Section 7 · The Best Card Trick
Hide the Key Card
You choose any five cards. The assistant covers one and lays the other four in a row. No card sits in a special place, nothing marks a key card, and yet those four cards, read left to right, name the fifth.
Try it
Pick any five cards, or deal five from a shuffled deck. This page plays both partners: the assistant who sees your hand and builds the row, and the magician who sees only the row.
Why it works
This is the Fitch Cheney five-card trick from Four cards name the fifth, with one change. A quick reminder of the basic method: five cards in four suits must include two of the same suit. The assistant hides one of them and shows the other, the anchor, which tells the magician the suit. On a clock of thirteen ranks (A, 2, …, K, then A again), one of the two directions from one card to the other takes at most six steps, so the assistant hides the card that sits 1 to 6 steps forward of the anchor. The remaining three cards send that number by their order: call them low, middle and high (rank first; equal ranks broken by ♣ < ♦ < ♥ < ♠), and use LMH = 1, LHM = 2, MLH = 3, MHL = 4, HLM = 5, HML = 6.
In the basic version the anchor always sits first, and an attentive viewer might notice that the first card always shares the hidden card’s suit. Here the anchor’s position is computed. Count ace as 1, jack 11, queen 12, king 13. Add the ranks of the four visible cards and divide by 4. A remainder of 1, 2 or 3 is the anchor’s position; a remainder of 0 means position 4. The other three cards go in the remaining slots, read from left to right, whether or not those slots are next to each other.
Could the rule demand a row the assistant cannot build? The assistant needs the position before arranging anything, while the magician computes it from the finished row. That works because a sum does not care about order: the same four cards have the same total in every one of their arrangements.
A sum ignores order, so the anchor’s slot is fixed the moment the hidden card is chosen: the assistant reserves it, and the magician finds the same slot in the finished row.
Step through the episode’s worked hand, the remainder-zero hand, or the last hand you chose above.
Why 24 orders are not the whole story
Four cards can be arranged in only 24 orders, yet 48 cards remain unseen. How can 24 arrangements name one of 48 cards? The puzzle quietly freezes something the assistant is free to choose: which four cards are shown. Hide a different card and the visible set changes, and the visible cards’ identities are part of the message.
So count the right objects. A hand is an unordered set of five cards. A message is four distinct cards in a definite order. A message together with the card it decodes to rebuilds the whole hand, so two different hands can never share a message. A working method therefore needs at least as many messages as hands.
Enough messages is necessary, not sufficient. Every displayed card must come from the hand, and that restriction could trap some group of hands with too few usable messages even when the totals balance. Imagine three hands and three messages, where the first two hands may only use the same single message: three messages exist, yet two hands are fighting over one of them. To see compatibility at work, shrink the problem.
A smaller miracle: three cards from eight
Take eight cards labelled 1 to 8. The spectator chooses three; the assistant hides one and shows the other two in order. There are 56 hands (8 × 7 × 6 ÷ 6) and exactly 56 messages (8 × 7), so a working method must use every message exactly once. Draw the messages as a table: the row is the first card shown, the column is the second, and the diagonal is empty because a card cannot be shown twice.
The complete table is a finished certificate for this one small puzzle, checked row by row. It is not yet a reason why such tables must exist for bigger decks. That reason is Hall’s marriage theorem, which the 124-card finale uses to push the trick to its true limit.
This table was not found by trial and error, either: it is exactly the finale’s arithmetic code for 124 cards, run with three chosen cards and two shown in place of five and four.
Challenge
Two tasks. First, be the magician: the assistant has laid out a row you have not seen before, and you must name the covered card, one stage at a time. Second, finish an eight-card table that a too-tidy assistant left stuck.
1. You be the magician
2. Finish the eight-card table
An assistant went through the 56 hands in dictionary order, giving each the first free message from its list: the three increasing pairs first, then the decreasing ones. Forty-eight hands got a message. Eight found all six of theirs taken. Place them without breaking any earlier hand: move a hand onto a taken message, and the hand you displace needs a new one.
Perform it yourself
Set-up
- An ordinary 52-card deck without jokers, something opaque to cover one card, and four places for the row, numbered 1 to 4 from the magician’s left.
