Puzzle #106: Cards and Two Magicians

This is a very interesting puzzle contributed by Pallav Pandey – thanks Pallav! I have not solved it yet, and looking forward to working on it!

Magician M1, calls an audience on stage who pulls random 5 cards from the deck.

M1 gives one card back to the guy from the audience; and lays out other 4 cards face-up and side-by-side on the table.

The Magician M2 comes and looks at the 4 open cards and is able to name the 5th card.

How ?

Please send your answers either directly on the blog site as comments, or to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

Enjoy!

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4 Responses to Puzzle #106: Cards and Two Magicians

  1. shreeram says:

    hey..i got the togglers puzzle right and yet you didn’t mention my name??

  2. Suman Saraf says:

    Uhh too bad this puzzle arrived while I was working on #105 🙂 The solution this time was posted a bit early.

    Anyways, for this there are two pieces to convey – the suit and the rank. According to Pigeon Hole Principle, at least 2 of the 5 will belong to the same suit. So he will return “One” such to the person from the audience and put the other in a pre-deterimed position among the 4. This conveys the suit.

    This leaves us with trying to represent 13 ranks using three cards. If the problem didn’t say all cards were face up, the software engineer in me would have used a binary scheme to encode 8 possible values i.e. face down = 0, face up 1

    However, for this problem we will have to rank order (which factors both rank and suit) all 52 cards in some pre-determined order which M1 and M2 understand. After that it is possible to arrange the 3 cards in 6 possible ways depending on the rank order i.e. Low,Mid,High Low,High,Mid Mid,High,Low Mid,Low,High High,Low,Mid High,Mid,Low

    There still is a small problem. Either of these schemes cannot encode the 13 values. They can do 6 or 8 (binary).

    Since M1 can decide which of the two cards to give back (in step 1), he can apply a modulo scheme i.e. the card given back to audience = (card left behind + encoded value) % 13 which would return 0-12

    So for example if the two cards were 1 and 12, he could keep 12, give 1 back and encode 2 with the remaining 3 cards. It is easy to see that the maximum number needed to be encoded would be 6.

    Suman

    >

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