Calculate the probability that the sum of 3 fair dice is at least 13 How would I approach this. I'm more concerned with method than answer. 
 A: $$\begin{align}
\mathsf P(A+B+C\geq 13)
 & = \sum_{a=1}^6 \frac{\mathsf P(B+C\geq 13-a)}{6}
\\[1ex] & = \sum_{a=1}^6 \sum_{b=7-a}^6 \frac{\mathsf P(C\geq 13-a-b)}{36} \tag{why?}
\\[1ex] & = \sum_{a=1}^6\; \sum_{b=7-a}^6 \;\sum_{c=13-a-b}^6\frac{1}{216}
\\[1ex] & = \sum_{x=1}^6\; \sum_{y=1}^x \;\sum_{z=1}^y\frac{1}{216} \tag{why?}
\\[2ex] & = \Box
\end{align}$$
Can you complete?
A: You can try to find the distribution of the output, or perhaps this problem is easier to solve using conditional probabilities. If first dice shows $x_1 \in [1;6]$, what is the probability of success? The other two dice showing at least $13-x_1$; in that way you can decompose the problem into simpler ones.
A: The most naive way is to count all the valid rolls. Although it is not fast solution and you can make an easy mistake it may be instructional to see what probability means in this case.
To ease the notation let's say that 
$$
3\circ5\circ2
$$
means that the first roll is $3$, the second $5$ and the last $2$. Then, since in a matter of the sum this is the same as $2\circ 5\circ 3$ we can write
$$
6\times (3\circ5\circ2),
$$
which means we count roll, where $3,5,2$ appears each once. Also $\sum(1\circ2\circ2) = 5$ denotes the sum of the rolls.
Let's start the counting. The number of all rolls is $216$ since we have three rolls and each roll may the dice have the number from $\{1,2,\ldots,6\}$. So
$$
\underline{6}\cdot \underline{6}\cdot \underline{6} = 216
$$
I won't show how to count all the valid rolls (mostly the easy ones). We can start with the rolls, where all numbers are the same, we get
\begin{align*}
6&\circ 6 \circ 6\\
5&\circ 5 \circ 5
\end{align*}
Each only once. $4\circ 4\circ4$ is not a valid since $\sum(4\circ4\circ4) = 12$. Then we count the rolls where two numbers are the same, let's start with $6$.
\begin{align*}
3\times&(1\circ 6\circ6)\\
3\times&(2\circ 6\circ6)\\
&\vdots\\
3\times&(5\circ 6 \circ6)
\end{align*}
Do not count $6\circ6\circ 6$ again! Each of those rolls we must count three times, because valid roll is $1\circ 6\circ6$, but also $6\circ 1\circ6$ and $6\circ6\circ1$. For $5$ we have 
\begin{align*}
3\times&(4\circ5\circ5)\\
3\times&(6\circ5\circ5)
\end{align*}
and for $4$
\begin{align*}
3\times &(5\circ4\circ4)\\
3\times &(6\circ4\circ4)
\end{align*}
There are no more valid rolls with two same numbers, the last bit (and the hardest I think ) is to count the rolls where each number is only once, but I hope this gave you general idea, how to do it.
So far we have 29 I think. So the result will be something like
$$
\frac{29+c}{216}
$$
where $c$ means the number of valid rolls without any repeting number.
Again this is not very nice approach, but I think it shows what we can imagine behind probability.
A: Can be done using generating functions which reduces down to polynomial multiplication.
$G(x)=\left(\frac{x}{6}+\frac{x^2}{6}+\frac{x^3}{6}+\frac{x^4}{6}+\frac{x^5}
{6}+\frac{x^6}{6}\right)^3 $
$G(x)$ models throwing 3 fair dice.
When that is expanded.
$\frac{x^3}{216}+\frac{x^4}{72}+\frac{x^5}{36}+\frac{5 \
x^6}{108}+\frac{5 x^7}{72}+\frac{7 x^8}{72}+\frac{25 \
x^9}{216}+\frac{x^{10}}{8}+\frac{x^{11}}{8}+\frac{25 \
x^{12}}{216}+\frac{7 x^{13}}{72}+\frac{5 x^{14}}{72}+\frac{5 \
x^{15}}{108}+\frac{x^{16}}{36}+\frac{x^{17}}{72}+\frac{x^{18}}{216}$
The ones that count are all the x's whose exponent are greater than or equal to 13.
$\frac{7 x^{13}}{72}+\frac{5 x^{14}}{72}+\frac{5 \
x^{15}}{108}+\frac{x^{16}}{36}+\frac{x^{17}}{72}+\frac{x^{18}}{216}$
since the x's are just placeholders and the coefficients are what counts we substitute $x = 1$
$P(\geq 13)=\frac{7}{72}+\frac{5}{72}+\frac{5}{108}+\frac{1}{36}+\frac{1}{72}+\frac{1}{216}=\frac{7}{27}$
