## 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??

• Alok Goyal says:

My sincere apologies Shreeram 😦

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 âOneâ 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 didnâ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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• Alok Goyal says:

Thanks Suman. I think the problem is solvable… Needs a but of lateral thinking 🙂