How do you do this style of dice problem: If your roll three dice what’s the probability that the sum is x? So effectively what I’m asking is how you do problems like with a number of dice y, what is the probability that the sub is x? Is there any way to do this besides memorizing the parabolas of amount of possibilities for each answer for each  dice. 
 A: Generalized Pascal's Triangle
One of the simpler ways is to use a generalized version of Pascal's Triangle.
In Pascal's triangle, we add the two entries just above the current one we want to calculate.  In the generalized version, we add more entries.  
In the case of six-sided dice, we add the six entry above each new one and get the following structure.
$$ 1 \\
1~~1~~1~~1~~1~~1 \\
1 ~~ 2 ~~ 3 ~~ 4 ~~ 5 ~~ 6 ~~ 5 ~~ 4 ~~ 3 ~~ 2 ~~ 1 \\
1 ~~ 3 ~~ 6 ~~ 10 ~~ 15 ~~ 21 ~~ 25 ~~ 27 ~~ 27 ~~ 25 ~~ 21 ~~ 15 ~~ 10 ~~ 6 ~~ 3~~ 1
$$
From this we can build a table.
$$
\begin{matrix}
\text{roll total} & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\  
\text{ways it can happen} &1 & 3 & 6 & 10 & 15 & 21 & 25 & 27 & 27 & 25 & 21 & 15 & 10 & 6 & 3 & 1
\end{matrix}
$$
So, for instance, if we want to find the probability of rolling a 7, we see that there are 15 ways this can happen.  Since there are $6^3=216$ total possibilities, we have 
$$P(7)=\frac{15}{216}=\frac{5}{72}$$
I usually use Excel to run the actual calculations if I go beyond 3 levels (it gets pretty tedious otherwise).
A: Suppose you are exploring $x=10$.  Start by imagining ghe dice to be distinct, call them red, green and blue.  To get that you need one of the following:
Red + green total is 4, blue gives 6 --> 3 permutations
Red + green total is 5, blue gives 5 --> 4 permutations
Red + green total is 6, blue gives 6 --> 5 permutations
Red + green total is 7, blue gives 5 --> 6 permutations
Red + green total is 8, blue gives 2 --> 5 permutations
Red + green total is 9, blue gives 1 --> 4 permutations
Add that up to get a total of $27$ out of $216$ so the probability for $x=10$ is one-eighth.
You should be able to figure out other values of $x$ similarly.
As an aside:  $10$ divided by $7$ gives a remainder of $3$.  If you add up the permutations for all sums giving this remainder ($3, 10, 17$) you get $31$ permutations.  You also get $31$ for each other remainder except $0$; that gives $30$ permutations.  Thus summing three standard dice modulo $7$ closely approximates a fair seven-sided die.
A: Encode using probability generating functions (pgfs). Let $X$ be the sum after performing the experiment. Then $X=\sum_{i=1}^3 X_i$ where $X_i$ are indepedent uniform $\{1,\dotsc, 6\}$ random variables. Let $g_X$ be the pgf of $X$. Since the $X_i$ are independent, it follows that 
$$
\sum_{i}P(X=i)t^i=g_X(t)=Et^{X}=\prod_i Et^{X_i}=\frac{1}{6^3}(t+t^2+\dotsb+t^6)^3=\frac{t^3}{6^3}\frac{(1-t^6)^3}{(1-t)^3}.
$$
Using the extended binomial theorem we can rewrite the previous line as
$$
\sum_{i}P(X=i)t^i=\frac{t^3(1-t^6)^3}{6^3}\left(\sum_{m=0}^\infty\binom{m+2}{2}t^m\right)
$$
Thus to find the probability $P(X=i)$ it sufficies to compute the coefficient of $t^i$ on the RHS.
