Completing a set is cheap until it is nearly done. Then every draw is a coupon you already own.
01 N·ln N, not N
To collect sixty distinct coupons, drawing uniformly at random, takes on average N·Hₙ ≈ 281 draws — not sixty. The harmonic pile-up of repeats is the entire cost, and it grows like N times the natural log of N.
02 The coupon tax
When you already hold k of the set, a fresh draw is new with probability (N−k)/N, so the next new coupon costs N/(N−k) draws on average. The first is free; the fifty-ninth costs thirty; the last one costs sixty draws all by itself.
03 A skewed finish
Completion times don't cluster around the mean — they trail into a long right tail (a Gumbel law). Most runs finish near 281, but a stubborn few run far past it. The histogram remembers every run and keeps its lopsided shape.