Simulation to Monty Hall problem (3 doors game)Simulating the Monty Hall Problem: Should You Stick or Switch?

I recently had this idea to simulate the famous Monty Hall Problem (also known as the 3 Doors Game) to see if reality matches the mathematical explanation. Here’s what I found.

Rules of the Game:

Let’s recap the setup of this intriguing brain teaser:

1. There are three identical doors. Behind one of them is a brand-new car (the grand prize), while the other two hide goats (no prize).
2. As the contestant, you pick one of the doors, guessing which one hides the car. For example, let’s say you choose Door 1.
3. After you make your choice, the host—who knows what’s behind each door—opens one of the two remaining doors, always revealing a goat.
4. Now, with two doors left (your chosen door and another unopened door), the host asks: “Do you want to stick with your original choice or switch to the other door?”

The goal is simple: maximize your chances of winning the car.

Stick or Switch?

At first, I thought whether to switch or stick had an equal chance of winning: 1 out of 3. But math tells a different story.

Let’s break it down:

• If you stick, your chances of winning are based on your initial choice, which had a 1/3 probability of being correct (one correct door out of three).
• If you switch, here’s what happens:
  • The door you initially chose had a 1/3 chance of hiding the car.
  • This means the other two doors collectively had a 2/3 chance of hiding the car.
  • When the host reveals a goat, the full 2/3 probability is transferred to the remaining unopened door.

This makes switching the better strategy because it increases your chances of winning from 1/3 to 2/3.

Why Does It Feel Counterintuitive?

At first glance, it might seem like after the host reveals a goat, the two remaining doors should each have an equal chance (1/2). But that’s not true because your initial choice locked in the probabilities.

• The door you picked still has a 1/3 chance.
• The unchosen, unopened door inherits the full 2/3 probability after the host eliminates one option.

Simulating the Monty Hall Problem

To test this concept, I wrote a simple simulation to play the game thousands of times. The goal was to measure the actual winning probabilities for both strategies—stick and switch—over a large number of rounds.

The Results:

Here’s a chart showing how the probabilities behave as the number of games increases:

• The x-axis represents the number of games played (from 1 to 10,000).
• The y-axis represents the winning probability (0.0 to 1.0).
• The blue line shows the probability of winning if you stick with your original choice.
• The orange line shows the probability of winning if you switch.

How to Read the Chart:

• With fewer games (left side of the chart): The probabilities vary wildly because outcomes are dominated by luck in the short term.
• As the number of games increases (right side of the chart): The probabilities stabilize, revealing a clear pattern:
  • Sticking hovers around 1/3 (blue line).
  • Switching stabilizes around 2/3 (orange line).

Conclusion:

If you’re playing the Monty Hall game once, the result might seem like pure luck. But if you get to play multiple times, trust the math—switching doors consistently gives you a better chance of winning the car. The simulation clearly demonstrates that 2/3 > 1/3 when it comes to maximizing your odds.

Give it a try yourself and see if you can trust your intuition—or if the math wins out in the end!