# Beyond Three Doors: Generalizing the Monty Hall Problem

## Introduction

The [Monty Hall problem](https://en.wikipedia.org/wiki/Monty_Hall_problem) is a classic probability puzzle. It involves three doors: one hides a car (the prize), while the other two conceal goats. The player chooses one door, the host opens another door to reveal a goat, and the player is then given the option to switch to the remaining door. This problem is intriguing because the counterintuitive optimal strategy is to always switch doors.

In this article, we will first revisit the original problem with three doors and then explore two generalized versions of the problem with $N$ doors.

## The Original Problem with Three Doors

### Analysis

To analyze the original problem, let us assume the player **chooses the first door** initially. This analysis applies equally if the second or third door were chosen, as the probabilities remain symmetric. The door with the car can be behind the:

1. First door (probability = $\frac{1}{3}$)
2. Second door (probability = $\frac{1}{3}$)
3. Third door (probability = $\frac{1}{3}$)

#### If the Player Does Not Switch

1. If the car is behind the first door ($\frac{1}{3}$), the player wins.
2. If the car is behind the second or third door ($\frac{2}{3}$), the player loses.

The winning probability for not switching is:

$$
P(\text{win if not switching}) = \frac{1}{3}
$$

#### If the Player Switches

1. If the car is behind the first door ($\frac{1}{3}$), the player loses.
2. If the car is behind the second or third door ($\frac{2}{3}$), the host reveals a goat behind one of the other doors, and switching leads to the door with the car.

The winning probability for switching is:

$$
P(\text{win if switching}) = \frac{2}{3}
$$

Switching doubles the player's chances of winning compared to not switching!

## Generalized Monty Hall Problem

We now explore two variations of the problem when $N$ doors are involved.
> Notes: In the original $N = 3$ case, the results of both variations align with the above analysis.

### First Variation: Host Opens $N-2$ Doors

#### Example: $N = 10$

The player chooses one door, and the host opens $N-2 = 10 - 2 = 8$ other doors with goats. For instance, if the car is behind the fifth door and the player initially chooses the first door, the host opens doors 2, 3, 4, 6, 7, 8, 9, and 10. By intuition, it’s very suspicious that the host skips the fifth door—it seems likely that the fifth door might hide the car. As a result, people might naturally lean toward switching to the fifth door. Can we verify that this intuition is indeed correct?

#### Analysis

- If the player does not switch:
1. The first door is correct ($\frac{1}{10}$): the player wins.
2. Any other door is correct ($\frac{9}{10}$): the player loses.

$$
P(\text{win if not switching}) = \frac{1}{10}
$$

- If the player switches:
1. The first door is correct ($\frac{1}{10}$): the player loses.
2. Any other door is correct ($\frac{9}{10}$): the player wins.

$$
P(\text{win if switching}) = \frac{9}{10}
$$

In general, for $N$ doors:

$$
P(\text{win if not switching}) = \frac{1}{N}
$$
$$
P(\text{win if switching}) = \frac{N-1}{N}
$$

Switching increases the probability of winning by $N-1$ times!


> Notes: The sum of the probabilities is 1.

### Second Variation: Host Opens Only One Door

#### Example: $N = 10$

The player chooses one door, and the host opens a single door with a goat. For instance, if the car is behind the fifth door and the player initially chooses the first door, the host opens one of the doors 2, 3, 4, 6, 7, 8, 9, or 10.

#### Analysis

- If the player does not switch:
1. The first door is correct ($\frac{1}{10}$): the player wins.
2. Any other door is correct ($\frac{9}{10}$): the player loses.

$$
P(\text{win if not switching}) = \frac{1}{10}
$$

- If the player switches:
1. The first door is correct ($\frac{1}{10}$): the player loses.
2. Any other door is correct ($\frac{9}{10}$), but the host opens a goat door, leaving $N-2 = 8$ doors. The probability of switching to the correct door is $\frac{1}{8}$.

$$
P(\text{win if switching}) = \frac{9}{10} \cdot \frac{1}{8} = 0.1125
$$

For $N$ doors:

$$
P(\text{win if not switching}) = \frac{1}{N}
$$
$$
P(\text{win if switching}) = \frac{N-1}{N} \cdot \frac{1}{N-2}
$$

Switching increases the probability of winning but less dramatically than in the $N-2$ doors opened scenario.


> Notes: The sum of the probabilities will not be 1 in this version, as there are cases where neither switching nor staying wins. Also, the winning probability for not switching remains the same as in the first variation.

## Discussion

The two generalized versions demonstrate that switching doors always improves your winning probability. However, the magnitude of this improvement depends on the number of doors the host opens.

- In the $N-2$ doors scenario, the advantage of switching increases dramatically.
- In the single-door scenario, the improvement is more modest.

If you want the best chance to win, **always switch!**

## References

- [Monty Hall Problem on Wikipedia](https://en.wikipedia.org/wiki/Monty_Hall_problem)