How to use the Coupon Collector Expected Draws calculator
If a set has n types and duplicates are unavoidable, how many draws does it take on average to own every type? That is the coupon collector problem, and the answer is n times the harmonic number H, where H equals 1 + 1/2 + ... + 1/n. With 10 types the expected number of draws is about 29.29. The variance is n² times the sum of 1/i² minus n times H, and the standard deviation is its square root.
The widely quoted approximation n(ln n + 0.5772) is shown as well, but only as a reference; the exact formula is the headline result. Every figure assumes each type is equally likely on every draw and that draws are independent. Enter a cost per draw and the page multiplies it by the expected draws, which is again a long-run average rather than a promise about one run.
Weighted rarity, where some types are scarcer than others, is not covered here, and neither are trade-in or duplicate-reward systems. Treat the output as a reference. Current as of October 2026.
Frequently asked questions
The approximation n(ln n + 0.5772) only closes in on the exact value once the number of types grows. With few types the gap is visible, so this page reports the exact n times H as the main result and keeps the approximation as a reference.
No. The expected number of draws is a long-run average across very many repeats, and the standard deviation is wide, so a real run can finish sooner or take far longer.
No. The formula here assumes every type appears with the same probability on every draw. Expected draws under weighted rarity tiers are not covered on this page.