Panini football stickers and the Coupon Collector Problem

Yesterday I was over at my sister’s, and her lad was excited because he had the new South Africa World Cup 2010 Panini sticker book. He had also bought four packets, each with five stickers in them, and I had the important job of unpeeling them so that he could put them in the album.

There are 638 stickers to collect in total and it made me wonder how many stickers you would expect to have to buy so that you had a complete set. I was interested in how many you would need to buy without doing swapsies with anybody and on the premise that there were an equal number of each sticker and that they were randomly distributed.


Be warned, there’s some maths up ahead.

This sort of problem has been modelled as the Coupon Collector’s problem by mathematicians and I’ll walk gently through the explanation to get to the answer to the problem applied to the World Cup football stickers album.

Step 1 – Probability of getting a new sticker

Now, the first sticker you take is absolutely guaranteed to not be one you have already got. The second sticker has a 637 in 638 chance of being a new sticker. Once you have this second sticker then the probability of the next sticker being different to the first two is 636 in 638 and so on . .. . . . . .until you have all but one of the stickers in your album to fill. Then, you have a 1 in 638 chance of any sticker being that last one you want.

Step 2 – Number of stickers we expect to buy to get a new one

So, we know we can work out the probability of getting a new sticker at any point. But we don’t want to know the probability of each sticker we open being a new one. What we actually want to know is how many stickers we should expect to open each time to get a new one.

This is actually quite easy. If you know the probability, call it p, of an event happening, then the expected number of times you should have to do something to get the outcome you want is 1/p.

To make sense of this then think of throwing a die. If you want to throw a 6 then you know that you have a 1 in 6 chance of throwing one. It makes sense that you should expect, on average, to throw a die 6 times until you get a 6.

So, to get the total number of stickers you should expect to buy altogether then you just add up the number of stickers you expect to buy to get the first, the second, the third . . . . . .right through to the 638th sticker.

Still with it?

Step 3 – Adding it all up

Turning this into numbers, this means that we want to work this sum out

1 + 638/637 + 638/636 + . . . . . . 638/2 + 638/1

And that, my friends, is an harmonic series, so it is. And, to estimate the sum of an harmonic series, and skipping some of the maths, we can use the equation

S = n ln(n)

In the equation above, applied to our example, S is the total number of stickers we should expect to have to buy, n is 638, the number of unique stickers we want to get, and ln(n) is the natural log of n. So

S = 638 x ln(638)

S = 638 x 6.46

S = 4120

And so, unless you swap, then you should expect to have to buy 4120 stickers, at a total cost of £412, to fill your Panini sticker album.

Which all goes to show, that it is far better to be sociable than rich.

  • http://www.demsoc.org Anthony Zacharzewski

    Good post. I think – at least it was the case in previous years – that if you needed 50 stickers or fewer, you could send off for them. So perhaps the target is not to get 638, but to get any 588 out of 638.

  • http://@ABlackCountryJo Jo Green

    Thanks for doing the maths on this…. I’ve posted it to my facebook to try and explain to our 7 yr old why we don’t want to buy into the sticker book collects. Even though they kindly gave a free book out to people…. err free.. I think NOT!

  • http://www.theaardvark.co.uk theaardvark

    Of course, this is skewed quite considerably because Panini release different amounts of each sticker. I believe.

  • http://siwhitehouse.co.uk Simon

    Anthony

    That would dramatically reduce the number of sticker packs you’d have to buy. If you think that you can expect to buy 638 stickers to get the last sticker, 319 to get the last but one, 159.5 to get the last but two . . . . . . 12.76 to get the last but 50, then that will take a hefty chunk out of the 4120.

    I think what you can do is take the equation

    S = n ln(n)

    and rewrite it as

    S = n.[ln(n) - ln(n-x)]

    with n = 638 and x = 588

    which gives you

    S = 638.[ln(638) - ln(50)]

    S = 638.[6.458 - 3.912]

    S = 638 x 2.546

    S = 1625

  • http://siwhitehouse.co.uk Simon

    Paul

    Yes, you’re right. If the stickers aren’t printed off in equal numbers, and if they aren’t randomly distributed then that would skew the figures.

    I’ve amended my post slightly to take account of that, so thank you.

  • http://jerometurner.posterous.com Jerome Turner

    You could also work out what was a fair price for buying new stickers, from a collector / comic shop for example. I remember when Pogs were very big and I completed my star wars pog set by buying the last few from a comic shop in Exeter.

