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Say we throw two dice. Firstly, we get 2 which is our set $A$. Secondly, we get 3 which is our set $B$. What does it mean for that these two sets are dependent?

$A = \{2\}$, $B = \{3\}$, sample space = $\{1, 2, 3, 4, 5, 6\}$

Because $P(A \cap B)\neq P(B)P(A)$ Therefore $A$ & $B$ are dependent.

user829347
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exer240
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    See Independence – Mauro ALLEGRANZA Jan 24 '22 at 14:45
  • @MauroALLEGRANZA As I understand is that two sets are independent if they do not affect the outcome of the other. So why does in this case A affect the outcome of B? (dependence, I guess(?)) – exer240 Jan 24 '22 at 14:49
  • Since it's a dice every side is available at every throw. – exer240 Jan 24 '22 at 14:54
  • "First we get 2 which is our set A"... This seems off. Do you mean to say you throw two dice in sequence and $A$ is the random variable corresponding to the result of the first die and $B$ is the random variable corresponding to the result of the second die? Then yes, the random variables $A$ and $B$ are independent. The sample space you refer to is not the sample space of the experiment as a whole but rather just the result of a single die. If you want the sample space of the experiment as a whole that would be ${(1,1),(1,2),(1,3),\dots,(1,6),(2,1),\dots,(6,5),(6,6)}$ – JMoravitz Jan 24 '22 at 15:01
  • And here, the event that $A$ took the value of $2$ would have corresponded to ${(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)}$ and similarly $B$ taking value three would have corresponded to ${(1,3),(2,3),(3,3),\dots,(6,3)}$. The intersection here would have been ${(2,3)}$. As for probabilities, $\frac{1}{6}\times\frac{1}{6}$ does indeed equal $\frac{1}{36}$ – JMoravitz Jan 24 '22 at 15:03
  • @JMoravitz Thank you, this cleared it up! So to summarize, it is a big difference in doing it in sequence and parallell? – exer240 Jan 24 '22 at 15:19
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    Not at all. Doing it parallel and doing it in sequence you still have the same description of the sample space as being pairs of die results. Note also that if using two indistinguishable dice and throwing them at the same time it is better to imagine that the dice were in fact distinguishable (e.g. a red die and a blue die) so that we can use counting techniques to find probabilities, otherwise some of the outcomes in our sample space would occur at different rates than others. – JMoravitz Jan 24 '22 at 15:23
  • As I understand is that two sets are independent if they do not affect the outcome of the other. Not quite: the correct question is whether the *probability* of the other's outcome is affected. I elaborate here: independence intuition. – ryang Jan 24 '22 at 15:29

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Before you threw the dice you had

$P(A=2)=1/ 6$ and $P(B=3) = 1 /6$. Also $P((A=2)\cap(B=3))=1/36$.

The events are independent. It is irrelevant whether you throw one die twice or two distinct dice at the same time.