Why $P(A)\cdot P(B \mid A) = P(B)\cdot P(A\mid B)$ for dependent events? Repeating the question

Why $P(A)\cdot P(B \mid A) = P(B)\cdot P(A\mid B)$?

It makes total sense to me when we consider independent events. However, I struggle to understand why it also holds for dependent events.
If we assume that $P(A)\cdot P(B \mid A) = P(B)\cdot P(A\mid B)$ holds, then we also assume that the order of the events happening does not matter. Why would we?
Let's consider an example. Let $A,B$ be arbitrary two events. Let $P(A) := \frac{1}{3}$ and $P(B) := \frac{1}{3}$.
Suppose that if $A$ happens, then probability of of $B$ happening is $\frac{1}{5}$, or in other words, $P(B|A) = \frac{1}{5}$
And suppose that if $B$ happens, then probability of $A$ happening is $\frac{1}{6}$, or $P(A|B) = \frac{1}{6}$
We have
$$P(A) \cdot P(B\mid A) = \frac{1}{3}\cdot \frac{1}{5} = \frac{1}{15} ≠ \frac{1}{18} = P(B) \cdot P(A\mid B)$$
So in this example, clearly, equality doesn't hold. 
I assume it would be possible to come up with some "real world" examples which would also show that equality might not necessarily hold.

So what am I missing?
 A: 
"Let's consider an example. Let $A,B$ be arbitrary two events. Let $P(A) := \frac{1}{3}$ and $P(B) := \frac{1}{3}$.
Suppose that if $A$ happens, then probability of of $B$ happening is $\frac{1}{5}$, or in other words, $P(B|A) = \frac{1}{5}$"

From that you can conclude that the probability that both events will occur is $P(A)P(B\mid A)=\frac13\frac15=\frac1{15}$.
With exactly the same reasoning we come to the conclusion that the probability that both events will occur equals $P(B)P(A\mid B)$ so our final conclusion is that:$$P(A)P(B\mid A)=P(A\cap B)=P(B)P(A\mid B)$$
A: note, that $P(A) P(B | A) = P(AB)$ is not a theorem, that holds by definition of conditional probability $P(B | A) = \frac{P(AB)}{P(A)}$. Similarly $P(A | B) = \frac{P(AB)}{P(B)}$ by definition, and that implies the bayes formula $P(AB) = P(A) P(B | A) = P(B) P(A | B)$
A: Suppose a box contains 5 white and 3 black balls. You take 2 balls without replacement. So, the second ball color depends on the first ball color. Hence:
$$P(W_1\cap B_2)=P(W_1)\cdot P(B_2|W_1)=\frac{5}{8}\cdot \frac{3}{7}=\frac{15}{56}\\
P(B_1\cap W_2)=P(B_1)\cdot P(W_2|B_1)=\frac38\cdot \frac57=\frac{15}{56}$$
