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Faulty Card Shuffler (Posted on 2013-07-24) Difficulty: 3 of 5
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)

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Some Thoughts Solution | Comment 2 of 10 |
After 13 shuffles, the cards will be back in their starting arrangement.  So running the shuffler 14 times will produce the same result as running it once.  Since we are given the result for two shuffles, applying that rearrangement six more times will be equivalent to 14 single shuffles.

First double shuffle (given): 10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7
Second double shuffle: J, 5, 2, A, 7, Q, 8, 10, K, 6, 3, 9, 4
Third double shuffle: 6, K, 9, 10, 4, 2, A, J, 7, 3, Q, 5, 8
Fourth double shuffle: 3, 7, 5, J, 8, 9, 10, 6, 4, Q, 2, K, A
Fifth double shuffle: Q, 4, K, 6, A, 5, J, 3, 8, 2, 9, 7, 10
Sixth double shuffle: 2, 8, 7, 3, 10, K, 6, Q, A, 9, 5, 4, J
Seventh double shuffle: 9, A, 4, Q, J, 7, 3, 2, 10, 5, K, 8, 6

The order the shuffler put the cards in from the first run is 9, A, 4, Q, J, 7, 3, 2, 10, 5, K, 8, 6.


  Posted by Brian Smith on 2013-07-24 16:28:59
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