This short note describes a simple non-Bayesian solution to the Monty Hall problem. A charming Bayesian analysis can be found in the book Bernoulli’s Fallacy.

The Monty Hall problem is stated as follows on Wikipedia:

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 No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

Solution

The probability of selecting the door with the car behind without switching is simply $\frac{1}{3}$.

The probability of getting the car after switching is the weighted average of two scenarios: the car was selected in the first place, and the car was not chosen in the first place. The associated probabilities are listed below.

Scenario Probability Probability of getting the car after switching Note
Car selected (car behind Door #1) $\frac{1}{3}$ 0  
Car NOT selected (Car not behind Door #1) $\frac{2}{3}$ 1 The probability of getting the car is $1$ because the car is behind the only remaining door (Door #2).

The overall probability of getting the car after switching is \(\frac{1}{3}\times0+\frac{2}{3}\times 1 = \frac{2}{3}. \notag\) Therefore, switching increases the probability of getting the car to $\frac{2}{3}$ from the probability of $\frac{1}{3}$ without switching.