Eight Is Not One More Than Seven

Two numbers about a deck of cards are famous enough to have escaped mathematics. Seven riffle shuffles randomise a deck. Eight perfect shuffles put it back exactly as it was. They sit one apart, they describe the same physical gesture, and they are not the same kind of quantity: the eight counts a set of eight arrangements, the seven counts steps across all 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 of them. Underneath four famous scrambling numbers there is a single curve, and this page computes it.

Ask how scrambled something is and you will be handed a number. A deck needs seven shuffles. A pocket cube can always be solved in eleven moves. A die rolled around a grid comes home in only twelve of its twenty-four orientations. A deck comes back after eight perfect shuffles. Each of those is true, each is on this site with its own working, and read together they invite an arithmetic that does not exist. Seven and eight are not a near miss. They are answers to different questions about different sets.

There is one object underneath all four. Take any finite set of states and any set of moves, and write B(k) for the number of states you can reach in at most k moves. That curve climbs at the branching rate of the moves and then flattens against a ceiling. Every number above is a reading of it: where the ceiling is, what fraction of the nominal space the ceiling covers, which step first touches the ceiling, and which step the spread over the ceiling goes flat. The first three are properties of a set. The fourth is a property of a distribution on that set, which is why it is never the smaller number.

The curve, for six systems

Pick a system. The vertical axis counts states, by powers of ten, from one up to the whole nominal space. The dotted line is everything the object could in principle be arranged into; the solid line is everything these moves can actually reach. Where the climbing curve meets the solid line is the diameter.

The two numbers everyone puts side by side

On the deck, the whole confusion is visible at once. A riffle is a random move: each shuffle picks one of an enormous number of interleavings, so the reachable set explodes, and after six riffles every arrangement of the deck is reachable. The deck is still not random. It takes a seventh shuffle for the spread over those arrangements to flatten. Covering the space and covering it evenly are different achievements, and on this object they are exactly one shuffle apart.

A perfect shuffle is not a random move. It is one fixed rearrangement, so its reachable set does not explode at all: repeating the out-shuffle reaches eight arrangements, and the eighth is the one you started with. That is the famous eight. It is the size of a reachable set of eight, out of a space of about 8.07 × 10⁶⁷. Comparing it to the seven is comparing the height of a doorway with the age of the door.

What the first four riffles cannot do, for reasons that are not probability

The famous shape of the seven-shuffles result is a cliff: the distance from random sits pinned at one for four shuffles, then falls away. The flat part needs no probability theory at all. Bayer and Diaconis proved that k riffles are exactly one 2^k-shuffle, and that such a shuffle can only produce arrangements with at most 2^k rising sequences. So counting alone fixes how much of the deck is even in play, and a distribution living on an S-sized fragment of an N-sized space cannot be closer to uniform than 1 − S/N.

Through the fourth riffle only 0.000000466568 of the deck's arrangements are reachable at all, so the distance from random cannot be below 0.999999533, and no probability was needed to say so. At the fifth, 0.998119142762 of them are reachable, the same argument yields only 0.001880857, and it never speaks again. The exact answer at that fifth shuffle is 0.923732, which counting cannot see at all. The cliff begins precisely where the counting argument expires. Below, the blue bar is what counting forces and the orange bar is the exact answer.

The ladder

Every system, every reading, in one place. Read across a row and the four columns are the four famous kinds of number. Read down a column and you can see that no two systems are even being asked the same thing.

What this page computes that its members do not

Four layers already on this ground hold the four numbers, one each. It would be too much to say none of them holds the curve, and two of them plainly do. Eleven Moves From Anywhere ships the pocket cube's complete depth histogram, whose running totals are B(k) exactly. Half the Ways Home ships its own reachable-in-at-most-n row, 1, 5, 17, 53, 153, 369, 684, 1060, and then says the thing this whole page is about: that the interesting invariant is not the count but the ceiling. What none of the four does is name the curve as the object the four numbers share, or notice that another layer's famous number is a different reading of the same shape. That is the portal's claim, and it is narrower than "nobody drew this curve" because two of them did. Four quantities here are new, and one of them is the reason the portal exists.

