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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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    $\begingroup$ In neither cases are they causally independent, which I think relates to your intuition. But in the latter case they are uncorrelated. $\endgroup$ – Lucas Dec 10 '12 at 22:38
  • $\begingroup$ @Lucas: You say that $F$ and $G$ are uncorrelated. What do you mean by that? I know that the correlation applies to random variables, but here it is about events. Also, $P(F\cap G) = P(F)P(G)$ means, by definition, that $F$ and $G$ are independent events. What do you mean that they are not "causally independent"? $\endgroup$ – digital-Ink Sep 17 '16 at 12:51
  • $\begingroup$ @digital-Ink good question. In that comment I was language to highlight the distinction between causation and correlation - and avoid using the terms dependent independent. So I was using correlation loosely here (though you could assign numerical values to the events and speak of correlation properly) I understood it to be implicit in the question that the first intuition would be that neither are be dependent - because neither dice affects (causally) the other. The question of what causation is is complex. Here, by causally independent I mean something not physically affecting something else $\endgroup$ – Lucas Sep 17 '16 at 20:28
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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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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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Think about it this way: when you roll a any number the first time, like a 4, there is only 1 out of 6 possibilities that the next number will give you a sum of seven, thus 1/6

Like Micah said, there are exactly 6 combinations of the two die that will give you seven, out of a total of 36 possible combos. 6/36=1/6

Note the probability of getting a sum of seven with two die, is equal to the probability of getting a sum of seven after rolling 4. That is the definition of independence.

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Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring

or P(B|A) = P(B)

Coming to our example,

As you had mentioned earlier,

𝐸 -> the sum of the dice is 6

F -> the sum of the dice is 7

G -> the first die rolled is a 4

F is a special case because we have 6 different ways of getting a sum of 7 unlike any other number.(Think of the sample space for any other sum.)

Lets look at the sample space of F

F = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

Notice that, irrespective of what the number is on the first or second die, the probability of getting a sum of seven is a constant.

This isn't the case for event E (the event that the sum of the dice is 6), let's confirm the sample space

E = {(1,5), (2,4), (3,3), (4,2), (5,1)}

Notice that, the event of getting a 6 on either die excludes the possibility of getting a sum of 6 and hence our probability is influenced by/dependent on that.

This, IMHO, is the intuitive reason for the independence of events F and G.

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