# Perfect riffle shuffles

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

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 $$\boxed{18\text{ and }35}$$.

Bonus:

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 $$\boxed{52\text{ shuffles}}$$.

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