- Both partners agree on every convention: ace 1, jack 11, queen 12, king 13; rank first with ♣ < ♦ < ♥ < ♠ for equal ranks; the table LMH 1, LHM 2, MLH 3, MHL 4, HLM 5, HML 6; and that the anchor’s position is computed, not fixed. If one partner uses the fixed rule and the other the computed one, the answer can be a plausible wrong card.
- A worksheet for the assistant’s arithmetic is fine. Let the audience see it, then hide it before the magician turns back.
Performance
- The magician turns away before any card is chosen. A spectator shuffles, chooses any five cards and hands them to the assistant, keeping the rest of the deck.
- The assistant finds two cards of one suit. For repeatable rehearsal, take the first suit in the order ♣, ♦, ♥, ♠ that has two cards, and its two lowest ranks.
- Count forward around the rank clock from one to the other. Hide the card that is 1 to 6 steps ahead and cover it; the other card is the anchor. Note the number of steps.
- Add the ranks of the four cards that will stay visible: the anchor and the other three. Never include the covered card. Divide by 4: remainder 1, 2 or 3 is the anchor’s position, remainder 0 is position 4. Place the anchor there.
- Rank the other three as low, middle and high, without the anchor. Put them in the remaining positions, left to right, in the pattern for your number of steps.
- The spectator confirms the row and the covered card. Nobody touches them again until the reveal.
- The magician turns back, adds the four ranks, takes the remainder (0 means 4) and finds the anchor. Reading the other three left to right, skipping the anchor, gives the pattern and so the number of steps. Count that many ranks forward from the anchor, keep its suit, name the card, and ask the spectator to turn it over.
The usual slips are worth rehearsing on purpose: including the hidden card in the sum, treating remainder 0 as the first position, ranking the anchor along with the three, and reading the three in a different order from the one they lie in.
Going deeper
What the computed anchor changes, and what it does not
The variation is presentational. Once the anchor is located there are still exactly six messages for the offset, and anyone who knows the rule can decode the row. What changes is that no single position always plays the key role; finding the role is itself a calculation over all four visible cards.
The order of dependencies is the whole design. The assistant goes hand → hidden card and anchor → visible set → sum → position → order of the other three. The magician runs it backwards: row → sum → position → anchor and ordered triple → offset → hidden card. A position rule that changed when the row was rearranged (say, “the anchor goes where the largest card is not”) would create a circular requirement and need its own proof that a compatible row always exists.
Over all 2,598,960 hands, the computed position lands on slots 1, 2, 3 and 4 almost equally often (649,900, 649,548, 649,924 and 649,588 times with the rehearsal pair rule).
The counting bound: at most 124 cards
With N distinct cards there are N(N − 1)(N − 2)(N − 3)(N − 4) ÷ 120 hands and N(N − 1)(N − 2)(N − 3) messages. Asking for at least as many messages as hands and cancelling the common positive product leaves (N − 4) ÷ 120 ≤ 1, that is, N ≤ 124. At 124 both counts are 225,150,024. At 125 there are 234,531,275 hands but only 232,593,000 messages, so two hands must share a message and no method can work, however clever.
The bound assumes the message is only the four identities and their order. Add face-up or face-down choices, spacing or spoken hints, and you have a different channel whose count must be redone.
From eight cards to 124
In the eight-card graph every hand has six legal messages and every message fits six hands. That equal degree on both sides is what rules out bottlenecks: any group of hands sends out six edges each, and those edges land on messages that can each absorb at most six, so the group always reaches at least as many messages as it has hands. Hall’s theorem turns that into a complete assignment. For 124 cards the same counting works with degree 120 on both sides; see the 124-card finale.
Sources
The five-card trick is due to William Fitch Cheney Jr. The computed-position variation follows Colm Mulcahy, Mathematical Card Magic: Fifty-Two New Effects, Chapter K, pp. 310–312 (the variable suit-giver principle); the position rule is his, while the first-position base version, the rehearsal pair rule and the rank-first card order are this course’s conventions.
The counting bound and the matching formulation follow Norman Do, “A magical card trick” (Mathellaneous), p. 11. The complete 56-row eight-card table comes from the course scripts, where it was generated by the same modular code as the 124-card finale (with three cards in place of five); it is a finite certificate for that puzzle, not a proof for larger decks.