Not sure if this directly answers your question, but it might help in further analysis.

Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$
 
|variable| $P(z^{0})$  | $P(z^{1})$  | $P(z^{2})$  | $P(z^{3})$  |
|--------|-------------|-------------|-------------|-------------|
|$x$        |$\frac{1}{4}$|$\frac{3}{8}$|$\frac{1}{8}$|$\frac{1}{4}$|

Which expands to the polynomial:
$\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$

If we were to introduce multiple variables $x_{i}$ then the resulting distribution is the product of those polynomials

|variable| $P(z^{0})$  | $P(z^{1})$  | $P(z^{2})$  | $P(z^{3})$    |
|--------|-------------|-------------|-------------|---------------|
|$x_{1}$  |$\frac{1}{4}$|$\frac{3}{8}$|$\frac{1}{8}$|$\frac{1}{4}$  |
|$x_{2}$  |$\frac{1}{3}$|$\frac{1}{5}$|$\frac{1}{9}$|$\frac{16}{45}$|

$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$

[link to plot](https://i.sstatic.net/jb0nn.png)