Monty Hall Problem

There are three doors, behind one is a car, behind the other two are goats. As a contestent, we pick a door. The host, who knows which door has the car, reveals one of the remaining doors that contain a goat and offers a choice: We can either stick with our initial choice or switch to the other unopened door.

Is switching a better stratagy?

Most people who encounter this problem decide to stick with their initial choice. The main reason is they think there are two remaining doors, and since the probabilities were initially were equal, nothing has changed after the host reveals one door.

They assume there is a 50-50 chance of winning. Psycologically, it also feels worse to switch and lose than to stick and lose.

This reasoning is incorrect. Switching wins with probability 2/3. Staying wins with probability 1/3.

We will show in three different ways that switcing is the better strategy:

1. Story Proof A story proof is a proof by interpretation. For counting problems, this often means counting the same thing in two different ways, rather than doing tedious algebra. A story proof often avoids messy calculations and goes further than an algebraic proof toward explaining why the result is true. The word “story” has several meanings, some more mathematical than others, but a story proof (in the sense in which we’re using the term) is a fully valid mathematical proof. - From Joe Blitzstein, Introduction To Probability, page 20.
2. Proof using the law of total probabilities
3. A computer simulation

Story Proof

If we were allowed to pick two doors at the start, our probability of winning would be 2/3. While the rules do not permit choosing two doors outright, host's elimination of one unselected door effectively gives us that opportunity. Switching is equivalant to having bet on the two doors we did not originally choose.

Mathematical Proof

Without loss of generality, assume we initially picked Door 1. Let $I_1$ denote this information.

Define:

• $S$: success by switching
• $D_j$: Door $j$ contains the car

Using the law of total probability:

Evaluate each term:

• If the car is behind Door 1, switching loses: $P(S \mid D_1, I_1) = 0$
• If the car is behind Door 2, switching wins: $P(S \mid D_2, I_1) = 1$
• If the car is behind Door 3, switching wins: $P(S \mid D_3, I_1) = 1$

Since each door initially has probability $\frac{1}{3}$:

Therefore, switching yields a winning probability of $\frac{2}{3}$, which is strictly greater than staying ($\frac{1}{3}$).

Computer Simulation

              
import numpy as np

N = 10_000  # number of trials

# Randomly pick the door with car behind
door_car = np.random.choice([1, 2, 3], size=N)

# Staying wins when car is behind door 1
stay_win_rate = np.mean(door_car == 1)
print(stay_win_rate)  # ~0.33

# Switching wins when car is behind door 2 or 3
switch_win_rate = np.mean((door_car == 2) | (door_car == 3))
print(switch_win_rate)  # ~0.66
              
            

This simulation captures the same logic: in about 1/3 of trials the car is behind the initially chosen door (stay wins), and in about 2/3 it is not (switch wins).