I once wondered if you could reach every permutation with "believable" shuffles. I define a "believable" shuffle as one where you cut the deck near the center, and then rejoin the two parts by alternately dropping cards from the two parts, with most drops being being one card and no drop being a lot of cards.

The answer is yes.

Let P be a perfect out riffle shuffle. Let S(n) be a shuffle that is almost a perfect out riffle except that it swaps the order of dropping cards n and n+1. Let O(X) be the number of times you have to consecutively do shuffle X to get back to the order you started with.

If you do O(S(n))-1 shuffles using S(n) followed by P, you end up with the original order except cards n and n+1 are swapped. This gives you a procedure for swapping any adjacent pair of cards, and that is sufficient to use bubble sort to achieve any desired permutation via believable shuffles.

It's not very efficient. Swapping 22 with 23 takes 120 shuffles, as does swapping 28 and 29. There are two pairs that take 72 shuffles to swap, two that take 56, 4 that take 40, and the rest take 16.

S(n), Px7, S(n+1), Px7, S(n), Px7 swaps n/2 with n/2+1 if n is even. If n is odd, it swaps that pair that is 26 past the pair that doing it for n-1 would have swapped. That gives a way to swap any adjacent pair in 24 shuffles, improving the prior bubble sort to 10 pairs that use 24 shuffles to swap, with the rest using 16.

S(n), Px7 swaps two cards, but they are not adjacent. If n is even, it swaps n/2 and n/2+26. If n is odd, it swaps (n+1)/2 and (n+1)/2+25. There's probably a way to reach an arbitrary permutation using that which is more efficient than what I've got in the prior paragraph.