Monty Hall problem

I first came across this problem in the film 21, where professor Micky Rosa puts it to Ben in class. It is a famous piece of conditional probability, and it is famous mostly because the correct answer feels wrong.

The problem

You are on a game show, choosing between three doors. Behind each door is either a car or a goat. You pick a door. The host, Monty Hall, then opens one of the other two doors — one he knows hides a goat — and shows you the goat. He asks whether you would like to switch to the remaining unopened door.

Assuming you would rather have the car, do you switch?

You should. Switching wins twice as often as staying.

Counting the outcomes

The quickest way to see it is to write down every case. Say you picked door 1. There are three equally likely arrangements:

Switching wins in two rows out of three.

The reason is easier to feel than the arithmetic suggests. Your first pick is wrong two times out of three. Whenever it is wrong, the car sits behind one of the other two doors, Monty is forced to open the goat one, and the car ends up behind the door he did not open. So "my first guess was wrong" and "switching wins" describe the same event — and that event has probability $\frac{2}{3}$.

Where the probability goes

graph LR
  A["You pick a door"] --> B["First pick correct, one third"]
  A --> C["First pick wrong, two thirds"]
  B --> D["Monty may open either goat door"]
  C --> E["Monty is forced to open the only goat door"]
  D --> F["Staying wins"]
  E --> G["Switching wins"]

The asymmetry lives on the lower branch. When your first pick is wrong, Monty has no choice about which door to open, and that absence of choice is exactly what leaks information to you.

The same answer from Bayes' theorem

Let $H$ be the hypothesis "the car is behind door 1" — the door you picked — and let $E$ be the evidence "Monty opened a door and revealed a goat." We want $P(H|E)$.

Bayes' theorem says

$P(H|E)=\frac{P(E|H)\times P(H)}{P(E)}$

where

$P(E)=P(E|H)\times P(H)+P(E|\text{not}H)\times P(\text{not}H)$

Taking each piece in turn:

• $P(H)=\frac{1}{3}$ — the prior that your door hides the car.
• $P(\text{not}H)=\frac{2}{3}$ — the prior that it does not.
• $P(E|H)=1$ — if the car is behind your door, both others hide goats, so Monty certainly reveals one.
• $P(E|\text{not}H)=1$ — if the car is elsewhere, Monty still knows which remaining door hides a goat, and opens that one.

Substituting:

$P(H|E)=\frac{1\times\frac{1}{3}}{1\times\frac{1}{3}+1\times\frac{2}{3}}=\frac{1}{3}$

Your door is still $\frac{1}{3}$. The evidence changed nothing, because Monty was always going to show you a goat regardless of where the car was — an outcome with probability $1$ carries no information. Since the car is behind either your door or the one still closed, that other door now carries $\frac{2}{3}$.

The variant where it really is 50–50

Now suppose Monty does not know where the car is. He opens one of the other two doors at random, and it happens to reveal a goat. Nothing about the view from your chair has changed — so why is the answer now different?

Because $P(E|H)$ and $P(E|\text{not}H)$ are no longer equal:

• $P(E|H)=1$ — if the car is behind your door, both remaining doors hide goats, so a random opening reveals a goat for certain.
• $P(E|\text{not}H)=\frac{1}{2}$ — if the car is elsewhere, one of the two doors he might open hides it, so he reveals a goat only half the time.

$P(H|E)=\frac{1\times\frac{1}{3}}{1\times\frac{1}{3}+\frac{1}{2}\times\frac{2}{3}}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$

Now it is a genuine coin flip. A goat from a random opening is evidence, because it was more likely to happen when your first pick was right, and that evidence pulls your door up from $\frac{1}{3}$ to $\frac{1}{2}$.

The two versions look identical from where you are standing. Same three doors, same revealed goat, same question. What differs is the process that produced the goat — and probability is a statement about processes, not about scenery.