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 $\frac{1}{2}$.