# Just How Expensive is it to Complete the Panini World Cup Album?

With the 2018 Football World Cup approaching amidst great excitement, in some quarters at least there is surely just as much enthusiasm for the accompanying Panini sticker album. There are 682 empty spots in the book waiting to be filled and the local newsagent will surely do a roaring trade in those pocket-money packets of stickers.

But how much does it really cost to fill the album completely? At a very minimum, you’d need 137 packets of 5 stickers at 80p each. That’s **£109.60**. And a swift back-of-the-envelope calculation will show that the chance of never getting a repeat sticker in all 137 packets is vanishingly small, much less likely than winning the lottery. So what’s a realistic estimate? What if we have friends to swap our spares with? And what would a good or a bad case scenario look like?

As actuaries, there’s really only one thing for it. Start by doing the maths and then, when that becomes too complex to handle neatly, build a model.

## The Question

**So what exactly is the question we’re trying to answer here?**

The initial problem is as follows:

- There are
**682**different stickers to collect. - Assume all stickers are
**independent**and**equally likely**. - A packet of five stickers costs
**80p**. - We continue to buy stickers until our album is full.

Suppose we start with an empty album and collect stickers one at a time. We know that the first sticker will be new, since we don’t have any yet. But after that, there’s a chance of getting a repeat. If we currently have k unique stickers, then the chance of the next sticker being a new one is (682–k)/682. On average, when we currently hold k stickers, we’ll therefore have to wait for 682/(682–k) more stickers before we see a new one.

If we assume all stickers are equally likely and independent of one another, we can simply sum these waiting times to get a total expected number of stickers as follows:

\(\begin{align}\sum\limits_{k=0}^{681} \frac{682}{(682-k)}\end{align}\)

That’s a total of around 4,844 stickers, or **£775.08**.

## Scenario 1 – the base case

**But we buy stickers in packets of five, not one at a time.**

This leads us on to consider the **base case**. This is the setup whereby one individual collector buys **packets** of stickers one at a time until their album is complete.

In contrast to the original assumptions, we actually know that stickers are not all independent of one another. In particular, we can add the following assumptions to our problem setup:

- The five stickers in any given pack will
**all be different**from one another. - Different packs are
**independent**of each other.

Now the problem becomes more complex, because at each point in the calculation we need to consider where we’ll be in five stickers’ time (one pack later), rather than just one sticker’s time.

To work this out, let’s start at the end and suppose for the moment that we have all but one of the stickers. Then in each new packet, we have a 5/682 probability of getting the one we’re missing. Therefore, on average, from this point we’ll need to buy 682/5 = 136.4 more packets. Let’s call that e_{681}, the expected number of packets we need given we currently hold 681 unique stickers.

We also know that e_{682} = 0, because at that point we already hold all the stickers.

The next step is to think about what happens if we currently hold 680 stickers. At this point, there are three possibilities for our next packet of stickers: we’ll get no new stickers, or just one new sticker or we’ll get both of the missing stickers. We can calculate the probabilities of these options fairly easily, knowing that all stickers are equally likely. Call p_{k,i} the probability of getting i new stickers in the next packet given we currently hold k unique stickers.

So from 680 stickers, our expected number of packets to completion can be expressed as:

\(\begin{align}e_{680} = 1 + p_{680, 0} \times e_{680} + p_{680,1} \times e_{681} + p_{680, 2} \times e_{682}\end{align}\)

Given what we know already, we can solve this for e_{680}.

In fact, we can repeat this approach for k = 679, 678, all the way back to k = 0 using the following recursive formula:

\(\begin{align}e_{k} = 1 + \sum\limits_{i=0}^{5} p_{k, i} \times e_{k+i}\end{align}\)

Or equivalently:

\(\begin{align}e_{k} = \frac{1}{1-p_{k,0}} \times \left( 1+\sum\limits_{i=1}^{5} p_{k,i} \times e_{k+i} \right)\end{align}\)

This gives us a total expected number of packets needed (e_{0}) of about 966.4, or a cost of **£773.12**. So the fact that all five stickers in a packet are different from one another doesn’t save us much!

We can use a similar method to calculate other features of the distribution in this base case.

Moment | No. packets | Total cost |
---|---|---|

Mean | 966.4 | £773.12 |

Standard deviation | 173.8 | £139.03 |

Skewness | 1.15 | 1.15 |

Excess kurtosis | 2.43 | 2.43 |

We can see from this that not only is the expected cost pretty high, but also that there’s a lot of variability around this expectation. The skewness of around 1 shows us that the distribution is asymmetric and positively skewed (a feature we also see clearly in the later graphs). The positive excess kurtosis tells us that the tails of this distribution are relatively heavy – meaning that we have quite a big chance of seeing extreme outcomes.

