You have a deck of 52 cards  for convenience, number them 1 through 52. You cut the cards into two equal halves and shuffle them perfectly. That is, the cards were in the order
1,2,3,...,52
and now they are
1,27,2,28,...,26,52. Let's call this a perfect inshuffle.
If you repeat this inshuffling process, how many inshuffles will it take for the deck to return to its initial ordering (taking for granted that the cards will eventually do so)?
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How does the solution change if you have a deck of 64 cards, or 10, or in general, n cards? For odd integer values of n, inshuffling will take 1,2,3,...,n to 1,(n+3)/2,2,(n+5)/2,...,n,(n+1)/2. For example, when n=5, the first inshuffle yields 1,4,2,5,3.
I was just shuffling my cards today, when I suddenly thought of this
particular puzzle. I realized that this puzzle is vulnerable to
cycle notation. I have no idea whether this came up in previous
comments.
Cycle notation is a way of notating different permutations. For
example, if 123456 changes to 214653, then the cycle notation would be
(12)(346)(5). I know I'm not explaining it very well, but you can
figure out the rest.
So anyway, in cycle notation, you can tell how many cycles are in the
permutation, and the length of each cycle. Naturally, the number
of inshuffles required is the least common multiple (LCM) of each of
the cycles' lengths.
So for a 52 card deck:
(1)(2,27,14,33,17,9,5,3)(4,28,40,46,49,25,13,7)
(6,29,15,8,30,41,21,11)(12,32,42,47,24,38,45,23)
(16,34,43,22,37,19,10,31)(18,35)(20,36,44,48,50,51,26,39)(52)
The lengths of each cycle are 1,8,8,8,8,8,2,8, and 1. The LCM is
8. It's funny how it all seems to reduce down to that.
I'm sure that someone with more time could derive a general equation from this, but I think it would get very messy very soon.

Posted by Tristan
on 20050323 00:33:01 