Posers and Puzzles

Posers and Puzzles

  1. Joined
    14 Dec '05
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    16 May '08 00:38
    I ask Alex to pick any 5 cards out of a deck with no Jokers.
    He can inspect then shuffle the deck before picking any five cards. He picks out 5 cards then hands them to me (Peter can't see any of this). I look at the cards and I pick 1 card out and give it back to Alex. I then arrange the other four cards in a special way, and give those 4 cards all face down, and in a neat pile, to Peter.
    Peter looks at the 4 cards i gave him, and says out loud which card Alex is holding (suit and number). How?
    The solution uses pure logic, not sleight of hand. All Peter needs to know is the order of the cards and what is on their face, nothing more.
  2. Joined
    12 Sep '07
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    16 May '08 09:38
    There are five cards. By the Pigeonhole principle, there are two cards of the same suit. The first card of the four you arrange determines the suit. Then the other three cards can be permuted in 6 ways of high, middle and low. Add this number to the first card that was arranged (and determined the suit) to get the secret card. This is possible because the two cards of the same suit can differ by at most 6 (m 13).
  3. Joined
    31 May '07
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    16 May '08 09:50
    I'm assuming you and Peter are in cahoots? In which case this is just a magic trick, and you could have any number of systems going.
  4. Joined
    17 Feb '08
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    16 May '08 09:53
    No you could have a spade (5th card) and then 4 of another suit.

    I was thinking the ordering of the highest card could note the suit.

    Highest card=1st could be diamonds
    Highest card=2nd could be spades
    etc. for the other two

    This would work for having all 4 of the same number, put the suit of that card first.
    (all jacks, put J of spades first if it is a spade)
    but that doesn't necessarily address when there are 2 or 3 of the highest order, especially when the 2 or 3 are not of the same suit as the card.

    That just semi-sorts out suit, but the actual number itself I do not know.
  5. Joined
    31 May '07
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    16 May '08 10:28
    dejection's method works fine.
  6. Joined
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    16 May '08 10:53
    If you draw 4 of the same value and the 5th card is something else I don't exactly see his method working.
  7. Joined
    31 May '07
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    16 May '08 10:59
    with same value you class the cards, as in spades highest, then clubs then hearts then diamonds lowest.
  8. Standard memberforkedknight
    Defend the Universe
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    16 May '08 13:121 edit
    Originally posted by doodinthemood
    with same value you class the cards, as in spades highest, then clubs then hearts then diamonds lowest.
    5 cards: 4s 2h 6h Jh Qh

    I choose the 4s.

    Please describe the 4s using: 2h 6h Jh Qh, and Dejection's method.
  9. Standard memberforkedknight
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    16 May '08 13:21
    Originally posted by Dejection
    Then the other three cards can be permuted in 6 ways of high, middle and low. Add this number to the first card that was arranged (and determined the suit) to get the secret card. This is possible because the two cards of the same suit can differ by at most 6 (m 13).
    Since they differ by a max of 6 only in mod 13, how do you describe the sign of the difference?

    If you have a 10 to work with, and have to describe the 2 as opposed to the 5, how would you account for the differences being equal with only 6 values.
  10. Standard memberforkedknight
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    16 May '08 13:382 edits
    A solution I would offer to this problem would be to use the side of the cards showing (back or front) as a binary 1 or 0 to describe the rank of the card, and use the ordering of the suits (if possible) or the ordering of the ranks to describe the suit of the card.

    You would then have a binary value from 1 to 13 for the rank using all four cards.

    A possible encoding scheme for the suit might be:

    If you have 3 distince suits available:
    order the suits highest to lowest: c d h s
    then use the 3 different suits as the first three cards and encode the desired suit with a possible permutation.

    If you have 3 distinct ranks available, but not 3 suits
    use the same encoding scheme as before, except ordering the ranks

    If you have 2 suits and 2 ranks:
    use the ordering of the ranks in each suit as a binary value to encode the desired suit.
  11. Standard memberforkedknight
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    16 May '08 13:451 edit
    Originally posted by Dejection
    There are five cards. By the Pigeonhole principle, there are two cards of the same suit. The first card of the four you arrange determines the suit. Then the other three cards can be permuted in 6 ways of high, middle and low. Add this number to the first card that was arranged (and determined the suit) to get the secret card. This is possible because the two cards of the same suit can differ by at most 6 (m 13).
    I just noticed that in the instructions, it is the person arranging the cards that also gets to choose which card to describe.

    Provided that constraint, Dejection's method, as well as a probably nearly infinite number of other methods, would work just fine.

    I think the problem is a little more interesting if someone other than the arranger gets to choose the card.
  12. Joined
    25 Apr '06
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    16 May '08 13:54
    Just wondering, how four '4's would be arranged to describe a 10 of spades?
  13. Joined
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    16 May '08 14:204 edits
    Originally posted by heinzkat
    Just wondering, how four '4's would be arranged to describe a 10 of spades?
    One of those 4's is a spade, so place that in the pre-arranged position for indicating the suit(say first position on the left). The other three cards have to code for a 'six'(the difference between 4 and 10).
    Dejection has already explained how to do that.
    You could have the convention used in bridge; which I think is;
  14. Joined
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    16 May '08 14:291 edit
    CDSH
  15. Joined
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    16 May '08 14:44
    Originally posted by luskin
    One of those 4's is a spade, so place that in the pre-arranged position for indicating the suit(say first position on the left). The other three cards have to code for a 'six'(the difference between 4 and 10).
    Dejection has already explained how to do that.
    You could have the convention used in bridge; which I think is;
    Interesting, thank you for the extra explanation (I didn't get it immediately)
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