Problem: Probability that that the left side and the right side are connected by a path
Graph
Consider a square lattice in a bond percolation. There are two sides which are connected together by a path with a probability $p$. I have a feeling that this is somehow related to sum of bernoulli variables. I want to get a polynomial in the single-bond probability $p$ for fixed $n$. In $p$, there is with integer coefficients and degree at most the number of bonds [RobertIsraeal].
There are finitely many members of the sample space and each has a probability that is a polynomial of that type. [RobertIsrael] This will be challenging polynomial which I want to understand. It is not only about independent random variables.
One attempt
Instead of trying to find the closed polynomial expression, I don't know how nice it is, you could approximate the connectness of the sides with the existence of infinite cluster
$$\theta_{\infty}(p)=P(|C|=\infty)$$
so given a cluster with $\theta_{1}(p)$: if $\theta_1(p)\approx \theta_{\infty}(p)$, then you could say that the sides are almost surely connected.
Random Networks for Communication (2008) book by Massimo Franceschetti et all explains different terms on convergences such as random variable converging almost surely, random variable converges in probability and random variable converges in distribution -- for finite graphs.
The almost surely convergence is the strongest such that
$$P(\lim_{n\rightarrow\infty} X_n=X)=1$$
where we say that $X_n$ converges almost surely to $X$.
Almost surely convergence implies convergence in probability which then again implies the convergence in distribution. More on the page 70.
You mean a polynomial in the single-bond probability $p$, for fixed $n$? Yes, this will be a polynomial in $p$ with integer coefficients and degree at most the number of bonds, simply because there are finitely many members of the sample space and each has a probability that is a polynomial of that type. But it will be an awfully complicated polynomial, not easy to compute. It's nowhere near as simple as a sum of independent random variables.