Is The Monty Hall Problem True?

You are presented with three doors, and there is a prize hidden behind only one of the doors. You chose door A, but before you can open the door, the host opens door B to show you that there is nothing behind it. Now you are given to chance to choose between sticking with your original plan (door A) or switching to door C.

What should you do?

It’s been a while since I last heard of The Monty Hall Problem until I came across Zack D. Film’s video while scrolling through YouTube Shorts a few days ago, and I thought it would be fun to prove this empirically with some good old Monte Carlo simulation.

Here we go!

To switch, or not to switch?

The optimal solution to the Monty Hall Problem, according to statistics, is to always switch. I’m not going to dive deep into the math behind it, but I’ll try to explain it in layman’s terms:

1. We start with doors A, B, and C, each with a 1/3 probability of having the prize.
2. The moment we chose door A to open, the probability then converges to 1/3 behind door A, and 2/3 not behind door A (thus behind B or C).
3. The host then reveals that door B is empty.
4. The previous 2/3 probability of the prize sitting behind door B or C then becomes the probability of the prize sitting behind door C, since B is eliminated
5. As a result, switching doors will now increase your probability of picking the right door from 1/3 to 2/3, therefore switching is always recommended.

Let’s try to validate this theory with some simulation.

Python Simulation

We’ll simulate 10000 such trials using the following script, and calculate the win rate of the two strategies:

1. Strategy 1: Do not switch doors.
2. Strategy 2: Always switch doors.

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import numpy as np


# experiment iteration
N = 10000

# randomly generate N prizes
games = np.random.randint(low=0, high=3, size=N)

# simulate player random picks
picks = np.random.randint(low=0, high=3, size=N)

# simulate revelation from host
reveal = []
for g, p in zip(games, picks):
    # if pick is the prize, randomly reveal another door
    choices = [0, 1, 2]
    if g == p:
        choices.remove(g)
        reveal.append(choices[np.random.randint(low=0, high=2)])
    else:
        # reveal the only non-empty door left
        choices.remove(g)
        choices.remove(p)
        reveal.append(choices[0])
reveal = np.array(reveal)

# strategy 1: do not switch
strat_1 = np.sum(picks == games) / N

# strategy 2: switch
new_pick = []
for r, p in zip(reveal, picks):
    choices = [0, 1, 2]
    choices.remove(r)
    choices.remove(p)
    new_pick.append(choices[0])
new_pick = np.array(new_pick)
strat_2 = np.sum(new_pick == games) / N

print(f'strategy 1: {strat_1:.2f}')
print(f'strategy 2: {strat_2:.2f}')

Here is the output:

1
2

strategy1 1: 0.33
strategy 2: 0.67

Strategy 1 and 2 have a win rate of approximately 1/3 and 2/3, exactly what we are expecting!

Final Thoughts

The Monty Hall Problem is a carefully crafted probability illusion designed to trick the audience into thinking that the probability doesn’t change after new information is introduced (host eliminating one option). Using a more obvious scenario could help people better understand why the probability changed:

Assume that instead of three doors, there are now 1,000,000 doors for you to choose from. After choosing a door, the host opened another 999,998 doors with nothing behind it. You are now left with two choices: sticking with your original hunch, knowing that it has a 1 in a million chance of winning, or switching doors (which now has a 99.9999% chance of winning)?

Until next time!