Conditional probability intuition.  Kolmogorov definition - Wikipedia
According to the Kolmogorov definition, given two events A and B, 
$P(A|B) = \frac{P(A \cap B)}{P(B)}$ .
But I dont really understand why this is true. Why isn't $ P(A|B) = P(A \cap B) $??

Somebody please help me get an intuition why the Kolmogorov definition is true??
 A: Check out this blog post. It's about a consequence of conditional probability, Bayes Theorem, but very helpful for visualizing how conditional probability works. 
You can think about the $P(B)$ in the denominator as reminding you that you're in the $B$ universe; you want to know the probability of $A$ knowing that $B$ has occurred. 
A: For one thing, it has to be a probability, so certainty must correspond to probability $1$, not $P(B)$.
An example may be helpful.  Consider choosing (with equal probabilities) a number from $1$ to $10$.  Let $B$ be the event that the number chosen is odd, $A$ that the number $\le 3$.
If you know that the number is odd, that cuts down the possibilities to $1, 3, 5, 7, 9$,
still equally likely, of which $1$ and $3$ are still in $A$.  So $P(A|B) = 2/5 = P(A \cap B)/P(B)$. 
A: $P(A|B)$ is the probability of $A$ happening, when you know that $B$ happened. This is not the same as $P(A\cap B)$, because then, you do not take into account that $B$ happened for sure. The probability of $B$ happening is $P(B)$, so you have to devide by that to obtain the probability of both of them happening assuming $B$ already happend, so $P(A|B)$.
EDIT
By definition, $P(A|B)$ is equal to
$$
P(A|B)=\frac{P(A\cap B)}{P(B)}
$$
because usually, $P(B)\neq 1$, they are not equal.
