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THE COUPON COLLECTOR

how many draws to collect the whole set
1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION
The coupon collector’s problem asks: if a cereal box holds one of n equally-likely coupons, how many boxes must you buy to collect them all? The exact expected number is n·Hn, where Hn = 1 + 1/2 + … + 1/n is the harmonic number. The reason is a beautiful use of linearity: once you hold i distinct coupons, each new box is new with probability (n-i)/n, so it takes n/(n-i) boxes on average to advance — and summing those independent waits gives n(1 + 1/2 + … + 1/n). Since Hn ≈ ln n + γ, collecting all n takes about n ln n boxes: the last few coupons dominate the wait.

LIT verified live: for n = 5, 10, 20 the exact formula n·Hn matches a Monte-Carlo simulation of tens of thousands of runs to within a fraction of a percent, and Hn tracks ln n + γ (window.__coupon). FIG no framing; the exact harmonic formula and the random simulation both run in-browser; the simulation is statistical, so its match is approximate by design.
2 HOW IT WAS WEAVED · AI + HUMAN
David (human) seated this at the-drop — the loot table where you keep rolling for drops until the whole set is complete, and the maths says exactly how many rolls that takes: n·Hn. AVAN (AI) built the instrument: the harmonic-number formula, the linearity-of-expectation derivation, and the Monte-Carlo simulation that confirms it.

Credit as content: the classical coupon collector problem (de Moivre, Laplace; Erdős–Rényi for the distribution). The weave: David names the drop; I confirm the expected number of rolls is n·Hn.
3 ONE DIMENSION
A single collection run: coupons light up as they first appear; the last few take by far the most draws.
4 TWO DIMENSIONS · INTERACTIVE
Run many collection trials; the empirical average draws converge onto the exact formula n·H_n.
5 THREE DIMENSIONS + AVAN’S INVERSE
The green forward object: the growing set of collected coupons.
AVAN’s addition (the inverse-companion): don’t count draws — sum the waits. The inverse of ‘how many draws to finish’ is ‘add the expected wait n/(n-i) at each stage’, and those independent waits sum to n·Hn. Magenta are the lengthening waits for each new coupon; green is the completing set. The tail dominates the hunt.
LIT Genuine coupon collector's problem (de Moivre, Laplace). Verified live: for n=5,10,20 the exact expectation n·Hₙ matches a Monte-Carlo simulation (15000 runs each) to within a fraction of a percent, and Hₙ tracks ln n+γ (window.__coupon.matches, .gammaErr).

FIG No framing; the exact harmonic formula and the random simulation both run in-browser. Honest scope: the simulation is statistical, so its match to n·Hₙ is approximate by design (within ~1%). The AVAN inverse is honest — instead of counting draws, sum the waits: the expected wait n/(n-i) at each stage, summing to n·Hₙ. Magenta are the lengthening waits for each new coupon; green is the completing set. The tail dominates the hunt.
◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE DROP · David Lee Wise (ROOT0), with AVAN