Math Squad

Bayes' Theorem
Example 1: Quality Control
Example 2: False Positives
Example 3: Monty Hall Problem

Example 3: Monty Hall

by Maliha Hossain, proud Member of the Math Squad.

keyword: probability, Monty Hall, Bayes' Theorem, Bayes' Rule 

The Monty Hall problem is based on the television game show Let's Make a Deal. The problem is named after the show's original host, Monty Hall. The problem was published by Marilyn vos Savant in Parade Magazine. It was also solved by the protagonist of Mark Haddon's A Curious Incident of the Dog in the Night-Time where he asks:

"You are on a game show on television. On this game show the idea is to win a car as a prize. The game show host shows you three doors. He says that there is a car behind one of the doors and there are goats behind the other two doors. He asks you to pick a door. You pick a door but the door is not opened. Then the game show host opens one of the doors you didn't pick to show a goat (because he knows what is behind the doors). Then he says that you have one final chance to change your mind before the doors are opened and you get a car or a goat. So he asks you if you want to change your mind and pick the other unopened door instead. What should you do?"

Many readers of vos Savant's column, including scholars holding PhD's, refused to believe that switching would be to the contestant's advantage even though it can be demonstrated using proofs and computer simulation.

We will use Bayesian inference to show that contestants have a 2/3 chance of winning the car by switching whereas contestants who do not switch only have a 1/3 chance of winning.

Let's assume you are the contestant on Let's Make a Deal and you initially choose door number 1.

Let $ a $ be the event that the car is behind door number 1.

Let $ b $ be the event that the car is behind door number 2.

Let $ c $ be the event that the car is behind door number 3.

This problem assumes that Monty knows where the car is hidden. From your point of view however, the prize is equally likely to be behind any one of the three doors. So

$ P[a] = P[b] = P[c] = \frac{1}{3} $

Let C be the event that Monty Hall opens door number 3 to reveal a goat after you have chosen door number 1.

Given that the car is behind door number 1, the door you have chosen, Monty has a 1/2 chance of picking door 3 from the remaining two doors to reveal a goat. So we have that

$ P[C|a] = \frac{1}{2} $

Given that the car is behind door number 2, and you have chosen door 1, that leaves Monty with only one choice. He can't reveal the door with the car and he also can't reveal what's behind the door you've chosen. So he has to pick door number 3.

$ P[C|b] = 1 \ $

Given that the car is behind door number 3, and you have chosen door 1, Monty will not choose door 3 since in doing so, he will reveal the car. Monty can not reveal the car before he's given you a chance to revise your decision. So

$ P[C|c] = 0 \ $

By the theorem of total probability, we have that

$ \begin{align} P[C] &= P[C|a]P[a] + P[C|b]P[b] + P[C|c]P[c] \\ &= \frac{1}{2} \times \frac{1}{3} + 1\times\frac{1}{3} + 0\times\frac{1}{3}\\ &= \frac{1}{2} \end{align} $

What is the probability that you've lost by switching to door number 2? In other words, what is $ P[a|C] $? By Bayes' Theorem, we have that

$ \begin{align} P[a|C] &= \frac{P[C|a]P[a]}{P[C]} \\ &= \frac{\frac{1}{2} \times \frac{1}{3}}{\frac{1}{2}}\\ &= \frac{1}{3} \end{align} $

Now what is the probability that you've won by switching? That is to say, what is $ P[b|C] $? By Bayes' Theorem, we have that

$ \begin{align} P[b|C] &= \frac{P[C|b]P[b]}{P[C]} \\ &= \frac{1 \times \frac{1}{3}}{\frac{1}{2}}\\ &= \frac{2}{3} \end{align} $

Since the assignment of door numbers that the you pick and the event C is arbitrary, you will arrive at the same results regardless of which door you pick first and which Monty opens later.

There are many other methods to solve the Monty Hall problem that seem more intuitive to some people. I however like to use Bayes' Theorem since the result can be derived analytically.


  • Mark Haddon, The Curious Incident of the Dog in the Night-Time

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