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2026-05-02
The men's World Cup is coming up, so it's time to work out the cost of filling the official Panini sticker book again.
Panini are boasting that this year's sticker book contains more stickers than ever, which can only mean one thing: you're going to have to buy more stickers than ever.
But how much more than last time is completing this sticker book going to cost you?
Buying stickers
This year, you can buy:
- A starter pack that includes a sticker book with 6 stickers (that are always the same and hence guaranteed to contain no duplicates) and 4 packs of 7 random stickers. This costs £4.99.
- Packs of 7 random stickers. Various sources claim these are £1.25 each, although our local Morrison's is selling them for £2.
- Multipacks of 6 or 8 packs of 7 random stickers. Both these multipacks work out at £1.25 per pack.
- Multipacks of stickers that come with a nice metal box to keep your swaps in (you'll need a few of these if you're planning to complete your collection). These are releasing later than the rest of the collection, but look like they'll work out at £1.25 per pack plus £1 for the metal box.
For the rest of this blog post, we'll assume that all
stickers are bought in multipacks.
![](javascript:showlimage("panini2026-1.jpg"))
![](javascript:showlimage("panini2026-2.jpg"))
Every album includes Declan Rice (England), Heechan Hwang (South Korea), Alexis Vega (Mexico), Mikel Merino (Spain), Danley Jean Jacques (Haiti) and Nicolas Jackon (Sengal)
How many stickers will I need?
After you've stuck your first 6 stickers in, you need to collect 974 more stickers.
Following the same method as I did in 2018, we can work out that if you want to get all of these from packs of random stickers, then
you should expect to buy 7265 stickers (including the 4 packs in your starter pack). Including the cost of the starter pack, buying all these stickers will cost
£1297.35.
But as always, you don't have to spend this much, as Panini let you order the last \(n\) stickers (for any value of \(n\) that you choose, although I have strong memories
of \(n\) being 50 when we had to send off paper sticker ordering forms when I was at school). On the Panini website, they're selling missing stickers
for other recent collections for 36p per sticker. The World Cup missing stickers aren't available yet: I'm guessing they'll be prices at 36p each, and will update this post once we know for certain.
If you decide to order the last 50 stickers, then you should expect 2883 stickers to get the rest of the stickers before ordering the last 50. In total this will
cost you £532.81. That's a big saving.
What is the best number of stickers to order?
If you want to minimise the amount that you have to spend to complete this sticker book, you should choose your value of \(n\) wisely. This plot shows the expected total cost
of completing the sticker book if you decide on different numbers of stickers to order:
![](javascript:showlimage("wc2026plot.png"))
The lowest expected cost is achieve when ordering 483 stickers: £295.78. That's an even bigger saving, although 483 does feel to me like an unreasonably large value for \(n\).
Swapping
As always, the cheapest way to complete the sticker book is to convince more of your friends to collect too so that you can swap your duplicate stickers with them. You could get
the price down as low as £174.99 if you manage to swap every single duplicate that you get.
If you decide to set \(n\) to 0, there's a good chance that going out now and buying 200 starter packs for £998 to give to your friends make them start collecting will mean spending less money in total
than trying to complete your sticker book on your own...
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2019-07-03
This year's Big Internet Math-Off
is now underway with 15 completely new contestants (plus one returning contender). As I'm not the returning contestant, I haven't been spending
my time preparing my pitches. Instead, I've spent my time making an
unofficial Big Internet Math-Off sticker book.
To complete the sticker book, you will need to collect 162 different stickers. Every day, you will be given a pack of 5 stickers; there are
also some bonus packs available if you can find them (Hint: keep reading).
How many stickers will I need?
Using the same method as I did for last year's World Cup sticker book,
you can work out that the expected number of stickers needed to finish the sticker book:
If you have already stuck \(n\) stickers into your album, then the probability that the next sticker you get is new is
$$\frac{162-n}{162}.$$
The probability that the second sticker you get is the next new sticker is
$$\mathbb{P}(\text{next sticker is not new})\times\mathbb{P}(\text{sticker after next is new})$$
$$=\frac{n}{162}\times\frac{162-n}{162}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next new sticker is
$$\left(\frac{n}{162}\right)^{i-1}\times\frac{162-n}{162}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find a new one:
$$\sum_{i=1}^{\infty}i \left(\frac{162-n}{162}\right) \left(\frac{n}{162}\right)^{i-1} = \frac{162}{162-n}$$
Therefore, to get all 162 stickers, you should expect to buy
$$\sum_{n=0}^{161}\frac{162}{162-n} = 918 \text{ stickers}.$$
Using just your daily packs, it will take you until the end of the year to collect this many stickers.
