For standard riffle shuffling, there is often a bias in pulling from the pile opposite from the last pulled card (so alternating pairs of left and right). The model saying “7 shuffles is sufficient” as people like to quote presumes you pull from a pile only proportional to the size of the pile (so you get many fewer alternating runs).
Suppose you're at a full table and 12 of the last 20 cards are aces or tens (rare but possible). These get shuffled and cut into the first three decks, giving you a true count of four (12/3) for the first half of the shoe, which is significant.
We never really put it into practice, though, since: 1) you have to track the count for the last 20 cards in addition to the regular count, 2) shuffles change, 3) dealers are inconsistent, 3) casinos use different shuffles, 4) the typical advantage is likely to be much smaller.
My knowledge on this is at least 20 years out of date, though, so who knows?
Let's call the two splits L for left and R for right and consider the resulting deck to be defined as a sequence of these letters indicating whether that card was taken from the left or right pile so in theory you can specify a shuffle uniquely like "LRRLLLLLLRRR...". This method of labeling shuffles has the advantage of being extremely easy to record and also makes the bias I found apparent where there are way too many pairs of "LR" and "RL" compared to "LL" and "RR" like the "7 shuffles is enough" model suggests there should be.
For some reason, when I did all this analysis last year, I was mostly thinking about how to ensure my MTG decks are sufficiently shuffled. I didn't really think about the implications for card counting. However, normal riffle shuffling seems to have less bias (but still present) than the mash shuffling I looked at and I think most casinos use machines for shuffling these days.
* Depending on how chunky your shuffle is. A pro card dealer can do almost a perfectly alternating riffle, which is obviously no good for national security.
• 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
Additionally, your analysis only considers when the permutation is reachable. You would also want to look at how likely each permutation is.
I think the bare minimum requirement for a good shuffler is "all hands are possible." And then getting some more bits beyond that is good to debias it.