TITLE: Independence intuition
QUESTION [13 upvotes]: Toss two fair dice. There are $36$ outcomes in the sample space $\Omega$, each with probability $\frac{1}{36}$. Let:

$A$ be the event '$4$ on first die'.
$B$ be the event 'sum of numbers is $7$'.
$C$ be the event 'sum of numbers is $8$'.

It says here $A$ and $B$ are independent. I don't understand why this is the case. What is the intuition behind this? Can someone offer an explanation to me that doesn't involve using the definition of $\mathbb{P}(A \cap B) = \mathbb{P}(A)\mathbb{P}(B)$?
My understanding is informally, an event is independent if the occurrence of one does not affect the probability of the other and vise versa. So if $A$ occurs, wouldn't that affect the probability of $B$? Since if I were to roll a $4$ on the first die, the sample space will be reduced and hence the probability of 'sum of numbers is $7$' will be affected?
It also says $A$ and $C$ are not independent and $B$ and $C$ are not independent. Why?
I think this is because I'm confusing independence with conditional probability?

REPLY [2 votes]: Here's an analogous example from my previous post:
Pairwise independence can be characterised as:

Let $P(A)\neq0.$
Events $A$ and $B$ are independent iff. the knowledge that $A$ happens doesn't change $B$'s probability.

When considering whether events $X$ and $Y$ are independent, the correct intuition is to ask whether the knowledge that $X$ happens changes the probability of $Y$—not whether it restricts how $Y$ can happen.
Consider this experiment: flip two fair coins, letting $H_1$ be the
event that the first coin lands on Heads, and $X$ be the event that
the coins land on different sides. Then $$ P\left(H_1 \cap
X\right)=\frac14=P(H_1)\:P(X);$$ i.e., Events $H_1$ and $X$ are, by definition, independent. Despite this, the knowledge that $H_1$ happens has reduced the number of ways that $X$ can happen: one of $X$'s outcomes, $TH,$ is no longer possible.