Of course, you'll only need to collect this many if you don't swap your duplicate stickers.
How many stickers will I need if I swap?
To work out the expected number of stickers stickers you'd need if you swap, let's first think about two people who want to complete
their stickerbooks together. If there are \(a\) stickers that both collectors need and \(b\) stickers that one collector has and the other one
needs, then let \(E_{a,b}\) be the expected number of stickers they need to finish their sticker books.
The next sticker they get could be one of three things:
- A sticker they both need (with probability \(\frac{a}{162}\));
- A sticker one of them needs (with probability \(\frac{b}{162}\));
- A sticker they both have (with probability \(\frac{162-a-b}{162}\)).
Therefore, the expected number of stickers they need to complete their sticker books is
$$E_{a,b}=1+\frac{a}{162}E_{a-1,b+1}+\frac{b}{162}E_{a,b-1}+\frac{162-a-b}{162}E_{a,b}.$$
This can be rearranged to give
$$E_{a,b}=
\frac{162}{a+b}+
\frac{a}{a+b}E_{a-1,b+1}
+\frac{b}{a+b}E_{a,b-1}
$$
We know that $E_{0,0}=0$ (as if \(a=0\) and \(b=0\), both collectors have already finished their sticker books). Using this and the
formula above, we can work out that
$$E_{0,1}=162+E_{0,0}=162$$
$$E_{1,0}=162+E_{0,1}=324$$
$$E_{0,2}=\frac{162}2+E_{0,1}=243$$
$$E_{1,1}=\frac{162}2+\frac12E_{0,2}+\frac12E_{1,0}=364.5$$
... and so on until we find that \(E_{162,0}=1269\), and so our collectors should expect to collect 634 stickers each to complete their
sticker books.
For three people, we can work out that if there are \(a\) stickers that all three need, \(b\) stickers that two need, and \(c\) stickers
that one needs, then
$$
E_{a,b,c}
= \frac{162}{a+b+c}+
\frac{a}{a+b+c}E_{a-1,b+1,c}
+\frac{b}{a+b+c}E_{a,b-1,c+1}
+\frac{c}{a+b+c}E_{a,b,c-1}.
$$
In the same way as for two people, we find that \(E_{162,0,0}=1572\), and so our collectors should expect to collect 524 stickers each
to complete their sticker books.
Doing the same thing for four people gives an expected 463 stickers required each.
After four people, however, the Python code I wrote to do these calculations takes too long to run, so instead I approximated the numbers
by simulating 500 groups of \(n\) people collecting stickers, and taking the average number of stickers they needed. The results are shown in
the graph below.
![](javascript:showlimage("mathoffstickerplot.png"))
The red dots are the expected values we calculated exactly, and the blue crosses are the simulated values.
It looks like you'll need to collect at least 250 stickers to finish the album: in order to get this many before the end of the Math-Off,
you'll need to find 20 bonus packs...
Of course, these are just the mean values and you could get lucky and need fewer stickers. The next graph shows box plots with the
quartiles of the data from the simulations.
![](javascript:showlimage("mathoffstickerboxplot.png"))
So if you're lucky, you could complete the album with fewer stickers or fewer friends.
As a thank you for reading to the end of this blog post, here's a link that
will give you two bonus packs and help you on your way to the 250 expected stickers...
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Link to sticker book, in the first paragraph, does not work. It points to mathoffstickbook.com
Pat Ashforth
minor typo for the 2 collector case
> and so our collectors should expect to collect 364 stickers
should be 634.
> and so our collectors should expect to collect 364 stickers
should be 634.
Road
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2018-07-07
So you've calculated how much you should expect the World Cup sticker book to cost
and recorded how much it actually cost. You might be wondering what else you can do with your sticker book.
