Suppose that I am given the graph $G = (V,E)$ where $V = \{ 1, 2, \dots 2N+1 \} \times \{ 1, 2, \dots 2N+1 \} $ and there is an edge between two vertices $(n,m)$ and $(n',m')$ if and only if $\vert n-n'\vert + \vert m-m'\vert = 1$.
Suppose that we remove some arbitrary edges between vertices $(n,m)$ and $(n',m')$ with $n, n' \leq N$.
Now, on the remaining graph, consider Bernoulli edge percolation with parameter $p \in [0,1]$. That is that we delete each of the remaining edges with probability $1-p$. Prove or disprove that, no matter what edges I deleted in the beginning, then the probability that there is a path from $(N+1,N+1)$ to a vertex of the form $(2N+1, m)$ for $m \in \{ 1, 2, \dots 2N+1 \}$ is larger or equal than the probability that there is a path to a vertex of the form $(1, m)$ for $m \in \{ 1, 2, \dots 2N+1 \}$.
See here for a related counterexample: Are there more paths exiting a box in $\mathbb{Z}^2$ to the right if I remove some edges to the left.
