The narrowing funnel

The coupon collector problem asks for the expected number of draws required to collect all n distinct items when draws are independent and equally likely. By linearity of expectation, total collection time decomposes into n separate geometric waiting stages. If you currently hold k distinct items, each subsequent draw finds a new item with probability p_k = (n - k) / n, requiring an average of n / (n - k) draws. Summing across stages gives n * H_n, where H_n is the n-th harmonic number. For a standard 6-sided die, the expected number of rolls to see all 6 faces is exactly 14.7.

Words you need

Expected value (EV): The average result if you could repeat the bet many times: each outcome times its probability, added up.e.g. A die pays its face in dollars: EV = 3.5.

Why it works

Each stage is an independent geometric trial whose mean is the reciprocal of its success probability.

E[T] = sum_{k=0}^{n-1} n / (n - k) = n * sum_{i=1}^n (1 / i) = n * H_n

Spot it: Calculating expected trials to see all outcomes of a random process.

Technique reference and sources

Collecting coupons is easy at first and increasingly difficult at the end. The first coupon arrives on draw 1. Once half the coupons are collected, the next new coupon takes 2 draws on average (for six coupons, reaching three distinct from two takes 6/4 = 1.5 draws; the draw after reaching halfway takes 6/3 = 2). The final missing coupon has only a 1/n chance on each draw, taking n full draws on its own.

When this method fits

Calculating expected trials to see all outcomes of a random process.
Evaluating card set completion, coverage testing, or collection games.
Pricing games of chance requiring full set completion.

Identify state k. Track how many distinct coupons out of n are already collected.
Calculate success probability p_k. Probability next draw is new: p_k = (n - k) / n.
Invert for stage wait time. Expected draws to find coupon k+1 is 1 / p_k = n / (n - k).
Sum across all stages. Add stage times from k = 0 to n - 1: E[T] = n * (1/1 + 1/2 + ... + 1/n).

Fair six-sided die: Expected rolls to see all 6 faces: 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 1 + 1.2 + 1.5 + 2 + 3 + 6 = 14.7 rolls.

Stage waits: 1, 1.2, 1.5, 2, 3, 6
Total rolls = 14.7

Four distinct items: Expected draws to collect 4 items: 4/4 + 4/3 + 4/2 + 4/1 = 1 + 4/3 + 2 + 4 = 25/3 ≈ 8.33 to 2 decimal places.

Total: 7 + 4/3 = 25/3 ≈ 8.33
Final item alone takes 4 draws

Three distinct items: Expected draws to collect 3 items: 3/3 + 3/2 + 3/1 = 1 + 1.5 + 3 = 5.5 draws.

Stage waits: 1, 1.5, 3
Total draws = 5.5

Remaining-coupon harmonic sum: With r unseen faces left on a d6, expected remaining time is the tail of the stage waits: 6/1 for one unseen, 6/2 + 6/1 for two unseen. Answer remaining-time questions directly without recomputing the full 14.7.

Two unseen faces: 6/2 + 6/1 = 3 + 6 = 9 remaining draws
One unseen face: 6/1 = 6 remaining draws

Common traps

Multiplying n by n (e.g. 6 * 6 = 36) to find total collection time. Total expected time is n * H_n, which is roughly n * ln(n), much smaller than n^2. For n = 6, it is 14.7, not 36.

Assuming every stage takes an equal number of draws. The first coupon takes 1 draw, while the final coupon takes n draws. Each stage wait is geometric with mean n / (n - k), and these means increase as the number of unseen coupons decreases. They form a reciprocal sequence (for d6: 1, 1.2, 1.5, 2, 3, 6), not a geometric sequence; only the waiting distributions are geometric.

Multiplying probabilities instead of inverting them for waiting time. For a geometric trial with success probability p, the expected number of trials until success is 1/p.

Another example: Fair coin: seeing both heads and tails

Flip 1: guaranteed to see a new face, wait = 2/2 = 1 flip
Flip 2: probability of the other face is 1/2, expected wait = 2/1 = 2 flips
Total expected flips: 1 + 2 = 3 flips

Source notes

These explanations and questions are written for QuantArc. Links show the underlying mathematics and learning design.

Harvard Stat 110: Probability (Joe Blitzstein)
Free course site with videos and the open companion book Introduction to Probability; the topic spine for counting, Bayes, and linearity. Lesson examples and questions are original.

Brainstellar: Probability interview puzzles
Free step-by-step walkthroughs of classic probability interview puzzles by difficulty; companion practice. Lesson examples and questions are original.

IES: Organizing Instruction and Study to Improve Student Learning
Supports worked examples, retrieval, explanations and spaced study. Does not establish QuantArc's exact lesson length, mastery threshold or review intervals.