# Blog

## Archive

Show me a random blog post**2018**

### Jun 2018

World Cup stickers 2018, pt. 2### May 2018

A bad Puzzle for Today### Apr 2018

Building MENACEs for other games### Mar 2018

A 20,000-to-1 baby?World Cup stickers 2018

### Jan 2018

*Origins of World War I*

Christmas (2017) is over

**2017**

**2016**

**2015**

**2014**

**2013**

**2012**

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## World Cup stickers 2018, pt. 2

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.

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.

The sticker I got most copies of was sticker 545, showing Panana player Armando Cooper.

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.

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...

### Similar posts

World Cup stickers 2018 | World Cup stickers | Euro 2016 stickers | A bad Puzzle for Today |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2018-03-23**

## A 20,000-to-1 baby?

This morning, I heard about Arnie Ellis on the Today programme. Arnie is the first baby boy to be born in his family in five generations, following ten girls. According to John Humphrys, there is a 20,000-to-1 chance of this happening. Pretty quickly, I started wondering where this number came from.

After a quick Google, I found that this news story had appeared in many of today's papers, including the Sun and the Daily Mail. They all featured this 20,000-to-1 figure, which according to The Sun originally came from Ladbrokes.

### What is the chance of this happening?

If someone is having a child, the probability of it being a girl is 0.5. The probability of it being a boy is also 0.5. So the probaility of having ten girls followed by a boy is

$$\left(\tfrac12\right)^{10}\times\tfrac12=\frac1{2048}.$$
If all 11 children were siblings, then this would be the chance of this happening—and it's a long way off the 20,000-to-1. But in Arnie's case, the situation is different. Luckily the Daily Mail article, there is an outline of Arnie's family tree.

Here, you can see that the ten girls are spread over five generations. So the question becomes: given a baby, what is the probability that the child is male and his most recently born ten relatives on their mother's side are all female?

Four of the ten relatives are certainly female—Arnie's mother, grandmother, great grandmother and great great grandmother are all definitely female. This only leaves six more relatives, so the probability of a baby being in Arnie's position is

$$\left(\tfrac12\right)^{6}\times\tfrac12=\frac1{128}.$$
This is now an awful lot lower than the 20,000-to-1 we were told. In fact, with around 700,000 births in the UK each year, we'd expect over 5,000 babies to be born in this situation every year. Maybe Arnie's not so rare after all.

This number is based on the assumption that the baby's last ten relatives are spread across five generations. But the probability will be different if the relatives are spread over a different number of generations. Calculating the probability for a baby with any arrangement of ancestors would require knowing the likelihood of each arrangement of relatives, which would require a lot of data that probably doesn't exist. But the actual anwer is probably not too far from 127-to-1.

### Where did 20,000-to-1 come from?

This morning, I emailed Ladbrokes to see if they could shed any light on the 20,000-to-1 figure. They haven't got back to me yet. (Although they did accidentally CC me when sending the query on to someone who might know the answer, so I'm hopeful.) I'll update this post with an explanaation if I do hear back.

Until then, there is one possible explanation for the figure: we have looked at the probability that a baby will be in this situation, but we could instead have started at the top of the family tree and looked at the probability that Beryl's next ten decendents were girls followed by a boy. The probability of this happening will be lower, as there is a reasonable chance that Beryl could have no female children, or no children at all. Looking at the problem this way, there are more ways for the situation to not happen, so the probability of it happening is lower.

But working the actually probability out in this way would again require data about how many children are likely in each generation, and would be a complicated calculation. It seems unlikely that this is what Ladbrokes did. Let's hope they shed some light on it...

### Similar posts

World Cup stickers 2018, pt. 2 | A bad Puzzle for Today | World Cup stickers 2018 | How much will I win on the new National Lottery? |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

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**2018-03-22**

## World Cup stickers 2018

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.

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.

### Similar posts

World Cup stickers 2018, pt. 2 | World Cup stickers | Euro 2016 stickers | A bad Puzzle for Today |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2018-03-24**

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

**2018-03-23**

RM

**Add a Comment**

**2017-03-27**

## The end of coins of constant width

Tomorrow, the new 12-sided one pound coin is released.

Although I'm excited about meeting this new coin, I am also a little sad,
as its release ends the era in which all British coins are shapes of constant
width.

### Shapes of constant width

A shape of constant width is a shape that is the same width in every direction,
so these shapes can roll without changing height. The most obvious such shape
is a circle. But there are others, including the shape of the seven-sided 50p
coin.