This suddenly stops looking like a pocket money hobby, and more like one you’d need a well-paid full-time job to sustain!

## The model

**Had enough of all this probability and fancy notation? Let’s build a model.**

Fortunately for all the enthusiastic collectors out there, there are ways of bringing the costs down. To investigate these, the next step is to build a model. The model used here is built in VBA, as this is used commonly across different areas of actuarial work, but it could easily be built using other software. Using this model we can look at a full range of outcomes over a large number of simulations.

The histogram below shows the results of a ‘base case’ run on 10,000 simulations. The height of each bar represents the number of simulations which fell within each cost category.

Moment | No. packets | Total cost |
---|---|---|

Mean | 965.97 | £772.76 |

Standard deviation | 174.49 | £139.59 |

Skewness | 1.16 | 1.16 |

Excess kurtosis | 2.58 | 2.58 |

Percentiles | No. packets | Total cost |
---|---|---|

0.1% | 627 | £501.60 |

0.5% | 659 | £527.20 |

1.0% | 678 | £542.40 |

2.5% | 709 | £567.20 |

5.0% | 740 | £592.00 |

10.0% | 774 | £619.20 |

25.0% | 842 | £673.60 |

50.0% | 938 | £750.40 |

75.0% | 1,058 | £846.40 |

90.0% | 1,192 | £953.60 |

95.0% | 1,285 | £1,028.00 |

97.5% | 1,382 | £1,105.60 |

99.0% | 1,510 | £1,208.00 |

99.5% | 1,621 | £1,296.80 |

99.9% | 1,907 | £1,525.60 |

From this we can see the result of every one of our 10,000 simulations plus some summary statistics. Reassuringly, the various moments of this distribution are pretty close to the ones calculated earlier!

You can see from the histogram that the total cost ranged from around £500 up to more than £1,500. In 10,000 simulations of filling the book, barely any cost the collector less than £500 – a far cry from the £109.60 that we might have hoped for at the very start. We can also look at the percentiles; half of the time, the cost was over £750 and in more than 1-in-20 cases the cost was over £1,000. Even somebody who was in the ‘luckiest’ 5% of collectors would be paying around £600.

## Scenario 2 – the swapsie

**Everyone in the office is collecting / all of the kids in your child’s class have an album. How much do ‘swapsies’ save you?**

The benefit of building the model is that it then becomes relatively easy to adapt and consider different setups and scenarios. In this case we can adjust the model to allow for sharing with other collectors and fortunately for you (or your child), this does translate to significant savings.

The output below is a result of a run which is as per the base case, but with **5 friends** swapping. It is based on the assumptions that everyone starts with an empty album and they continue buying packets until they have 5 complete albums between them. This shows the mean cost per person. In this run there were 1,000 simulations.

Notice the change of scale on the x-axis.

Moment | No. packets | Total cost |
---|---|---|

Mean | 411.75 | £329.40 |

Standard deviation | 48.31 | £38.65 |

Percentiles | No. packets | Total cost |
---|---|---|

0.5% | 323 | £258.74 |

5.0% | 351 | £280.80 |

10.0% | 361 | £288.50 |

50.0% | 404 | £322.96 |

90.0% | 469 | £375.30 |

95.0% | 501 | £400.48 |

99.5% | 624 | £499.31 |

We can see here that the average cost is less than half of what it was in the base case. In this case, collectors have a decent chance of spending less than £300 and none of our 1,000 simulations saw the cost rise above £550. Not only do we see a significant saving in costs, but also a much smaller standard deviation of costs, so we can be more certain that our actual cost will be close to our expected cost.

## Scenario 3 – the cheat

**You can buy 50 stickers direct from Panini – does this save you much money?**

The next scenario to consider is that it is possible to buy the last few stickers to complete the album direct from Panini. Starting again from the base case, we here add the following assumptions:

- Each person can buy
**up to 50**direct. - They cost
**22p**each. - Each simulation will run until the album has
**at least 632**unique stickers (bearing in mind that the packet that gives the 632^{nd}sticker may also give the 633^{rd}, 634^{th}etc) and then the packet-buying stops, and the remaining stickers bought directly.

Here, the ‘Number of packs’ column tells us how many packets were purchased, whereas the ‘Cost’ column is the total cost of packets plus individual stickers bought directly.

