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How do you shuffle cards?



Given a regular deck of cards (diamonds, clubs, hearts, spades, with ace,2,3,4,5,6,7,8,9,10,Jack,Queen,King in each) minus the jokers, is there a way of stacking the deck in such a way that no card is on top of or below any card of the same value or the same suit? If so, how many ways are there?

Sorry for bad wording, but ask questions!

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HA DK SQ CJ H10 D9 S8 C7 H6 D5 S4 C3 H2 DA SK CQ HJ D10 S9 C8 H7 D6 S5 C4 H3 D2 SA CK HQ DJ S10 C9 H8 D7 S6 C5 H4 D3 S2 CA HK DQ SJ C10 H9 D8 S7 C6 H5 D4 S3 C2

first letter is suite, second symbol is card value.

I think you can rotate the suite giving at least four with the current symbol orientation. Of course I can put the first card the HA at the end and due the same rotation to get a new orientation. so i am thinking there are 4*52 or 208 orientations?

to be clear, i simply counted down from A to 2 and rotated the suits {H, D, S, C}

Edited by BMAD
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There are many, many more ways though.

If we stick with a "length 4"-cycle suit sequence (for example HSDC...), there are 4! = 24 ways to make such a cycle. If we stick with a "length 13"-cycle rank sequence (for example (AQ58923T47JK6...), there are 13! = 6,227,020,800 ways to make such a cycle. Any suit cycle of length 4 combined with any rank cycle of length 13 will work just like BMAD's example sequence did. This is true because all such sequences are algebraically isomorphic (they are the same except for choice of symbols). So we get 4!*13! = 149,448,499,200 ways to stack the deck in this fashion. But we aren't done yet.

Suppose we still stick with the ranks cycling with length 13, but explore other ways to arrange the suits. We already know that no two cards of the same rank touch each other. As long as no two cards of the same suit touch, and no two cards of the same suit are a multiple of 13 apart from each other, we will be fine. Since all cards of the same rank are multiples of 13 apart from each other, if two cards of the same suit were also a multiple of 13 apart from each other, we would have two identical cards in the deck. One way that works is to split the suits into two "length 2"-cycles and add them together at the end, for example HDHDHDHDHDHDHDHDHDHDHDHDHDSCSCSCSCSCSCSCSCSCSCSCSCSC. There are 4! ways to do this as well, so after multiplying with the 13! ways to cycle the suits, we get another 149,448,499,200 possible stackings of the deck.

Now I have found almost 300 billion solutions. I think there are still more of them, but I'm not completely sure. I can't think of any more easily verifiable solutions right now, and I don't know (or at least I can't remember) enough algebra to solve it completely.

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