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Tr a d i n g c o u p o n s : c o m p l e t i n g t h e Tr a d i n g c o u p o n s : c o m p l e t i n g t h e Wo r l d C u p f o o t b a l l s t i c ke r a l b u mWo r l d C u p f o o t b a l l s t i c ke r a l b u m

The agonies and the ecstasies of the World Cup are over for another four years. England supporters may not wish to

be reminded of Ronaldo’s spot-kick, and we are loath to add to pain. But one set of players who probably did very

well from the competition was Panini, the company which produced the offi cial 2006 World Cup football sticker

album—and, more importantly, the collectable stickers which go inside it. Kevin Hayes and Ailish Hannigan tried

to fi ll up their albums.

The World Cup stickers are pictures of up to 17 players from each of the 32 teams playing in the fi nals in Germany, plus a team photograph and a badge, and photographs of the various stadia where the matches are played. There are 598 different stickers in all. Young fans—generally young fans, we assume—buy sealed packets of

stickers in the hope of completing their col-lection of stickers and fi lling their album. The stickers are randomly distributed in packets of 5. Packets cost 35p and may have multiple copies of an individual sticker.

With more stickers than ever to collect this year, the National Consumer Council has com-plained about the cost and irritation for parents whose children are trying to complete the al-

bum. The Times newspaper reported “it could cost more than £100 to complete one album” and “the law of probabilities means that, the nearer the collectors get to completing the al-bums, the more diffi cult it is to fi nd those elu-sive, last few players1”! Other newspapers took a more experimental approach to determine the cost of completing one album. The Daily Mail reported that journalists from their sister paper The London Evening Standard bought “nearly 150 packets” of stickers, but “still had 100 empty spaces in their album and more than 200 duplicate stickers” and “still had not completed a single country’s team”2.

So can we predict how much, on average, it will cost to complete the album? Ignore (initially anyway) the fact that stickers come in packets of 5. Instead, consider examining stickers one at a time. The fi rst sticker examined will be new to our album with a probability of 1, denoted here by p1 = 1. The waiting time T1 (measured in number of stickers examined) until this event takes place has to satisfy p1 × T1 = 1 and so T1 = 1. The probability of fi nding a second sticker suitable for inclusion in our album—rather than a dupli-cate that we have already—is p2 = 597/598. The

average waiting time T2 until this event takes place satisfi es p2 × T2 = 1, so that T2 = 598/597. The same argument can be used to establish the waiting times T3 = 598/596, T4 = 598/595 and T5 = 598/594, and so on. The average waiting time (measured in number of stickers required) to complete the album is

Since the stickers come in packets of 5, the aver-age number of packets required to fi ll the album is 4169.042/5 = 833.8, or rounding up we get 834. At a cost of 35p a packet this implies a total average cost of 834 × 35p = £291.90 per album.

So buying 835 packets, and spending just short of £300, will on average fi ll your plastic-

covered ringbinder album. £300 would also buy you, say, the complete works of Shakespeare in leather-bound hardback some 10 times over, but let that pass.

Filling the World Cup football sticker album is an example of the well-known coupon collector’s problem and appears in classic references such as Feller (1968)3. Suppose that independent tri-als, each of which results in any of n possible outcomes (number of footall stickers to collect for our album) with probability n–1, are continu-ally performed. Let W denote the number of trials needed (stickers examined) until each outcome has occurred at least once. For w ≥ n, the density function

The nearer the collector gets to completing the album,

the more diffi cult it is to fi nd those last few players

The average cost to fi ll an album is £291.90

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more than 919 packets to complete their album (estimated cost greater than £321.65), but the luckiest 25% of collectors will purchase fewer than 724 packets to complete their album (es-timated cost less than £253.40). The median equals 815 packets, that is, half the fans will be expected to spend less than this amount to complete their album and half will spend more. The density function of packets sold is posi-tively skewed. Consequently the median (815 packets) will be a more appropriate measure of centre than the mean (834.3 packets). The estimated cost based on the median equals £285.25, but of course, this lower estimate may serve the needs of attention-grabbing headlines almost as well as the mean.

So far we have assumed that collectors of the stickers operate in isolation, but in the real world this is a naïve assumption. Collectors can swap stickers with other collectors. In earlier World Cups this would have happened in the school playground; this time round there has been a fl ourishing market in the stickers on ebay, and even Panini, the company who makes the World Cup stickers, offers a swap area and a facility for buying missing stickers on their web-site (www.paninigroup.com/2006fi faworldcup). A collector can buy missing stickers for 8p each with a minimum order of 20 stickers and a maxi-mum order of 50 stickers, though it does take 35 to 40 days for the missing stickers to arrive.

