Friday, 13 April 2018

Friday factoid- Friday 13th occurs far too frequently.

Friday 13th is an unlucky day in the UK. British people stay at home, don't make eye contact in public, and spend their day complaining, mostly about the weather.
Did you know, however, that the 13th is more likely to be a Friday than any other day? And also, coincidentally, Friday the 13th is the equally-most likely day number/day name combination to occur?

Why? We know the year for the Earth (the time it takes us to do one orbit of the Sun) is 365.242189 days. This is annoying for calendar makers, as we just can't put an extra 0.256 days into each year. So what do we do? Well, we try and even it out over a 400 year cycle. We fix the year to have 365 days normally (remember the old rubbish rhyme: "30 days has September, April, June and November, all the rest have 31, except February...")

February has 28 days, except in a leap year, where it has 29. A leap year is defined as
  • any year which is divisible by 4 except
  • years divisible by 100 are not leap years except
  • years divisible by 400 are leap years.
If we repeat this on a 400 year cycle, then this means that the average year length is 365.2425 days (or very close to it)
If we work out the number of days in 400 years, then this is 146097 days; by chance this is divisible by 7, so it means that if a day is a Monday, the same day 400 years later will be a Monday. In other words, the minimum period the calendar cycles over is 400 years. Note that 146097/400=365.2425, very close to Earth's year.

We can then run a simple computer script to count the number of times that each number/day of week combination occurs in a 400 year cycle, and graph the results

This shows us that the 13th is a Friday 688 days out of every 400 year cycle, more than any other day. Also, we can see that Friday 13th is the joint most common number/day of week combination (all the yellow squares in the diagram). I have left off the 29th,30th and 31st just because it's a nicer pattern this way.

So is this important? No! But next time you have bad luck on Friday 13th remember, this day will happen 688 days in the next 400 years so you'd better get used to it...


Thursday, 5 April 2018

So what have we learned?

Although we provided a fun marking scheme at the end of every episode we decided to actually plot our opinions in the following quality graphs. Horizontal axes provide a measure of quality of the film, whilst vertical axes provide a a measure of quality of the maths.

Our first graph illustrates all of our differing opinions. Each film has a different marker and the colour of each marker links to a particular person's tastes. 
We can clearly see that Interstellar was a favourite all round, whilst The Oxford Murders took a battering from all sides.

Equally, although Thomas and Ben agreed that Pi wasn't a good Liz seemed to enjoy it more than them (still not good, just not as bad!). Proof, was perhaps the most divisive amongst the team.

A couple of interesting points is that there are very few, if any, points in the Bad Film/Good Maths category, or the Good Film/ Bad Maths. This is, of course easy to explain. The Bad Film/Good Maths section would probably be inhabited by videos of mathematical lectures. Whilst, (good) non-science films would inhabit the Good Film/Bad maths quadrant.

Below we plot the average values of the above data in order to visualise the trends better.
What is most striking about this representation is that most of the films tend to cluster around The Line of Equal Quality. This means that a film portraying good maths is also likely to be a good film overall, equally, a film presenting poor maths is likely to be poor overall.

Again, this makes sense, as we have been focusing on such mathematical films, the mathematical content will be central to its subject matter. Thus, a lot of the films quality will rest upon the mathematics representation.

Of course the outlier from this theory is The Imitation Game. A gripping, if slightly embellished, story with actually very little mathematics presented.

So, ten movies later and what have we learned?

A good maths film has to present both good maths and a good film.

Yes, mathematicians are often known for stating the obvious!

Monday, 2 April 2018

Answer to the Interstellar puzzle

In our Interstellar podcast, Thomas posed the following puzzle:

A group of people with assorted eye colors (say 10 blue and 10 brown) live on an island.

Everyone can see the eye colour of everyone else, but they can't communicate to each other to tell each other their eye colour.

Every night at midnight, a ferry stops at the island. Any islanders who have figured out the colour of their own eyes can leave the island.

One day a sailor from the ferry gets off the boat and says:

"I can see someone who has blue eyes".

Everyone hears and understand the statement, but the sailor is immediately shot dead for communicating with the islanders and no-one ever speaks again. However, given this information some people are able to figure out their eye colour.

Who leaves the island, and on what night?


The simple answer is that all 10 blue-eyed people leave on the tenth day. However, the thinking behind this answer is probably more important than the answer itself. Critically, this is a difficult puzzle. You need to have an extended chain of thinking, but, before we make a chain, let's start with a single link.