If so, look no further: this post contains 5 mathematical things involvolving your sticker book and stickers.
Test the birthday paradox
![](javascript:showlimage("birthdaysbrazil.jpg"))
Stickers 354 and 369: Alisson and Roberto Firmino
In a group of 23 people, there is a more than 50% chance that two of them will share a birthday. This is often called the birthday paradox, as the number 23 is surprisingly small.
Back in 2014 when Alex Bellos suggested testing the birthday paradox on World Cup squads, as there are 23 players in a World Cup squad. I recently discovered that even further back in 2012, James Grime made a video about the birthday paradox in football games, using the players on both teams plus the referee to make 23 people.
In this year's sticker book, each player's date of birth is given above their name, so you can use your sticker book to test it out yourself.
Kaliningrad
![](javascript:showlimage("kaliningrad.jpg"))
Sticker 022: Kaliningrad
One of the cities in which games are taking place in this World Cup is Kaliningrad. Before 1945, Kaliningrad was called Königsberg. In Königsburg, there were seven bridges connecting four islands. The arrangement of these bridges is shown below.
![](javascript:showlimage("7bridges.png"))
The people of Königsburg would try to walk around the city in a route that crossed each bridge exactly one. If you've not tried this puzzle before, try to find such a route now before reading on...
In 1736, mathematician Leonhard Euler proved that it is in fact impossible to find such a route. He realised that for such a route to exist, you need to be able to pair up the bridges on each island so that you can enter the island on one of each pair and leave on the other. The islands in Königsburg all have an odd number of bridges, so there cannot be a route crossing each bridge only once.
In Kaliningrad, however, there are eight bridges: two of the original bridges were destroyed during World War II, and three more have been built. Because of this, it's now possible to walk around the city crossing each bridge exactly once.
![](javascript:showlimage("kaliningrad-path.png"))
A route around Kaliningrad crossing each bridge exactly once.
I wrote more about this puzzle, and using similar ideas to find the shortest possible route to complete a level of Pac-Man, in this blog post.
Sorting algorithms
If you didn't convince many of your friends to join you in collecting stickers, you'll have lots of swaps. You can use these to practice performing your favourite sorting algorithms.
Bubble sort
In the bubble sort, you work from left to right comparing pairs of stickers. If the stickers are in the wrong order, you swap them. After a few passes along the line of stickers, they will be in order.
![](javascript:showlimage("bubble-sort-stickers.gif"))
Bubble sort
Insertion sort
In the insertion sort, you take the next sticker in the line and insert it into its correct position in the list.
![](javascript:showlimage("insertion-sort-stickers.gif"))
Insertion sort
Quick sort
In the quick sort, you pick the middle sticker of the group and put the other stickers on the correct side of it. You then repeat the process with the smaller groups of stickers you have just formed.
![](javascript:showlimage("quick-sort-stickers.gif"))
Quick sort
The football
![](javascript:showlimage("correct-football.jpg"))
Sticker 007: The official ball
Sticker 007 shows the official tournament ball. If you look closely (click to enlarge), you can see that the ball is made of a mixture of pentagons and hexagons. The ball is not made of only hexagons, as road signs in the UK show.
Stand up mathematician Matt Parker started a petition to get the symbol on the signs changed, but the idea was rejected.
If you have a swap of sticker 007, why not stick it to a letter to your MP about the incorrect signs as an example of what an actual football looks like.
Psychic pets
Speaking of Matt Parker, during this World Cup, he's looking for psychic pets that are able to predict World Cup results. Why not use your swaps to label two pieces of food that your pet can choose between to predict the results of the remaining matches?
![](javascript:showlimage("timber-football1.jpg"))
![](javascript:showlimage("timber-football2.jpg"))
Timber using my swaps to wrongly predict the first match
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@Matthew: Thank you for the calculations. Good job I ordered the stickers I wanted #IRN. 2453 stickers - that's more than the number you bought (1781) to collect all stickers!
