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2014-03-29
Following my last post, I wrote to my MP (click to enlarge):
![](javascript:showlimage("mp-coin-letter.png"))
Today I received this reply (click to enlarge):
![](javascript:showlimage("mp-reply.jpg"))
I'm excited about hearing what the Treasury has to say about it...
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2014-03-19
Vending machines identify coins by measuring their width. Circular coins have the same width in every direction, so designers of vending machines do not need to worry about incorrectly rotated coins causing a blockage or being misidentified. But what about seven-sided 20p and 50p coins?
Perhaps surprisingly, 20p and 50p coins also have a constant width, as show by this video. In fact, the sides of any regular shape with an odd number of sides can be curved to give the shape a constant width.
![](javascript:showlimage("Reuleaux_polygons.png"))
3, 5, 7 and 9 sided shapes of constant width.
Today, a new 12-sided £1 coin was unveiled. One reason for the number of sides was to make the coin easily identified by touch. However, as only polygons with an odd number of sides can be made into shapes of constant width, this new coin will have a different width when measured corner to corner or side to side. This could lead to vending machines not recognising coins unless a new mechanism is added to correctly align the coin before measuring.
Perhaps an 11-sided or 13-sided design would be a better idea, as this would be easily distinguishable from other coins by touch which being a constant width to allow machines to identify it.
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2013-12-23
Here is a collection of Christmas relates mathematical activities.
Flexagons
I first encountered flexagons sometime around October 2012. Soon after, we used this template to make them at school with year 11 classes who had just taken GCSE papers as a fun but mathematical activity. The students loved them. This lead me to adapt the template for Christmas:
![](javascript:showlimage("xmas-flex.png"))
And here is an uncoloured version of the template on that site if you'd like to colour it yourself and a blank one if you'd like to make your own patterns:
![](javascript:showlimage("flex.png"))
![](javascript:showlimage("blank-flex.png")){kind=spacer}
The excitement of flexagons does not end there. There are templates around for six faced flexagons and while writing this piece, I found this page with templates for a great number of flexagons. In addition, there is a fantastic article by Martin Gardner and a two part video by Vi Hart.
Fröbel stars
I discovered the Fröbel star while searching for a picture to be the Wikipedia Maths Portal picture of the month for December 2013. I quickly found these very good instructions for making the star, although it proved very fiddly to make with paper I had cut myself. I bought some 5mm quilling paper which made their construction much easier. With a piece of thread through the middle, Fröbel starts make brilliant tree decorations.
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2013-12-15
Last week, I was watching Pointless and began wondering how likely it is that a show features four new teams.
On the show, teams are given two chances to get to the final—if they are knocked out before the final round on their first appearance, then they return the following episode. In all the following, I assumed that there was an equal chance of all teams winning.
If there are four new teams on a episode, then one of these will win and not return and the other three will return. Therefore the next episode will have one new team (with probability 1). If there are three new teams on an episode: one of the new teams could win, meaning two teams return and two new teams on the next episode (with probability 3/4); or the returning team could win, meaning that there would only one new team on the next episode. These probabilities, and those for other numbers of teams are shown in the table below:
| No of new teams today | |||||
Noof new teams tomorrow
| 1 | 2 | 3 | 4 | |
| 1 | 0 | 0 | \(\frac{1}{4}\) | 1 | |
| 2 | 0 | \(\frac{1}{2}\) | \(\frac{3}{4}\) | 0 | |
| 3 | \(\frac{3}{4}\) | \(\frac{1}{2}\) | 0 | 0 | |
| 4 | \(\frac{1}{4}\) | 0 | 0 | 0 | |
Call the probability of an episode having one, two, three or four new teams \(P_1\), \(P_2\), \(P_3\) and \(P_4\) respectively. After a few episodes, the following must be satisfied:
$$P_1=\frac{1}{4}P_3+P_4$$
$$P_2=\frac{1}{2}P_2+\frac{3}{4}P_3$$
$$P_3=\frac{3}{4}P_3+\frac{1}{2}P_4$$
$$P_4=\frac{1}{4}P_1$$
And the total probability must be one:
$$P_1+P_2+P_3+P_4=1$$
These simultaneous equations can be solved to find that:
$$P_1=\frac{4}{35}$$
$$P_2=\frac{18}{35}$$
$$P_3=\frac{12}{35}$$
$$P_4=\frac{1}{35}$$
So the probability that all the teams on an episode of Pointless are new is one in 35, meaning that once in every 35 episodes we should expect to see all new teams.
Edit: This blog answered the same question in a slightly different way before I got here.
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2013-07-24
A news story on the BBC Website caught my eye this morning. It reported the following "uncanny coincidence" between a Northern Irish baby and a Royal baby:
But both new mothers share the name Catherine, the same birthday - 9 January - and now their sons also share the same birth date.
I decided to work out just how uncanny this is.
The Office for National Statistics states that 729,674 babies are born every year in the UK. This works out at 1,999 babies born each day, assuming that births are uniformly distributed, so there will be approximately 1,998 babies who share Price Nameless's birthday.
So, what is the chance of the mother of one of these babies having the same birthday as Princess Kate? To work this out I used a method similar to that which is used in the birthday "paradox", which tells us that in a group of 23 people there is a more than 50% chance of two people sharing a birthday, but that's another story.
First, we look at one of our 1,998 mothers. The chance that she shares Princess Kate's birthday is 1/365 (ignoring leap days). The chance that she does not share Princess Kate's Birthday is 364/365.
Next we work out the probability that none of our 1,998 mothers shares Princess Kate's birthday. As our mothers' birthdays are independent we can multiply the probabilities together to do this (this is why we are looking at the probability of not sharing a birthday instead of sharing a birthday). Our probability therefore is \(\left(\frac{364}{365}\right)^{1998} = 0.00416314317\).
Back to the original question, we wanted to know the probability that one of our mothers shares Princess Kate's birthday. To calculate this we do take 0.00416314317 away from 1. This gives 0.99583685682 or 99.6%.
There is a 99.6% chance that there is a resident of the UK who shares the same birthday as Princess Kate and had a child on the same day.
Uncanny.
But let's be fair. The mother in our story is also called Kate. So what are the chances of that? In fact, the same method can be followed, working with the probability of having neither the same birthday or name as Princess Kate.
I think it is safe to assume that this would still be considered news-worthy if our non-princess was called Katie, Cate, Cathryn, Katie-Rose or any other name which is commonly shortened to Kate, so I included a number of variations and used this fantastic tool to find the probability of a mother being called Kate. The data only goes back to 1996, but as the name is dropping in popularity, we can assume that before 1996 at least 1.5% of babies were called Kate. Disregarding males, we can estimate that 3% of mothers are called Kate.
If anyone would like the details of the rest of the calculation, please comment on this post and I will include it here. For anyone who trusts me and isn't curious, I eventually found that the probability of none of our 1,998 mothers share the same name and birthday as Princess Kate is 0.84855028964. So the probability of another Kate having a child on the same day and sharing Princess Kate's birthday is 0.15144971035 or 15.1%. Just over one in seven.
So this is as uncanny as anything else which has a probability of one in seven, such as the Royal baby being born on a Monday (uncanny!).
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