Number of ways to throw at most 14 with 4 dice - generating functions

Question

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.

Answer 1

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.

Answer 2

As mentioned in Karl's answer, one surefire way of doing it is by adding together all coefficients of $(x+x^2+x^3+x^4+x^5+x^6)$ on the terms of power less than or equal to $14$.

You have included in your proposed solution a very nice way to get around the difficulty of having to either personally sum or word to the computer how to sum all of those coefficients by multiplying another polynomial which takes care of the summing for you. There is but one problem: as $x'$ is allowed to be zero, you are missing a term, $x^0=1$, in the right term.

The correct generating function will be:

$$(x^1+x^2+x^3+x^4+x^5+x^6)^4(0+\color{red}{x^0}+x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10})$$

With the generating function, you can plug it into your favorite CAS and read the coefficient to the power in question. In this case wolfram reads the coefficient is $721$, agreeing with Karl's earlier arithmetic.

It is common as well to see this written instead as $(x^1+x^2+x^3+x^4+x^5+x^6)^4(0+1+x+x^2+\dots)$ where the right term is an infinite series. We don't actually mind that the series doesn't converge in combinatorics as we look at partial sums anyways. This allows us to reuse the same generating function for other sums as well.