If one deck of 52 cards can have 52! possible arrangements, how many can you get when you have two decks ?


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  • $\begingroup$ Hint: how many cards are in two decks? $\endgroup$ – Sean Roberson Oct 13 '17 at 0:56
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    $\begingroup$ Do you consider cards of identical rank and suit to be indistinguishable? $\endgroup$ – Graham Kemp Oct 13 '17 at 0:59
  • $\begingroup$ Yes, so an ace of spades looks exactly the same in either deck, and so on and so forth for every card. $\endgroup$ – Darth Ratus Oct 13 '17 at 1:23

You have now 104 cards, but they are paired. Assuming the two decks are identical, you have $\frac {104!}{2^{52}}$ combinations. Indeed there are $104!$ possible shuffles if the two decks are different, but since they are identical, for each pair of identical cards, you can swap them and you have the same shuffle.

  • $\begingroup$ I am still trying to wrap my head around the 2^52 denominator. I know is correct, i worked it out by hand for the cases of two identical decks each with 2 cards (4!/2^2 = 6), and two identical decks with three cards each (6!/2^3 =90). $\endgroup$ – Darth Ratus Oct 13 '17 at 1:48
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    $\begingroup$ @DarthRatus Think of it this way. Assume that both the ace of spades are different and calculate. You get $104!$, but in every arrangement you can swap the two ace of spades. So you only get $\frac{104!}{2}$ arrangements. But you have every card that's a duplicate, so you need to divide by two 52 times. So there you go. $\endgroup$ – Pritt Balagopal Oct 13 '17 at 1:54
  • $\begingroup$ That makes a lot more sense. Thank you. $\endgroup$ – Darth Ratus Oct 13 '17 at 2:05

If the decks must stay separate, then you have 52! arrangements in deck 2 for each of the 52! arrangements in deck 1, so there are $52! \times 52!$ total arrangements.

If the decks are combined, you just increase the operand of the factorial to the new card count, so there are $(52 + 52)!$ total arrangements.

  • $\begingroup$ Decks do not stay separate, and the cards maintain their identities, that is, there are 13 cards of 4 different suits. $\endgroup$ – Darth Ratus Oct 13 '17 at 1:22

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