I am very confused by the Monty Hall Problem. See Wikipedia for the set up.
Suppose we have the three doors $A,B,C$. Suppose $t$ indicates rounds of choosing.
Round 1
Prior to round one, no doors have been opened. Thus, $$P_{t=1} (A)= P_{t=1} (B)=P_{t=1} (C)=\frac{1}{3}$$
The contestant chooses a door.
Round 2
Suppose the contestant choose door $A$. We now learn the prize is not behind door $C$. Therefore, we know $$P_{t=2} (C) = 0$$
But what about A and B.
My Question
Would this problem be less paradoxical if we defined "equally probable" as merely $P(A^c) = P(B^c) = P(C^c) = \frac{2}{3}$?
My Reasoning
In other words, let's suppose we do not know $P(A),P(B),P(C)$ but do know the probabilities for their complements. Then if the contestant chose door A and is shown door C is empty, there would be no surprise since all this has told us is now we have $P(C)$. So we can find $P(B)$ since $P(B) = P(A^c) - P(C)$. This logic makes sense and avoids confusion and leads to the $\frac 2 3$ answer cleanly because we never knew $P(A)$, just $P(A^c)$. So $P(C) = 0$ implies $P(B) = \frac 2 3$ since $P(A^c) = \frac 2 3$.
To me, the Monty Hall solution doesn't make sense if you define the probabilities to the doors explicitly. If you define the probabilities, the logic is "I have $n$ doors, so each should have a $\frac{1}{n}$ chance." But this procedure shouldn't change in the second round because there is no difference in probability between $P(\text{revealed empty door})$ and $P(\emptyset)$, meaning we could have just not included $C$ to begin with. In which case, there are just $2$ doors. So the probabilities for each should be $\frac{1}{2}$.
This is a clever way to think about it and it does work. One can clean up the reasoning you give with probability. In particular, essentially, the important statement is the following:
This is because the last step of your argument where you write $$P(B)=P(A^C)-P(C)$$ needs all those probabilities to be calculated given that the host revealed door $C$ and if $P(A^C)$ weren't independent of that event, your proof fails. Formally, if $R_C$ is the event that the host reveals door $C$, we should be writing: $$P(B|R_C)=P(A^C|R_C)-P(C|R_C).$$
You can see that the host revealing door $C$ is independent from the prize being behind door $A$ intuitively by noting that, if the prize isn't behind door $A$, then the host has to reveal the door the prize isn't behind - so either $B$ or $C$ with $50\%$ probability. Similarly, if the prize is behind door $A$, then the host chooses between doors $B$ and $C$ arbitrarily with $50\%$ probability*. This essentially says $P(R_C|A)=P(R_C|A^C)$ which suffices to show independence.
With that technicality out of the way, we trivially have $P(A^C|R_C)=P(A^C)=\frac{2}3$ and easily achieve the desired result that $P(B)=\frac{2}3$. It's worthy of note that this also means that $A$ is independent of $R_C$, meaning that we, in a way, "lock in" the probability of $A$ being right to $\frac{1}3$ when we choose it, and the host's actions do not affect it since, from our point of view, they act the same regardless of whether the prize is behind our door or not.
If the host doesn't follow this randomized procedure - e.g. if they always chose $C$ when the prize is behind $A$, then the argument fails and $P(B|R_C)$ is in fact less than $\frac{2}3$. This is a subtlety not captured in the argument you give. Moreover, this reveals an important feature: the event that the host reveals door $C$ is different from the event that the prize isn't behind door $C$. That is to say, the naive argument of having a probability behind each door essentially says correctly that $P(B|C^C)=\frac{1}2$, but then incorrectly assumes this is the same as $P(B|R_C)$. This is what defeats the idea of treating door $C$ like it was never even there.