Circular conditional probability Is it true that:
$$
P(A \mid P(A) = p) = p
$$
It sounds logical but is there a way to derive it nicely?
 A: By just filling in the definition of conditional probability, we see that $$P(A|P(A)=p)=\frac{P(A\mathrm{\ and\ } P(A)=p) }{P(P(A)=p)}=\frac{P(A)\cdot P(P(A)=p)}{P(P(A)=p)}=P(A)=p.$$ Note that $P(P(A)=p)=1$ if we assume '$P(A)=p$' is true. We assume '$P(A)=p$' is true and we see that a random event ($A$ or $A^c$) is independent from an event that is true with probability one.
A: The issue you encounter here is that a conditional probability statement must be conditional on an event in the probability space.  You are trying to define an event by specification of a probability condition on an event; that is going to lead you to problems, since there is the probability condition is actually referring to a subset of the sigma-field of events.
To see this, consider a probability space $(\Omega, \mathscr{G}, \mathbb{P})$ and suppose we interpret the conditioning statement $\mathbb{P}(A) = p$ as referring formally to the following class of events:
$$ \mathscr{A}_p = \{ A \in \mathscr{G} | \mathbb{P}(A) = p \} \subseteq \mathscr{G}.$$
The classes $\{ \mathscr{A}_p | 0 \leqslant p \leqslant 1 \}$ form a partition of $\mathscr{G}$.  The problem here is that each $\mathscr{A}_p$ is a subset of $\mathscr{G}$ rather than being an event (i.e., an element of $\mathscr{G}$).  This means it does not make sense to specify $\mathscr{A}_p$ as a conditioning event; it is not an event in the probability space!
Okay, so how do you deal with this properly?  Well, you can specify what you want to do here by defining a random variable in the probability space to measure the conditional probability of the event of interest.  Given an event of interst $A \in \mathscr{G}$ we define a random variable $\theta: \Omega \rightarrow [0,1]$ that is $\mathscr{G}$-measurable and has the following conditional probability:
$$\mathbb{P}(A|\theta) = \theta.$$
This random variable effectively measures the probability of the underlying event $A$ conditional on itself.  Now, we clearly have $\mathbb{P}(A|\theta=p) = p$, which is a reasonable translation of your statement, so we can achieve the rule you want by defining a random variable measuring the probability of an event.  Notice that this form is slightly different from what you are trying to do, since the random variable $\theta$ represents the probability of $A$ occurring conditional on knowledge of $\theta$.  The marginal probability of $A$ is obtained via the law-of-total probability as $\mathbb{P}(A) = \int \mathbb{P}(A|\theta) \pi(\theta)$.