What perfect shuffles can reach at all, and why so little. The eight-perfect-shuffles layer establishes the order of a single shuffle, and it does ask the group question, so this portal must be exact about what is left. It asks it of a 24-card packet, it names Diaconis, Graham and Kantor for the structure, and it is scrupulous about the seam: in its own words it checks the resulting order, not the theorem that names the group. What it does not do is put the question to a full deck at all. Running a stabiliser chain over the group the two shuffles generate on 52 cards gives exactly 2²⁶ × 26!, which is 27,064,431,817,106,664,380,040,216,576,000,000 arrangements: one ordering in 2,980,227,913,743,310,874,726,229,193,921,875. A lifetime of flawless faro shuffles cannot bring a deck to all but a vanishing sliver of its own arrangements.

The reason is a single invariant, and it is visible once you go looking for it. A perfect shuffle never separates a card from its mirror partner. The top card and the bottom card stay a pair; the second from the top and the second from the bottom stay a pair; all twenty-six such pairs survive every shuffle, in or out, forever. That partition is computed here rather than assumed, by searching for the smallest block system the two shuffles permute: it comes back as the twenty-six pairs {i, 51 − i} and nothing finer. The largest group that can permute 52 objects while keeping such a pairing intact has order exactly 2²⁶ × 26!, the same number, so the shuffles generate the whole of that group and not one element more. The vanishing sliver has a name: it is everything a deck can do without ever breaking a mirror pair.

Checked three further ways: against brute-force enumeration of the whole group on decks of 8, 10 and 12 cards; against the order 194,641,920 that our own 24-card layer quotes from the literature, which this chain now derives rather than takes; and by testing the stabiliser-chain routine on nine groups whose orders are not in doubt before it was trusted on this one. The general classification of shuffle groups is due to Diaconis, Graham and Kantor (1983), and nothing above is taken from it: every route to that paper returned 403 tonight, so it is cited as the prior work it is and not as a source read.

The riffle's diameter, next to its mixing time. Six and seven, on the same object with the same moves, so covering a space and covering it evenly differ here by a single shuffle. The sixth riffle is not a formality: 151,706,512,551,444,814,120,700,127,652,998,499,269,204,008,764,705,163,930,790,422,064 arrangements cannot be produced by five, however the cards fall. That is 0.001880857 of the deck, and it is the same number as the counting floor in the table above, necessarily: the share counting cannot rule out is exactly the share still out of reach. The extreme case is the one you would guess, and it is unique. The deck in exact reverse order is the only arrangement of the 52! with the maximum fifty-two rising sequences.

The counting floor, everywhere. The gap between what counting forces and what is true is itself a measurement, and it differs by system: two shuffles on the deck, two moves on the cube in half turns, four in quarter turns.

The four layers this walks

• How Many Shuffles Until It's Random? The mixing time. Exact total-variation distance under the Gilbert, Shannon and Reeds model, and the cutoff cliff. It shows the flat part; the explanation of the flat part is here.
• Eight Perfect Shuffles The order. Its own card states plainly that the two shuffles are chance against control, which is the nearest anything here comes to this portal's claim. What it does not say is that the eight is a count of a set, and how small that set is.
• Eleven Moves From Anywhere The diameter, with the full depth histogram of the pocket cube's 3,674,160 states. That histogram is the curve on this page, spelled out layer by layer; the portal's contribution is naming it as the same object the other three read differently, and putting the counting floor underneath it.
• Half the Ways Home Reachability, as a fraction. Twelve orientations of twenty-four, welded to the colour of the square by a parity law.

Two layers originally proposed for this spine, on juggling notation and on leapers on a torus, are not members. Both count arrangements rather than walk a state graph, so they have no reachable-set curve to read, and putting them here would have been a resemblance rather than a join.