Intuitively grasp why these two events are independent

Question

Two well constructed dice are tossed. Let $A$ denote the event that the first die shows $2$ or $5$ and $B$ denote the event that the sum of the values of the numbers shown on the two tossed dice is $\ge7$.

Are the events $A$ and $B$ independent or dependent?

$P(A) = g/m = 2/6 = 1/3$

$P(B) = g/m = 21/36$

$P(A \cap B) = g/m = 7/36$

$P(A)P(B) = (1/3)(21/36) = 7/36$

Since $P(A)P(B) = P(A\cap B)$ we can conclude that the events $A$ and $B$ are independent.

If we would have chosen the numbers $3$ and $5$ instead of $2$ and $5$ in event $A$ then $P(A\cap B) \neq P(A)P(B)$. However, when choosing numbers whose sum equals $7$, like $1$ and $6$, we get that the events are independent.

The math clearly shows that the events are independent when certain numbers are picked, but is there any intuitive way whatsoever of grasping why this is the case?

I find the result very counter intuitive.

Any explanation of this is greatly appreciated!

Answer 1

It may be less surprising when phrased as the following equivalent problem:

Suppose you roll two six sided dice whose faces are numbered $\pm 1, \pm 3, \pm 5$.

Let $A$ be the event that the first die has magnitude $3$.

Let $B$ be the event that the sum is positive.

Are these events independent?

Answer 2

What is the probability of getting $7$ or more when the first die shows $2$ or $5$?

What is the probability of getting $7$ otherwise?

Are they the same? Yes! Then, no matter if the first die shows $2$ or $5$ or it does not.

So, a more intuitive formula for independence would be

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$$p(A\mid B)=p(A\mid B)$$ That is $$\frac{p(AB)}{p(B)}=\frac{p(A\bar B)}{p(\bar B)}$$

But $p(A\mid B)=p(A)-p(AB)$ because $AB$ and $A\bar B$ are disjoint and their union is $A$. Also, $p(\mid B)=1-p(B)$. Therefore, the last equality can be written as:

$$\frac{p(AB)}{p(B)}=\frac{p(A)-p(AB)}{1-p(B)}$$ or $$p(AB)-p(B)p(AB)=p(A)p(B)-p(AB)p(B)$$ and this is $$p(AB)=p(A)p(B),$$

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the very definition of probabilistic independence.

Note that the formulas between the lines can be derived in reverse order, assuming that no involved probability is $0$.

Answer 3

In terms of the first die $X$ given the event $B$ the conditional probability of $P(X=2|B)$ has obviously diminished compared with $P(X=2)$ but $P(X=5\mid B)$ has augmented precisely as much as the other diminished. So $$P(X\in \{2,5\}\mid B)=P(X\in \{2,5\})$$