The upper and lower bunk graphs are symmetrical in their probabilities, so it would intuitively seem like connecting to the same bunk's point would be easier/more likely than connecting to the other bunk's point. The first would require zero (or an even number of) crossings, the latter would require one (or an odd number of) crossings. Every crossing would seem to only decrease the odds of the path existing, or at best leave it the same.

To me it smells like a trap, though. This feels like exactly the kind of problem where some algorithmically chosen infinite graph would break the conjecture, possibly through some kind of NP-hardness-proof-style transformation of the problem into another form, and who knows, maybe the Axiom of Choice gets involved too (but probably not based on the headline!).