My question is this: Will there always be one more edge swap, always one less, or 50/50 depending on cycle breaks? Statistically, if we count a dedicated parity fix as equal to any other swap, is there a swap count advantage to fixing parity by exchanging targets?

(My guess is that it's 50/50, so in half of parity solves I'd be replacing parity fix with an extra swap, and in the other half I'd be saving two steps, the fix and a swap.)