As shown below, each side of a 50p is part of a circle centred around the opposite corner.
As a 50p rolls, its height is always the distance between one of the corners and
the side opposite, or in other words the radius of this circle. As these circles
are all the same size, the 50p is a shape of constant width.

Shapes of constant width can be created from any regular polygon with an
odd number of sides, by replacing the sides by parts of circles centred at the
opposite corner. The first few are shown below.

It's also possible to create shapes of constant width from irregular polygons with an odd number,
but it's not possible to create them from polygons with an even number of sides.
Therefore, the new 12-sided pound coin will be the first non-constant width British coin since
the (also 12-sided) threepenny bit was phased out in 1971.

Back in 2014, I wrote to my MP in an attempt to find
out why the new coin was not of a constant width. He forwarded my letter to
the Treasury, but I never heard back from them.

### Pizza cutting

When cutting a pizza into equal shaped pieces, the usual approach is to
cut along a few diameters to make triangles. There are other ways to fairly
share pizza, including the following (that has appeared here before as an answer to this puzzle):

The slices in this solution are closely related to a triangle of constant
width. Solutions can be made using other shapes of constant width,
including the following, made using a constant width pentagon and heptagon (50p):

There are many more ways to cut a pizza into equal pieces. You can find them in

*Infinite families of monohedral disk tilings*by Joel Haddley and Stephen Worsley [1].You can't use the shape of a new pound coin to cut a pizza though.

Edit: Speaking of new £1 coins, I made this stupid video with Adam "Frownsend" Townsend about them earlier today:

#### References

[1]

**by***Infinite families of monohedral disk tilings***Joel Haddley and Stephen Worsley**. December 2015. [link]### Similar posts

New machine unfriendly £1 coin, pt. 2 | New machine unfriendly £1 coin | World Cup stickers 2018, pt. 2 | A bad Puzzle for Today |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

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**2016-05-04**

## Euro 2016 stickers

Back in 2014, I calculated the expected cost of
filling Panini world cup sticker album. I found that you should expect to buy
4505 stickers, or 1285 if you order the last 100 from the Panini website (this
includes the last 100). This would cost £413.24 or £133.99
respectively.

Euro 16 is getting close, so it's sticker time again. For the Euro 16
album there are 680 stickers to collect, 40 more than 2014's 640 stickers.
Using the same calculation method as before,
to fill the Euro 16 album, you should expect to buy 4828
stickers (£442.72), or 1400 (£134.32) if you order the last 100.

This, however, does not tell the whole story. Anyone who has collected
stickers as a child or an adult will know that half the fun comes from
swapping your doubles with friends. Getting stickers this way is not
taken into account in the above numbers.

### Simulating a sticker collection

Including swaps makes the situation more complicated: too complicated
to easily calculate the expected cost of a full album. Instead, a different
method is needed. The cost of filling an album can be estimated by
simulating the collection lots of times and taking the average of the cost of
filling the album in each simulation. With enough simulations, this estimate
will be very close the the expected cost.

To get an accurate estimation, simulations are run,
calculating the running average as they go, until the running averages after recent simulations
are close together. (In the examples, I look for the four most recent running averages to be within 0.01.)
The plot below shows how the running average changes as more simulations are performed.

The simulations estimate the number of stickers needed as 4500. This is
very close to the 4505 I calculated last year.

Now that the simulations are set up, they can be used to see what happens if you have friends to swap with.

### What should I do?

The plots below shows how the number of stickers you need to buy each changes based on how many friends you have.

In both these cases, having friends reduces the number of stickers you need to buy significantly, with your first few friends
making the most difference.

Ordering the last 100 stickers looks to be a better idea than ordering no stickers. But how many stickers should you order to
minimise the cost? When you order stickers, you are guaranteed to get those that you need, but they cost more: ordered stickers cost 14p
each, while stickers in 6 pack multipacks come out at just 9.2p each. The next plot shows how the cost changes based on how many you order.

Each of the coloured curves represents a group of a different size. For each group, ordering no stickers works out the most
expensive—this is expected as so many stickers must be bought to find the last few stickers—and ordering all the stickers also works
out as not the best option. The best number to order
is somewhere in the middle, where the curve reaches its lowest point. The minimum points on each of these curves are summarised in the
next plots:

Again, having friends to swap with dramatically reduces the cost of filling an album. In fact, it will almost definitely pay off in future
swaps if you go out right now and buy starter packs for all your friends...

### Similar posts

World Cup stickers 2018, pt. 2 | World Cup stickers 2018 | World Cup stickers | How to kick a conversion |

### Comments

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