Abstract
When $n$ kinds of prize come out one at a time with equal probability, how many draws are needed to collect them all (the coupon collector’s problem)? Where those draws are spent is counted. No new theorem or law is claimed.
First, collecting $50$ kinds takes $\mathbf{224.960267}$ draws on average——holding $j$ kinds, a new kind takes $50/(50-j)$ draws on average, and the sum is $50$ times $1+1/2+\cdots+1/50$. Collecting $20000$ times with random numbers gave a mean of $225.598950$, within $1.5$ standard errors of the formula (Section 2).
Second, and this is the core. More than a fifth of it is spent waiting for the last kind——once one kind remains, it takes $50$ draws on average, $\mathbf{0.222261}$ of the whole. With $10$ kinds the share is $0.341417$, with $100$ kinds $0.192776$; it falls only slowly as the kinds increase (Section 2).
Third, the whole and the last kind spread differently——the median of the whole is $215$ draws, $\mathbf{0.955724}$ times the mean, close to it, but the median wait for the last kind alone is $35$ draws, $0.70$ times its mean of $50$. For a single wait the median is about $\ln2$ times the mean; adding many waits brings the ratio near $1$ (Section 3).
Fourth, the separator is the least likely kind——with $49$ kinds of weight $1$ and one of weight $0.1$, that kind has probability $0.002037$, waiting for it alone takes $491$ draws on average, and collecting everything takes $536.096370$, $\mathbf{2.383071}$ times the equal case. The integral formula and random collection agreed for weights $1$, $0.5$, $0.2$ and $0.1$ (Section 4).
Keywords: coupon collector’s problem, harmonic numbers, expected value, last coupon, median.
Scope (what is not claimed)
No new theorem or law is claimed——that the mean number of draws is the number of kinds times the harmonic number, that its distribution approaches a fixed shape for many kinds (Erdős and Rényi, $1961$), and that the mean for unequal probabilities is given by an integral (Flajolet, Gardy and Thimonier, $1992$) are all known. The design or sale of lotteries is not discussed——the weight of the rare kind is an example for seeing the mechanism. Limits of the distribution are not treated——only the means for $10$, $50$ and $100$ kinds and the random distribution for $50$ kinds are looked at. Relation to earlier papers: Paper 554 showed that although a control chart runs $370$ points on average before a false alarm, half alarm within $256$ points (the median is close to $\ln2$ times the mean)——the last kind here is a wait of that type, and the whole is the sum of $50$ of them, so the ratio of median to mean approaches $1$. What is added is giving, for each number of kinds, the share of the draws taken by the last kind, setting the ratio of median to mean side by side for the whole and the last kind, and placing the separator on the least likely kind and checking it with the formula and random collection for several weights.
1. Notation and the Question
Setting: $n$ kinds of prize come out one per draw, each with probability $1/n$. Let $T$ be the number of draws until every kind has appeared at least once. Holding $j$ kinds, a new kind appears with probability $(n-j)/n$, so the mean number of draws to the next new kind is $n/(n-j)$. Adding these gives
$$E[T]=n\left(1+\frac12+\frac13+\cdots+\frac1n\right).$$The bracket is called the harmonic number $H_n$.
Question: how much of $E[T]$ is spent waiting for the last kind, what is the shape of the spread, and what happens if one kind is less likely?
2. The Mean and the Last Kind
Computation 2.1: from the harmonic numbers the mean and standard deviation were found, together with the share of the whole taken by the $n$ draws spent on the last kind.
| kinds $n$ | mean $E[T]$ | standard deviation | share of the last kind |
|---|---|---|---|
| $10$ | $29.289683$ | $11.211025$ | $0.341417$ |
| $50$ | $\mathbf{224.960267}$ | $61.950557$ | $\mathbf{0.222261}$ |
| $100$ | $518.737752$ | $125.821704$ | $0.192776$ |
Computation 2.2: independently of the formula, $50$ kinds were collected $20000$ times with random numbers. The mean was $225.598950$ draws; its difference $0.638683$ from the formula is within $1.5$ times the standard error $0.439149$.
Observation 2.3 this is the core: more than a fifth of the draws are spent waiting for the last kind. The first kind comes in $1$ draw, but with one kind left, only one draw in $50$ brings something new——those $50$ draws are $0.222261$ of the whole $224.960267$. The share is $1/H_n$, falling only from $0.341417$ to $0.192776$ when $n$ grows tenfold.
3. The Shape of the Spread
Computation 3.1: the distribution of the $20000$ random collections was examined. The median was $215$ draws; $0.568800$ finished at or below the mean and $0.005800$ took more than twice the mean. The ratio of median to mean is $\mathbf{0.955724}$.
Computation 3.2: the wait for the last kind alone is the number of draws until an event of probability $1/50$ occurs. Its median is $35$ draws, $0.70$ times the mean of $50$, close to $\ln2=0.693147$.