Let's make the puzzle as simple as possible. Suppose on the island there is only one person with blue eyes and a load of people with other colour eyes (it doesn't matter how many, or what colour, as long as they're not blue). This blue-eyed person would immediately realize that they had blue eyes because they could see no one else with blue eyes. Therefore, that person would leave on the first night.

Suppose, now, there are two people with blue eyes. They would both leave on the second night, because they would each look at the other blue-eyed person on the second morning and realize that the only reason the other blue-eyed person wouldn't leave on the first night is because they see another person with blue eyes. Seeing no one else with blue eyes, each of these two people realize it must be them.

Carrying on this argument inductively we see that n blue-eyed people would leave on night n because on the n-1 previous night they cannot deduce that the other blue-eyed people are not leaving because of them.

This is a slightly hand-wavy proof, but can be made more rigorous, indeed Reddit has a very formal proof.


One part that still blows our mind is that everyone can always see everyone else's eyes. So in the case of 10 blue-eyed people on the island everyone knows that there are blue-eyed people on the island, so what information have they gained from the sailor?

The answer is common knowledge, but we'll let Wikipedia explain that.

Wednesday, 14 March 2018

Maths at: pi day

Happy pi day everyone. 

We figure that pi day is the like Hallowe'en for mathematicians. It's a day they get to cut loose and throw away their inhibitions.

As such Lorraine's intro pertaining to strong language is particularly pertinent for today's podcast. It starts out quite coarse.

So sit back, grab a slice of your favourite pie and enjoy as Ben sings you the song of his people.


Further reading links:
Subscribe via iTunes.

Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Saturday, 10 March 2018

Puzzle from Interstellar

In our Interstellar podcast, Thomas posed the following puzzle:

A group of people with assorted eye colors (say 10 blue and 10 brown) live on an island.

No one knows the color of their eyes as there are no mirrors on the island and the water is muddy so you can't use the reflection. For all each person knows, they could have green eyes!

However, everyone can see the eye colour of everyone else, but they can't communicate to each other to tell each other their eye colour.

Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes can leave the island.

One day a sailor from the ferry gets off the boat and says:

"I can see someone who has blue eyes".

Everyone hears and understand the statement, but the sailor is immediately shot dead for communicating with the islanders and no-one ever speaks again. However, given this information some people are able to figure out their eye colour.

Who leaves the island, and on what night?



It's difficult, but possible.

If you think you have the answer comment below, tweet it to us @PodcastMathsAt, or email us at podcastmaths@gmail.com.

The answer will be posted next week.

Wednesday, 7 March 2018

Solution to The Oxford Murders puzzle

In our Oxford Murders podcast, Ben posed the following puzzle:

Suppose you are going to play chess against two people: one person is really good, one person is quite bad. You are going to play three games and you always have to alternate your opponents. Namely, you can either choose to play the opponents in the order
Good, Bad, Good,
or you can play the opponents in the order
Bad, Good, Bad.

Which of these two play sequences gives you the optimal chance of winning two consecutive games?

You can approach this problem using probability and tree diagrams. However a little logic goes a long way.

Specifically, in order to win two consecutive games you have to win the middle game. Thus, it is best to put your weaker opponent in the middle. Thus, Good, Bad, Good is the best strategy.

An alternative way of also seeing this answer it that you're probably going to lose against the good player, so the Good, Bad, Good play order gives you two chances to win against the good player, rather than just one.

Simple no?

If you want a bit more rigor then Ben has created a YouTube video solution.
Alternatively, you could try three player chess and team up with the weak player to beat the good player. But that might be considered cheating...


Friday, 2 March 2018

Puzzle from The Oxford Murders

In our Oxford Murders podcast, Ben posed the following puzzle:

Suppose you are going to play chess against two people: one person is really good, one person is quite bad. You are going to play three games and you always have to alternate your opponents. Namely, you can either choose to play the opponents in the order
Good, Bad, Good,
or you can play the opponents in the order
Bad, Good, Bad.

Which of these two play sequences gives you the optimal chance of winning two consecutive games?


If you think you have the answer comment below, tweet it to us @PodcastMathsAt, or email us at podcastmaths@gmail.com.

The answer will be posted next week, or you can listen to the answer in our Interstellar podcast

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