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Untangle the permutation as a big cycle green-blue-green-blue, and choose an orientation. We're going to count the posible shapes for this cycle, with the vertices labeled by their original position. We first have an alternating cyclic permutation of all the green wires. There are $(n/2)!\cdot (n/2)!/(n/2)$ ways to do this. Then we can choose an orientation of each of the green wires. There are $2^n$ ways to do this. Finally we note that two big cycles with reversed orientation correspond to the same permutation, so we divide by $2$. The total number of ways to get one big cycle is:
$\frac{2^n \left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n}$
$\frac{2^n \left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n\cdot n!}$
which if I remember the estimates correctly as Noam Elkies points out is asymptotically proportional to $o(2^{-n})$.n^{-1/2}$. 1 Untangle the permutation as a big cycle green-blue-green-blue, and choose an orientation. We're going to count the posible shapes for this cycle, with the vertices labeled by their original position. We first have an alternating cyclic permutation of all the green wires. There are$(n/2)!\cdot (n/2)!/(n/2)$ways to do this. Then we can choose an orientation of each of the green wires. There are$2^n$ways to do this. Finally we note that two big cycles with reversed orientation correspond to the same permutation, so we divide by$2$. The total number of ways to get one big cycle is:$\frac{2^n \left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n}$and the probability is:$\frac{2^n \left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n\cdot n!}$which if I remember the estimates correctly is$o(2^{-n})\$.