Observation 3.3 the whole and the last kind spread differently. A single wait stretches far to the right, pulling the median down to about $\ln2$ times the mean. The whole is the sum of $50$ waits, which evens out the stretch and brings the ratio of median to mean close to $1$——still, the median $215$ of the whole is below the mean $224.960267$: more than half of collectors finish sooner than the mean.
Observation 3.4 beside Paper 554. Paper 554 showed that although a control chart runs $370$ points on average before a false alarm, half alarm within $256$——the last kind here is a wait of that type, with a median $0.70$ times its mean. The whole number of draws is the sum of $50$ waits of that type.
4. The Separator
Computation 4.1: with $49$ kinds of weight $1$ and one kind of weight $w$, the mean number of draws was found numerically from the integral formula
$$E[T]=\int_0^\infty\left(1-\prod_{i}\left(1-e^{-p_i t}\right)\right)dt$$and compared with random collection ($4000$ times each). The integral reproduced $224.960267$ in the equal case.
| weight $w$ | probability of the rare kind | mean wait for it alone | mean to collect all | random |
|---|---|---|---|---|
| $1$ | $0.020000$ | $50$ | $224.960267$ | $223.740500$ |
| $0.5$ | $0.010101$ | $99$ | $234.159532$ | $232.313250$ |
| $0.2$ | $0.004065$ | $246$ | $323.833899$ | $320.603000$ |
| $0.1$ | $0.002037$ | $491$ | $\mathbf{536.096370}$ | $530.662250$ |
For every weight, the integral and random collection differed by less than $4$ standard errors of the random mean. The mean for $w=0.1$ is $\mathbf{2.383071}$ times the equal case.
Observation 4.2 the separator is the least likely kind. With weight $0.1$, waiting for that kind alone takes $491$ draws on average, most of the $536.096370$ needed for everything——the other $49$ kinds are usually complete while one waits for it. The number of draws comes to be decided less by the number of kinds than by the probability of the least likely one.
Observation 4.3 what is not said. The design or sale of lotteries is not discussed, and the limiting shape of the distribution is not entered——what is shown is the share of the mean taken by the last kind, the different spreads of the whole and the last kind, and what dominates when one kind is rarer.
5. Roll Call and Closing
- mean for $50$ kinds $\mathbf{224.960267}$ (Computation 2.1): $50H_{50}$.
- random mean $225.598950$ (Computation 2.2): within $1.5$ standard errors.
- the last kind takes $\mathbf{0.222261}$ (Computation 2.1): the core.
- $0.341417$ for $10$ kinds, $0.192776$ for $100$ (Computation 2.1): falling only slowly.
- median of the whole $215$, ratio $\mathbf{0.955724}$ (Computation 3.1): close to the mean.
- the last kind alone: median $35$, ratio $0.70$ (Computation 3.2): close to $\ln2$.
- a single wait stretches; a sum evens out (Observation 3.3): the type of Paper 554.
- one kind at weight $0.1$: mean $536.096370$ (Computation 4.1): $2.383071$ times.
- the $491$ draws for that kind are most of it (Observation 4.2): the separator.
- integral and random agree for four weights (Computation 4.1): two routes.
Closing: Collecting all of $50$ equally likely kinds takes $224.960267$ draws on average. More than a fifth of that is spent waiting for the last kind——with one kind left, only one draw in $50$ brings something new, and those $50$ draws are $0.222261$ of the whole. The share is the reciprocal of the harmonic number and is still $0.192776$ for $100$ kinds. The whole and the last kind spread differently——for the last kind alone the median is $0.70$ times the mean, while for the sum of $50$ waits it is $0.955724$ times. The separator is the least likely kind——making one kind ten times less likely raises the mean to $536.096370$, $2.383071$-fold, and the $491$ draws spent waiting for it take up most of it. Placed among the earlier papers——it looks at what is evened out, and what remains, when $50$ waits of the type seen in Paper 554 are added. To be honest——the limiting shape of the distribution is not entered, and the design of lotteries is not discussed.
References
- Erdős, P., & Rényi, A. (1961). On a classical problem of probability theory. Magyar Tudományos Akadémia Matematikai Kutató Intézetének Közleményei.
- Flajolet, P., Gardy, D., & Thimonier, L. (1992). Birthday paradox, coupon collectors, caching algorithms and self-organizing search. Discrete Applied Mathematics.
- Yamagishi, Y. (2026). Half the Charts That “Last 370 Points on Average” Alarm by 256 [Paper 554]. Zenodo. 10.5281/zenodo.23014204
Reference
Yamagishi, Y. (2026). More Than a Fifth of the Draws Needed to Collect Fifty Kinds Are Spent Waiting for the Last One [Paper 609]. YOI: https://www.yamagishiyuuki.com/yoi/609
DOI: 10.5281/zenodo.23014464
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 609. This paper is a preprint and has not been peer reviewed.