This run was completed on 10,000 simulations. See the output below (notice again the change in scale):

Moment | No. packets | Total cost |
---|---|---|

Mean | 354.51 | £294.58 |

Standard deviation | 16.21 | £12.97 |

Percentiles | No. packets | Total cost |
---|---|---|

0.5% | 316 | £263.80 |

5.0% | 329 | £273.98 |

10.0% | 334 | £278.20 |

50.0% | 354 | £294.20 |

90.0% | 376 | £311.58 |

95.0% | 382 | £316.60 |

99.5% | 399 | £330.20 |

Again, you can see the expected cost falling enormously, to significantly less than half the base case. The variance of outcomes is also very narrow.

This might seem counter-intuitive, given that 50 stickers represent far less than a tenth of the book, and yet buying these directly saves over 60% on expected cost. This happens because the more stickers you have, the less and less likely it becomes that the next sticker will be new. As we saw in the first section of this article, the last sticker alone takes us an average of 136 packets to find (in the base case). Cutting off that tail end of the search saves an enormous amount of money.

## Scenario 4 – the shiny

**What if you’re convinced there’s a conspiracy, and that you’re only half as likely to get a ‘shiny’ as any other sticker?**

Now Panini stress (see http://collectibles.panini.co.uk/company/curiosity.html#c606) that all stickers are printed in the same quantities, and we certainly aren’t disputing this claim – so our base case assumes that all 682 stickers are equally likely to appear in any packet.

But a quick internet search suggests that many collectors insist that this isn’t the case and that some stickers – perhaps the most popular players or the team photos – are rarer than others. This scenario looks at how, if that were the case, the costs would be impacted.

This scenario has the same setup as the base but with the following additional assumptions:

- There are
**50 ‘shiny’**stickers (this is how many are in the World Cup 2018 album). - You believe that each ‘shiny’ sticker is
**half as likely**to appear as each ‘plain’ sticker.

Of course, there are any number of possible combinations here. Maybe you think that the Germany squad stickers are only a quarter as likely as any other, in which case we could set the model up with 18 ‘rare’ stickers and a ‘rare probability’ of 25%. However, this example should give an idea of the impact.

This scenario has been run over 5,000 simulations. Again, note the scale.

Moment | No. packets | Total cost |
---|---|---|

Mean | 1,222.70 | £978.16 |

Standard deviation | 303.38 | £242.70 |

Percentiles | No. packets | Total cost |
---|---|---|

0.5% | 745 | £596.00 |

5.0% | 845 | £676.00 |

10.0% | 900 | £720.08 |

50.0% | 1,159 | £927.20 |

90.0% | 1,624 | £1,299.20 |

95.0% | 1,808 | £1,446.36 |

99.5% | 2,327 | £1,861.60 |

It might not be much of a surprise to learn that introducing rare stickers makes the whole project more expensive – now close to £1,000 on average. For 1-in-100 collectors that cost is now over £1,750. The variance of outcomes has also increased.

## Scenario 5 – the game

**So now let’s put this all together**

Finally, how do all of these features interact? This scenario combines the ones considered earlier. It’s as per the base setup but with:

**5 collectors**all sharing their ‘swapsies’.- The final
**50**stickers purchased**directly**from Panini. **50 rare**stickers (the ‘shinies’) each with a relative probability of**0.5**as compared to the rest of the stickers.- Run on 2,000 simulations.

Moment | No. packets | Total cost |
---|---|---|

Mean | 188.33 | £161.65 |

Standard deviation | 3.71 | £2.97 |

Percentiles | No. packets | Total cost |
---|---|---|

0.5% | 179 | £154.36 |

5.0% | 182 | £156.88 |

10.0% | 184 | £157.88 |

50.0% | 188 | £161.56 |

90.0% | 193 | £165.40 |

95.0% | 195 | £166.68 |

99.5% | 199 | £170.04 |

## And finally…

This article ends with apologies to anyone that has been put off of collecting stickers having found out how costly it really is (similarly to anyone who feels this level of mathematical analysis takes the fun out of it all)! However, armed with this information, those still determined to continue should now be able to go ahead and complete their album in the most cost-efficient way possible. Best of luck to you – you will need it!

The Excel model used in these projections can be downloaded here. Please do play around, try out your own scenarios and let us know of any interesting observations or feedback on the model. On the published version we have protected the VBA code, but if you would like a copy of the unprotected version do get in touch with us via our contact page and we’d be happy to send you a copy.

##### Sammy Ford

June 2018