Concept

Probability

A number between nought and one measuring how much of a space of outcomes an event takes up, adding over disjoint events. It is a measure normalised to total one, and everything about it follows from that.

Named by 15 essays across 5 fields — each of them below, with the objects they name alongside it.

The share of arrangements that fix nothing, up to 8 objects. A bar per number of objects, giving the proportion of its arrangements that leave nothing in place, against the horizontal line at 1/e.

The constant that counts what does not happen

Nothing grows in a shuffled pack of cards, and nothing grows in a factorial. Yet e sits in the middle of both — as the chance that a shuffle leaves nothing in place, and as the base that makes n! nearly a power.

analysis · The exponential
A game that stops, over totals 0 to 5. States in a row with arrows up and down between them and the two ends absorbing, above a table of the expected number of steps and the chance of ending at the top from each start.

The chain that stops

Give a chain a state it cannot leave and there is no long run to find — every walk ends. What is worth computing instead is how long it lasts and where it finishes, and both are exact answers to a linear system rather than limits of anything.

probability · Markov chains
A walk on a weighted graph, and a cycle whose traffic goes one way. A weighted graph with the long-run share of each state read off its total weight, beside a three-state cycle whose shares are equal and whose traffic circulates.

The chain that runs the same backwards

Put weights on the edges of a graph, step to a neighbour in proportion to them, and the long-run share of a state is its own weight over the total — read straight off the picture, with nothing to solve. The condition that makes that work is strictly stronger than being stationary.

probability · Markov chains
The time a single walk spends in each state, against the share it should hold. Paired bars for each state, one the fraction of a long run's time spent there and one the computed stationary share, above a table of expected return times.

The time spent and the share held

Stationary shares are a limit of distributions — where the walk probably is after many steps. Here the question is about a single walk: the fraction of its time spent in each state is that state's share, and the expected wait between visits is exactly the reciprocal.

probability · Markov chains
One walk on the whole numbers, three chances, three different fates. The relative weight of each state for three step-up chances, drawn as bars, with the running total of those weights and what each case means beneath.

Where the shares have nowhere to go

On finitely many states, a chain that can reach everywhere and is not forced into a rhythm settles down. Give it infinitely many and both conditions can hold while the walk leaves and never returns — or returns with certainty and takes an unbounded average time about it.

probability · Markov chains
One table of shares, two different lotteries. A doubly stochastic table decomposed into whole assignments twice, by two different orders, giving two mixtures that reconstruct the same shares.

One table, two lotteries

A table of shares says what fraction of each task each person does. It does not say how — the same table is a mixture of whole assignments in many different ways, and the differences are exactly what the people being assigned would care about.

applied · Assignment
Three properties that leave the middle, and one that cannot. Four measured curves of the share of random graphs having a property, plotted against the number of points: three first-order properties running to zero or one, and the parity of the edge count sitting on a half throughout.

Nearly always, or nearly never

Toss a coin for every pair of points and ask whether the graph that results has some property. For a property a first-order sentence can state, the answer in the limit is never a genuine probability — it is zero or it is one, and the game is what proves it.

logic · Ehrenfeucht–Fraïssé games
Infinitely many guessers, finitely many wrong. Three rows over the first 40 places: the hats worn, the chosen representative of their class, and a row of marks showing each guess right or wrong. The 5 wrong guesses all fall within the first 14 places, up to a marked place; every later guess is right.

Infinitely many guessers, finitely many wrong

An infinite line of people each wears a black or white hat, sees every hat in front and none of their own, and must guess their own colour. With a finite line, each guesser is right half the time whatever they agree in advance. With an infinite line and the axiom of choice, they can agree a strategy under which all but finitely many are right — and nobody can carry it out.

logic · Axiom of choice
Birthdays within k days: a half at 23, 14, 9, 7 people. Curves of the probability that two of m people have birthdays within k days, for k = 0, 1, 3, 6, with the group size at which each passes one half marked.

