I want to illustrate this problem using a causal diagram.

Here is the rule of the game:

Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, open another door, say #3, which has a goat. He says to you, ‘Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?

There are basically two groups of answers to this question:

- It doesn’t matter. There are only two door left in the game, the probability that the car is behind the door is 1/2.
- Your should switch. The probabiliy of winning is two times more than not switching

The correct answer is the second one. The intuition is that since the host knows where is the car and he cannot open the door with a car behind it, it increases the probability that there is a car behind door 2. There is extra information when the host opens a door with a goat. If on the other hand, the host just randomly opens any remaining door, the probability of winning remains the same no matter you switch or not.

So now you have the intuition. We need to show it in a more systematic way.

# You Should Swtich

Door 1 (Your first choice) | Door 2 | Door 3 | Host Open | Outcome if you switch | Outcome if you stay |

Auto | Goat | Goat | Door 2/3 | Lose | Win |

Goat | Auto | Goat | Door 3 | Win | Loss |

Goat | Goat | Auto | Door 2 | Win | Loss |

66.67% | 33.33% |

There are 3 possible cases if the host will open the door without the car instead of random. For example, in row 2, the host must open door 3 since behind door 2, there is a car. Using this simple table, we calculate the probability of having the car is 66.67% if you switch which is two times the probability if you do not switch.

To further show it mathematically, we calculate the conditional probability using Bayes Theorem.

The P(Door2=auto) = 1/3. The denominator = 1/3 since if you know Door2 = auto, the host will definitely open door 3, and the probability of you open door 1 is 1/3. The nominator is 1/6 since the probability for you to choose door 1 is 1/3 and after that, the host has 1/2 probability to open door 3. The updated probability from P(Door2=auto) to P(Door2=auto|HostOpen = Door3, YourChose = Door1) is therefore 2/3, increasing from 1/3.

# If Host Opens Door Randomly

Door 1 (Your first choice) | Door 2 | Door 3 | Host Open | Outcome if you switch | Outcome if you stay |

Auto | Goat | Goat | Door 2 | Lose | Win |

Auto | Goat | Goat | Door 3 | Loss | Win |

Goat | Auto | Goat | Door 2 | Loss | Loss |

Goat | Auto | Goat | Door 3 | Win | Loss |

Goat | Goat | Auto | Door 2 | Win | Loss |

Goat | Goat | Auto | Door 3 | Loss | Loss |

The host completely opens the door at random. If he opens the door with a car behind it, you lose. As you can see, the probability of winning is the same.

Look at the same formula in the above session, the nominator is the same, 1/6. However, the denominator now changes from 1/3 to 1/6 since knowing Door2 = auto does not affect the probability of the host choosing which door to open. As a result, the updated probability is the same as the old know. No information is passed.

# Why It is So Confusing?

Our brain is wired to do causal problems, and this produces systematic probabilistic mistakes. There is no causal link between “Location of the Car” and “The Door You open” directly or a common cause. We find it difficult to comprehend why the “Location of the Car” somehow affects which door I should open, Door 1 (stay) or Door 2 (switch).

However, they are actually linked by a collider, “Which door will the host open”. The information starts to flow if we are conditioned on this collider.