I can't find or think of an intuitive way to think about P(A∩B) for non-independent events A and B. I get that P(A∩B) = P(A) * P(B) for independent events.  That makes complete intuitive sense for me.  But I am struggling to think about P(A∩B) for events that overlap (that is, when B happening implies A happened or vice versa).  
I was reading Introduction To Algorithms and it mentioned the following question: if you flipped a coin twice, what is the probability you flipped two heads given that you know at least one flip was a head. 
The breakdown made total sense: there are three scenarios that have at least one head, HH, HT, TH.  Of those three, one is HH.  P(A|B) = 1/3.  
But then it showed the formula for P(A|B) = P(A∩B)/P(B)
So in this case it's P(both were heads ∩ at least one was heads) / (3/4)
Which means that P(A∩B) = 1/4.  Which makes sense to me still (every single time A happens B happens so it's just the probability of A).  But I can't think of a general way to think about this in cases where A doesn't always imply B.  Can anyone help me out?  
 A: The best way to understand it in simple situation is in terms of set theory:
$$
P(A\cap B)=\frac{\text{ the number of elements common to set $A$ and set $B$}}{\text{ the total number of possibilities}}.
$$
A: You have the formula for conditional probability. Now just rearrange it to solve for the probability of the intersection: $P(A\cap B) = P(B)P(A \mid B)$. The right-hand side can be interpreted as applying the ``rule of product''. Following successive branches of the probability tree into finer and finer partitions of the sample space (not only does event B happen, but so does event A, or perhaps event Z, etc.), you get the probability of your intersection.
A: An example where $A, B$ are not independent but neither implies the other might run as follows: suppose you roll two "three-sided" dice, which can roll 1,2, or 3 uniformly and independently. Let $A$ be the event that you roll at least one $2$ and let $B$ be the event that you roll no $3$s. As there are only $9$ outcomes, you can still list them all fairly easily, and you should still have $P(A|B) > P(A)$, but neither $A$ nor $B$ implies the other.
A: Suppose that we roll a fair die twice and add the results. We are interested in the probability that the sum is an even number given that the sum is larger than seven. We have that
\begin{align*}
&P(\text{sum is an even number}\mid\text{sum is larger than seven})=\\
&\quad=\frac{P(\text{sum is an even number}\cap\text{sum is larger than seven})}{P(\text{sum is larger that seven})}\\
&\quad=\frac{9/36}{15/36}=\frac 35.
\end{align*}
If the sum is an even number, it does not imply that the sum is larger than seven. If the sum is larger than seven, it does not imply that it is even.
I hope this example is useful.
