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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...

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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?

• 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 4814 more stickers after this.
The cheapest way to buy these stickers is to buy them in multipacks for 10p per stickers. This gives a total expected cost of filling the sticker album of £485.39... On this plus side, I've just saved you £288.84 by pointing out that multipacks exist!

### 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.
Panini'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 £187.39. I've just saved you another £298.

### 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
• £$$\displaystyle0.1\left(\sum_{n=1}^{681-a}\frac{682}{682-n}-31\right)$$ 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 400 stickers are ordered, giving a total expected cost of £109.14. That's another £78.25 I've saved you. You're welcome.
Still, ordering nearly half the stickers to minimise the cost doesn't sound like the most fun way to complete the sticker book, so maybe you might want to order a few less than this and spend a little more money.

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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2018-03-23
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 of just it's very unlikely... Follow up post looking into this coming soon
Matthew
2018-03-23
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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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

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

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 New machine unfriendly £1 coin, pt. 2 New machine unfriendly £1 coin A 20,000-to-1 baby? World Cup stickers 2018

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2017-02-25

## The importance of estimation error

Recently, I've noticed a few great examples of misleading uses of numbers in news articles.
On 15 Feb, BBC News published a breaking news article with the headline "UK unemployment falls by 7,000 to 1.6m". This fall of 7,000 sounds big; but when compared to the total of 1.6m, it is insignificant. The change could more accurately be described as a fall from 1.6m to 1.6m.
But there is a greater problem with this figure. In the original Office of National Statistics (ONS) report, the fall of 7,000 was accompanied by a 95% confidence interval of ±80,000. When calculating figures about large populations (such as unemployment levels), it is impossible to ask every person in the UK whether they are employed or not. Instead, data is gathered from a sample and this is used to estimate the total number. The 95% confidence interval gives an idea of the accuracy of this estimation: 95% of the time, the true number will lie of the confidence interval. Therefore, we can think of the 95% confidence interval as being a range in which the figure lies (although this is not true, it is a helpful way to think about it).
Compared to the size of its confidence interval (±80,000), the fall of 7,000 is almost indistinguishable from zero. This means that it cannot be said with any confidence whether the unemployment level rose or fell. This is demonstrated in the following diagram.
A fall of 7,000 ± 80,000. The orange line shows no change.
To be fair to the BBC, the headline of the article changed to "UK wage growth outpaces inflation" once the article was upgraded from breaking news to a complete article, and a mention of the lack of confidence in the change was added.
On 23 Feb, I noticed another BBC News with misleading figures: Net migration to UK falls by 49,000. This 49,000 is the difference between 322,000 (net migration for the year ending 2015) and 273,000 (net migration for the year ending 2016). However both these figures are estimates: in the original ONS report, they were placed in 95% confidence intervals of ±37,000 and ±41,000 respectively. As can be seen in the diagram below, there is a significant portion where these intervals overlap, so it cannot be said with any confidence whether or not net immigration actually fell.
Net migration in 2014-15 and 2015-16.
Perhaps the blame for this questionable figure lies with the ONS, as it appeared prominently in their report while the discussion of its accuracy was fairly well hidden. Although I can't shift all blame from the journalists: they should really be investigating the quality of these figures, however well advertised their accuracy is.
Both articles criticised here appeared on BBC News. This is not due to the BBC being especially bad with figures, but simply due to the fact that I spend more time reading news on the BBC than in other places, so noticed these figures there. I quick Google search reveals that the unemployment figure was also reported, with little to no discussion of accuracy, by The Guardian, the Financial Times, and Sky News.

### Similar posts

 A 20,000-to-1 baby? World Cup stickers 2018 The end of coins of constant width Euro 2016 stickers

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2017-12-26
I've seen archaeologists claiming proof that event A happened before event B because the radiocarbon date of A was 50 years before B. Except the standard error on both dates was 100 years. They even showed the error bars in their own graphics, but seemed to not understand what it meant.

My favorite species of ignoring the measurement error is the metric conversion taken to way too many decimal places. The hike was 50 miles (80.467 kilometers) long.
Perry Ramsey

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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.
Stickers needed if you and your friends order no stickers.
Stickers needed if you and your friends all order the last 100 stickers. The last 100 are not counted.
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.
The expected cost of filling an album based on number of people in group and number of stickers ordered.
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:
How the number you should order changes with the number of people in the group.
How the cost changes with the number of people in the group.
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...

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