Monty hall problem extended. 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.
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 $k &gt; 0$, $\frac{n-1}{n-k-1} &gt; 1$ 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.
@DwB

Your explanation would work if you chose your door AFTER you were shown a door that had a goat behind it.

Since you make your choice while there are 3 choices, the probability you choose the right door is 1/3 no matter if he shows you a goat or not.

If you choose one of the goats at first, you have a probability of 2/3 of choosing one of those right?  So the beauty of this problem is that if you choose a goat first, and switch you will always get the car.

Think about it, if you choose one of the goats first, then the other goat will always get shown.  If you switch then you will always switch to the car.

You want to choose a goat first to win this game by SWITCHING.(2/3)
You want to choose a car fiRst to win this game by NOT SWITCHING.(1/3)

4 doors...

You choose a car at first(1/4)
You choose a goat at first(3/4)

He opens 2 doors with a goat behind them, so your first choice and the last door is left.

If you chose the car first and switched now, you would lose. 

If you chose any of the 3 goats initially, and now switched, you would win the car.

It just comes down to the fact you have a better chance of picking one of the goats initially, so assuming he shows all the other "bad choices" to you, it is better to switch your answer you chose originally.

The probabilities you win by switching your choice, assuming that Monty opens all the other doors accept for the original one you picked and one other one.

3 doors (2/3)
4 doors (3/4)
5 doors (4/5)
6 doors (5/6)
7 doors (6/7)
8 doors (7/8)

n doors [(n-1)/n] = 1 - 1/n

as n approaches infinity 1/n becomes 0, and the general probability = 1 - 0 = 1.

So if we were to play this game with an infinite amount of doors, and switch our choice, we would win 100% of the time!