Poisson approximation
Named by 6 essays across 3 fields — each of them below, with the objects they name alongside it.
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.
A room where nobody is alone
Twenty-three people probably include two who share a birthday. How many are needed before every single person shares a birthday with somebody else in the room? The answer is 3,064 — more than it takes for every day of the year to be somebody's birthday — and the reason is a count of loners, which rises as the room fills, peaks at 134 when the room is the size of the year, and then falls so slowly that the last loner lingers for thousands of arrivals.
A random tree is one part in e leaves
Choose a labelled tree on n points uniformly at random. A point is a leaf exactly when its label never appears in the tree's Prüfer code, so the share of leaves is (1 − 1/n)^(n − 2) — half the points for a tree on four, 36.8% for a large one, the reciprocal of e. The whole degree distribution follows the same way: one plus a Poisson count with mean one.
Charged for the variance, not the range
Hoeffding's inequality knows one thing about each term of a sum: the interval it lies in. For ten thousand coins that each land heads once in a thousand, that makes it promise almost nothing — a 92% chance of thirty heads, when the truth is two in ten million. Tell the bound each term's variance as well and it changes character: Bernstein's and Bennett's inequalities decay like the normal curve while the deviation is small and like a Poisson tail beyond, and for rare events they are millions of times sharper.
A third of random games have no pure equilibrium
Fill a game's payoff table with random numbers and ask whether some cell is each player's best reply to the other. Exactly one cell is expected to be, at every size, and the chance that none is climbs from one in eight to 1/e — the same constant that counts the shuffles in which nobody gets their own hat back, and for the same reason.
How many new kinds the next sample will show
Count how many kinds were seen once, twice, three times. Good and Toulmin's alternating sum f₁t − f₂t² + f₃t³ − … then predicts, with no bias at all, how many new kinds a further sample t times as large will reveal. It works perfectly up to t = 1 and is worthless just beyond: past that point the kinds seen most often are multiplied by t to a high power, and the spread of the estimate passes 10⁴⁰ by t = 1.5. Truncating the sum at a random point rescues it, and each fourfold increase in the first sample buys about half a unit more of future.
Named alongside it
The objects these essays reach for when they reach for this one.
Birthday problemSimulationVarianceAlternating seriesConcentration inequalityCounting argumentCoupon collectorDerangemente, the numberEstimatorEstimator biasExhaustive search