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?

Okay, now I've looked at everyone else's ideas..there were a lot of things suggested that I like that I didn't think of.

The 24-permutation thing by Brian is nifty, but man! You have to mentally take out the four cards, and sort out the remaining 48, as well as determine the relative values of the four you have, remembering them in order, and pick out the hidden card. As a couple people said, that's pretty intese.

Charlie's suggestion with the 'clock' is pretty cool, and it uses just the three cards to determine the value of the hidden one from the fourth. Pretty neat, and the giveaway card doesn't have to be the first card, though that is probably easiest since you are adding to/subtracting from that value. The only thing with this, as with Brian's, is that you have to remember the cards in order as they are handed to you. That's doable, with some practice. Also, cards with the same value could be troublesome, but ranking the suits works (I think the standard convention for that is ♠-♥-♣-◊ from high to low).

I also liked the idea of the 'camouflage suit' (I forget who said it) so that the giveaway is less obvious, but, what if all five cards are the same suit? You could make up some way to indicate that special exception, but that would be rather clumsy. Better might be just making the giveway card second or third instead of first, and therefore less consipicuous.

For Jon's last proposition, having the cards face down would obviously make the problem more difficult. My solution ignores everything but the suit of one card, but without that you have four times as much work to do. A ternary system with 4 bits yields 81 combinations, much more than are necessary. Three ternary bits and one binary bit gives 54 combinations, very close to what we want (heck, throw in two different jokers).
How about this:
Number the cards 1-52: A-K, Diamonds, Clubs, Hearts, Spades, again omitting zero for intuitivity (is that a word?).
For example; 1 = A◊, 2 = 2◊, 3 = 3◊, ... , 14 = A♣, all the way up to 52 = K♠.

For the first card, if the assistant is holding on the the corner/edge nearest you, do nothing (add 0), and for the corner nearest him, add 26.
For the second card, if he is holding the corner nearest you, add 0, the middle of the card, add 9, and the corner nearest himself, add 18.
For the third card, add 0 for the corner nearest you, 3 for the middle of the card, and 6 for the corner nearest him.
Finally, the last card has similarly placed values of 0, 1, and 2.
This method can be used to determine any value between 0 and 53. Determining the hidden card from this number is relatively simple, especially if you think of the numbers in groups of 13. Determine the suit from its range:
1-13 = ◊
14-26 = ♣
27-39 = ♥
40-52 = ♠
Then just subtract the base (0, 13, 26, or 39) to get the value 1-13 that corresponds to A-K.
That whole process is a lot simpler than I made it sound. The hardest part, really, is for the assistant to quickly figure out the value of the card and convert it to this ternary-binary system mentally. That is still very much doable with a little practice (and remember, the assitant gets to choose the card, so he can pick a simpler one if that is possible).
The reason I am still saying to go by what corner or edge of the card is held, rather than the number of fingers, is because it is a heck of a lot easier to see where someone's thumb is on the top of the card than to try to count the number of fingers on the bottom (and being blocked by the thumb). Any way of distinguishing three methods, of course, will work.

Posted by DJ on 2003-05-11 15:44:23