When should we consider objects as distinguishable in probability? Example : Why is the probability of getting a sum of 12 when we roll two fair dices is 1/6 whereas probability of getting  5  is 2/6 . When we labeling the dice  red and green in our head , isn't (6 of red dice , 6 of green dice ) different from (6 of green dice ,6 of red dice ) ? Aren't we using the same thing to differentiate between [ 3 (Red) , 2 (Green) ] and [ 3 (Green) , 2 (Red) ]?
Also a follow up question that confused me : what is the probability of getting at-least one 6 when we throw 6 fair dices . Here we calculate the complement first and get the answer as 1 - (5/6) Power (6) . But in this answer aren't we over-counting these cases (5,5,5,5,5,5) , (2,2,2,2,2,2) and so on ?
 A: Getting a sum of $12$ means there is only one kind of combination: $(6,6)$, therefore the probability is $\frac16\times\frac16=\frac{1}{36}$ 
Getting a sum of $5$ means these possible combinations: $(1,4),(2,3),(3,2),(4,1)$, the possibility is therefore $\frac{4}{36}$.
As for your follow-up question, $(\frac56)^6$ is denoting the possibility where there is no $6$ when we throw 6 dices. It is interpreted as: For the first dice, the numbers could be $(1,2,3,4,5)$, which are $5$ out of $6$ possible outcomes;
The second dice can also be one of the number among  $(1,2,3,4,5)$, same for the third, fourth, fifth, and sixth dice.
So for example, you have $1$ on the first dice and again $1$ on the second dice, has in fact the same possibility with having $1$ on the first dice and $2$ on the second dice. So the possibility of having $(2,2,2,2,2)$ is the same with having $(1,2,3,4,5)$. You are simply choosing from a pool of numbers each with the same possibility of being chosen.
A: I used to say, take the identical dice, get some paint, draw a tiny red dot on one of them, draw a tiny green dot on the other. Now, draw all 36 outcomes, along the left write the number 1,2,3,4,5,6 in a column, that will be the number that the red die shows. Across the top row, 1,2,3,4,5,6, that will be the green die. Fill in the 36 squares with the sum. 
Maybe I can figure out how to draw that.
A: The probability of getting a sum of $12$ when we roll two fair dice is $1/36$.  You can think of this experiment as being a sequence of two experiments: first you roll the red die, and then you roll the green die. Or if the two dice are indistinguishable, you can still roll them one after another rather than together.  It is now clear that the sample space has $36$ elements.  Only one element in this sample space - the $2$-tuple $(6,6)$ - has a sum of $12$.