  • http://pigsonthewing.org.uk Andy Mabbett

    When I was a child, and such stickers were given away with petrol, my uncle was a trade rep for a bakery. He used to do “swapsies” of his trade samples, with a rep for a petrol company – a carrier-bag full at a time

    Clearly, your nephew has a deficiency in the uncle department. Just sayin’

  • Paul Vincent

    An additional complication is that you don’t buy the stickers individually, you buy them in packets of five. Presumably each packet of five contains five DIFFERENT stickers. So the first 5 terms of the series are all 1.

  • http://siwhitehouse.co.uk Simon

    Paul

    I’ve not presumed that all the stickers in a pack are different. But you’re right, if they are then the first five terms are all 1. It’s also not a harmonic series any more and I ended up with a sum looking like this

    S = 5n[1/n + 1/(n-5) + 1/(n-10) + . . . . . . 1/n-635)] + 2n/(n-635)

    Which left me a bit stumped as it’s been a while since I studied series, and my books are all in the loft.

    But I’m pretty sure there’s something obvious to be done with that series. Hmmmph.

  • http://www.nickparkhouse.com Nick Parkhouse

    I know it’s not the sexiest way of collecting the stickers (and I have started the 2010 World Cup album myself) but eBay is very much your friend when looking to collect the stickers you’re missing.

  • http://www.twitter.com/liencam Liencam

    Sorry to add another adjustment, but there are actually 640 stickers to collect. Numbers “000″ & “00″ are on the inside cover.

    So I’ve been told….

  • siwhitehouse

    Well if it is 640 then the equation changes slightly to

    S = 640 ln(640)

    S = 4135

    So, an increase of 15 for the extra 2 stickers. Could you should say a thank you on my behalf to whoever told you about this, please?

  • Laurent et Julie

    Simon. Excel or computer can be used to calculate exact number with no need of approx function. (By the way, in wikipedia article the approx function is n.ln(n)+Euler times n + 1/2 and I believe this is correct). I calculated in Excel a staggering 4505 stickers to buy, meaning 4505-640=3865 spare stickers. To get to 540 stickers (and buy the remaining 100 from Panini, 100 being max number to make an order), 1185 must be purchased. At 12 stickers to a sterling pound, it’s £99 +£8=£107. Cost of 640 stickers is £53.
    Swapping is key, in’t ?
    The above assumes duplicates can be found in a pack of 5. I don’t understand the maths in otherwise situation, so could not calculate. Maybe it’s good news for the overall price to pay ?

  • siwhitehouse

    I made a mistake in my working I think. I took the total to be

    S = n ln(n) + γ

    where γ = the Euler number. And so, for simplicity’s sake, I left it out. But, it should really be

    S = n [ln(n) + γ]

    which makes the γ more significant. Thank you for pointing it out.

    And yes, I have made the assumption that you can have duplicates in a pack of 5. If you can’t then you end up with a series that looks something like this

    S = (640/640) + (640/640) + (640/640) + (640/640) + (640/640) + (640/635) + (640/635) +(640/635) +(640/635) +(640/635) + (640/630) + . . . . . . + (640/5) + (640/5) + (640/5) + (640/5) + (640/5)]

    which can be factorised to produce

    S = 640 [5/640 + 5/635 + 5/630 + . . . . . . . 5/5]

    S = 640 [1 + 1/2 + 1/3 + . . . . .1/128]

    S = 640 [ln(128) + γ]

    S = 640 [4.8520 + .5772157]

    S = 640 [5.42925]

    S = 3474.72

    So, roughly 3475 stickers need to be bought if you can’t have duplicates within a pack. If my maths is correct.

  • Ciaran

    i have 178 stickers and 3 doulds

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  • Greg Newman

    Panini claim all stickers are equally printed and distributed and Swiss TV allagedly did an experiment was done with 6000 packets that backs up this theory, see comments at end of this article. http://www.guardian.co.uk/football/2010/may/23/panini-world-cup-football-sticker

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  • Mike

    Here’s a more difficult problem:

    Suppose the total stickers are a multiple of 4, eg 640. Suppose the’re sold in packs of 4 . each pack has 4 different random players. Each time you buy a pack , you throw the pack away if you already have one of the 4 players.
    How many packs do you need to buy to complete the set?

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  • mancity

    i just opened my second pack and got number 00

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