The Sticker Collector's Problem

An open box of Panini 2026 World Cup stickers with rows of packets inside, and a spread of opened stickers in front showing many duplicates of Mbappé, Messi, Haaland, Yamal and Ronaldo.

I’ve collected Panini World Cup stickers for as long as I’ve watched the World Cup. If you like football, the album is part of the run-up to the tournament. You learn the squads and the badges, you meet names you’ve never heard of, and you slowly fill the pages before the first game kicks off.

This year I’m doing it again, but the album is bigger than it has ever been. The 2026 World Cup now has 48 teams, and the sticker set grew to match. There are 980 stickers in the full album, and that changes the economics quite a bit. A bigger album is more difficult to finish, and the last stretch drags on longer and costs more.

It gets most difficult near the end. Say you’re down to your last 12. You buy a packet, open it, and all seven stickers are ones you already have. So you buy another one and get roughly the same thing. At that point it’s easy to start thinking some stickers must be rationed. But this happens even when Panini prints every sticker in the same numbers.

If you fill a 980-sticker album by buying random packets, you’ll open about 7,300 stickers on average. That works out to a bit more than 1,000 packets of seven. At $1 a packet you can expect to spend about $1,000 to fill 980 slots. You end up buying over seven albums’ worth of stickers, and most of what you open is a duplicate.

Mathematicians have studied this for a long time. They call it the coupon collector’s problem. The question is how long it takes to get a full set when items show up at random, one at a time. Swap the coupons for World Cup stickers and nothing about the logic changes.

The math also tells you how to buy more sensibly. Random packets are good value at the start and poor value at the end. What you really need to know is the point where you should switch.

Open Table of contents

Every new sticker is harder than the last
Your rare sticker isn’t actually rare
When the rarity is real: McDonald’s Monopoly
A collector is really a buyer
Trading is the cheapest strategy
The strategy: buy in bulk early, trade towards the end
Under the hood
What to do with all this

Every new sticker is harder than the last

When you tear open a packet, the only thing you care about is whether any of the stickers are ones you’re missing. So let’s start there.

It’s easiest to think about the stickers in a packet one by one. Say the album has $N$ spaces and you’ve filled $k$ of them already. Any new sticker is equally likely to be any of the $N$ in the set, so the chance it matches one of your $k$ filled spaces is $k/N$ . Put simply, your duplicate rate is the same as how full your album is.

So if the album is half full, about half your stickers will be doubles. If it’s 90 percent full, about 90 percent of them will be, and by then most packets do very little for you.

You can also put it like this. If a fraction $f$ of the album is done, a random sticker is new with probability $1-f$ . That means the average number of stickers you’ll open before you hit the next new one is:

$\frac{1}{1-f}.$

Show the math: why this formula works

Let $p = 1-f$ be the chance that any one sticker is new, and let $m$ be the average number of stickers you open until you get the first new one.

Think about the very first sticker you open. With probability $p$ it’s new, and you stop after one sticker. With probability $1-p$ it’s a duplicate. You’ve used up one sticker and you’re back where you began, still looking at an average of $m$ more. Combining the two cases:

$m = 1 + (1-p)\,m.$

Solve for $m$ and you get $p\,m = 1$ , so

$m = \frac{1}{p} = \frac{1}{1-f}.$

When the album is 50 percent complete, you open about two stickers for every new one you get. At 90 percent it’s about 10 stickers. At 99 percent it’s about 100.

Now take a 980-sticker album with one space left. Each sticker you open has a $1/980$ chance of being the one. On average you’ll open 980 stickers before you find it, which is around 140 packets to fill a single space.

Add up all of those waits, from the first sticker through to the last, and you get:

$\frac{N}{N} + \frac{N}{N-1} + \frac{N}{N-2} + \cdots + \frac{N}{1} = N\left(\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{N}\right).$

The part in brackets is the harmonic series, $1 + \tfrac12 + \tfrac13 + \cdots$ , cut off after the $N$ -th term. It grows slowly, roughly like $\ln N + 0.58$ . That 0.58 is the Euler-Mascheroni constant. Plug in 980 stickers and you get about 7,300, the same number from the start of the post.

Each of these waits is independent of the others. Once you own a certain number of stickers, the wait for the next new one only depends on how full the album is. It doesn’t matter which stickers you got or how long the earlier waits were. That’s why the expected total is simply the expected waits added together.

Show the math: why those waits add up

Say you already have $k$ of the $N$ stickers. The next one you open is new with probability $\frac{N-k}{N}$ , so using the rule from the box above, the average wait for it is one over that, $\frac{N}{N-k}$ .

