Can independence go one way? I.e., so that P(A|B) = P(A), but P(B|A) ≠ P(B) As I understand it, independence of A and B can be informally established by asking whether learning something about one of those events tells you something new about the other. This must be borne out mathematically, however. For example:
If P(A|B) = P(A), then A and B are independent.
And if P(A & B) equals P(A) x P(B), then A and B are independent.
The above imply that P(B|A) = P(B)
It’s this last statement that confuses me, at least in application to certain cases. For example:
You devise a way to randomly choose a number from all real numbers, uniformly distributed. The probability of the chosen number being prime is 0, given that 0% of the reals are prime. Similarly, choosing the number 2 from the set of all real numbers has a probability of 0. Likewise, the probability of choosing a 2 from the set of all prime numbers has a probability of 0. Given that 2 is a prime number, it seems that choosing a 2 and choosing a prime number must be dependent events, at least in one direction (namely, if I know I’ve chosen a 2, then I’m certain I’ve chosen a prime number). Here’s what I mean:
P(2|prime number) = P(2) = 0 (passes for independence)
P(prime number|2) = 1 (i.e., not 0, or P(prime number), and so fails for independence)
But can also test as follows:
P(2 & prime number) = P(2) x P(prime number) = 0
P(2 & prime number) = P(2) x P(prime number|2) = 0
P(2 & prime number) = p(prime number) x P(2|prime number) = 0
Everything here comes out to 0, as I suppose it should. This also aligns with my understanding that anything with probability 0 is independent from any other event. (Right?) And yet, I’m stuck with the intuition that:
If I learn I got a 2, I know I got a prime number, wherein learning I got a prime is insufficient for updating my beliefs about getting a 2 (provided I really do believe that the probability of pulling a 2 from the primes is 0), and the same goes for learning I got an even, a natural, an integer, and so on. Yet I learn I got all those things if I learn I got a 2.
I’ve thought of other examples, though all of them deal with some single event occurring out of a set of infinite possible outcomes. E.g., pulling from the natural numbers: P(2|even) = P(2) = 0; but P(even|2) = 1 (rather than the P(even) = 1/2). So I imagine there’s something I’m naive about in the domain of infinite possible outcomes.
What am I missing?
 A: *

*The symmetry of independence is manifest in its definition $P(A\cap B) = P(A)P(B)$ which makes no reference to conditional probability.

*Conditional probability is problematic when the conditioning event has probability zero, as evidenced by the division by zero that would occur in the definition $P(A\mid B) = \frac{P(A\cap B)}{P(B)}.$ For it to have meaning, we must be in a situation where a limit of $B$ approaching a null event is understood.

*Your example is problematic in other ways. There is no uniform distribution on the real line, nor is there a uniform distribution on the prime numbers. It does not make sense to say the probability that a prime is even is zero. There is a related concept of density, but there is a good reason why density is not regarded as the same thing as probability.

*A less problematic variant of your question would be to look at something like a standard normal $X$ and then consider the events $X>0$ and $X=2.$ It does seem a little strange that we have independence since $$P(X>0,X=2) = P(X=2) = 0 =P(X>0)P(X=2) $$ whereas clearly $X=2$ implies $X>0,$ and we want to write something like $$P(X>0\mid X=2)=1$$ even though it's undefined as a conditional probability (and indeed, we can write something like it... see the caveat to point (2) above regarding there being a well-defined limit.) This is probably best viewed as a counterintuitive fact about independence when events have probability zero or one. An event with probability zero or one is always independent of any other event. As a particular extreme case, note this means an event with probability zero or one is independent of itself!

A: From the definition of conditional probability, we have that $$P(A\mid B)P(B) = P(A\cap B) = P(B\mid A)P(A).$$ Rearranging a bit, we see that 
$$
\frac{P(A\mid B)}{P(A)} = \frac{P(B\mid A)}{P(B)}.
$$
$A$ is independent from $B$ iff the left-hand side is equal to $1$, and $B$ is independent from $A$ iff the right-hand side is equal to $1$. They're equal, so independence is symmetric.
A: First problem: Choose a number from all real numbers, uniformly distributed? There is no such distribution on $\Bbb R$.
Second problem: If you have any probability space $(\Omega,\mathcal F,P)$ and an event $A\in\mathcal F$, then  a probability space $(A,\mathcal F|_A,P')$ is only induced (via $P'(S)=\frac{P(S)}{P(A)}$) if $P(A)>0$.
