Determining number of combinations on odd dice Suppose we had four dice with four sides. Two sides have a $\frac{1}{10}$ chance of being rolled, two sides have a $\frac{4}{10}$ chance of being rolled. The dice score points equal to 1, 3, 4, and 6, with the 1 and 6 sides having the $\frac{1}{10}$ chance.
I'm trying to determine the number of combinations for different scores, but it's much more complex than using regular dice, especially when you usually calculate for only two. I've seen websites that give chances of rolling specific numbers of standard types of dice, but nothing that would allow me to customize the numbers. And writing out all the combinations by hand would take a book's worth of work. 
Any suggestions?
 A: One way to do this is using generating functions.  Note that
$$
f(x)=\frac{1}{10}\left(x + 4x^3 + 4x^4 + x^6\right)
$$
is the generating function for a single such die, in that the coefficient of $x^k$ is equal to the probability of rolling exactly $k$.  The generating function for the sum of four dice is then $f(x)^4$.  You can play around and try to find a nice way to evaluate that polynomial, or just ask WolframAlpha (omitting the $1/10$ prefactor for simplicity), like this; the result is

So the most likely sum is $14$, for instance, and the probability of rolling a sum of exactly $14$ is $1926/10^4=0.1926$.
A: Writing out all the combinations by hand would take a while, but assessing all combinations in Excel takes about 10 minutes. Including the probabilities.
$$
\begin{array}{c} \\
\text{Total}&\text{Combinations}&\text{Probability}\times 10^4\\ \hline
4 & 1 & 1\\
5 & 0 & 0\\
6 & 4 & 16\\
7 & 4 & 16\\
8 & 6 & 96\\
9 & 16 & 196\\
10 & 10 & 352\\
11 & 24 & 816\\
12 & 25 & 1072\\
13 & 20 & 1472\\
14 & 36 & 1926\\
15 & 20 & 1472\\
16 & 25 & 1072\\
17 & 24 & 816\\
18 & 10 & 352\\
19 & 16 & 196\\
20 & 6 & 96\\
21 & 4 & 16\\
22 & 4 & 16\\
23 & 0 & 0\\
24 & 1 & 1\\
\end{array}$$
A: If you didn't have a convenient tool such as Excel or
Wolfram Alpha handy to crunch the numbers for you
(or if you wonder how you could solve this without them),
you could use a kind of "Pascal's triangle" approach.
The idea of Pascal's triangle is that the entries are in rows and you
can compute any given entry by combining nearby entries from the row above it.
So you can fill in the triangle one row at a time;
each time you finish a row you have enough data to (relatively) easily
compute the next row.
In the usual Pascal's triangle, 
if we use $f(n,m)$ to denote entry number $m$ in row number $n$,
we assume that $f(0,0) = 1$ and $f(0,m) = 0$ whenever $m \neq 0$,
which gives us row number $0$.
Then for $n = 1, 2, 3, \ldots$ in turn, continuing as far as we need to,
we compute row $n$ by this formula:
$$f(n,m) = f(n-1,m-1) + f(n-1,m).$$
If instead you use the formula
$$f(n,m) = f(n-1,m-6) + 4f(n-1,m-4) + 4f(n-1,m-3) + f(n-1,m-1)$$
to compute the $m$th entry in the $n+1$st row,
then $f(n,m)$ will be $10^{n}$ times the probability of
rolling a total of $m$ on $n$ dice.
You are interested in row $4$, so you can write the entire portion of
the triangle you need in a grid of five rows and $26$ columns,
in which much fewer than half the entries actually need to be computed at all.
(You can fill in the non-zero entries of rows $0$ and $1$ immediately; 
in fact it's almost not worth even making space for row $0$ or column $0$.)
If you imagine that you are working with dice whose faces are labeled
$0,2,3,5$ instead of $1,3,4,6$ then you can use a grid of even fewer columns
(though you will have to add $4$ to each sum in order to get results
that match the original set of dice).
