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?

  • $\begingroup$ Try proving that $P(E \mid F) = P(E)$. $\endgroup$ – Nate Eldredge Oct 19 '15 at 16:34
  • $\begingroup$ 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? $\endgroup$ – Carlos Mendoza Oct 19 '15 at 18:06

Since there is the same amount of even and odd numbers on a dice, it is independant. No mather what's rolled on the first dice, you have a 50% chance of having an even sum after you roll the second dice.


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.

  • $\begingroup$ What do you mean by statistical and casual independence ? $\endgroup$ – John Oct 19 '15 at 16:52
  • $\begingroup$ Statistical independence is condition of the equality in my answer holding. When I refer to causal independence, I just mean that in a particular instance, the value of one has an effect on the process that results in the value of the other being determined. $\endgroup$ – Eric Brooks Oct 19 '15 at 21:31
  • $\begingroup$ 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. $\endgroup$ – jeremy radcliff Jun 30 '16 at 18:23

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