Roll a die $2$ times. What's the probability that the rolled sum is at least $6$? 
Roll a die / cube $2$ times. What's the probability that the rolled
  sum is at least $6$?

I'm not quite sure how this is solved. So we got $2$ cubes. In total we have $6^2=36$ different possibilities. But we want the reverse order too, so we need to multiply $36$ by $2$, we have $72$. This is the denominator of the fraction. Now we need to look for the enumerator; amount of sums which is at least $6$.
Now I would just write all possibilities (without reverse order) on a paper and then multiply by $2$. But this is very inefficient.. Is there a better way of doing this? And is my approach correct at all? :s
So I got $26 \cdot 2 = 52$
In the end we have $$\frac{52}{72}= \frac{13}{18}\approx  72,2 \text{%}$$
Result seems realistic for me at least.
 A: Your multiplication by $2$ is not correct.  The order of throws is already accounted for in the $6^2=36$.  You might then notice that there is one way to get a sum of $2$, two ways to get a sum of $3$, etc. for a total of $10$ ways to get a sum less than $6$.  The chance of getting at least six is then $1-\frac {10}{36}=\frac{13}{18}$
A: This isn't the most efficient solution, but it will help you to understand what's going on.
Let's start by making a table of the possibilities:
\begin{array}{cccccc}
(1,1)& (1,2)& (1,3)& (1,4)& (1,5)& (1,6)\\
(2,1)& (2,2)& (2,3)& (2,4)& (2,5)& (2,6)\\
(3,1)& (3,2)& (3,3)& (3,4)& (3,5)& (3,6)\\
(4,1)& (4,2)& (4,3)& (4,4)& (4,5)& (4,6)\\
(5,1)& (5,2)& (5,3)& (5,4)& (5,5)& (5,6)\\
(6,1)& (6,2)& (6,3)& (6,4)& (6,5)& (6,6)
\end{array}
There are $36$ possibilities, and the sum is constant along diagonals going from bottom-left to top-right. We see that the $(5,1)-(1,5)$ diagonal is the first with sum equal to $6,$ and there are $26$ possible pairs that are on or below this diagonal. Therefore the probability is equal to $$\frac{26}{36}=\frac{13}{18}.$$
