independent events in rolling a balanced die once. I understand well the concept of independence if it is rolling a balance die twice. of course, the first roll and the second roll are independent, because the first roll does not impact the second roll.
A:{ 6 appears in the first roll } P(A)= 1/6
B:{ 6 appears in the second roll } P(B)= 1/6
A ∩ B: {both rolls show 6} P(A ∩ B) = 1/36
P(A ∩ B)= P(A)P(B)

Now, I am rolling the die once. Can I say A and B are independent in the following case:
A = {1, 2}; P(A) = 2/6
B = {2, 3, 4} ; P(B) = 3/6
A∩B = {2}; P(A∩B) = 1/6   
P(A∩B) = P(A)P(B)

How to understand the concept of independance that event A = {1, 2} does not affect event B = {2, 3, 4} . I am confused.
 A: In a way it is of course just coincidence that it works out that way. Imagine if you extend $A$ to $\{1,2,5\}$. Then it no longer is the same.
But, it still is true that the chance of $A$ happening is the same whether $B$ happens or not, since if $B$ happens (so we rolled a 2,3, or 4), then we have a 1/3 chance that $A$ happens (that we got the 2 out of those three), and if $B$ does not happen (so we got a 1,5,6) we again have a 1/3 chance.
So: $P(A|B)=P(A)$ (and hence $P(A \cap B) = P(A|B)*P(B)=P(A)*P(B)$) ... So while the presence of $B$ will effect the possible ways $A$ can happen, the chance of $A$ happening remains the same... and in that sense they are still said to be independent.
But you are right to be confused: the way in which these two events are independent is quite different from the way in which the outcomes of two dies are independent:  wih two dice, the chance of getting a 6 is 1 out of six, no matter what the other die comes up with. (And this is how we typically think about independence). But with your example we go from 2 out of 6 to 1 out of 3 .... So the probability stays the same, even as the possible outcomes are effected!
A: Let´s take the example of Bram28 first.
$A=\{1,2,5\}$ and  $B=\{2,3,4\}$ 
Consequently the corresponding probabilities are $P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{2}$. For independency  $P(A| B)=P(A)$ must be true.  If you notice that Event $B$ has occurred then the probability that event $A$ has occured  is $P(A|B)=\frac{1}{3}$. Thus $P(A| B)\neq P(A)$. Or the other way round. If you notice that event $A$ has occurred then the probability that event $B$ has occured  is $P(B|A)=\frac{1}{3}$.
Thus the events are not independent and  $P(A\cap B)=\frac16$

Now have a quick look at the example of your question.

A = {1, 2}; P(A) = 2/6
B = {2, 3, 4} ; P(B) = 3/6


$P(A|B)=\frac{1}{3}$ Thus $P(A|B)=P(A)$. Here the events $A$ and $B$ are independent as you have already investigated. If you notice that event $B$ has occurred then the probability that event $A$ has occured  is the same  (unconditional) probability that event $A$ occcur.
