The Coupon Collector Problem: How Many Tries to Collect Every Item?
October 29, 2026
If there are n equally likely different items, such as the n different toys in cereal boxes or the six faces of a die, and each purchase or roll gives one at random, how many do you expect to need to collect all n? The answer is E = n × (1 + 1/2 + 1/3 + ... + 1/n) = n × Hₙ, where Hₙ is the nth harmonic number. For a six-sided die, E = 6 × (1 + 0.5 + 0.333 + 0.25 + 0.2 + 0.167) = 6 × 2.45 = 14.7 rolls.
The result grows a bit faster than n: for large n, E ≈ n(ln n + 0.5772). Collecting 50 different items takes on average about 225 purchases, or 4.5 per item; collecting 151 takes about 845, or 5.6 per item.
Skip the math and get your answer instantly:
Open the Permutation & Combination Calculator →Why the Last Items Take the Longest
The calculation adds the expected waiting time for each new item. When you have none, the first purchase is certainly new (1 try). With one item in hand, the chance that the next is new is (n−1)/n, so you expect n/(n−1) tries. With k items collected, you expect n/(n−k) tries to find the next new one. The final item takes n tries on average, since only one of n possible outcomes helps. For n = 50: the first 25 items take about 34 tries in total, while the last 5 items alone take about 50 × (1/5 + 1/4 + 1/3 + 1/2 + 1) = 114 tries, half of the total.
This explains the frustration of collecting: you finish most of the set quickly, and the last few cards take forever.
| Set size n | Expected purchases | Per item | Example |
|---|---|---|---|
| 6 | 14.7 | 2.45 | Faces of a die |
| 12 | 37.2 | 3.1 | Zodiac signs or months |
| 26 | 101.2 | 3.9 | Letters of the alphabet |
| 50 | 225.0 | 4.5 | Toy set |
| 151 | 845.3 | 5.6 | Large collectible set |
Variability and Confidence
The expected value is a mean, and the actual number needed varies widely. The standard deviation for n = 50 is about 62 purchases, with a long right tail. To be 95% sure of completing a set of 50, plan on roughly 350 purchases or more. Planning for a deadline requires adding a margin. The variance is roughly n² × π²/6 for large n, so the standard deviation is about 1.28n.
Trading and Unequal Probabilities
The formula assumes all items are equally likely. When some are rare, as in many card packs, the expected time is dominated by the rarest item: a card with a 1% chance takes an average of 100 purchases to appear. Trading with others reduces the effective number of purchases: if k collectors trade duplicates perfectly, the group needs far fewer purchases than k times the solo amount. Packs that contain several distinct items at once reduce the time, by roughly dividing the expected number of packs by the pack size when drawing without replacement within a pack.
Other Applications
The same math appears in computer science (hashing, load balancing), testing (how many random tests cover all code paths), biology (how many reads cover the whole genome), and sampling (how many people must be surveyed to see all categories). If you want to be sure that every one of 30 equal groups appears in a random sample, you need roughly 30 × H₃₀ = 120 observations on average.
Frequently Asked Questions
What is the coupon collector problem?
How many random draws are needed to collect every one of n different items.
What is the formula?
n × (1 + 1/2 + ... + 1/n) expected draws.
How many rolls to see every face of a die?
About 14.7 on average.
Why does the last item take so long?
Because only one outcome out of n helps, so it takes n tries on average.
Does trading help?
Yes, it dramatically reduces the purchases needed per collector.