The idea is to take the complement of the dual of the graph, see it's $(n+1)\times n$, and see that there is a left to right path in the graph iff there isn't a top to bottom path in the complement of the dual of the graph

This leads to $P(\text{path}) = P(\text{no path}) = 1/2$

Dual graph = graph made by the faces?

Isn't it $n\times(n-1)$?

Yeah kinda

Lemme do a drawing

sketchtoy.com/68171227

sketchtoy.com/68171233

Here's an example

Black are actual graph, blue complement of the graph, red dual graph and green complement of dual graph

I guess the edges on the left and right of the actual graph (and the top and bottom of the ghost one) don't matter

That's right

OK, I see

So, ignoring those, you're essentially turning all edges of the complement around 90 degrees

Or, perhaps, turning all edges 90 degrees and then flipping colors

And that's clearly a bijection then

Cool!

Like, 90 degrees around each edge's individual midpoint

Yup, very short proof as promised :p

Noice