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 $10$.
 A: It's more likely to get 10 than 13, although your reasoning is incorrect: would you say that 3 is the most likely outcome then?
For this particular use case: every die, on average, gives a value of $\frac{1+2+3+4+5+6}{6} = 3.5$. So on average, 3 dice give a sum of 10.5, which is closer to 10 than 13.
Note that it's quite specific to this use case: if you had to compare, say, 10 and 11, you'd probably need to look more closely. Also, this method wouldn't work for bimodal distributions, e.g. if the dice were more likely to be a 1 or 6 than a 3 or 4.
A: 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
$${\small{\binom{12}{2}}} - {\small{\binom{3}{1}}}{\small{\binom{6}{2}}} = 21$$
