Section 5 · Count Out and Transfer
Nine Cards, One Named
Shuffle nine cards and remember the third. Name any card in the deck, even one that is not among the nine, and spell its name twice, one card per letter. Your card is third again.
Watch the lectureLecture 45 · Nine cards and a freely named card
Try it
Peek at a card, name any card you like, and deal along with the spelling. You make every choice; the cards follow the real moves.
Why it works
Each word is one count out and transfer, or COAT. You deal k cards one at a time onto the table, so each lands on the one before and the counted cards come out reversed; then the rest of the packet goes on top. Follow one card: if it was dealt (its position i is at most k) it moves to position 10 − i; if it stayed in your hand it moves up to i − k.
What the words mean never matters, only how long they are. Ace, Two, Six and Ten have three letters; Four, Five, Nine, Jack and King have four; Three, Seven, Eight and Queen have five. The plural suits have five letters (Clubs), six (Hearts, Spades) or eight (Diamonds). So fifty-two names give only nine pairs of counts. Call them s for the value and t for the suit: every name deals s, t, s, t.
When s + t is at least 9, split the packet into three blocks: T, the top 9 − t cards; B, the bottom 9 − s; and M, the s + t − 9 cards between. The first count deals exactly T and M, so the packet becomes B, Mᴿ, Tᴿ (ᴿ for reversed). The second count deals exactly B and Mᴿ, giving Tᴿ, M, Bᴿ: every block is home, the outside two reversed. Spelling the pair again reverses them back, and the whole packet is restored.
Eight of the nine pairs reach 9. The ninth, 3 and 5, belongs to the Ace, Two, Six and Ten of Clubs, and adds up to only 8. For those four names the whole packet does not come back: it ends 4, 2, 3, 9, 5, 6, 7, 8, 1. Position 3 does, by a route of its own: 3 → 7 → 2 → 8 → 3.
Eight pairs restore every position and the ninth restores six positions, 3 among them, so position 3 is safe whatever card is named.
Step through it below with the nine cards labelled by where they start. Try Queen of Hearts, where the blocks do the work, then Ace of Clubs, where they cannot.
Take the positions that every one of the nine pairs fixes, and you get 2, 3, 5, 6, 7 and 8. Positions 1, 4 and 9 work for 48 names and fail for the four exceptional Clubs. That is why the performer asks for the third card, and not for any card the spectator likes: one simple procedure then works for every name.
Challenge
Two tasks. First, predict two different promises: does one card come home, and does the whole packet? Then find the positions a performer may safely promise for a whole menu of names, including one you design.
Perform it yourself
Set-up
- Hand any nine ordinary cards to the spectator to shuffle. Their order does not matter. The packet stays face down.
- Ask them to remember the third card: lift the top two together as a little block, look privately at the next card, then put the two-card block back exactly as it was. The remembered card never moves. Look away while they do it.
Performance
- Ask them to name any card from an ordinary deck. It can be the card they remembered, or a different one, and it need not be among the nine. Say: “We will use that value and that suit, with the word of left out.”
- Spell the value first. Deal one card for each letter into a single pile, so each new card lands on top. When the word ends, place all the cards still in hand on top of the pile. That is one operation.
- Spell the suit the same way, using the plural: Clubs, Diamonds, Hearts or Spades. Remainder on top.
- Repeat the same value, then the same suit, with the same spelling and the same dealing. Four operations in all: value, suit, value, suit.
- Square the packet and keep it face down. There is no final cut.
- Count the top two cards aside one at a time, without turning them over. Ask which card they remembered, then have them turn over the next card.
Handling notes
- The spellings are fixed: Queen has five letters, Ten three, and the suits are plural. No spaces, no punctuation, no of. Putting of back in adds a count; spelling the suit first, or changing words halfway, are different procedures that need their own check.
- Squaring the packet is harmless. A cut is not: moving even one card from top to bottom puts the remembered card second, and you would reveal the wrong card.
- If someone names a Joker: “For this version, choose one of the fifty-two ordinary card names.” No claim covers Jokers or other spellings.
- If a deal goes wrong: “Let us reset and follow those four words carefully.” Start again openly from a fresh shuffle and a new peek. Never hide a reset.
- Rehearse with cards numbered 1 to 9 in order. Say each word aloud as you deal. For Ace, Two, Six or Ten of Clubs the packet should end 4, 2, 3, 9, 5, 6, 7, 8, 1; for every other name it should end exactly as it began.
Going deeper
Unequal counts can still cancel
The theorem behind the eight ordinary pairs holds for any packet of n cards: if both counts lie between 1 and n and s + t ≥ n, the schedule s, t, s, t restores everything. The blocks have lengths n − t, s + t − n and n − s; T and M together make s cards, and M and B together make t. With thirteen cards and counts 6, 9, 6, 9, the first pair leaves 4, 3, 2, 1, 5, 6, 13, 12, … , 7, and the second pair restores the lot.
The ordered pair matters. Call the two moves together G. Then G reverses T and B and leaves M alone, so doing G twice changes nothing. The single moves need not undo themselves; only the pair does. Run the pair the other way round (9, 6 instead of 6, 9) and it still squares to nothing, but it protects a different middle region.
You can also work backwards. To keep positions a to b exactly in place after each pair, choose s = b and t = n − a + 1. For twelve cards and positions 4 to 8, that is 8 and 9.
Four equal transfers restore the whole packet
The special case s = t = k needs 2k ≥ n: count at least half the packet four times and every card returns. Writing F for one transfer, F⁴ is the identity. That does not make four the first return. With ten cards counted seven at a time, the positions form the cycles (1 10 3 8)(2 9)(4 7)(5 6): the four cards on (1 10 3 8) need four moves; the other six return after two. Every cycle length divides four, so a three-cycle can never occur.
The named-card trick asks for something weaker than full restoration, and that is what lets it survive the one pair outside the theorem. Promising less, and choosing what to promise carefully, made the effect work for every name.
Other languages, other promises
The English length table does not carry over to another language. To translate the effect, list the new value and suit spellings, find their lengths, and work out which positions every pair protects. Even an English variation changes the answer: with singular suits (Club, Heart…) the safe positions shrink to 3, 6 and 7. Position 3 survives that one too. The menu designer in the challenge will compute the safe set for any words you give it.
Sources
Colm Mulcahy, Mathematical Card Magic: Fifty-Two New Effects (CRC Press, 2013). The COAT move and its three- and four-transfer properties are in the COAT material at printed pp. 30–42; unequal counts and the nine-card named-card spelling are in chapter 5, printed pp. 105–118. The spoken presentation, the safe-position analysis as presented here and the challenges are teaching adaptations for this course.