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I know that after "ruffle shuffling" 1 deck (54 cards) 7 times, every configuration is nearly equally likely. How should I shuffle (for example) 3 decks (162 cards) assuming I can shuffle together only 1 deck (54 cards) at a time? I'd like an answer that can be easily adapted to other amounts of decks.

jkoop
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    First, you'll need to define what a 'shuffle' is in mathematical terms ... same for a 'perfect shuffling' – Bram28 Oct 29 '18 at 14:57
  • I think the method of shuffling is called the "ruffle shuffle". And by "perfect shuffle" I mean that every configuration is nearly equally likely. – jkoop Oct 29 '18 at 15:02
  • The devil is in the details ... the basic idea of the ruffle shuffle is clear ... but again, you'll need to have some kind of mathematical description of it in order to work with it (for example: split the deck into two, and keep picking the top card from each deck randomly to be put into a new pile until both half decks are empty) Same for 'perfect shuffle': how do you operationalize 'nearly equally likely'? – Bram28 Oct 29 '18 at 15:30
  • I just found this online: "In 1992, Bayer and Diaconis showed that after seven random riffle shuffles of a deck of 52 cards, every configuration is nearly equally likely. Shuffling more than this does not significantly increase the 'randomness'; shuffle less than this and the deck is 'far' from random." It just needs to good enough for gameplay. – jkoop Oct 29 '18 at 15:32
  • OK, that's a good start. So I would go to that paper, and see how they defined their terms. – Bram28 Oct 29 '18 at 15:34
  • I took a quick look at that paper myself ... the riffle shuffle is defined easily enough: after splitting the deck into two, keep picking the top (bottom) card from a deck with a probability proportional to how many cards that deck has proportional to the total cards of both decks. E.g. if in the middle of the shuffle you have 13 cards left in one deck, and 17 in the other, then the probablity that you pick the next card from the deck with 13 cards is 13/30. Unfortunately, the math involved in the next step (from single shuffle to 'perfect shuffle') is a hell of a lot harder! – Bram28 Oct 29 '18 at 15:45
  • I've edited my question. It doesn't need to be a perfect shuffle, just so that every configuration is nearly equally likely as defined by that paper. – jkoop Oct 29 '18 at 20:21

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