Please explain how to derive the PGF of a 6 sided dice

Question

I understand the PGF is $$ P(s) = \frac{s}{6} \frac{1-s^6}{1-s} $$ Can someone explain all the steps to get there

Answer 1

We have $p(1)=...=p(6)=\frac16$, $p(0)=0$ and for $k>6$, $p(k)=0$

The PGF is

$G(s)=\sum_{k=0}^{+\infty} p(k) s^k=\sum_{k=1}^{6} p(k) s^k$

So $G(s)=\frac16 \sum_{k=1}^{6} s^k$

We recognize a geometric sum:

$G(s)=\frac{s}{6} \frac{1-s^6}{1-s}$

Answer 2

We want

$$f(x)=x+x^2+x^3+\cdots +x^6.$$

By long division we can prove that

$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$

If we multiply the above expression by $1-x^7$ we are going to see a pattern where all the $x^n$ exponents will cancel out after $x^6$:

$$\begin{align}&(1-\color{red}{x^7})+(x-\color{blue}{x^8}) +(x^2-x^9)+\cdots+(\color{red}{x^7}-x^{14})+(\color{blue}{x^8}-x^{15})+\cdots\\[2ex]&=1+x+x^2+\cdots+x^6 \end{align}$$

Now we have to get rid of the $1$ in front by multiplying the series by $x.$ That will require reducing the exponent in the numerator exponent from $1-x^7$ to $1-x^6.$

This leaves us with $x\frac{1-x^6}{1-x}.$ And since the probability of each outcome is $1/6,$ the expression is derived:

$$f(x)=\frac 1 6 \frac{x(1-x^6)}{1-x}$$