Milad
@Matthew: Here is how I calculated it:
You want a specific set of 20 stickers. Imagine you have already \(n\) of these. The probability that the next sticker you buy is one that you want is
$$\frac{20-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not wanted})\times\mathbb{P}(\text{sticker after next is wanted})$$
$$=\frac{662+n}{682}\times\frac{20-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next wanted sticker is
$$\left(\frac{662+n}{682}\right)^{i-1}\times\frac{20-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find the next wanted one:
$$\sum_{i=1}^{\infty}i \left(\frac{20-n}{682}\right) \left(\frac{662+n}{682}\right)^{i-1} = \frac{682}{20-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{19}\frac{682}{20-n} = 2453 \text{ stickers}.$$
You want a specific set of 20 stickers. Imagine you have already \(n\) of these. The probability that the next sticker you buy is one that you want is
$$\frac{20-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not wanted})\times\mathbb{P}(\text{sticker after next is wanted})$$
$$=\frac{662+n}{682}\times\frac{20-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next wanted sticker is
$$\left(\frac{662+n}{682}\right)^{i-1}\times\frac{20-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find the next wanted one:
$$\sum_{i=1}^{\infty}i \left(\frac{20-n}{682}\right) \left(\frac{662+n}{682}\right)^{i-1} = \frac{682}{20-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{19}\frac{682}{20-n} = 2453 \text{ stickers}.$$
Matthew
@Matthew: Yes, I would like to know how you work it out please. I believe I have left my email address in my comment. It seems like a lot of stickers if you are just interested in one team.
Milad
@Milad: Following a similiar method to this blog post, I reckon you'd expect to buy 2453 stickers (491 packs) to get a fixed set of 20 stickers. Drop me an email if you want me to explain how I worked this out.
Matthew
Thanks for the maths, but I have one probability question. How many packs would I have to buy, on average, to obtain a fixed set of 20 stickers?
Milad
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2018-06-16
This year, like every World Cup year, I've been collecting stickers to fill the official Panini World Cup sticker album.
Back in March, I calculated that I should expect it to cost £268.99 to fill this year's album (if I order the last 50 stickers).
As of 6pm yesterday, I need 47 stickers to complete the album (and have placed an order on the Panini website for these).
So... How much did it cost?
In total, I have bought 1781 stickers (including the 47 I ordered) at a cost of £275.93. The plot below shows
the money spent against the number of stickers stuck in, compared with the what I predicted in March.
![](javascript:showlimage("panini2018plot1.png"))
To create this plot, I've been keeping track of exactly which stickers were in each pack I bought. Using this data, we can
look for a few more things. If you want to play with the data yourself, there's a link at the bottom to download it.
Swaps
The bar chart below shows the number of copies of each sticker I got (excluding the 47 that I ordered). Unsurprisingly, it looks a lot like
random noise.
![](javascript:showlimage("panini2018plot3.png"))
The sticker I got most copies of was sticker 545, showing Panana player Armando Cooper.
![](javascript:showlimage("sticker545.jpg"))
Armando Cooper on sticker 545
I got swaps of 513 different stickers, meaning I'm only 169 stickers short of filling a second album.
First pack of all swaps
Everyone who has every done a sticker book will remember the awful feeling you get when you first get a pack of all swaps.
For me, the first time this happened was the 50th pack. The plot below shows when the first pack of all swaps occurred in 500,000 simulations.
![](javascript:showlimage("panini2018plot2.png"))
Looks like I was really quite unlucky to get a pack of all swaps so soon.
Duplicates in a pack
In all the 345 packs that I bought, there wasn't a single pack that contained two copies of the same sticker.
In fact, I don't remember ever getting two of the same sticker in a pack. For a while I've been wondering if this is because Panini
ensure that packs don't contain duplicates, or if it's simply very unlikely that they do.
If it was down to unlikeliness, the probability of having no duplicates in one pack would be:
\begin{align}
\mathbb{P}(\text{no duplicates in a pack}) &= 1 \times\frac{681}{682}\times\frac{680}{682}\times\frac{679}{682}\times\frac{678}{682}\\
&= 0.985
\end{align}
and the probability of none of my 345 containing a duplicate would be:
\begin{align}
\mathbb{P}(\text{no duplicates in 345 packs})
&= 0.985^{345}\\
&= 0.00628
\end{align}
This is very very small, so it's safe to conclude that Panini do indeed ensure that packs do not contain duplicates.