To get the total wait to finish, add these up from your first new sticker ( $k=0$ ) to your last ( $k=N-1$ ):

$\frac{N}{N} + \frac{N}{N-1} + \cdots + \frac{N}{1} = N\left(1 + \frac{1}{2} + \cdots + \frac{1}{N}\right).$

The bracket is the harmonic sum, written $H_N$ , and it’s close to $\ln N + \gamma$ where $\gamma \approx 0.58$ . So the expected total is

$E[T] = N\,H_N \approx N(\ln N + \gamma).$

With $N = 980$ that’s about $7{,}316$ stickers, or roughly $1{,}045$ packets of seven.

The average doesn’t tell you everything, though. For one thing, the variance is big. Two people can buy the same number of packets and be in very different spots, and the typical spread around the average is well over 1,000 stickers (I work this out at the bottom of the post). For another, most of the effort is at the end. The last one percent of the album, around 10 stickers here, eats up a huge share of all the stickers you open.

That’s why people remember one sticker that seemed impossible to find, even when the print run treated it like every other sticker.

You’ll find the coupon collector’s problem in lots of other places too, like toys in cereal boxes, random loot rewards in games, and algorithms that need to visit every state. A sticker album is just the version most of us have actually lived through.

Your rare sticker isn’t actually rare

A lot of collectors are convinced that some stickers get printed in smaller numbers. After hundreds of stickers without seeing one particular badge or star player, “bad luck” starts to sound like a weak excuse.

In a normal sticker album, though, bad luck and a nearly full album usually explain it fine.

Researchers have gathered big samples of real stickers and checked whether certain numbers turn up less often than others. Mostly they haven’t found that. The stickers look close to evenly spread, which is what we’ve been assuming all along. The feeling that a sticker is rare is real enough. It just doesn’t have to come from an uneven print run. Late in the album, every sticker you’re missing is hard to find, and that’s where the feeling comes from.

You see something similar in the birthday problem. Put just 23 people in a room and the odds are already better than even that two of them share a birthday. It surprises people because they compare 23 people to 365 days. What you should really compare is the number of possible pairs of people, and there are a lot of pairs. Collisions happen sooner than our gut expects.

A sticker album is a slow-motion version of this. Equal odds don’t stop duplicates from happening. If anything, they guarantee them. Doubles start showing up well before the album looks anywhere near full, and by the end they’re most of what you get.

Show the math: the birthday problem in disguise

In the birthday problem, with $d$ equally likely birthdays, the chance that $n$ people all have different birthdays falls off roughly like $e^{-n^2/2d}$ . It drops under one half once $n \approx 1.2\sqrt{d}$ . Sticker collisions follow the same counting. With $N$ different stickers, the expected number of stickers you open before your first duplicate is on the order of $\sqrt{N}$ , and for $N = 980$ that’s only about 30.

Even odds alone are enough to bury you in duplicates. You don’t need any secret scarcity in the print run.

So Panini probably isn’t sitting on the sticker you need. Once you’re down to a handful of specific stickers, buying random packets is just a pricey way of hunting for them.

When the rarity is real: McDonald’s Monopoly

McDonald’s Monopoly makes a good comparison. It looks like a collecting game, but it works in nearly the reverse way, and it shows you what real scarcity looks like.

With a sticker album, every sticker is roughly as likely as any other. McDonald’s Monopoly makes some pieces scarce on purpose. Each color group has a few common properties and a single scarce piece. You can pick up the common ones over and over and get no closer to a prize if the rare one never turns up.

McDonald's Monopoly piece: New York Avenue, $5,000, needs B09, B10 and B11. — Two real game pieces. Each lists the properties you still need for the set, and in practice there is usually one piece that matters much more than the others.

McDonald's Monopoly piece: States Avenue, $1,000, needs B06, B07 and B08. — Two real game pieces. Each lists the properties you still need for the set, and in practice there is usually one piece that matters much more than the others.

It’s still a coupon collector’s problem, only now the probabilities aren’t equal. When they’re lopsided enough, the rarest item is what counts, and the common pieces barely matter next to that one bottleneck.

The bottleneck was worth enough that someone stole it. For over ten years Jerome Jacobson, who handled security for the promotion, took rare winning pieces and handed them to a network of associates. He rigged more than $20 million in prizes before the FBI worked out what was going on. The story was later turned into the documentary McMillions.

The scam worked for the same reason the game did. The rare pieces really were rare, and the common ones were close to worthless by themselves.

So the two games behave differently. In a sticker album, a sticker usually only seems rare because of even odds and a long tail at the end. In McDonald’s Monopoly, the rarity is built in on purpose. With stickers, the print run is probably not your problem. The bigger question is how long you keep buying random packets.

A collector is really a buyer

You could treat all this as a question of luck. I find it more useful to treat it as a question of cost.