Fourteen people within a day

Twenty-three people probably include two with the same birthday. Fourteen probably include two whose birthdays are at most a day apart, and seven, two within a week. The near miss has an exact formula, found by a trick that takes k days away after every birthday and turns the question back into the plain one on a shorter year, and the pattern behind every threshold is a single square root: a window of k days either side makes each pair 2k + 1 times as likely to collide.

probability · Birthday problem
The ground a 1,500-step walk covers. A random walk on a square grid drawn as a path, with every grid square it visited shaded and its start and end marked.

The ground a walk covers

A random walk of a thousand steps visits far fewer than a thousand places: in one dimension about fifty, in the plane about four hundred, in space about six hundred and sixty. The share of steps that land on new ground is exactly the chance of never coming home — so the number that decides whether a walker returns also decides how much of the world it sees.

probability · Random walk
Four stages of the four-corner Cantor dust. Four panels showing stages 1 to 4 of the four-corner Cantor set: 4, 16, 64 and 256 squares kept at the corners.

A dust that almost every line misses

Keep the four corner squares of a square, then the four corners of each of those, and so on. What is left has a length, in the sense that measures length, and yet almost every straight line misses it entirely. Stage by stage the chance that a random line hits the dust falls — to 48% by the eighth stage — and how fast it falls to nothing is one of the few questions about a simple picture that is still open.

analysis · Arc length
Which pairs of the square's symmetries commute: 40 of 64. A 8-by-8 grid over the elements of the 8 symmetries of a 4-gon, rows and columns grouped into 5 conjugacy classes, with a filled square wherever the two elements commute: 40 filled squares, and each row's count of filled squares written at its end.

Five-eighths of the pairs, and no more

Pick two symmetries of a square at random and do them in both orders: forty times in sixty-four the result is the same. No group that fails to commute does better. The reason is a count of pairs that turns into a count of conjugacy classes, and a two-line argument about the centre that caps the answer at five-eighths — reached by the square and the quaternions, approached from above by nothing, and approached from below by groups that commute a little more than half the time.

algebra · Symmetry groups
Every pair of rearrangements of four letters, and what the two generate. A 24 by 24 grid of ordered pairs of permutations of four letters: 216 generate S4, 96 generate A4, 264 generate a smaller subgroup.

Two random shuffles reach every shuffle

Pick two ways of rearranging n letters at random and allow them to be repeated in any order. Almost always the two together reach every rearrangement there is, or every even one — and when they fail, the usual reason is a coincidence about a single letter that both happen to leave alone. Eugen Netto guessed it in 1882 and John Dixon proved it in 1969.

algebra · Symmetry groups
The 3,600 pairs of icosahedral rotations, and what each pair generates. A 60 by 60 grid of ordered pairs of elements of A5 shaded by the order of the subgroup they generate; 2280 generate A5.

Nineteen copies and no more

Two of the icosahedron's rotations, chosen at random, generate all sixty of them nineteen times in thirty. Philip Hall found the number in 1936 by inclusion and exclusion over every subgroup at once, weighted by the Möbius function of the subgroups' order — and the same count says that nineteen copies of the group, side by side, can still be generated by two elements, while twenty cannot.

algebra · Symmetry groups
How good each judge must be, as the premises multiply. k = 1: 0.5000; k = 2: 0.7071; k = 3: 0.7937; k = 4: 0.8409; k = 5: 0.8706; k = 6: 0.8909; k = 7: 0.9057; k = 8: 0.9170; k = 9: 0.9259; k = 10: 0.9330.

Every premise makes the verdict vote stricter

A court that must find several things true before it finds liability can vote on each of them or on the verdict. With two elements, judges right seven times in ten are enough for either. With five, the verdict vote needs judges right eighty-seven times in a hundred, or a larger court makes it worse — while the premise vote still needs only better than even. And when any one ground is enough instead of all, the two procedures swap their failures.

applied · Judgement aggregation

Named alongside it

The objects these essays reach for when they reach for this one.

Random walkExpectationMarkov chainInvariantLimitStationary distributionAlternating groupCounting argumentExhaustive searchGenerating setInclusion exclusionParity

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