chance that you got the one you need. This is wrong.

First lets look at our three doors:

Set 1 | Set 2 | Set 3 |
---|---|---|

Car A | Valet A | Car C |

Car B | Valet B | Valet C |

Now we can see all of our options:

Revealed | Hidden |
---|---|

Car A | Car B |

Car B | Car A |

Valet A | Valet B |

Valet B | Valet A |

Car C | Valet C |

Valet C | Car C |

Since we saw that the revealed option was a car we can remove all the other options, which leaves us with the following:

Revealed | Hidden | Getting Home |
---|---|---|

Car A | Car B | Bus |

Car B | Car A | Bus |

Car C | Valet C | Driving |

We can see that there is a 2/3 chance that you will be taking the bus home. This is actually a re-skin of another famous puzzle called the Bertrand’s Box paradox.