There are two ways to get stickers. One is buying packets at the shop. The other is buying singles, which means the exact stickers you want, either on eBay or, for a limited time, from Panini’s official website. Packets are cheap per sticker but you don’t pick what’s inside. Singles cost more each, but you get exactly what you ordered. At the start, the randomness doesn’t hurt because nearly everything is new. Near the end, that same randomness is the reason packets get so expensive.

Let $c$ be the cost of one sticker from a packet. If a packet of seven costs $1, $c$ is about 14 cents. Let $s$ be what it costs to buy one specific missing sticker as a single.

When you’re at completion level $f$ , only a fraction $1-f$ of what you open is new to you. So one useful new sticker from packets really costs:

$\frac{c}{1-f}.$

With an empty album that’s just the price of a packet sticker. At half full it’s double. At 90 percent full, your 14-cent sticker costs about $1.40 for each new one you actually need. At 99 percent full it’s about $14.

The marginal cost, meaning the cost of your next new sticker, keeps going up for packets as the album fills. For singles it doesn’t move. You should switch at the point where the two costs are equal:

$f^{*} = 1 - \frac{c}{s}.$

If a single costs twice what a packet sticker does, you should switch around 50 percent complete. If it costs four times as much, switch around 75 percent. The pricier singles are compared to packets, the longer packets stay worth it.

Show the math: where the switch point comes from

Let $c$ be the price of a packet sticker and $s$ the price of a single. At completion level $f$ , a random sticker is new with probability $1-f$ , so one new sticker from packets costs $\frac{c}{1-f}$ on average. A single costs $s$ no matter what.

Packets are the cheaper option while

$\frac{c}{1-f} < s \quad\Longleftrightarrow\quad f < 1 - \frac{c}{s}.$

Set the two equal and you get the switch level $f^{*} = 1 - \frac{c}{s}$ . Under that level, packets are cheaper on the margin. Over it, singles win. If singles don’t cost any more than packet stickers, so $s \le c$ , then $f^{*} \le 0$ and you should just buy singles from day one.

You can use the calculator below to compare the two costs and find the switch point for your own album. Put in the album size, how many stickers you have, how many come in a packet, and the two prices. It shows you where packets stop being the cheaper choice, and it compares what it would cost to finish from where you are with each strategy.

In practice there’s a catch. The official singles service often doesn’t open until late in the campaign, and it may limit how many stickers you can order. So you can’t always follow the formula to the letter. Think of it as a marker for when packets have stopped paying off, even if the market won’t fully let you act on it yet.

Trading is the cheapest strategy

If you collect on your own, the hobby costs more, since every duplicate you get just sits in your pile.

To you, a duplicate is worth almost nothing. To someone else it might be the exact sticker they’re after. Trade it for one of their doubles that fills a gap in your album and you both get closer to finishing without anyone buying another packet. It’s the standard argument for why trade helps both sides, just with stickers.

Paper collage of two collectors exchanging duplicate stickers above albums with complementary gaps.

The effect gets stronger with more people. When a few collectors pool their duplicates, the sticker you never seem to pull is more likely to be in somebody’s spare pile. A bigger group makes that more likely still. Trading won’t get rid of the long tail, but it makes it cheaper to get through.

It only goes so far, though. A group that has opened thousands of stickers can still be stuck on one last sticker. That final gap stays hard because the probability problem underneath is still there. It’s just spread across more people now. So trading helps most in the middle stretch, and the last few stickers are usually easier to buy directly.

Your spares might also be worth some money, especially when a campaign is popular. Selling them won’t change the overall economics of the album, but you can get some of your money back. What the album really costs you is whatever you spend on packets and singles, minus what you get back from swaps and sales.

The strategy: buy in bulk early, trade towards the end

The sensible plan is to buy random packets for as long as they make sense and then switch to buying exactly what you need.

I’d break it into three stages. Start by buying in bulk, since almost every sticker you open early on is useful. A box of packets, or a box that promises no repeats, gets you through the easy part of the album cheaply.

Next, trade to shrink the list of gaps. You might open a few more packets to give yourself more to swap, but you’re not really buying for volume anymore. What you want now is to turn your duplicates into stickers you’re missing.

When you’re down to the last few, order the exact stickers you need. Packets are at their most expensive here, in real terms, and this is where singles give you the most value.

It can make a big difference. In the 980-sticker example, buying packets right to the end costs a bit over $1,000 on average. If you bulk buy, trade, and then put in one singles order at the end, you can finish the same album for much less. All you’ve changed is the order you buy things in.

Once you think of the album as a spending problem, you stop paying random prices for specific stickers.

Under the hood

Here are a few extra details if you want to see how the numbers come out.

How spread out is the total, really?

The expected total $N\,H_N$ tells you where the middle of the distribution is, but it doesn’t say how far from it a typical collector ends up. For that you need the variance.

