Does The Monty Hall Problem Still Apply With Infinite Doors? Here's been a bunch of questions on the Monty Hall problem, so I'll assume people know the basics. 
This answer helped clarify a few things for me, but talking with some colleagues yesterday, someone brought up the idea that as you increase the number of doors, the probability of winning the car by switching approaches 1. Intuitively, this makes sense, but I also know that infinite sets aren't always intuitive. 
Is it really the case that if the number of doors increases to infinity, the probability of winning approaches 1? If not, why not? 
EDIT: To clarify, the infinite case would look as follows: 


*

*There are an infinite number of doors. One has a car behind it, the others have goats.

*The host knows which door has the car behind it.

*The contestant selects a door

*All doors except the selected door and an additional one are opened revealing goats.

*The contestant is asked if they would like to switch doors. 


The question is: in this case, does P(winning by switching doors) = 1? Or is this even well defined? What changes when the number of doors is large but finite?
 A: You can only validly state this problem in a decent probabilistic way if you either consider only finite versions and take the limit for the probability and the finite number of doors tends to infinity, or when considering one problem with infinitely many doors you specify a probability distribution that gives the a priori probability for each door that the car is behind it. You cannot say that infinitely many doors all have the same probability, because (given there is no more than one car) that probability would have to be$~0$, and it would imply that the probability that there was any car at all would be$~0$ as well, which it is not supposed to be. There is no such thing as a uniform probability on a countably infinite set.
Given the problem statement, the chances you win by switching is exactly the probability that you did not choose the door with the car in the first place. So in the limit of finite problems, the success probability for switching, $1-\frac1n$ for $n$ doors, indeed tends to$~1$.
In an infinite version with a given probability distribution, you can make your chances of success as close to$~1$ as you like by choosing a very improbable door in the first place (one situated way beyond the furthest known galaxy might be fine, but it really depends on the given probability distribution), but quite possibly you cannot make it $1$ exactly (namely if there are no doors that have been assigned probability$~0$).
I cannot help feeling sorry for those infinitely many undesired goats.
A: I've always found the Monty Hall problem interesting in that it's used to illustrate conditional probability, but there is a much simpler approach that I've always used when discussing this problem and its variants:
In essence, the Monty Hall problem boils down to two events: 


*

*The car is behind the door the contestant selected (S)

*The car is behind the door Monty Hall has left closed (M)


Note that S and M are exhaustive, disjoint events, so $P(S) + P(M) = 1$. If the contestant selects from among N doors at random, the probability of selecting the door with the car is $P(S_N)=\frac{1}{N}$. From the above observation that the events S and M are disjoint-exhaustive, we can conclude $P(M_N)=1-P(S_N) = \frac{N-1}{N}$.
Taking the limit as $N\rightarrow \infty$ we get $P(S_{\infty})=\lim\limits_{N\rightarrow \infty} \frac{1}{N} = 0$ hence, as the number of doors increases, the success probability of switching does indeed approach 1.
