Two thirds of the work is the last tenth
An album has 640 spaces and the stickers come one at a time, each equally likely, with no swapping and no ordering the ones you are missing. Filling it takes — stickers on average. That is the famous answer and it is the least interesting thing about it: half the album costs — of those, and the last sixty-four spaces cost —. The work is nearly all at the end, and the reason is one line of arithmetic.
What it costs to fill
When you already have k of the n stickers, a new one is any of the n − k you are missing, so each sticker you buy has chance (n − k) ⁄ n of being new and you wait n ⁄ (n − k) of them, on average, for the next one. Add those waits up from an empty album to a full one and the whole thing costs
n ⁄ n + n ⁄ (n−1) + … + n ⁄ 2 + n ⁄ 1, which is n × Hn,
where Hn is the sum of the reciprocals up to n. The last term is the whole of the trouble. The first sticker is new for certain and costs one; the last one is a single space in n, and costs n stickers on its own —
The amber line is a single run of the thing being averaged: stickers drawn one at a time until the album is full. It is the first run from seed 1, not a picked one — the page opens on it so that every reader starts on the same collector — and the button draws the next seed along. One run is not evidence of anything; the checks at the foot of the page average ten thousand of them.
Each new sticker, one at a time
The same arithmetic drawn as it is paid. Each bar is what the next new sticker costs on average, from the first at the left to the last at the right, and the bars are a hyperbola: nothing at all for most of the album, then a wall.
How long it really takes
An average is a poor summary of this. The total is a sum of waits, most of them tiny and the last few enormous, so the distribution leans: there is a floor under it — you cannot fill n spaces in fewer than n stickers — and a tail that runs a long way to the right. This is the exact distribution, not a simulation: the chance of holding each number of stickers after each purchase, carried forward one purchase at a time.
The dashed line is the average and the claim it makes is that most people are done before it — A distribution with a tail puts its mean above its middle, and the tail here is the last sticker refusing to turn up.
A die, a deck, an album
The six-sided die is the whole thing in miniature and the one you can check by rolling: 14.7 rolls on average to see all six faces, and it is exactly 14.7 — six times (1 + 1⁄2 + 1⁄3 + 1⁄4 + 1⁄5 + 1⁄6) is 147/10, which the checks below confirm in whole numbers rather than in decimals. The commonest single answer is 13 rolls, and half of all attempts are done by then.
What the tail looks like from far away
For a large album the share still collecting after n × (log n + c) stickers settles down to a shape that does not depend on n at all: the chance of being finished by then goes to e−e−c. Put c = 0.5772, which is Euler's constant and the place the average sits, and it gives — — which is where the share finishing before the average comes from, and why it hardly moves as the album grows. The page computes the exact share for each size in the table above and compares.