Card Trick - Answer

It seems as though this trick should be impossible. The four revealed cards can be revealed in any of 24 different orders, each of which could be encoded to indicate a card. But, once four cards are removed from the deck, 48 remain, not 24! So the 24 different possible orders don't seem like enough -- it seems as though Magician B could specify that the last card is one of two possibilities, but couldn't nail down which of the two it is. But a little trick gets around this.

The spectators have chosen five cards. There are only four suits in the deck. So at least two of the five cards must be in the same suit. Therefore, the first thing Magician A does is to note two cards of the five that are of the same suit. One of these will be the target card for Magician B to name and the other will be the first card that Magician A will reveal.

Moreover, if one envisions the 13 cards in the noted suit arranged in a circle, with the Ace at the top and the other 12 cards arranged clockwise, then, of the two cards Magician A has selected in the same suit, one will be no more than six places clockwise from the other in the circle. (E.g., if the two cards are the 3 and 10 of spades, then, if one imagines the spades laid out in a circle and starts with the 10, it is only necessary to count six places around the circle to get to the 3.)

So, Magician A selects as the first card to reveal one of the two cards in the same suit, such that the other is no more than six places around the circle from the first. The other card in the same suit is the target card that Magician B will name. (E.g, in the previous example, the first card revealed is the 10 of spades and the 3 of spades is the target card.)

So as soon as Magician B sees the first card, Magician B knows that the target card is one of six cards: the six cards, in the same suit as the first card, counting up from the first card and going around the circle if necessary. (E.g., if the first card is the 10 of spades, Magician B knows the target card is the Jack, Queen, King, Ace, Two, or Three of spades.)

Now the remaining three cards that Magician A reveals tell Magician B which of the six possible cards the target card is. To do this, just imagine the whole deck laid out in a specific order: say, we consider the spades to be highest, then the hearts, then the diamonds, then the clubs, and, within each suit, the cards are ordered in the normal way, with Ace highest and 2 lowest. The three remaining cards are then "high (H)," "medium (M)," and "low (L)" within this ordering. The magicians have agreed that the each of the six possible orders of revealing the three cards corresponds to one of the numbers one through six, as follows:

HML = 1, HLM = 2, MHL = 3, MLH = 4, LHM = 5, and LMH = 6

So Magician B looks at the three last cards revealed, notes the corresponding number from the above code, and counts up from the first card revealed to reach the target card!

Not the most stunning trick to actually perform, perhaps, but mathematically very satisfying.