The formula $P(E\cap F)=P (E)\times P (F)$ states that these events are independent. How is it that these events do not affect each others outcomes?

A fair six sided dice is rolled twice.

$E=\{ \text{2 appears on the first roll}\}$

$F=\{\text{the sum on the two rolls is an even number}\}$

Using the formula $P(E\cap F)=P(E)\times P(F)$ the event are independent. I cannot understand this however as surely $E$ will affect the outcome of $F$. Could anyone explain how these are independent and how $E$ does not affect $F$?

$P(E)= \frac{1}{6}$ and $P(F)= \frac{18}{36}$. If $E$ occurred would it not affect $F$ outcome?

• Try proving that $P(E \mid F) = P(E)$. – Nate Eldredge Oct 19 '15 at 16:34
• What do you mean by "as surely $E$ will affect the outcome of $F$"? In which way do you think that $E$ affects $F$? Do you think that knowing $E$ modifies the probability of getting $F$? If that is the case, what is your reasoning behind that argument? – Carlos Mendoza Oct 19 '15 at 18:06

The answer is that there is a distinction between statistical independence and causal independence. Yes, whether the first roll is $2$, given what the second roll is, affects the value of $F$. But the question of statistical independence asks, does $P(F | E) = P(F|\neg E)$? The answer to that question is yes, and for the formula to hold true, all we need is statistical independence.
• This is very helpful, as it clarifies the idea very common in math that there usually isn't a complete overlap between mathematical definitions and physical intuition (something I've struggled with many many times, and still do). In this case, all we care about is comparing the likelihood of $F$ happening if $E$ has already happened with the likelihood of only $F$ happening, regardless of whether or not $E$ has an impact on the sum of the dice. – jeremy radcliff Jun 30 '16 at 18:23