What is the Monty Hall problem?

The Monty Hall problem is a famous illustration of the concept of conditional probability and has been the subject of much debate and discussion in the field of probability theory. The Monty Hall problem is a probability puzzle named after the host of the game show Let’s Make a Deal, Monty Hall. The problem goes as follows:

You are a contestant on a game show, and there are three doors in front of you. Behind one door is a valuable prize, such as a car, and behind the other two doors are goats. Our goal is to choose the door that has the prize behind it. Here’s how the game works:

1. Initial choice: We start by selecting one door, let’s say Door 1, without knowing what’s behind it.

2. Host’s intervention: Now, the host, who knows what’s behind each door, intervenes. They’re forced to open another door that we did not choose, revealing a goat. For instance, if we initially chose Door 1 and the prize is behind Door 2, the host would open Door 3 to reveal a goat.

3. Reevaluation: At this point, we’re given a choice. We can either stick with our initial choice (Door 1) or switch to the remaining unopened door (Door 2). What should we do?

Let’s play

You’re now a contestant on our very own game show. First, pick a door. The game will reveal a door that has a goat behind it. We can then choose to stick to our original choice or switch to the other door. Let’s see if we can win a car.

Is switching better?

The surprising answer is that it’s statistically advantageous to switch doors. Switching gives us a $2/3$ probability of winning the car while sticking with our original choice only has a $1/3$ probability. Let’s try to understand why with some illustrations.

Initial choice

• When the contestant makes their initial choice, they have a 1 in 3 chance of picking the car and a 2 in 3 chance of picking a goat.

• So, the probability distribution for the initial choice is:

  • Car: $1/3$

  • Goat: $2/3$

The host opens a door

• Regardless of the contestant’s initial choice, Monty Hall, who knows what’s behind each door, will always open a door with a goat behind it.

• If the contestant initially chose a door with a goat (which happens with a $2/3$ probability), Monty has no choice—he must open the other door with a goat.

• If the contestant initially chose the door with the car (which happens with a $1/3$ probability), Monty can choose to open either of the two other doors with a goat behind it.

• So, after Monty opens a door, the probability distribution remains the same:

  • Car: $1/3$ (if the contestant initially chose the car)

  • Goat: $2/3$ (if the contestant initially chose a goat)

Switching

• Now, the contestant has the option to either stick with their initial choice or switch to the remaining unopened door.

• If the contestant chooses to stick to their initial choice, the probability of them winning a car remains the same, which is $1/3$.

• If the contestant chooses to switch, and:

  • If they initially chose the car (with a $1/3$ probability), they’ll lose because the car is behind their initial choice, and switching means picking a door with a goat.

  • If they initially chose a goat (with a $2/3$ probability), they’ll win because their initial choice was a goat, and switching means picking the door with the car. We can be certain about the win condition here because the host has already revealed the door with the goat. The only door left is the one with the car.

This result is counterintuitive to many people because it seems like both remaining doors should have an equal chance ($1/2$ probability) of hiding the prize after Monty reveals a goat. However, the crucial point is that Monty’s actions provide additional information that changes the probabilities. The act of Monty opening a door with a goat doesn’t change the probabilities associated with the contestant’s initial choice. If they switch, they essentially transfer the probability of choosing a goat initially to the probability of winning the car. The result is that switching increases the chance of winning from $1/3$ to $2/3$ .

Scaling the problem

Here’s another way to look at a similar problem. What if we had more than three doors and the rules of the game stayed the same? If the host had to reveal all the doors with goats except one, would we still stick with our initial choice? Try to increase the number of doors to see how the probabilities change. Do keep in mind that with switching, we’re essentially transferring the probability of choosing a goat initially to the probability of winning the car.

Conclusion

The Monty Hall problem is a fascinating blend of probability, strategy, and intuition. Despite its seemingly counterintuitive solution, the math backs the wisdom of switching doors to enhance the winning odds. This puzzling scenario dives into mathematical intricacies and challenges our instincts, highlighting how unexpected choices can often lead to success.