Number of ways to throw at most 14 with 4 dice - generating functions Determine the chance to throw at most 14 with 4 normal dice. I will set up the right generating function to determine the number of ways tot thow at most 14 with 4 normal dice and I need some help. I know that:
\begin{align}
x_1 + x_2 + x_3 + x_4 \leq 14 \qquad \text{with} \qquad 1 \leq x_i \leq 6. 
\end{align}
\begin{align}
x_1 + x_2 + x_3 + x_4 + x' = 14 \qquad \text{with} \qquad 0 \leq x' \leq 10
\end{align}
So we have:
\begin{align}(x + x^2 + x^3 + x^4 + x^5 + x^6)^4 \cdot (0 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10}).
\end{align} 
Does this make sense? I do not have to determine the coefficient of $x^{14}$ only the right generating function.

The probability generating function for rolling a normal die is:
$G(t)=\frac{t}{6} + \frac{t^2}{6} + \frac{t^3}{6} + \frac{t^4}{6} + \frac{t^5}{6} + \frac{t^6}{6}$
To find the probability generating function for the sum of four independent dice then you have
$G_S(t)=\left(\frac{t}{6} + \frac{t^2}{6} + \frac{t^3}{6} + \frac{t^4}{6} + \frac{t^5}{6} + \frac{t^6}{6}\right)^4$
Where $S=X_1+X_2+X_3+X_4$
Since you want the probability of getting at most $14$ you want the sum of the coefficients of $t^4$ up to and including $t^{14}$ in the expansion.  This is because the terms in the expansion represent mutually exclusive events.
I used Wolfram and got the probability to be $\frac{721}{1296}$
Hopefully I haven't made and error and have understood your question correctly.
