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 vertice $(n,m)$ 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 $n, n' \leq N$. Prove or disprove that there are more edge-selfavoiding paths from $(N+1,N+1)$ to a vertex of the form $(2N+1, m)$ for $m \in \{ 1, 2, \dots 2N+1 \}$ than paths to a vertex of the form $(1, m)$ for $m \in \{ 1, 2, \dots 2N+1 \}$.



Are there more paths exiting a box in $\mathbb{Z}^2$ to the right if I remove some edges to the left
Expanding on Anthony Quas' comments above, it is indeed possible to show that there are cases when there are more paths to the left than to the right.


Let $H$ be the graph obtained from $G$ by removing all edges from $(N-1,i)$ to $(N,i)$ except one (which then clearly is a bridge in $H$).


Any walk in $H$ that ends on the right cannot cross the bridge and thus uses at most $2N^2 + O(N)$ vertices. One way to encode such a walk is to store its length and remember the behavior (turn left/turn right/continue straight) every time we encounter a previously unvisited vertex; the behavior at previously visited vertices is determined by this encoding since we are not allowed to use any edge twice. This shows that there are at most $3^{2N^2 + O(N)}$ walks that end on the right.


For a lower bound on the number of walks in $H$ ending on the left, we observe that there is a connected subgraph of $H$ with $4N^2-O(N)$ vertices of degree $4$ in which $(N+1,N+1)$ and $(1,1)$ are the only vertices of odd degree. Starting with a Eulerian walk from $(N+1,N+1)$ to $(1,1)$ we can obtain $2^{4N^2-O(N)}$ different such Eulerian walks by iteratively performing local modifications at each vertex. Indeed, observe that if we fix how the walk traverses every vertex apart from one vertex of degree $4$, then there are always two options at this last vertex completing it to a Eulerian walk.


