The algebra may not be associative, is that OK with you? The trouble is that you can have $a$,$b$,$a+b$,$c$,$a+b+c$ vertices, but $b+c$ not be a vertex.

The fiber product you write is typically singular.

In fact, high degree rational functions (at least over complex numbers) will generally lead to curves with self-intersection. This is because a generic high-degree rational curve in $\mathbb P^2$ has many nodes. Thus to avoid the intersections in the affine patch, you would need to have all of the singularities at the line at infinity.

Thank you for the comment, I will edit the answer accordingly.

This is somewhat stronger than Goldbach's conjecture for even numbers of the form $12n+10$.

Yes, you are right. Thank you for the correction!

They are toric.

Out of curiosity, wouldn't you want to weigh each group $G$ by $1/Aut(G)$ in the count of probabilities?

I don't see any obvious maps to a $\mathbb P^1$ here. It would be more reasonable to ask whether this variety is a $K$-equivalent to a $\mathbb P^2$-bundle over $(\mathbb P^1)^3$.

Are your coefficients exact or is everything approximate? What sort of equations are they?