One Rank, One Suit
Cut a full deck as often as you like and deal three cards in a row. Tell me only the value of the first card and the suit of the second. I will name all three, including the one you said nothing about.
Watch the lectureLecture 15 · One rank, one suit, three revelations
Try it
You play all three spectators. Cut as often as you like, deal three cards, then give me the two clues. The deck follows the real rules: straight cuts only, and no peeking on my side.
Why it works
The deck is not shuffled. It is built from two short lists that repeat. The values follow a cycle of thirteen: 8, K, 3, 10, 2, 7, 9, 5, Q, 4, A, 6, J, and then 8 again. The suits follow a cycle of four: clubs, hearts, spades, diamonds, and then clubs again. Each card takes the next value and the next suit.
Could a card turn up twice? Two equal cards would need a gap that is a multiple of 13 (to repeat the value) and a multiple of 4 (to repeat the suit). Since 13 and 4 share no common factor, the first gap that does both is 13 × 4 = 52. So the 52 positions hold 52 different cards: an ordinary full deck.
A cut only changes where we start reading the circle. The three dealt cards are still neighbours. The first clue gives the first value. The second clue gives the second suit, and stepping back one place on the suit cycle gives the first suit. Now the first card is known, and stepping both cycles forward names the other two.
A value and the next suit pin down where you are on both clocks at once, and the clocks then tell you every card that follows.
Step through it below. Watch the two cycles build the deck, then decode the worked example from the script: eight, then hearts.
Suits as an alphabet
Earlier tricks in this act asked for colours: an alphabet of two letters, red and black. Four colours in a row give 2 × 2 × 2 × 2 = 16 possible reports. A suit is a richer letter. It has four possible answers, so just two suits in a row already give 4 × 4 = 16 ordered reports: clubs then hearts is different from hearts then clubs, and clubs then clubs counts too.
This trick mixes two alphabets. The value has 13 possible answers and the next suit has 4, which gives 13 × 4 = 52 possible clue pairs, one for each place the deck could have been cut. Counting tells us that 52 clue pairs could be enough. The two cycles are the construction that makes every clue pair appear exactly once. A count on its own is not an arrangement: the Challenge below asks you to build one.
Challenge
Two tasks. First, be the magician: decode three cards from a value and a suit. Then be the designer: arrange sixteen suits in a circle so that every ordered pair of suits appears exactly once.
Perform it yourself
Set-up
- Arrange the full deck, face down, from the top: 8♣ K♥ 3♠ 10♦ 2♣ 7♥ 9♠ 5♦ Q♣ 4♥ A♠ 6♦ J♣ 8♥ K♠ 3♦ 10♣ 2♥ 7♠ 9♦ 5♣ Q♥ 4♠ A♦ 6♣ J♥ 8♠ K♦ 3♣ 10♥ 2♠ 7♦ 9♣ 5♥ Q♠ 4♦ A♣ 6♥ J♠ 8♦ K♣ 3♥ 10♠ 2♦ 7♣ 9♥ 5♠ Q♦ 4♣ A♥ 6♠ J♦
- You do not need to memorise 52 cards. Learn the two cycles. Values: 8, K, 3, 10, 2, 7, 9, 5, Q, 4, A, 6, J. A traditional sentence helps: “Eight kings threatened to save ninety-five queens for one sick jack.” Suits: clubs, hearts, spades, diamonds (“CHaSeD”).
- Set out three numbered places, so that the three cards cannot be confused or reordered. A printed reference sheet of the two cycles is entirely fine while you are learning.
Performance
- Hand over the deck. Ask for a straight cut: move a block from the top to the bottom, keeping the order inside each block. Several cuts, by several people, are fine. No shuffling.
- Ask for the top three cards to be dealt one to each place: the first to place 1, the next to place 2, the next to place 3. Each person looks privately at their own card.
- Place 1 names only the value (“ace, king, queen or jack, otherwise the number”). Repeat it back so you agree. Place 2 names only the suit. Place 3 says nothing.
- Decode. Step back one suit from the second suit to get the first suit, and combine it with the named value: that is card 1. Step value and suit forward once for card 2, and once more for card 3.
- Name the cards in order, and ask each person to turn their card over only after you have named it.
Handling warnings
- Deal to three separate places. If the three cards are dealt into one face-down pile and read from its top, their order is reversed and the decoding fails.
- Ask for rank names (“jack”), not numbers, to avoid confusion over whether a jack is eleven.
- Wraparound is not a special case: before clubs comes diamonds, and after jack comes eight.
- If the deck is shuffled, reset it. There is no decoder for a deck whose neighbour relations have been destroyed.
Going deeper
A count is not an arrangement
The counting argument (13 values × 4 suits = 52 clue pairs) says the trick is possible. It does not make any arrangement work. Swap just the top two cards of the stack, so it begins K♥ 8♣ 3♠ … Now the clue “king, then clubs” fits two different places: the start K♥ 8♣ 3♠ and the start K♦ 3♣ 10♥, further down. Three different clue pairs collide this way after that one swap. The two cycles are what guarantee that no clue pair is used twice.
On the other hand, the particular value order is only a memory aid. Any order of the thirteen values, used as a repeating cycle together with any order of the four suits, gives a working deck, provided you decode with the same cycles.
How far can a suit alphabet go?
With an alphabet of q letters and a window of k positions there are qk possible reports. A cycle that uses every one of them exactly once is a de Bruijn cycle. For suits (q = 4) and two reports (k = 2) that is a circle of 16, built in the Challenge. The same Euler-tour argument that builds binary cycles works for any alphabet: four suits as vertices, sixteen ordered pairs as arrows, four arrows in and four out of every suit.
Three suit reports give 4 × 4 × 4 = 64 words, more than 52. But a full 64-position suit cycle uses each suit 16 times, and a deck has only 13 cards of each suit. The mathematics allows it; one ordinary deck does not. A 52-card arrangement with distinct three-suit windows would need its own construction and check. This trick avoids the problem by asking for a value, from an alphabet of 13, and a suit, from an alphabet of 4.
Sources
The three-person effect is Ronald Wohl’s, as described in Persi Diaconis and Ron Graham, Magical Mathematics (Princeton, 2012), pp. 52–55, which also discusses the “Eight Kings” value order and the CHaSeD suit order. Larger alphabets appear there on p. 25. The 16-position suit cycle 0010203112132233 (clubs 0, hearts 1, spades 2, diamonds 3) is a checked teaching construction from episode 20 of this course.