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Roll $2$ dice.
Let $E$ be the event that the sum of the dice is $6$
Let $F$ be the event that the sum of the dice is $7$
Let $G$ be the event that the first die rolled is a $4$

$E$ and $G$ are dependent (since $P(E\cap G) \neq P(E)P(G)$ )
$F$ and $G$ are independent (since $P(F\cap G) = P(F)P(G)$ )

Intuitively, why is this true?

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In neither cases are they causally independent, which I think relates to your intuition. But in the latter case they are uncorrelated. –  Lucas Dec 10 '12 at 22:38

2 Answers 2

It's slightly more intuitive to think about this problem in terms of conditional probabilities.

We want to understand why P(F | G) = P(F). There are exactly 6 pairs of dice values whose sum is 7: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. That is for each possible outcome of the first roll, there is exactly one (unique) outcome of the second roll which would yield a sum of 7. In other words, given the value of the first die, the probability that the sum of the dice roll is 7 is 1/6.

P(E|G) != P(E) because there are values of the first roll which make it impossible to have a sum of 6 - in particular, if the value of the first roll is 6. Therefore, knowledge of the value of the first die changes the probability that the sum of the two dice is 7.

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Roughly speaking, if you know the sum of the dice is $6$, that gives you some information about what the first die rolled: namely, that it couldn't possibly have rolled a $6$. On the other hand, if you know the sum of the dice is $7$, that doesn't tell you anything about the result of the first die. For any number $n$ it might have rolled, the second die could have rolled the number $7-n$. So events relating to the result of the first die are independent of whether you rolled a $7$, but not of whether you rolled a $6$.

You can make this a little more rigorous by thinking in terms of conditional probabilities: if $A$ and $B$ are events with nonzero probability, then they're independent if and only if $P(A|B)=P(A)$ (or $P(B|A)=P(B)$). For your events, a bit of case analysis will show you that $P(G|F)=1/6=P(G)$, but $P(G|E)=1/5 \neq P(G)$.

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