Dealing Twins

Not that dissimilar to Mark Henning's solution. Apply a random permutation to a random permutation yields another random permutation. Therefore, this question is the same as the probability of at least one singleton cycle in a random permutation, which in the limiting case is $1-\frac{1}{\mathbb{e}} \approx 0.632121$ . 52 is sufficient to approximate the limiting case. I did a Monte Carlo simulation with $10^8$ and the first four digits matched.

Imagine turning over all cards in both packs, and then rearranging the card pairs so that the cards in the left-hand pack come in "ascending order" (A-K of Clubs, Diamonds, Hearts, Spades in that order). The desired probability is now the probability that at least one of the cards in the second pack occurs in its "ascending order" position, and hence is $p \; = \;1 - \frac{d(52)}{52!}$ where $d(n)$ is the number of derangements of $n$ symbols. Now $\frac{d(n)}{n!} \; = \; \sum_{j=0}^n \frac{(-1)^j}{j!}$ and hence $p \; = \; \boxed{0.632121} \approx 1 - e^{-1}$