Take a deck of cards (with indexed position from 1 (top) to 52 (bottom)) and perform a perfect riffle shuffle, such that the top card (1) is still on top and the bottom (52) is still on the bottom.

Amazingly, if you perform 8 such riffle shuffles you will return to where you started.

Obviously, cards 1 and 52 do not change position. Most of the cards will go through some cycle and land back where they started only after 8 riffles.

But two cards will simply swap position back and forth each shuffle.

What are they?

Bonus question:

If you throw in the two jokers, you will have 54 cards. How many riffle shuffles will it take to get this deck back to the starting positions?


1 Answer 1


Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.

If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=k\iff3k=54\iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $\boxed{18\text{ and }35}$.


It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $\boxed{52\text{ shuffles}}$.

  • 1
    $\begingroup$ These answers are the reason i feel guilty for not going back to continues learning. $\endgroup$
    – Alex
    Mar 25, 2019 at 19:32
  • 2
    $\begingroup$ I got a PhD and I still can't put together answers like these! xD $\endgroup$
    – Somebody
    Mar 26, 2019 at 16:30

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