We shall ignore the facility for buying miss-ing stickers and consider only the situation where stickers are traded between collectors. Generalising the coupon collector’s problem to this (more) real-life situation is diffi cult math-

ematically. Programming a computer simulation to calculate the cost-saving effect of a formal sticker trading arrangement between two par-ties is, however, relatively straightforward. Sup-pose collectors A and B each purchase a packet of fi ve stickers and arrange to trade stickers ac-cording to the following protocol. If a sticker is required for A’s album and is available in the surplus held by B, then a trade takes place if and

only if A’s surplus contains a sticker required by B. We are assuming that the only motivation for trades is the desire to complete the album and no money is changing hands. Finally, when one collector has completed their album, there is no incentive to continue trading, so the other col-lector will subsequently have to behave as an individual collector in order to complete their album.

We simulated 20 000 pairs of collectors trad-ing stickers according to the protocol described above. Figure 2 depicts the empirical bivariate density of the number of packets required by the fi rst collector to complete their album (fi rst fi n-ished) and the number of packets required by the second collector to complete their album (sec-ond fi nished). The colours red, orange and yel-low represent increasing levels of density, while the colour white represents the region of highest density. In other words, the most frequent joint result is that the fi rst collector has to buy around 469 packets and the second around 500—the co-ordinates of the white top of the ‘mountain.’ Black contour lines have been included to help to visualise the overall shape of the bivariate density. The median number of packets required to complete an album by the fi rst fi nisher is 469 (illustrated in Figure 2 by the vertical arrow) at a cost of £164.15 and the median number of pack-ets required by the second fi nisher is 607 (il-lustrated in Figure 2 by the horizontal arrow) at a cost of £212.45. These costs are approximately 60% and 75% of the median cost of £285.25 calculated earlier for an individual collector to complete the album.

The simulations show that modelling the pos-sibility of a trading arrangement with just one other collector produces important downward adjustments in the estimated costs. Clearly, the more traders there are, the cheaper the costs be-come. So, although the theory of the coupon col-lector’s problem is often used by statisticians to answer queries from newspapers about the cost of fi lling sticker albums, perhaps the real answer

and distribution function

are available from Feller (1968)3. Dawkins (1991)4 showed that W has expected value

(calculated for our example to be 4169.042 stick-ers) and approximated the distribution function using Pr(W ≤ w) ≈ exp(–n exp(–w/n)). These results have been used on other occasions by statisticians to answer queries from newspapers about the cost of fi lling similar sticker albums.

A computer simulation approach gives ap-proximately the same results. The number of packets required in order to complete 1000 simulated sticker albums was calculated using the statistical software package R5. Essentially, the computer bypasses the mathematics, and just pretends to buy sticker after sticker until it has fi lled its album a thousand times over—the approach used by the Evening Standard journal-ist earlier, but buying virtual stickers not real ones, and without spending money. The results are shown in Figure 1, along with theoretical values. The simulation shows that an unlucky 25% of collectors will be expected to purchase

Figure 1. Empirical probability (black) of 1000 simulated World Cup sticker albums, simulation summary statistics (red) and theoretical distribution function (blue)

Three hundred pounds would also buy the complete works of Shakespeare several times over

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0.0

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0.6

0.8

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Number of packets sold

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Lower quartile = 723.8

Median = 815

Mean = 834.3

Upper quartile = 919.2

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can only be obtained by modelling the trading environment of the school playground!

The authors’ own nation, Ireland, sadly did not qualify for the World Cup, which perhaps allows us a more dispassionate analysis. Our interest in this problem arose when one of the authors was asked by a newspaper to comment

on the cost of fi lling the offi cial 2006 Gaelic Athletic Association Players Stickers Album (376 stickers to collect, sold in packets of eight stickers at a cost of €0.99 per packet). The text of the newspaper article reported the costs based on the mean and upper quartile for the single collector problem and a plot similar

to Figure 1 was used. The article also warned that the costs arrived at excluded the possibil-ity of sticker trading. However, the headline of the article was based on the cost calculated using the upper quartile6. Newspaper headlines are not quite the same thing as good statis-tics.

References1. The Times (2006) The £100 World Cup

penalty. The Times, May 31, 2006.2. The Daily Mail (2006) Football stickers

sting young fans for £100. The Daily Mail, May 31, 2006.

3. Feller, W. (1968) An Introduction to Probability Theory and Its Applications, Vol. 1. Chichester: Wiley.

4. Dawkins, B. (1991) Siobhan’s problem: the coupon collector’s problem revisited. The American Statistician, 45, 76–82.

5. R Development Core Team (2003). R: a Language and Environment for Statistical Computing. Vienna: R Foundation for Statistical Computing. (See www.r-project.org.)

6. Ireland on Sunday (2006) €335, the cost of fi lling new GAA sticker album. Ireland on Sunday, June 11th.

Kevin Hayes’s main research is in statistical analysis of biomedical signals, images and shapes. Ailish Han-nigan’s interest is in multivariate survival analysis. Both are Lecturers in the Department of Mathematics and Statistics at the University of Limerick, Ireland.

Figure 2. Empirical bivariate density of the number of packets purchased by two collectors trading stickers with each other

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450

500

550

600

650

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Packets purchased by first finished

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median = 469

median = 607

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