It's not even obvious to me that repeated shuffles of this form

• Given a deck of 52 cards (C1, C2, ... C52) pick some k near 26 and split the deck into two piles, (C1, C2, ..., Ck) and (Ck+1, ..., C52). Call these piles P1 and P2,

• Make a new empty pile, and then repeatedly take cards from the front of P1 or P2 and append those cards to the new pile. Cards from the same source pile should be in the same relative order in the new pile. Continue until all the cards are in the new pile,

• When taking cards from P1 or P2 to append to the new pile, take about the same number each time,

are capable of resulting in all possible 52! permutations. That kind of shuffle preserves a lot of order so I wondered if there might be some order that it cannot remove.

It turns out that you can in fact reach all permutations. I don't know of any elegant way to prove this though. I have an ugly way to do so.

Let R be a perfect out riffle shuffle, where the deck is split exactly in half (k = 26), and cards are merged by pulling one at a time from alternate piles starting with P1. In other words the deck after one shuffle is (C1, C26, C2, C27, ..., C26, C52).

Let S(n) be a shuffle just like R with one exception: when the cards that would end up at positions n and n+1 from the bottom of the shuffled deck are the next two cards to be pulled from P1 and P2 switch the order you pull them.

The resulting shuffle is the same as if you did R and then swapped the cards at positions n and n+1 from the bottom.

Let O(X) be the "order" of a shuffle X. The order of a shuffle is how many times you have to do apply the shuffle consecutively to get back to where you started. O(R) for example is 8. Start with a new deck and do 8 perfect out riffle shuffles and you will be back to the original order.

It turns out that if you take a deck and do O(S(n))-1 shuffles using S(n) and then do an R you get back to where you started except the cards n and n+1 from the bottom are swapped.

A swap of any two cards can be accomplished using a series of swaps of adjacent cards and so this gives us a method to swap any two cards in the deck using only R and S shuffles. Since any permutation can be generated using only pair swaps this means any of the 52! possible permutations can be reached using only R and S shuffles.

This isn't very efficient. Just swapping n and n+1 in the deck this way takes at least 16 shuffles, and many more for some values of n. Here's a table:

      n           # of shuffles
   0, 50            72
   1, 49            56
  16, 17, 33, 34    40
  22, 28           120
   other            16

Remember, adjacent swaps are just one small step in getting to a given permutation. Actually reaching an arbitrary permutation using R and S shuffles this particular way would take tens of thousands or more shuffles. But it does show that all permutations are reachable using reasonably normal shuffles.