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 | $P(z^{0})$ | $P(z^{1})$ | $P(z^{2})$ | $P(z^{3})$ |
|---|---|---|---|---|
| $\frac{1}{4}$ | $\frac{3}{8}$ | $\frac{1}{8}$ | $\frac{1}{4}$ |
The distribution of outcomes can be directly read off the polynomial expansion: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$
If we were to add multiple variables x, then the resulting distribution is the product of those polynomials
| variable | $P(z^{0})$ | $P(z^{1})$ | $P(z^{2})$ | $P(z^{3})$ |
|---|---|---|---|---|
| $x_{1}$ | $\frac{1}{4}$ | $\frac{3}{8}$ | $\frac{1}{8}$ | $\frac{1}{4}$ |
| $x_{2}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\frac{1}{9}$ | $\frac{16}{45}$ |
$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$