Don't think that this is my homework. Also, I do not have my graduation in mathematics. I was looking at different exams and their sample papers and came across this question...

If a fair die (with 6 faces) is cast twice, what is the probability that the two numbers obtained differ by 2?
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> 
> (A) 1/12
> 
> 
> (B) 1/6
> 
> 
> (C) 2/9
> 
> 
> (D) 1/2
> 
> 
> 

I believe the answer is 2/9 and I want to find out if I am correct.

For the first time rolling the dice, probability would be 1/6

(as per my knowledge) there are 8 possible combinations for rolling the dice for the second time and difference between two numbers is equal to 2

```
1 -> 3

2 -> 4

   |-> 1
3 -|
   |-> 5

   |-> 2
4 -|
   |-> 6

5 -> 3

6 -> 4

```

And thus, final probability as per my knowledge should be (1/6)(8/6) = 2/9

Am I correct?

Also I would welcome better approaches that can solve this problem more quicker.