TITLE: Monty hall problem extended.
QUESTION [160 upvotes]: 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.

REPLY [3 votes]: This is an adaptation of an extremely short and simple solution to the Monty Hall problem which I posted here. I think it vastly simplifies matters in this case as well.
Of course the probability of winning if the contestant does not switch doors is $1/n$. So let's suppose the contestant does switch doors and let's break it into two cases: 1. the contenstant chooses the correct door initially, 2. he doesn't. The probability in each case respectively is $1/n$ and $\frac{n-1}{n}$. In the first case, switching doors loses. In the second case since another door is now eliminated the probability of winning by switching doors is $\frac{1}{n-2}$. Therefore the probability of winning by switching doors is $\frac{n-1}{n}\cdot\frac{1}{n-2}={\bf \frac{n-1}{n(n-2)}}$. Lastly, observe $\frac{n-1}{n(n-2)}>\frac{n}{n^2}=\frac{1}{n}$.
In the example where $n=4$, the probability of winning by switching is $\frac{4-1}{4(4-2)}=\frac{3}{8}$.