Intuition behind independent events in the pebble world In the Pebble World, the definition says that probability behaves like mass: the mass of an empty pile of pebbles is $0$, the total mass of all the pebbles is $1$, and if we have non-overlapping piles of pebbles, we can get their combined mass by adding the masses of the individual piles. The pebbles can be of differing masses and we can also have a countably infinite number of pebbles as long as their total mass is $1$.

In the above image, suppose each pebble weighs $\frac{1}{9}$ (for brevity). The pebbles inside the red box denote event $A$ and the pebbles inside the green box denote the event $B$. Note that the two boxes are intersecting.
If I am to go by the formula, then $P(A \cap B) = \frac{2}{9} = \frac{3}{9} \cdot \frac{6}{9} = P(A) \cdot P(B)$. This means $A$ and $B$ are independent events. Independent events are such that the occurrence of one event says nothing about the occurrence of the other event. How is that exactly happening in this pebble world? I don't understand the independence of events intuitively.
 A: Addendum added to respond to comment question.

My take on the posted question is geometric.  The red box of pebbles is row specific (1st row only) but column neutral.  This means that the chance that a pebble is in the red box is independent of which column the pebble is in.
The blue box is the opposite: column specific (2nd and 3rd columns only), but row neutral.  This means that the chance that a pebble is in the blue box is independent of which row the pebble is in.
So, when examining whether a pebble is in the blue box, the added the information that the pebble is in the red box can not affect the chance that the pebble is in the blue box.
This is because the information that the pebble is in the red box is row specific and column neutral.  Since the blue box is row neutral, specifying that the pebble is in a particular row can not possibly affect the probability of whether the pebble is in the blue box.

Addendum
Response to the comment question of dictatemetokcus.
Throughout this addendum, it is assumed that
$p(A) \neq 0 \neq p(B).$
Using Bayes Theorem, you have that regardless of whether events $A,B$ are independent:
$$p(A|B) ~p(B) = p(A \cap B) = p(B|A) ~p(A). \tag1$$
Using (1) above, and also using the definition of independent events, you have that:
Events $A,B$ are independent 
$\iff~ p(A) ~p(B) = p(A \cap B)$ 
$\iff~ p(A|B) = p(A)$ 
$\iff~ p(B|A) = p(B)$.
Let $C$ denote the event that a pebble is in the  blue box. 
Let $D$ denote the event that a pebble is in the  red box.
Then, based on the above analysis, with $p(C) \neq 0 \neq p(D)$, 
you have that events $C,D$ are independent 
$\iff ~p(C|D) = p(C)$.
The portion of my answer before the Addendum demonstrates that $p(C|D) = p(C).$
