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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)

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 The Hard Way? (spoiler) | Comment 1 of 10
Well, on a single shuffle, the A which winds up in the original 8 position after 2 shuffles, must go into one of the 11 positions after one shuffle.  It cannot got into the A or 8 position, so it must go into original position 2,3,4,5,6,7,9,10,J,Q or K after 1 shuffle.

I started by trying position 2.
Since the second shuffle carried the A from position 2 to 8, this implies that the first shuffle carried the 2 to position 8.
Since the second shuffle carried the 2 from position 8 to Q, this implies that the first shuffle carried the 8 to position Q.
Similarly,
the second shuffle carried the 8 from position Q to 4, so
the second shuffle carried the Q from position 4 to 3, so
the second shuffle carried the 4 from position 3 to 7, so
the second shuffle carried the 3 from position 7 to 6, so
the second shuffle carried the 7 from position 6 to K, so
the second shuffle carried the 6 from position K to J, so
the second shuffle carried the K from position J to 5, so
the second shuffle carried the J from position 5 to 10, so
the second shuffle carried the 5 from position 10 to 9, so
the second shuffle carried the 10 from position 9 to A, so
the second shuffle carried the 9 from position A to 2, which agrees with our first guess.

After one shuffle, the order was 9 A 4 Q J 7 3 2 10 5 K 8 6

Was this just a lucky guess?  It appears so.  If I had initially guessed that the A moved to position 3 on the first shuffle, then

Since the second shuffle carried the A from position 3 to 8, this implies that the first shuffle carried the 3 to position 8.
Since the second shuffle carried the 3 from position 8 to 6, this implies that the first shuffle carried the 8 to position Q.
Similarly,
the second shuffle carried the 8 from position 6 to 4, so
the second shuffle carried the 6 from position 4 to J, so
the second shuffle carried the 4 from position J to 7, so
the second shuffle carried the J from position 7 to 10, so
the second shuffle carried the 7 from position 10 to K, so
the second shuffle carried the 10 from position K to A, so
the second shuffle carried the K from position A to 5, but this contradicts our guess, so the A did not move to position 3 on the first shuffle

I stopped working at this point, without trying out the remaining 9 possibilities.

 Posted by Steve Herman on 2013-07-24 14:37:37

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