Last Cards Match
Two face-down packets: ace to four of hearts, and ace to four of spades. Eleven times a round, you pick a packet and move its top card to the bottom. Then a pair is set aside. After three rounds, the last two cards match, and so does every pair you set aside.
Try it
You choose the packet for every move: all on one side, split down the middle, or switch every time. The page only counts.
Why it works
Start with one packet. A move takes its top card to the bottom, so four moves on a four-card packet bring every card back where it began. The packet behaves like a clock with four positions: only the remainder of its move count matters.
Now put the four ranks on a ring: ace, two, three, four, clockwise. The hearts packet reads that ring clockwise from its top card. The spades were prepared in the opposite order, four, three, two, ace, so they read the same ring anticlockwise.
Mark each packet’s top card with a pointer. A move on the hearts steps the ♥ pointer one place clockwise. A move on the spades steps the ♠ pointer one place anticlockwise. Either way, the two pointers come one step closer, which is why your choice of packet never mattered.
Every move, on either packet, closes the gap between the two top cards by one step, so only the total counts.
The pointers start three steps apart, one short of a full turn. After a total of 11 moves they are 3 − 11 = −8 steps apart, and −8 on a four-step clock is 0: the tops match. Set that pair aside, and the three ranks left still form a ring read in opposite directions, now two steps apart. Eleven closes that gap too, because 11 + 1 = 12 is a multiple of 3, and once more with two cards, because 12 is a multiple of 2.
In symbols: with m cards in each packet, a moves on the hearts and b on the spades, the tops match exactly when a + b + 1 is a multiple of m. You chose a and b; the routine fixed a + b = 11.
Challenge
Two tasks. Set the packet size, the total and how the moves are split, run every round, and read why each pair does or does not match.
Perform it yourself
Set-up
- Take the ace, two, three and four of hearts, and the same four ranks in spades.
- Hearts, face down, top to bottom: A♥, 2♥, 3♥, 4♥ (the ace on top).
- Spades in the opposite order, face down, top to bottom: 4♠, 3♠, 2♠, A♠.
- Put the two packets side by side with room for three discarded pairs. No cuts and no shuffles.
Performance
- Explain a move: take the top card of either packet and put it underneath that same packet. It is not turned over, and it never goes to the other packet.
- The spectator makes eleven moves in all, choosing the packet each time. All eleven on one side is allowed. Count aloud, but only after each completed move: one, two, … eleven.
- After the eleventh, take the top card of each packet and put the two together, face down, as the first discarded pair. Keep the order of the cards still in each packet.
- Three cards remain on each side. Repeat: eleven moves, counting from one again, then set the top pair aside.
- Repeat once more with two cards on each side.
- One card is left in each packet. Predict that they have the same rank, and have them turned over together.
- Then turn over the discarded pairs, one pair at a time. Every pair matches in rank.
Handling warnings
- Moves are one card at a time, top to bottom. Do not deal a group or reverse a packet: this is a rotation, not a count-out-and-transfer.
- If anyone miscounts, drops a card, turns a card over or moves one to the other packet, say so: “That changes the procedure. Let us reset the two packets and start again.” Never claim the guarantee after an untracked move.
- Promise only that the cards match. The final rank depends on the spectator’s choices.
Going deeper
Why eleven, and which other totals work
At a stage with m cards on each side, the total must be one less than a multiple of m. The stages have 4, 3 and 2 cards, so the total plus one must be divisible by all three. Their least common multiple is 12, so 11 is the smallest total that works, followed by 23, 35, and every 11 + 12q.
Why 12 and not 4 × 3 × 2 = 24? Divisibility by 4 already includes divisibility by 2. A product gives a common multiple, but not always the least one.
Zero fails: the starting tops are an ace and a four. Five works for the three-card and two-card stages, since 6 is divisible by 3 and 2, but fails at four cards. With five ranks on each side, the stages 5, 4, 3, 2 have least common multiple 60, so the total would be 59 moves a round: valid, but long for a performance. You could also change the total every round to m − 1 (3, then 2, then 1). That works too, but loses the single repeated number.
Why the packets must be in opposite order
If both packets started in the same order, the top cards would match when the two move counts agree on the clock, that is, when a − b is a multiple of m. A fixed total says nothing about a difference, so the spectator could break the match. Reversing one packet turns the relevant relation into a sum, and a sum is exactly what the count of eleven controls.
Clocks that disagree: the Chinese Remainder Theorem
In this trick every clock asks for the same thing, one short of a full turn, so a common multiple solves everything at once. Other targets need more care. Suppose a total must leave remainder 2 on a 3-clock and remainder 1 on a 4-clock. The candidates for the first clock are 2, 5, 8, 11, …; on the 4-clock they read 2, 1, 0, 3. The first match is 5, and every solution is 5 plus a multiple of 12.
Sometimes there is no solution. Remainder 1 on a 4-clock forces an odd number; remainder 2 on a 6-clock forces an even one. Change the second target to 3 and both ask for odd numbers; 9 works, and so does every 9 + 12q.
The general rule: conditions “remainder a modulo m” and “remainder b modulo n” can both hold exactly when a and b agree modulo the greatest common divisor of m and n; then the answer is unique modulo their least common multiple. These clock puzzles are mathematics in their own right, not new card tricks: packets of unequal size do not keep the pair-by-pair match.
Sources
The opposite-order, last-cards-match mechanism is inspired by Tadashi Tokieda’s Chinese Remainder Theorem and Cards (Numberphile). The exact four-rank, eleven-move routine here, with rotations only and no count-out-and-transfer, is a checked teaching adaptation of that idea, not a transcription of the video.
Episodes 50 and 51 of Magic & Math with Cards develop the performance and the remainder arithmetic behind it.