The total wait is a sum of independent pieces, one for each new sticker, and when things are independent their variances add. So the variance for the whole album is the sum of the variances of each wait. A geometric wait with success probability $p$ has variance $\frac{1-p}{p^2}$ , and when $p$ is small that’s close to $\frac{1}{p^2}$ . Put another way, the rarer a success is, the bigger the variance, and it grows like the square of the average wait.

That’s why the very end of the album drives most of the spread. When the album is nearly full, every new sticker has a small $p$ , a long average wait of $\frac{1}{p}$ , and an even bigger variance of $\frac{1}{p^2}$ . Nearly all the uncertainty comes from the last handful of stickers. If you add those variances up and keep only the part that matters for a big album, you get

$\operatorname{Var}(T) \approx \frac{\pi^2}{6}\,N^2.$

The $\frac{\pi^2}{6}$ comes from the sum $1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots$ , which is what the squared waits at the end of the album turn into. Take the square root and you get the standard deviation:

$\operatorname{sd}(T) \approx \frac{\pi}{\sqrt{6}}\,N \approx 1.28\,N.$

For a 980-sticker album that comes to about 1,260 stickers, or roughly 180 packets. The average, $N\,H_N \approx 7{,}300$ , grows like $N\ln N$ , but the standard deviation only grows like $N$ . So relative to their size, bigger albums are actually a little more predictable. The absolute spread is still big, though. With a standard deviation of 180 packets, two collectors using the same strategy can easily finish hundreds of dollars apart purely through luck.

Packs of seven, not one at a time

So far I’ve treated stickers as if they come one at a time. Real packets hold several at once, usually seven, and newer packets often promise no repeats inside a single packet. In principle, either of these could change the count.

In practice they hardly do. A packet is only worse than seven separate draws if two stickers in it would have been the same number, so one gets wasted. But you’re pulling seven stickers out of a set of almost a thousand, and the chance any two of them match is tiny (it’s the birthday problem again, with seven “people” and 980 “birthdays”). Nearly every packet acts like seven independent draws, so bundling them in sevens barely changes anything.

The no-repeats promise pushes the other way. It removes the few collisions inside a packet that would have happened anyway, which makes a packet very slightly better than seven independent draws. The effect is small again, since there were hardly any collisions to begin with.

Both corrections are tiny and they pull in opposite directions, so the packet count stays near the simple estimate you get by dividing the sticker total by seven: about $7{,}316 / 7$ , or 1,045 packets.

When stickers aren't equally likely (the Monopoly case)

Up to now every sticker has had the same chance, $1/N$ . McDonald’s Monopoly doesn’t work like that. Sticker $i$ shows up with its own probability $p_i$ , and some of those are kept tiny on purpose. When the probabilities aren’t equal, the expected time to finish the set comes from an integral worked out by Flajolet and colleagues:

$E[T] = \int_{0}^{\infty}\!\left(1 - \prod_{i=1}^{N}\bigl(1 - e^{-p_i t}\bigr)\right)dt.$

It looks scary, but each part means something simple. Treat $t$ as the number of stickers you’ve opened. For one sticker $i$ , the chance you still haven’t seen it after $t$ stickers shrinks like $e^{-p_i t}$ . A common sticker ( $p_i$ large) drops off your “still missing” list fast, and a rare one ( $p_i$ small) hangs around. So $1 - e^{-p_i t}$ is the chance you have it by time $t$ , the product over all $i$ is the chance you have the full set, and $1$ minus that product is the chance you’re still missing something. Add up that “still missing” chance over all $t$ and you get the expected time to finish.

The reason a single item can take over is in how fast it decays. Say one sticker is ten times less likely than the others. Its $e^{-p_i t}$ term takes roughly ten times longer to shrink, so well after you’ve almost certainly got every common sticker, that one rare sticker is still dragging the product down. Whichever sticker is slowest to show up sets your finishing time, and when the odds are lopsided that’s nearly always the rarest one. This is the formal version of the McDonald’s Monopoly point. Once one piece is far rarer than all the rest, the whole game comes down to waiting for that piece.

As $N$ gets large, the finishing time also settles into a known shape. Rescale it the right way and it approaches a Gumbel distribution, the same extreme-value distribution you see in problems driven by the maximum of lots of random quantities. That suits the sticker problem, since finishing an album has more to do with the last sticker than with the average one.

What to do with all this

The album probably isn’t rigged, and it wouldn’t need to be. Fair odds are enough on their own to make the end slow and expensive. The good news is you can plan for it. Buy packets while most stickers are still new, trade while your duplicates are still useful to other people, and order directly once randomness has stopped doing much for you.

So next time you open a packet and all seven are doubles, it’s probably not Panini working against you. More likely you’ve just kept buying random packets for longer than you should have.