What is the intuition behind the product rule of probability? Let's say that I have two independent events $A$ &amp; $B$ from two different experiments. 

Assume that $P(A) = 3/13$ i.e. Out of the total 13 ways that an experiment can be performed, 3 ways make $A$ happen.

Assume that $P(B) = 7/13$ i.e. Out of the total 13 ways that another experiment can be performed, 7 ways yield $B$.

The product rule says $P(A \cap B) = P(A) * P(B)$. In this case the combined probability is $$P(A \cap B) = P(A) * P(B) = \frac {3}{13} * \frac {7}{13}$$ This makes sense to me that $13 * 13$ is actually the total number of ways that both experiments can be performed together &amp; $3*7$ is the number of ways that $A$ and $B$ together can happen (by the fundamental rule of counting).

But this intuition quickly falls apart if $A$ &amp; $B$ were dependent events of the same experiment. Let's say that experiment can be performed in 13 ways. $P(A) = \frac {3}{13}$ &amp; $P(B) = \frac {7}{13}$ and lastly, $P(B|A) = \frac 13$. Now ,
$$P(A \cap B) = P(A) * P(B|A) =\frac {3}{13} * \frac {1}{3} = \frac {3}{39}$$
So $\frac {3}{39}$ shows 3 ways out of "39" ways the two events of the experiment happen. But the original experiment could only be done in "13" ways, what does 39 here represent?

I can understand if its unclear what I'm trying to convey. In essence, what is the intuition behind the product rule of probability?
Well, you've sort of found out that if your configuration space is $\Omega=\{1,2,...,n\},$ your probability distribution is 
uniform (say you're rolling with a fair $n$-sided die as opposed to considering the sum of eyes on two fair six-sided dies, for instance), then any event $A$ such that $\mathbb{P}(A)=\frac{k}{n}$ with $k$ and $n$ co-prime can never be independent of any other non-trivial event - if $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)=\frac{kj}{n^2},$  then there needs to be a set in $\Omega$ with probability $\frac{kj}{n^2}$, which can only happen in the above set-up if $n$ divides $kj$.

However, if $A$ and $B$ are events corresponding to different trials of experiments that are essentially the same, then you're back to the situation with a "natural" configuration space of cardinality $n^2$, and thus, you get independent events. This should be somewhat intuitively satisfying: It's not all that clear how many independent sub-events of a single experiment there really should be, unless the experiment involves independent components - say you're rolling two dice, or making measurements on two people and so on and so forth.
You can define indipendence in an other way
$$A \perp B \iff P(A|B)=P(A)$$
Which means that che event B doesn't affect A.
Now if you agree that the definition of
$$P(A|B)=P(A,B)/P(B)$$
Is intuitive, It becomes trivial that
$$A \perp B \iff P(A,B)=P(A)P(B)$$