The problem

Imagine you’re a contestant on a popular game show. The host shows you three closed doors: behind one is a brand-new luxury car, and behind the other two are goats (which, unless you’re in the market for a goat, are considerably less valuable prizes).

You select Door #1, hoping it contains the car. Before revealing what’s behind your chosen door, the host—who knows what’s behind each door—opens Door #3 to reveal a goat. The host then gives you a choice: stick with your original selection (Door #1) or switch to the remaining unopened door (Door #2).

The question: Should you stay with your original choice, or should you switch doors? Does it even matter?

Many people instinctively feel it makes no difference. After all, there are now two doors, one with a car and one with a goat—surely it’s a 50-50 chance either way? But is that correct? Think carefully about the probabilities involved in this famous mathematical puzzle known as “The Monty Hall Problem.”

Mathematical solution

This problem is one of the most famous probability puzzles in mathematics, and its solution is often counter-intuitive.

Let’s analyze the probabilities:

Initial Setup

• The car is randomly placed behind one of the three doors (1/3 probability for each door)
• You randomly select one door, giving you a 1/3 chance of selecting the car
• This means there’s a 2/3 chance the car is behind one of the doors you didn’t select

When the Host Reveals a Goat

Here’s where the magic happens. The host (who knows where the car is) must reveal a goat, and they must choose from the doors you didn’t select.

If you stay with your original choice:

• Your probability of winning remains 1/3 (unchanged from your initial selection)

If you switch to the other unopened door:

• If your original door contained the car (1/3 chance), switching loses
• If your original door didn’t contain the car (2/3 chance), switching wins
• Therefore, switching gives you a 2/3 chance of winning!

Counterintuitively, switching doubles your chances of winning from 1/3 to 2/3.

Why Switching Works Better

When you initially choose a door, there’s a 1/3 chance it contains the car and a 2/3 chance the car is behind one of the other doors. When the host reveals a goat behind one of those other doors, they’re essentially giving you information. The unopened door you didn’t choose now has a 2/3 probability of containing the car.

Discussion Questions

1. If there were 100 doors, with 99 goats and 1 car, and after you chose a door the host opened 98 doors (all showing goats, leaving your original door and one other), would you switch? How does this change your intuition about the problem?
2. Does the host’s knowledge matter? What if the host randomly opened doors without knowing where the car was, and happened to reveal a goat?
3. How would the probabilities change if sometimes the host opened a door to reveal the car, ending the game immediately with a loss?
4. Can you think of real-life scenarios where understanding the Monty Hall Problem might help you make better decisions?

Run the simulation multiple times and see the results for yourself! I’ll share the best reader solutions and insights next week.