The data
If you'd like to have a play with the data yourself, you can download it here. Let me know if
you do anything with it...
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2018-03-22
Back in 2014, I worked out the cost of filling an official Panini World Cup 2014 sticker book. Today, the 2018 sticker book was relased: compared to four years ago, there are more stickers to collect and the prices have all changed. So how much should we expect it to cost us to fill the album this time round?
How many stickers will I need?
There are 682 stickers to collect. Imagine you have already stuck \(n\) stickers into your album. The probability that the next sticker you buy is new is
$$\frac{682-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not new})\times\mathbb{P}(\text{sticker after next is new})$$
$$=\frac{n}{682}\times\frac{682-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next new sticker is
$$\left(\frac{n}{682}\right)^{i-1}\times\frac{682-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find a new one:
$$\sum_{i=1}^{\infty}i \left(\frac{682-n}{682}\right) \left(\frac{n}{682}\right)^{i-1} = \frac{682}{682-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{681}\frac{682}{682-n} = 4844 \text{ stickers}.$$
How much will this cost?
You can buy the following [source]:
- Starter pack (an album and 31 stickers) for £3.99
- Sticker packs (5 stickers) for 80p
- Sticker multipacks (30 stickers) for £4.50
First of all you'll need to buy the starter pack, as you need an album to stick everything in. This comes with 31 stickers; we should expect to buy 4813 more stickers after this.
The cheapest way to buy these stickers is to buy them in multipacks for 15p per sticker. This gives a total expected cost of filling the sticker album of £725.94. (Although if your local newsagent doesn't stock the multipacks, buying 80p packs to get these stickers will cost you £774.07.)
What if I order the last 50 stickers?
If you'd like to spend a bit less on the sticker book, Panini lets you order the last 50 stickers to complete your album. This is very helpful as these last 50 stickers are the most expensive.
You can order missing stickers from the Panini website for 22p per sticker's sticker ordering service for the 2018 World Cup doesn't appear to be online yet; but based on other recent collections, it looks like ordered stickers will cost 16p each, with £1 postage per order.
Ordering the last 50 stickers reduces the expected number of other stickers you need to buy to
$$\sum_{n=0}^{631}\frac{682}{682-n} = 1775 \text{ stickers}.$$
This reduces the expected overall cost to £291.03. So I've just saved you £434.91.
What if I order more stickers?
Of course, if you're willing to completely give up on your morals, you could order more than one batch of 50 stickers from Panini. This raises the question: how many should you order to minimise the expected cost.
If you order the last \(a\) stickers, then you should expect to pay:
- £3.99 for the album and first 31 stickers
- £\(\displaystyle0.15\left(\sum_{n=1}^{681-a}\frac{682}{682-n}-31\right)\) for other stickers you need
- £\(\displaystyle \left(0.16a+\left\lceil \frac{a}{50}\right\rceil\right)\) to order the last \(a\) stickers
The total expected cost of filling your album for different values of \(a\) is shown in the graph below.
![](javascript:showlimage("worldcup2018graph.png"))
The red cross shows the point at which the album is cheapest: this is when the last 550 stickers are ordered, giving a total expected cost of £120.18. That's another £170.85 I've saved you. You're welcome.
Still, ordering nearly all stickers to minimise the cost doesn't sound like the most fun way to complete the sticker book, so you probably need to order a few less than this to maximise your fun.
What about swaps?
Of course, you can also get the cost of filling the book down by swapping your spare stickers with friends. In 2016, I had a go at simulating filling a sticker book with swapping and came to the possibly obvious conclusion that the more friends you have to swap with, the cheaper filling the book will become.
My best advice for you, therefore, is to get out there right now and start convinding your friends to join you in collecting stickers.
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@RM: It would slightly reduce the expected cost I think, but I need to do some calculations to make sure.
I'm not certain whether they guarantee that there are no duplicates in a pack or just it's very unlikely... Follow up post looking into this coming soon
I'm not certain whether they guarantee that there are no duplicates in a pack or just it's very unlikely... Follow up post looking into this coming soon
Matthew
Helpful article!! My head always go into a spine with probability, but does the fact that you would not get a double within a pack affect the overall strategy or cost? My head says this is an another advantage of buying 30 packs over the 5.
RM
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