Always Five Face Up
Ten cards, all face down. You cut them and turn pairs over for as long as you like, then turn over every second card. A prediction, sealed before you began, says exactly five will face up.
Watch the lectureLecture 06 · Ten cards, always five face up
Try it
You choose every cut, every turn and when to stop. The cards follow the real moves.
Why it works
Number the places 1 to 10 from the top and split them into two lanes: the odd places 1, 3, 5, 7, 9 and the even places 2, 4, 6, 8, 10. Each lane has five places.
Suppose that just before the finish each lane holds the same number of face-up cards — say two. The finish leaves the odd lane alone, so it still has two. It turns over every card in the even lane, so that lane’s two face-up cards go down and its three face-down cards come up. Two plus three is five. If the common number is a, the total is a + (5 − a) = 5, whatever a was. We never need to know a.
So everything rests on the balance: the two lanes always hold equally many face-up cards. At the start both hold none. A one-card cut moves every card to a place of the other parity — even the card that wraps from place 1 to place 10, because ten is even — so the two counts simply swap, and equal counts stay equal. A bigger cut is several one-card cuts.
Turning the top two together touches only places 1 and 2: one odd place and one even place. Remember that the pair swaps as well as turning over.
| Top pair | After turning together | Odd lane | Even lane |
|---|---|---|---|
| down, down | up, up | +1 | +1 |
| down, up | down, up (cards swapped) | 0 | 0 |
| up, down | up, down (cards swapped) | 0 | 0 |
| up, up | down, down | −1 | −1 |
In every case both lanes change by the same amount. Every legal move keeps the balance, so it is there whenever you decide to stop, however many moves you made.
Every legal move keeps the face-up counts of the odd and even places equal, and turning one lane of five over changes a balance of a and a into a + (5 − a) = 5.
Step through the run from the episode: ten cards, all face down, labelled A to J so that letters name cards and numbers name places, cuts of 3, 6 and 1 with a turn of the top two after each, then the finish.
Challenge
Each version below changes one rule. Find a history that breaks the promise, then name the step of the proof that failed. These are changed rules, kept apart from the real trick.
Perform it yourself
Set-up
- Write “Exactly five face up” on a card or slip and set it aside, face down.
- Take any ten different cards. The spectator may shuffle them, as long as all ten end face down.
Performance
- Show the two moves. A cut: any number of cards from the top to the bottom, order unchanged. A turn: lift the top two together, turn that small packet over, and replace it on top. Demonstrate once, then put all ten face down again.
- The spectator cuts and turns the top two together, round after round, choosing a different cut each time. They may look at the cards. They stop whenever they like, with a final straight cut if they wish.
- Deal the cards one at a time into a row of ten places, left to right, keeping each card facing as it is — except at places 2, 4, 6, 8 and 10, where you turn the card over as you put it down.
- Count the face-up cards, then the face-down cards separately: five and five. Turn over the prediction.
“Turn over” means reverse its current direction. A face-up card at place 4 goes face down. Do not make every second card face up — that agrees with the rule only when the cards there were all face down, which mixing does not guarantee.
Turn the two together, never one at a time. And before a repeat, turn all ten face down again: the all-down start is what makes the lanes begin balanced. If the count ever comes out wrong, check the start and the alternating flips openly, and start again.
Going deeper
Any even packet works
With 2n cards each lane has n places, and the same proof gives exactly n face up: three for six cards, six for twelve. The start need not be all face down — any pattern with equal lane counts works. Six cards up, up, down, down, up, up have two face up in each lane, so they also finish with three.
When the lanes start unequal
Call the odd-lane count a, the even-lane count b, and their difference D = a − b. A turn of the top two leaves D alone; a one-card cut swaps the lanes, so D becomes −D. The size of D never changes, but its sign can, and the finish gives n + D face up. Six cards with only the top one up have D = 1 and finish with four; cut one card first and D = −1, so the finish gives two.
Odd packets and bigger turns
With five cards, a one-card cut sends the top card from place 1 to place 5 — odd to odd — so the cut no longer exchanges the lanes. Start all down, turn the top two (up, up, down, down, down), cut one (up, down, down, down, up), and turning over places 2 and 4 gives four face up; without the cut it gives two.
Turning any even number of cards together is safe. Inside a turned block of 2m cards every card changes lane and every face flips: if the block had x face up in odd places and y in even places, afterwards it has m − y and m − x, and the difference (m − y) − (m − x) = x − y is unchanged. Turning three is not safe, as the Challenge shows.
How much freedom does the spectator have?
A computer search from ten face-down cards finds that cuts and two-card turns reach exactly 252 face patterns. That is every pattern whose lanes are balanced: the spectator can reach any of them, and none of the others.
Sources
Persi Diaconis and Ron Graham, Magical Mathematics, Chapter 1 (pp. 4–8, 12–14): Bob Hummer’s ten-card effect, the balance proof and the discussion of odd packets; the row-dealing finish is a teaching handling of “turn over every second card”. Colm Mulcahy, Mathematical Card Magic, Chapter 9 (p. 212): equal face-up counts in odd and even positions after cut-and-turn-two mixing (CATO). Course episodes 06 (performance) and 07 (a proof in two lanes).