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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)

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Some Thoughts Lots of Little Somethings | Comment 15 of 20 |
Since there are five cards, and only four suits, there must be two cards with the same suit. So, have the assistant pick two cards with the same suit and keep the lower of the two. Always consider the ace to be low (or high, whatever) for convention. The other card would be laid down first. Now you know the suit, and you have at most 12 cards to work with, and at best, you already know what the hidden card is (if there happen to be a three and a deuce of the same suit in there). The remaining three cards are used to determine the value of the last card. Using a binary system, that only yields eight different possibilities. With a ternary system, there are 27 combinations of 3 bits. You only need, 12, or 13 to make the trick a little neater, so that gives room for some play. Actually, it would probably be most convenient to work out an arbitrary mechanism that would be inconspicuous to the audience and yet fast and obvious to the magician.
..Ignore everything I just said, cause there are so many ways to do it that a comprehensive analytic solution is pretty much impossible.

Here's what I would do (I am sitting here with a deck of cards, mulling over the easiest ways to put the cards down):

The first card is used to determine the suit. You could make it the lowest or highest of the flushed cards, but probably to have some sort of absolute method of determining the value only from the last three cards would be that much simpler. Or put down the card with the same suit last (just do it the same all the time). It doesn't matter, as long as you pick a convention and adhere to it.
The second card is used to determine the range. The simplest convention I came with is, hold on to the corner closest to the magician to indicate a 'high' card, that is, 10-K. Hold the middle or side of the card to indicate 6-9, and the corner nearest yourself to indicate 2-5. Maybe, put all the cards down with your left hand instead of your right, if the card is an ace.
At this point, either you already know the hidden card (assuming that the 'suit' card is laid first), or it is narrowed down to four cards. The remaining cards could be used in a binary system, say, hold the corner toward the magician for a '1' and the corner nearest you for a '0'.
00 = 2, 6, or 10
01 = 3, 7, or J
10 = 4, 8, or Q
11 = 5, 9, or K
and if it's an ace you already know because of the aforementioned 'left-hand rule'. That requires a little thought, but with a few minutes of practice for both people it can go very quickly and smoothly.

PS: Sorry if I said anything redundant to the other comments; I haven't been around but I still wanted a chance for a fresh look at the problems. I'll check out the other comments now...
  Posted by DJ on 2003-05-11 14:27:30
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