Simple question: When viewing the Monty Hall Problem by Bayes' Theorem, one of the key steps involves calculating the conditional probability of $O$, where $O$ denotes the event of Monty Hall opening the curtain b. The probability is as follows:
$Pr(O \mid A) = \frac{1}{2}$, if the prize is behind $a$, Monty Hall can open either $b$ or $c$.
$Pr(O \mid B) = 0$, if the prize is behind $b$, Monty Hall cannot open curtain $b$.
$Pr(O \mid C) = 1$, if the prize is behind $c$, Monty Hall can only open curtain $b$.
Why isn't $Pr(O \mid C)=Pr(O \mid A)$ as $B$ and another curtain are always free for Monty Hall's choosing?
There seems to be a lot of undefined notation here. However, it looks like you are assuming that the player always chooses curtain A for his initial guess. If event A occurs (which seems to be that the prize is behind curtain A), then Monty Hall chooses one of B and C uniformly at random because neither one hides the prize. Therefore, $\mathrm{prob}(O \mid A) = \frac{1}{2}$. If event B occurs, then Monty Hall must choose curtain C in order to avoid revealing the prize, so $\mathrm{prob}(O \mid B) = 0$. Finally, if event C occurs, then Monty Hall must choose curtain B in order to avoid revealing the prize, so $\mathrm{prob}(O \mid C) = 1$. Thus, $\mathrm{prob}(O \mid A) \neq \mathrm{prob}(O \mid c)$ due to the (unstated) assumption that the contestant always chooses curtain A for his initial guess.