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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).

  Submitted by Charlie    
Rating: 4.0000 (2 votes)
Solution: (Hide)
9, A, 4, Q, J, 7, 3, 2, 10, 5, K, 8, 6

After two "shuffles", the first postion has gone from holding an Ace to holding a 10. The 10 position has gone from holding a 10 to holding a Jack, the Jack (11th) position has gone from holding a Jack to holding a 6, etc.

That sequence of "position is replaced by" can be represented as follows:

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

The "square root" of this transformation can be obtained by going forward 7 or backward 6 in the cycle, so that the first position would go from Ace to 9, and the 9 to the 10, and the 10 to the 5, etc.:

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

So after one iteration instead of two, the Ace has been replaced by a 9, the deuce by an ace, the trey by a 4, etc:

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

which is the solution.

To verify:

The 9, in the A position after one "shuffle" is moved to the 2 position as required.
The A, in the 2 position after one "shuffle" is moved to the 8 position as required.
The 4, in the 3 position after one "shuffle" is moved to the 7 position as required.
The Q, in the 4 position after one "shuffle" is moved to the 3 position as required.
The J, in the 5 position after one "shuffle" is moved to the 10 position as required.
The 7, in the 6 position after one "shuffle" is moved to the K position as required.
The 3, in the 7 position after one "shuffle" is moved to the 6 position as required.
The 2, in the 8 position after one "shuffle" is moved to the Q position as required.
The 10, in the 9 position after one "shuffle" is moved to the A position as required.
The 5, in the 10 position after one "shuffle" is moved to the 9 position as required.
The K, in the J position after one "shuffle" is moved to the 5 position as required.
The 8, in the Q position after one "shuffle" is moved to the 4 position as required.
The 6, in the K position after one "shuffle" is moved to the J position as required.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(3): ? Solution for any set of subcycles?Steve Herman2013-07-29 23:37:59
re(2): ? Solution for any set of subcycles?Larry2013-07-29 21:42:52
re: ? Solution for any set of subcycles?Charlie2013-07-29 10:05:25
? Solution for any set of subcycles?Larry2013-07-28 15:03:18
Some ThoughtsCycle length determinedSteve Herman2013-07-25 09:13:57
Brians solutionbroll2013-07-25 03:09:11
Some ThoughtsCycle length is not obviousSteve Herman2013-07-25 00:38:57
re: SolutionAdy TZIDON2013-07-24 17:00:42
Some ThoughtsSolutionBrian Smith2013-07-24 16:28:59
Some ThoughtsThe Hard Way? (spoiler)Steve Herman2013-07-24 14:37:37
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