Section 7 · The Best Card Trick
Four Cards Name the Fifth
You choose any five cards from a full deck. The assistant hides one and lays the other four in a row. From those four cards alone, the magician names the fifth.
Try it
Pick any five cards, or let the site deal you five. The assistant decides which one to hide; the magician sees only the row.
Why it works
The four cards in the row carry two kinds of information: which four cards they are, and the order they lie in. The magician and the assistant agreed on how to use both before you chose anything. Three small facts do the work.
A suit repeats. Five cards, four suits: if every suit held at most one card, there would be at most four cards. So two of your cards share a suit. One of them goes first in the row as the anchor, and the other is hidden. The anchor tells the magician the hidden card’s suit.
The short way round. Put the thirteen ranks on a clock, ace after king. Go forward from one rank to the other, then forward again back to the first: together that is one full lap, so the two forward distances add to 13 and one of them is at most 6. The assistant shows the card at the start of the short trip and hides the one at its end. Now the hidden card is one of just six cards: 1 to 6 steps forward from the anchor, same suit.
Three cards, six orders. Rank the other three cards as low, middle and high (L, M, H): by rank first, with equal ranks split by suit ♣ < ♦ < ♥ < ♠. Three different cards can be laid out in exactly six orders, one for each distance:
1 LMH 2 LHM 3 MLH 4 MHL 5 HLM 6 HML
Take the hand 2♣ 10♠ 2♥ J♣ 8♦. The jack and two of clubs are four steps apart going jack, queen, king, ace, two. So the jack is shown first, the two is hidden, and the other three go down as M H L: the row is J♣, 8♦, 10♠, 2♥.
Choosing which card to hide is the assistant’s real power: it turns a twelve-way question about rank into a six-way one, and three cards have exactly six orders.
Step through it below. The three views are deliberately separate: what the assistant knows (the hand), what is actually sent (the row), and what the magician does with the row alone.
Challenge
Play both roles. As the magician, decode a fresh row and name the hidden card. As the assistant, build a row the magician can read. Every mistake is explained, never just marked wrong.
Perform it yourself
Set-up
- An ordinary 52-card deck with no jokers, a partner who knows the method, and an envelope or card to cover the hidden card.
- Agree on every convention in advance, and rehearse until neither of you hesitates: ranks A < 2 < … < K with equal ranks split ♣ < ♦ < ♥ < ♠; the table 1 LMH, 2 LHM, 3 MLH, 4 MHL, 5 HLM, 6 HML; the anchor in position 1; the row read left to right from the magician’s seat.
The assistant’s job
- Find the first suit, in the order ♣ ♦ ♥ ♠, with at least two of the five cards, and take its two lowest ranks. (Any same-suit pair works; a fixed rule makes rehearsal repeatable.)
- Count forward around the clock from one to the other. One direction is 1 to 6 steps. Show the card where that short trip starts; hide the card where it ends.
- Rank the other three as L, M, H, rank first and suits only for ties.
- Lay the anchor in position 1 and the other three in the order that sends the distance. Cover the hidden card.
The magician’s job
- The first card is the anchor: it gives the suit and the place to start counting.
- Rank the other three, read their order in the row, and turn it into a number from 1 to 6.
- Count that many steps forward from the anchor’s rank, ace following king, and keep the anchor’s suit. Name the card before it is uncovered.
Performance
- Let the spectator check and shuffle the deck, then choose any five cards. Turn away before any face is shown, and stay turned away while your partner prepares the row.
- Be honest about the roles: the spectator chooses the hand; the assistant chooses which card to hide. That freedom is essential. If the spectator insists on choosing the hidden card, this method no longer applies.
- Once the row is laid, it must not be touched. The four identities and their order are the whole message: no spacing, gestures or timing. The row could be read out to you as four card names and the trick would still work.
- Turn back, read the row without moving any card, commit to one card, then ask for the covered card to be turned over.
Watch for: counting the anchor itself as step one (from J, the first step lands on Q); calling the hidden card “the lower one” (with 2 and J, the two is hidden, because the short trip crosses king to ace); and sorting the row before reading it, which erases the message.
Going deeper
Why five cards, and why five is the threshold
Drop the five cards into four boxes labelled ♣, ♦, ♥, ♠. If every box held at most one card there would be at most four cards, so some box holds two. This is the pigeonhole principle; it does not depend on how well the deck was shuffled or on a spectator trying to be difficult. The suits of five cards can split as 2-1-1-1, 2-2-1, 3-1-1, 3-2, 4-1 or 5, and even the least generous split offers one pair.
Try to build a counterexample: one card of each suit uses only four cards, and the fifth must land in an occupied box. With five suits, or with duplicate cards allowed, the argument would need repair. Distinct cards in one suit always have different ranks, so the clock distance between them is never zero.
Odd clocks never tie
On a clock with 2r + 1 positions, the two forward distances between different positions add to 2r + 1, so one is at most r, and because the total is odd they can never be equal. Our ranks give 13 = 2 × 6 + 1, so r = 6, and three cards have exactly 3 × 2 × 1 = 6 orders. The numbers fit exactly. On an even clock, opposite positions tie and a protocol would need an extra rule.
The rehearsal rule picks one pair, but a hand may have several. That ambiguity is harmless: the assistant may choose any pair and orientation that works, as long as the row the magician receives names exactly one card. The extreme hand A♣ A♦ A♥ A♠ 2♣ works too: the only pair is A♣ and 2♣, one step apart, and the three other aces are told apart by suit, A♦ A♥ A♠ reading L M H, which sends 1.
Same row, different convention
A code is a shared agreement, not a property of the cards. Take the three cards 8♣, 10♥, 2♠ in that order. With our rank-first order, 2♠ is L, 8♣ is M and 10♥ is H, so the row reads M H L and sends 4. With a suit-first order (all clubs, then diamonds, hearts, spades), 8♣ is L, 10♥ is M and 2♠ is H, and the same row reads L M H, sending 1.
Reading direction matters in the same way: M H L read backwards is L H M, which turns 4 into 2. Because all six orders are used, a mismatch does not produce an obvious error; it produces a plausible wrong card. Both partners must use the same order, the same table and the same anchor position. Some published versions use other conventions; any of them works when both partners share it.
Twenty-four orders, forty-eight unseen cards?
Four cards can be arranged in only 24 orders, yet 48 cards are not on the table. How can 24 arrangements pick out one of 48? Because the assistant also chooses which four cards are shown, and the identities are part of the message. Across the whole deck there are 2,598,960 possible five-card hands but 6,497,400 ordered rows of four cards, so there is room to spare. The same count shows that no deck of more than 124 cards can support a four-card message, and a cleverer method works for exactly 124; that is where this act ends.
Sources
The trick is due to William Fitch Cheney Jr.; it first appeared in print in Wallace Lee’s Math Miracles (1950). Michael Kleber’s article “The Best Card Trick” (The Mathematical Intelligencer, 2002) made it famous among mathematicians and gave it its nickname.
Colm Mulcahy, Mathematical Card Magic: Fifty-Two New Effects (CRC Press, 2013), Chapter K, pp. 305–310, explains the mechanism and the pigeonhole and short-arc principles. This course deliberately uses its own conventions: the anchor in the first position and a rank-first order with suits breaking ties. The book uses a different, equally consistent, convention.
Every one of the 2,598,960 possible hands was checked for this page: the assistant’s rule always produces a row, and the magician’s rule always names the hidden card.