The Monty Hall problem is an interesting problem having stumped experienced mathematicians, despite the seemingly simple problem statement. The problem is considered a paradox of the veridical type, because the solution is so counterintuitive that some believe the conclusion is absurd. [1]

The purpose of this note is to consider a derivative of the original problem statement to build further intuition. I was recently introduced to a similar iteration that helped to cement the intuition behind the logic to solve the problem.

### Monty Hall

The original problem statement had the following constraints: [1]

- The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999).
- The host must always open a door to reveal a goat and never the car.
- The host must always offer the chance to switch between the originally chosen door and the remaining closed door. The question is should the player switch given the host has revealed a goat? The answer is yes, the optimal strategy is to switch, as the remaining door has a probability of \(\frac{2}{3}\) to contain the car. Given that any door has a probability of \(\frac{1}{3}\) to contain the car prior to any information via action by player, and the host, the probability of \(\frac{2}{3}\) is somewhat counterintuitive.

### Monty Hall with a Twist

Adding a bit of a complexity to the original problem, actually helped in my understanding and I hope will have a similar impact on the reader.

In addition to the original constraints, assume the following constraints.

- The player must select the number of doors.
- The number of doors can range from \(3\) to \(15\) inclusive.
- At each turn, the player has the option to switch.
- If there are \(2\) doors remaining, the player is forced to make a final decision on switching and the game ends.

Now rephrase the problem statement to *what is the optimal strategy for this
game?*.

### Solution

#### Lets try by select \(3\) doors, which gives us the original Monty Hall

problem.

The probability that an given door has the car is \(\frac{1}{3}\). After
player selection, and host action to reveal goat, the probability that the other
door contains a car is \(\frac{2}{3}\). **Thinking about probability as an
object with mass helps with building intuition for this problem.** Since
\(\frac{1}{3}\) of mass is contained in the player selected door, the
remaining \(\frac{2}{3}\) has to be contained by the remaining doors. When
there are two doors remaining, it is evenly split. When the number of doors is
reduced by host action, the probability mass of \(\frac{2}{3}\) is contained
by the remaining door. Thus when given the choice, of \(\frac{2}{3}\) or
\(\frac{1}{3}\), of probability mass, it is logical to select the door with
more mass. Its ok to be greedy here. Thus we reach the solution that the player
should switch.

#### Lets try selecting \(7\) doors.

The player selects a door, and the host reveals a goat. The probability that any given door has the car is \(\frac{1}{7}\), and thus the probability mass contained in the player selected door is \(\frac{1}{7}\), while the remaining \(5\) doors, remember the host has removed a door, equally share the remaining probability mass of \(\frac{6}{7}\), thus each door has a probability of \(\frac{6}{35}\), or \(0.1714\) vs. the \(0.1428\) of probability mass currently held by the player. Should the player switch? Well if this was the end of the game then yes, but the player still doesnt know what the result of other doors being revealed will have on the probability masses contained in each remaining door.

The host reveals another goat leaving \(4\) doors plus the \(1\) door the player selected originally. Keep in mind that the probability mass of \(\frac{1}{7}\) that the player currently has remains constant, thus the remaining \(\frac{6}{7}\) of probability mass needs to be shared by the remaining \(4\) doors. This equates to each door having \( \frac{\frac{6}{7}}{4}\), or \(\frac{6}{28}\), \(0.2143\) in decimals. Ok now we see that by waiting and staying put, the available probability mass by switching has increased from \(0.1714\) to \(0.2143\). This is nice discovery.

Lets wait until there are \(2\) doors remaining in addition to the door the originally selected by the player. Again, the same logic applies. The remaining \(\frac{6}{7}\) of probability mass needs to be split between the \(2\) doors, leaving \(\frac{6}{14}\), or \(0.4286\) of mass per door.

I would guess that this pattern is consistent, in other words, the per door probability mass contained by the other doors increases as more doors are revealed.

If this is true, we can easily show by computation that the optimal strategy given an upper bound of \(n\) doors is to select the n-door game, and wait until there is a total of \(2\) doors remaining before switching, as the remaining \(1\) door will have \(\frac{n-1}{n}\) of probability mass.

### Summary

Thinking about probabilities as an object with mass was really the key for me to build intuition. I hope this walk through helps to build similar intuition for others. In completing this note, the next question, is what happens when \(n\) goes to infinity?

### References

- https://en.wikipedia.org/wiki/Monty_Hall_problem