I just learned about the Monty Hall problem and found it quite amazing. So I thought about extending the problem a bit to understand more about it.
In this modification of the Monty Hall Problem, instead of three doors, we have four (or maybe $n$) doors, one with a car and the other three (or $n-1$) with a goat each (I want the car).
We need to choose any one of the doors. After we have chosen the door, Monty deliberately reveals one of the doors that has a goat and asks us if we wish to change our choice.
So should we switch the door we have chosen, or does it not matter if we switch or stay with our choice?
It would be even better if we knew the probability of winning upon switching given that Monty opens $k$ doors.
Best Answer
I decided to make an answer out of my comment, just for the heck of it.
$n$ doors, $k$ revealed
Suppose we have $n$ doors, with a car behind $1$ of them. The probability of choosing the door with the car behind it on your first pick, is $\frac{1}{n}$.
Monty then opens $k$ doors, where $0\leq k\leq n-2$ (he has to leave your original door and at least one other door closed).
The probability of picking the car if you choose a different door, is the chance of not having picked the car in the first place, which is $\frac{n-1}{n}$, times the probability of picking it now, which is $\frac{1}{n-k-1}$. This gives us a total probability of $$ \frac{n-1}{n}\cdot \frac{1}{n-k-1} = \frac{1}{n} \cdot \frac{n-1}{n-k-1} \geq \frac{1}{n} $$
No doors revealed
If Monty opens no doors, $k = 0$ and that reduces to $\frac{1}{n}$, which means your odds remain the same.
At least one door revealed
For all $k > 0$, $\frac{n-1}{n-k-1} > 1$ and so the probabilty of picking the car on your second guess is greater than $\frac{1}{n}$.
Maximum number of doors revealed
If $k$ is at its maximum value of $n-2$, the probability of picking a car after switching becomes $$\frac{1}{n}\cdot \frac{n-1}{n-(n-2)-1} = \frac{1}{n}\cdot \frac{n-1}{1} = \frac{n-1}{n}$$ For $n=3$, this is the solution to the original Monty Hall problem.
Switch.