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 Faulty Card Shuffler (Posted on 2013-07-24)
This card shuffler, instead of randomizing the positions of cards in the deck, always rearranges any given set of cards into the same new order.

If for example one puts all the hearts into the shuffler, in the order A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, and runs that through the shuffler first once, and then, without rearranging the cards, runs it through again, the resulting sequence after the second shuffle is always:

10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7

The question is:

What was the order of the cards after the first run through this non-randomizing shuffler?

From the Skeptics' Guide to the Universe podcast episode of June 22, 2013 (episode 414).

 See The Solution Submitted by Charlie Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): ? Solution for any set of subcycles? | Comment 9 of 10 |
(In reply to re: ? Solution for any set of subcycles? by Charlie)

Ah...
I get it.  So similarly, if the problem had given us results after zero shuffles and 3 shuffles then any sub-cycles divisible by 3 would have a similar blind spot

 Posted by Larry on 2013-07-29 21:42:52
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