Imagine there is a ring network $G$ with $N$ (odd) nodes. Nodes are initially $iid$, $a_i\overset{\text{iid}}\sim Ber(p)$. However, once the value is assigned, a node with $a_i=0$ will switch to $1$ if both of its neighbours are $1$. Define $f(G)=\sum_{i=1}^Na_i$ (total number of $1$'s in the network).
Then how can we express $P(f(G)\geq(N+1)/2)$ (majority of $1$'s) as a function of $N$ and $p$?
For example, if $N=3$ then $P(f(G)\geq2)=p^3+3p^2(1-p)$.
Without the conversion rule it would basically boil down to a Binomial distribution (which is now instead a lower bound) but the conversion messes up things.
This probability will always be of the form:
$$a_{0} p^{0} (1 - p)^{N} + a_{1} p^{1} (1 - p)^{N - 1} + \dots + a_{N - 1} p^{N -1} (1 - p)^{1} + a_{N} p^{N} (1 - p)^{0}$$ For a particular value of $N$ we can just go through all possibilities and compute them. I've tabulated all values below $16$ in the table below:
Please notice that each of these rows has its second half in common with the Pascal's triangle (the lower bound you have observed). But we have the equality only for $N = 1$ and $N = 3$. The first half are half zeros and half some positive noise which I unfortunately cannot decipher (the fact that first $\frac{1}{4}$ is obvious though - the number of ones cannot double in this process).
Maybe you (or anyone else) can make sense of that. I was searching a bit on OEIS but haven't found anything of value.