Cut a deck into two exact halves and interlace them one card at a time. Do that eight times and the deck is in precisely the order it started in.
8
perfect out-faros · 52 cards · back where it began
On the rest of this site the mathematics is what makes a trick work. Here it does the opposite: a perfect shuffle does not shuffle anything. It moves every card to a place decided entirely by where it already was, so nothing is lost, and eight repetitions walk the whole deck around a circle and back to the start.
Every card in the deck, in order. Press a single shuffle and the deck cuts into two halves and the cards come down one at a time into the row underneath, alternately from each: green edges came from the top half, red from the bottom. Watch the two packets go down while the woven row builds. A card is ringed in brass when it is back where it started.
Pick a position from 1 to 52. The deck resets and the shuffles are worked out from the number written in binary, which is explained below.
Number the positions 0 to 51 from the top. An out-faro takes the top 26 cards and the bottom 26 and lays them alternately, top half first, so the card at position 0 stays at 0 and the card at position 51 stays at 51. Every other card lands at:
new position = 2 × old position, modulo 51
The doubling is the whole mechanism. A card at 3 goes to 6, then 12, then 24, then 48, then 96 − 51 = 45, and so on. Do the shuffle k times and the card that began at i is at:
position after k shuffles = 2k × i, modulo 51
So the deck is back exactly when 2k = 1 modulo 51, whatever i was. That is a question about the number 51 and nothing else:
21=2 22=4 23=8 24=16
25=32 26=13 27=26
28 = 256 = 5 × 51 + 1 →
1
Eight. Not a fact about cards at all, but a fact about 51, which is the number of positions that are allowed to move.
An in-faro is the same move with the other half first, so the top card goes second instead of staying on top. Now no card is fixed, the deck behaves as though it had 53 positions, and the rule becomes 2i + 1 modulo 53.
53 is prime, and the smallest power of 2 that comes back to 1 modulo 53 is the fifty-second. One card of difference in how you interlace, and the answer goes from 8 to 52.
The count is the order of 2 modulo (size − 1), so it jumps about rather than growing. Fifty-two is a good number to have been handed.
| Deck size | Out-faros to return it |
|---|---|
| 8 cards | 3 |
| 12 cards | 10 |
| 24 cards | 11 |
| 32 cards | 5 |
| 40 cards | 12 |
| 50 cards | 21 |
| 52 cards | 8 |
| 54 cards, with two jokers | 52 |
| 64 cards | 6 |
Adding two jokers takes it from eight shuffles to fifty-two, because 53 is prime and 51 is 3 × 17.
An out-faro doubles a position. An in-faro doubles it and adds one. Those are the two operations you need to build any number in binary, one digit at a time, which means the two shuffles together can put the top card anywhere you name.
Write the position in binary and read the digits as shuffles: a 1 is an in-faro, a 0 is an out-faro, most significant digit first. Position 26 is 25 in binary, 11001, so that is in, in, out, out, in.
At most six shuffles reach any of the 52 positions, because 52 needs six binary digits. That is the practical reason magicians learn this: not to return a deck to its order, but to move one card a known distance without looking at it. The control for it is with the deck above, so you can watch it happen.
A riffle shuffle destroys order precisely because it is sloppy. Cards fall in clumps of unpredictable size, and after about seven of them the deck is close enough to random that no arrangement is more likely than another.
A faro removes the sloppiness, and with it the randomness. Nothing is mixed, because every card's new position is a function of its old one, and a function you can undo is a function that lost nothing. Eight repetitions and the deck has been nowhere.
The two numbers sit either side of the same fact: seven imperfect shuffles to destroy the order of a deck, eight perfect ones to leave it untouched. The size of what is being destroyed is on the 52 factorial page.
Eight, if every shuffle is an out-faro on a full 52-card deck.
An out-faro sends the card at position i to 2i modulo 51, so after k shuffles it is at 2ki, and the deck is back the moment 2k = 1 modulo 51. That happens at eight, because 256 is 5 × 51 + 1. In-faros take 52.
Cut the deck into two halves of exactly 26, hold one in each hand, and press the corners together so the cards interlace one at a time rather than in clumps.
It is a difficult sleight and it takes months. Nothing on this page needs you to be able to do it, because the arithmetic is the same whoever is holding the cards.
One card. In an out-faro the top card stays on top and the bottom card stays on the bottom. In an in-faro the top card goes second.
That single change moves the arithmetic from modulo 51 to modulo 53, and the number of shuffles needed to return the deck goes from 8 to 52.
No, and that is the point of it. A perfect faro moves every card to a position decided entirely by where it already was, so nothing is lost.
A riffle shuffle destroys order because it is imperfect, and it takes about seven of those to reach something close to random.
Yes, and the instruction is the position written in binary. To bring the top card to position n, write n − 1 in binary and read the digits as shuffles: a 1 is an in-faro and a 0 is an out-faro, most significant digit first.
At most six shuffles reach any of the 52 positions, and the control is with the deck above so the card can be watched arriving.
This sits under mathematical card tricks, which collects the tricks that run on arithmetic rather than on sleight of hand. The number a faro refuses to touch is on the 52 factorial page.