I am considering a generalized Monty Hall problem.
Let $D$ be a set of doors, $|D| \geq 2$. A car is behind one door, and goats are behind all the others. Instead of having to choose a single door as in the original Monty Hall problem, you choose a nonempty proper subset $A$ of $D$ out of which you would eventually like to choose one. (There are multiple doors that you suspect might lead to the car, but you cannot decide on one, so you say, “I’ll choose from this one, this one, ... and that one, but haven’t decided which.”) Upon your declaration, Monty, who knows which door has the car, reveals some doors as having goats. Let $B$ be the set of the doors revealed. Monty chooses $B$ so that $A \backslash B \neq \varnothing$ and $\overline{A} \backslash B \neq \varnothing$, where $\overline{A} = D \backslash A$ is the complement of $A$ in $D$. Now, you can maintain your original intention and make your final choice from (the remaining doors in) $A$, or change your mind and choose a door from outside $A$. What is your chance of winning the car in each case? The original Monty Hall problem arises with $|A| = 1$.
Here is my attempt.
Originally, the chance that the car is behind some door in $A$ was $\frac{|A|}{|D|}$. After Monty’s revelation, you have $|A \backslash B|$ doors left in $A$ to choose from, but the chance that the car is behind some door in $A$ remains the same. So the chance of winning the car by choosing a door from $A$ is now $\frac{|A|}{|A \backslash B|\cdot |D|}$.
On the other hand, the chance that the car is behind some door in $\overline{A}$ was originally $\frac{|\overline{A}|}{|D|}$. After Monty’s revelation, there are $|\overline{A} \backslash B|$ doors left in $\overline{A}$, but the chance that the car is behind some door in $\overline{A}$ remains the same. So the chance of winning the car by choosing a door from $\overline{A}$ is now $\frac{|\overline{A}|}{|\overline{A} \backslash B| \cdot |D|}$.
Is this correct? Intuitively, this conclusion is somewhat hard to accept.
EDIT: Since I was asked to clarify how Monty chooses which doors to reveal, let's say we assume either Condition ($\alpha$) or Condition ($\beta$) given below (though they might mean the same). Note that $B$ could be empty. I would appreciate explanations as to how we end up with different probabilities without such conditions or under different conditions.
($\alpha$) $B$ is randomly selected from the set $$ \{X \subseteq D \mid \text{ all doors in $X$ have goats}, A \backslash X \neq \varnothing \text{ and } \overline{A} \backslash X \neq \varnothing \}. $$ That is, all elements of this set have an equal chance of being selected by Monty.
($\beta$) All revealable doors have an equal chance of being included in $B$, where a revealable door is a door $x$ such that (i) $x$ has a goat, (ii) $A \neq \{x\}$ and (iii) $\overline{A} \neq \{x\}$.
Let's consider a game under Condition ($\alpha$) with four numbered doors, $D = \{1, 2, 3, 4\}.$
Suppose you initially choose $A = \{1, 2, 3\}.$ Let's repeat this game many times, making the same choice of $A$ each time.
In the long run, in $3/4$ of the games the car will be behind a door in set $A,$ and Monty has three choices for set $B.$ For example, if the car is behind door $1,$ then $B \in \{\{2\}, \{3\}, \{2,3\}\}.$ In that particular case Monty chooses to open one door with probability $2/3,$ two doors with probability $1/3.$ The same holds when the car is behind another door in $A,$ so altogether, in $1/4$ of all games Monty will open two doors and the car will be behind the only remaining unopened door in set $A.$ Any set of two doors in set $A$ is equally likely to be opened.
Also in the long run, in $1/4$ of the games the car will be behind door $4.$ Then $B \in \{\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}\}.$ Monty will open two doors with probability $1/2$ in that case, so in $1/8$ of all games Monty will open two doors and the car will be behind door $4$. Again, any set of two doors in set $A$ is equally likely to be opened.
In the other $5/8$ of the games Monty opens only one door.
Therefore, if you have gotten to the stage where Monty has opened the doors in set $B$ and you can still make a final choice, and if in this particular game $\lvert B\rvert = 2,$ by Bayes' Theorem the conditional probability that the car is behind the only remaining door in $A \setminus B$ (given all these circumstances) is $2/3.$ But
$$ \frac{\lvert A\rvert}{\lvert A \setminus B\rvert\cdot \lvert D\rvert} = \frac{3}{1\cdot 4} = \frac34, $$
so the alleged formula does not hold for this decision point; it overestimates the conditional probability of winning by picking a door from set $A.$
On the other hand, a similar analysis of the cases where Monty opens only one door shows that when $\lvert B\rvert = 1,$ the conditional probability that the car is behind one of the two remaining doors in $A \setminus B$ (given all these circumstances) is $4/5.$ That is, $2/5$ for one door in $A \setminus B$ and $2/5$ for the other door in $A \setminus B$, whereas
$$ \frac{\lvert A\rvert}{\lvert A \setminus B\rvert\cdot \lvert D\rvert} = \frac{3}{2\cdot 4} = \frac38. $$
In this case, this formula underestimates the conditional probability of winning by picking a door from set $A.$
If $B$ can be the empty set with the same probability as any other permitted set, then (continuing the example above) if the car is in $A$ then Monty opens two doors with probability $1/4,$ so this occurs in $3/16$ of all games. If the car is behind door $4$ then Monty opens two doors with probability $3/7,$ and this happens in $3/28$ of all games. So given that Monty opened two doors, the car is in $A$ with probability $$ \frac{3/16}{3/16 + 3/28} = \frac{7}{11} < \frac34. $$ The case where the car is in $A$ and Monty opens no doors occurs in $3/16$ of all games and the case where the car is behind door $4$ and Monty opens no doors occurs in $1/28$ of all games, and $$ \frac{3/16}{3/16 + 1/28} = \frac{21}{25} > \frac34, $$ so the fact that Monty "does nothing at all" is actually information about where the car is.
It is possible to make the formula work by changing how Monty chooses the doors to open. For example, if you are playing with four doors, you choose $\lvert A\rvert=3,$ and he always opens two doors, the formula gives the correct probability $3/4$ that the car is behind the remaining unopened door in set $A.$
I think (though I have not proved) that if Monty always opens exactly $n$ doors, where $n$ is some fixed number such that $0 \leq n \leq \lvert D\rvert - 2,$ that the formula in the question will then give the correct probabilities. The intuitive reasoning behind this is that the reason for the different probabilities in the examples earlier in this question is that the probability of choosing $n$ doors (for some particular $n$) is different when the car is in $A$ than when the car is in $\bar A.$ If the probabilities are both $1$ then this effect should disappear.