All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Games
Magic trick (Posted on 2003-05-08) Difficulty: 4 of 5
We have a normal deck of 52 cards. We want to do the following magic trick:

A person from the audience chooses 5 random cards. The magician's assistant looks at the 5 cards, chooses 4 of them, hands them to the magician one by one face up and keeps the other one hidden. The magician then guesses the fifth card (the one that the assistant kept hidden) just by looking at the 4 cards he was handed in.

Is it possible to devise a strategy, so that no matter what the original 5 cards were, the trick always works?

See The Solution Submitted by Fernando    
Rating: 3.8571 (14 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution: (I don't remember where I saw this but it was in the past year.) | Comment 7 of 20 |
With 5 cards, at least two must be in the same suit. The first card the assistant hands the magician will be one of these two cards and the one kept hidden will be the other (or one of the others, if more than one). The way to decide which, and the communication of the value depend on the following:

Consider the Ace low and King high arranged around a 13-position circle, like a clockface except with 13 positions. Two cards from the same suit will have different positions on this imaginary circle. Choose to hand to the magician the one that is shorter to go counterclockwise from the other, which will be the card kept hidden. This way, the hidden card can be 1 to 6 clockwise positions from the one presented.

Then the remaining three cards can be presented in one of 6 orders. You had to decide on a hierarchy of suits, say spades higher than clubs higher than hearts higher than diamonds, and within these suits, king is high and ace low. This way each of the remaining 3 cards to be presented to the magician can be evaluated as being the high, middle or low card of the set of three. If the order of presentation is LMH, then the hidden card's denomination is 1 position counterclockwise from that first presented; LHM means 2 positions; MLH, 3 positions; MHL, 4 positions; HLM, 5 positions; and HML, 6 positions. And of course the suit is the same as the first card presented.

This may have been in one of Ivars Peterson's MathLand (or is it MathTreks) sites for Science News.
  Posted by Charlie on 2003-05-08 07:59:47
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information