Monty Hall Game (audience selected to open a door at random) The classic Monty Hall game is as follows: There are 3 doors. 1 door contains a prize. The other 2 doors do not. The prize has an equal chance of being randomly placed behind a given door. The contestant chooses a door to open. Prior to opening the door, the Host will open 1 of the 2 remaining doors, and it is guaranteed that there is nothing behind this door. If you initially chose a door that has the prize, the host will choose to open 1 of the 2 remaining doors with equal probability. You are then given the choice of whether to switch your choice or not switch after the host reveals which door is empty. 
The brute force way to prove this is using Bayes Theorem. But intuition allows for a quick solution by examining the 3 possible placements of the prize:
\begin{array}{|c|c|c|c|}
\hline
Door: & 3 & 2 & 1 \\ \hline
Content: & Prize & Empty & Empty\\ \hline
Content: & Empty & Prize & Empty\\ \hline
Content: &  Empty & Empty & Prize \\ \hline
\end{array}
Assume that we initially choose door 3. If the prize is in door 3, we will lose the prize by switching regardless if the Host opens door 1 or 2. If the prize is in door 2, then the Host is guaranteed to open door 1, so switching guarantees we win the prize. Same logic for if prize is in door 1. So for the 3 possibilities, we see that switching is advantageous $\frac{2}{3}$ of the time for repeated trials, so switching is advantageous. For $N$ games, the expected number of prizes we get should equal $2/3 N$. 
Now we want to answer the same question, but instead of the host choosing an audience member is chosen to select the door to open for a SINGLE round. This audience just so happens to choose the door that doesn't contain the prize. Should you switch? 
The solution from the book "Heard on the Street" states 
"Suppose instead that the host of this week’s show has, just for a one-off special
occasion (his 60th birthday, say), decided to let an audience member (who is
ignorant of the prize’s location) reveal another door. Suppose it just happens
to turn out to be empty. We cannot now talk about a frequentist approach and
repeated plays of the game because in repeated plays, the audience member
would reveal the prize one third of the time, and that is not the situation
we find ourselves in. The ignorant audience member has, on this occasion,
just happened to show us an empty door. The audience member has blindly
removed one door from the sample space, and we have a 50/50 chance of
winning (and also of losing) if we switch. We are therefore indifferent."
I cannot comprehend the author's logic behind removing one door from the sample space on the basis that this just a single trial. Can someone explain this with more detail (dumb it down), or perhaps reword it for me? 
 A: A direct computation is simple if you make a small modification to your "brute force" way for the original problem.
Suppose you initially chose door 3.


*

*Prize is behind door 3.


*

*If the audience member opens door 1, then you lose if you switch.

*If the audience member opens door 2, then you lose if you switch.


*Prize is behind door 1.


*

*Audience member must have opened door 2. You win if you switch.


*Prize is behind door 2.


*

*Audience member must have opened door 1. You win if you switch.



These four outcomes are equally likely, so the chance of winning if you switch is $1/2$.
A: The $50/50$ result is correct, but the part about this being a single trial and not amenable to a frequentist analysis is nonsense. (Full disclosure: I tend towards a Bayesian view of probabilities, so I’m not just defending my favourite philosophy here.) Of course you can do this experiment many times; sometimes the audience member will open the door with the prize, sometimes not; and you can count the relative frequency with which you win in the cases where the prize was not behind the opened door. The result will be that no matter whether you switch the relative frequency of winning tends to $\frac12$ in the limit of infinitely many trials. That’s a perfectly normal frequentist way of looking at things.
Edit in response to the comments (both under this answer and under @angryavian’s):
Things can be “guaranteed to happen” in very different ways. For a conditional probability, we only count certain events. The conditional probability of you winning, conditioned on the audience member having opened a door without the prize, is $\frac12$. So you could say that by our choice of cases for the conditional probability, we are “guaranteeing” that the audience member opens a door without the prize.
But you should be focusing instead on which results are equiprobable; that’s what makes the two scenarios different.
In the original Monty Hall problem, three events are equiprobable: That the prize is behind door $1$, $2$ or $3$. As you correctly analyzed, in two of these cases you win if you switch and in the third you don’t, so your winning probability is $\frac23$. 
In the audience version, there are $6$ equiprobable events: The prize is equally likely to be behind any door, and the audience member is equally likely to open either door you didn’t pick. In $2$ out of those $6$ cases the game ends because the prize has been revealed. If we condition on this not happening and the game continuing, then that leaves $4$ equiprobable events: the four pairs of prize door and opened door where the prize door is neither the opened door nor the one you picked. Say you picked door $1$, then these four are $(1,2)$, $(1,3)$, $(2,3)$ and $(3,2)$ (where the first number represents the prize door and the second represents the opened door). In two of these you win if you switch and in the other two you win if you don’t. Since these four are equiprobable, that gives you a winning probability of $\frac12$.
If you now compare this to the original Monty Hall problem, note that the two events $(1,2)$ and $(1,3)$ that are equiprobable in the audience scenario together form one of the three equiprobable events in the original scenario. Of course you can distinguish them as possibilities just like you do in the audience scenario; but that doesn’t make them equiprobable. 
The reason for the different conditions of equiprobability in the two scenarios is that in the audience scenario there’s a second random choice by the audience member that makes all $6$ outcomes equiprobable, whereas in the original scenario there’s no random choice in two of the cases; the host only makes a random choice which door to open if you pick the prize door, and this random choice in only one of the three equiprobable cases doesn’t lead to a new set of equiprobable cases as in the audience scenario.
Another way of looking at it: Imagine that the host always flips a coin, no matter which door you choose. If you open the prize door, she uses that coin flip to decide which door to open; otherwise, she ignores it. Now we have $6$ equiprobable events, as in the audience scenario; and the corresponding door pairs are $(1,2)$, $(1,3)$, $(2,3)$, $(2,3)$, $(3,2)$ and $(3,2)$ (again, the first number represents the prize door, the second the opened door, and you picked door $1$). You win if you switch in $4$ out of these $6$ cases.
So in the audience scenario, the six equiprobable states are $(1,2)$, $(1,3)$, $(2,2)$, $(2,3)$, $(3,2)$, $(3,3)$, and the “guarantee” that the prize door isn’t opened comes from our conditioning, which removes the two states $(2,2)$ and $(3,3)$ from consideration and leaves $4$ equiprobable states. By contrast, in the original scenario, the six equiprobable states are $(1,2)$, $(1,3)$, $(2,3)$, $(2,3)$, $(3,2)$ and $(3,2)$, none of which are removed, and the “guarantee” that the prize door isn’t opened comes from the way the host picks which door to open based on her knowledge of the prize door.
