We play with $3$ dice. Which sum of pips on the dice is more probable, $10$ or $13$? (Question isn't as long as it looks.)

We play with $3$ dice. Which sum of pips on the dice is more
  probable, $10$ or $13$?

Without doing any calculation, I would say that it is more probable to have $10$ as sum of pips because we only have numbers between $1$ and $6$ and because $10$ is lower than $13$, it might be easier to get it. I don't know if this makes sense but that's the very first thing that came to my mind. To not just make a wild guess, I tried to calculate it.
Did I do it correctly and is there a faster / better way of doing it? :)

Here are the combinations to get $10$ as sum ($25$ in total):
$$(1,4,5)$$ $$(1,5,4)$$ $$(1,6,3)$$ $$(2,2,6)$$ $$(2,3,5)$$ $$(2,4,4)$$ $$(2,5,3)$$ $$(2,6,2)$$ $$(3,1,6)$$ $$(3,2,5)$$ $$(3,3,4)$$ $$(3,4,3)$$ $$(3,5,2)$$ $$(3,6,1)$$ $$(4,1,5)$$ $$(4,2,4)$$ $$(4,3,3)$$ $$(4,4,2)$$ $$(4,5,1)$$ $$(5,1,4)$$ $$(5,2,3)$$ $$(5,3,2)$$ $$(6,1,3)$$ $$(6,2,2)$$ $$(6,3,1)$$
Combinations for sum $13$ (in total $21$):
$$(1,6,6)$$ $$(2,5,6)$$ $$(2,6,5)$$ $$(3,4,6)$$ $$(3,5,5)$$ $$(3,6,4)$$ $$(4,3,6)$$ $$(4,4,5)$$ $$(4,5,4)$$ $$(4,6,3)$$ $$(5,2,6)$$ $$(5,3,5)$$ $$(5,4,4)$$ $$(5,5,3)$$ $$(5,6,2)$$ $$(6,1,6)$$ $$(6,2,5)$$ $$(6,3,4)$$ $$(6,4,3)$$ $$(6,5,2)$$ $$(6,6,1)$$

So for sum $10$ we have probability of $p=\frac{25}{6^3} \approx 11.6$ %
For sum $13$ we have probability of $p=\frac{21}{6^3} \approx 9.72$ %
Thus, it is indeed more probable to get sum $10$.

By the stars-and-bars formula, if $n,r$ are positive integers, the equation
$$x_1 + \cdots + x_n = r$$
has exactly ${\large{\binom{r-1}{n-1}}}$ solutions in positive integers $x_1,...,x_n$. 

Let the values of the dice rolls be represented as an ordered triple $(d_1,d_2,d_3)$.

First count the number of triples $(d_1,d_2,d_3)$ for the case $d_1+d_2+d_3=10$ .  . .

If we temporarily ignore the condition $d_1,d_2,d_3 \le 6$, the stars-and-bars formula for the equation
$$d_1+d_2+d_3=10$$
would yield ${\large{\binom{10-1}{3-1}}} = {\large{\binom{9}{2}}}$ positive integer triples $(d_1,d_2,d_3)$.

We need to subtract the count of the triples where at least one of $d_1,d_2,d_3$ exceeds $6$.

Given that $d_1+d_2+d_3=10$, at most one of $d_1,d_2,d_3$ exceeds $6$.

Consider the case $d_3 > 6$.

The number of positive integer triples $(d_1,d_2,d_3)$ such that $$d_1+d_2+d_3=10\;\;\text{and}\;\;d_3 > 6$$ 
is the same as the number of positive integer pairs $(d_1,d_2)$  such that $$d_1+d_2<4$$
which is the same, using $x_3$ as a dummy variable, as the number of positive integer triples $(d_1,d_2,x_3)$ such that 
$$d_1+d_2+x_3=4$$
which, by the stars-and-bars formula, has ${\large{\binom{4-1}{3-1}}} = {\large{\binom{3}{2}}}$ solutions.

Hence the corrected count is
$${\small{\binom{9}{2}}} - {\small{\binom{3}{1}}}{\small{\binom{3}{2}}} = 27$$
where the factor ${\large{\binom{3}{1}}}$ accounts for the choice of which of $d_1,d_2,d_3$ exceeds $6$.

By analogous reasoning, the number of triples $(d_1,d_2,d_3)$ for the case $d_1+d_2+d_3=13$ is
$${\small{\binom{12}{2}}} - {\small{\binom{3}{1}}}{\small{\binom{6}{2}}} = 21$$
