In many online sources I've read a statement similar to:
Probability of B happening given A is equal to the probability of A and B both happening divided by B happening
or
$p(A | B) = p(A \cap B) / p(B)$
But what is the difference between "B happening given A" and "A and B happening at the same time" ? If B happens given A, didn't A and B happen at the same time ? Why is that chance smaller than the intersection ?
On
The difference is that in one case, you already know B has happened. This is not subject to chance anymore, it's a fact. So you compute the probability of A happening, knowing for a fact that B happened
For the probability in the intersection, on the other hand, you don't know that B has happened and have to calculate the probability that both A and B occur
"If B happens given A, didn't A and B happen at the same time?" Yes that is true (or it could be true--it could also be considering sequential events--not necessarily simultaneous ones). However, A and B happening together might be "unlikely" (really it just means they might not always happen together).
So consider that you have three outcomes and exactly two always occur together. That is there are three possible outcomes: A and B, A and C, or B and C. Now what is the probability that A occurs given that B occurs? Well the probability that they both occur is $\frac{1}{3}$ (since A and B is only one possible outcome out of three). But what is the probability that given that B occurs that A occurs? Well there are only two events such that B occurs: either A and B or B and C. Therefore given that B occurs the probability that A also occurs is $\frac{1}{2}$ (because you have excluded the possibility that A and C occur since you assumed B occurred).
This is indeed inline with Bayes' Theorem. The probability of B occurring is $\frac{2}{3}$ and the probability of A and B occurring is $\frac{1}{3}$ therefore the probability of A given B is: $\frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$. (this is probably a poor example of Bayes' Theorem because, in this case, if you chose any two events, you'd always get $p(x|y \neq x) = \frac{1}{2}$)
Second Example: Independent Events
Consider that you flip a (fair) coin twice, sequentially. What are the possible outcomes? You can have $\{HH, TT, HT, TH\}$ (since it's a fair coin, each of these possibilities are equally likely--that's important here although we could extend to situations where they are not equally likely--but then we start getting more into the Bernoulli random variable rather than conditional probabilities). What is the probability that you get a heads on the first flip? Clearly from the above we get $\frac{2}{4} = \frac{1}{2}$.
Now, what is the probability that you flip a heads given that the first flip is heads? There is only a single outcome that satisfies this (i.e. that you get two heads in a row). However there is only one other event, $HT$, such that heads was rolled first, therefore the probability that you get a heads given that the first flip was heads is $\frac{1}{2}$. Again, this is the same as the probability of two heads (heads both times), $\frac{1}{4}$, divided by the probability of getting heads the first time, $\frac{2}{4}$: $\frac{\frac{1}{4}}{\frac{2}{4}} = \frac{1}{2}$.
Indeed we see by the above that the probability of getting heads given an initial flip of heads is the same as the probability of getting heads on the second roll--both are $\frac{1}{2}$. This tells us that each flip is independent of the other since the probability of getting heads isn't affected by the previous flip.