Applied mathematics at its… best? After predicting statistically that it would take 400-500 packets of Skittles before you’d expect to find the same permutation of colours, an
experiment finds empirical backing for this answer at pack number 464.
I show that for the same reason that the golden ratio, ϕ=1.6180334.., can be considered the most irrational number, that 1 + √2 can be considered the 2nd most irrational number, and
indeed why (9 + √221)/10 can be considered the 3rd most irrational number.
Let us imagine a game between two kids, Emily and Sam – both strong and determined in their own way who spend their entire lives trying to outwit each other, instead of doing their
homework. (A real life Generative Adversial Network…)
Emily, proudly reminds us that she simultaneously bears the same first name as Emily Davison, the most famous of British
suffragettes; Emily Balch, Nobel Peace Prize laureate; Emilie du Chatelet, who wrote the first French translation and commentary of Isaac Newton’s
“Principia”; Emily Roebling, Chief Engineer of New York’s iconic
Brooklyn Bridge; Emily Bronte author of Wuthering Heights;
Emily Wilson, the first female editor of ‘New Scientist‘ publication; and also Emmy Noether, who revolutionized the field of theoretical physics.
On the other side we have Sam (and all his minion friends, who are aptly called Sam-002, Sam-003, Sam-004) who is part human / part robot and plays Minecraft or watches Youtube, 24/7.
They agree to play a game where Emily thinks of a number, and then Sam (with the possible help of his minions) has 60 seconds to find any fractions that are equal to Emily’s number.
And so the game begins…
Emily says “8.5″.
Sam & friends quickly reply with “85/10″,… “34/4″,… “17/2″,… “425/50″,…
They soon realize that all these answers are equally valid because they are all equivalent
fractions. Being competitive they want to pick a single winner, so they all agree that the best answer is the one with the lowest denominator. And so, 17/2
is deemed the best answer.
This time, Emily tries to make it harder by picking ‘0.123456‘. After only a slight pause, Sam slyly says “123456/1000000“.
Emily’s annoyed with this answer. She knows that although the best answer would be the irreducible
fraction1929/15625, Sam’s answer is still valid answer, and furthermore he will always be able to instantly answer like this if she picks any number
with a terminating decimal expansion.
So this time, Emily picks “π”.
…
Delightful exploration of the idea that while all irrational numbers are irrational, some can be considered more-irrational than others if you consider the complexity of the convergent
series of fractions necessary to refine the representation of it. Some of this feels to me like the intersection between meta-mathematics and magic, but it’s well-written enough that I
was able to follow along all the way to the end and I think that you should give it a go, too.
Summary: if an idealised weight slides into another, bouncing it off a wall then back into itself, how many times will the two collide? If the two weights are the same then the answer
is 3: the first collision imparts all of the force of the first into the second, the second collision is the second bouncing off the wall, and the third imparts the force from the
second back into the first. If the second weight weighs ten times as much as the first, the answer turns out to be 31. One hundred times as much, and there are 314 bounces. One thousand
times, and there are 3,141. Ten thousand times, and there are 31,415… spot the pattern? The number of bounces are the digits of pi.
Why? This is mindblowing. And this video doesn’t answer the question (completely): it only poses it. But I’ll be looking forward to the next episode’s explanation…
Here’s a thought: what’s the minimum number of votes your party would need to attract in order to be able to secure a majority of seats in the House of Commons and form a government?
Let’s try to work it out.
The 2017 general election reportedly enjoyed a 68.8% turnout. If we assume for simplicity’s sake that each constituency had the same turnout and that votes for candidates other than
yours are equally-divided amongst your opposition, that means that the number of votes you need to attract in a given constituency is:
68.8% × the size of its electorate ÷ the number of candidates (rounded up)
For example, if there was a constituency of 1,000 people, 688 (68.8%) would have voted. If there were 3 candidates in that constituency you’d need 688 ÷ 3 = 229⅓, which rounds up to 230 (because you need the plurality of the ballots) to vote for your candidate in order to secure the seat. If there are
only 2, you need 345 of them.
It would later turn out that Barry and Linda Johnson of 14 West Street had both indented to vote for the other candidate but got confused and voted for your candidate instead. In
response, 89% of the nation blame the pair of them for throwing the election.
The minimum number of votes you’d need would therefore be this number for each of the smallest 326 constituencies (326 is the fewest number of seats you can hold in the
650-seat House of Commons and guarantee a strict majority; in reality, a minority government can sometimes form a government but let’s not get into that right now). Constituencies vary
significantly in size, from only 21,769 registered voters in Na h-Eileanan an Iar (the Western Isles of Scotland, an SNP/Labour marginal) to 110,697 in the Isle of Wight (which
flip-flops between the Conservatives and the Liberals), but each is awarded exactly one seat, so if we’re talking about the minimum number of votes you need we can take the
smallest 326.
Win these constituencies and no others and you control the Commons, even though they’ve tiny populations. In other news, I think this is how we end up with a SNP/Plaid coalition
government.
By my calculation, with a voter turnout of 68.8% and assuming two parties field candidates, one can win a general election with only 7,375,016 votes; that’s 15.76% of
the electorate (or 11.23% of the total population). That’s right: you could win a general election with the support of a little over 1 in 10 of the population, so long as it’s the
right 1 in 10.
I used a spreadsheet and everything; that’s how you know you can trust me. And you can download it, below, and try it for yourself.
I’ll leave you to decide how you feel about that. In the meantime, here’s my
working (and you can tweak the turnout and number-of-parties fields to see how that affects things). My data comes from the following Wikipedia/Wikidata sources: [1], [2], [3], [4],
[5] mostly because the Office of National Statistics’ search engine is terrible.
You are at a party and overhear a conversation between Lucy and her friend.
In the conversation, Lucy mentions she has a secret number that is less than 100.
She also confesses the following information:
“The number is uniquely describable by the answers to the following four questions:”
Q1) Is the number divisible by two?
Q2) Is the number divisible by three?
Q3) Is the number divisible by five?
Q4) Is the number divisible by seven?
…
I loved this puzzle. I first solved it a brute-force way, with Excel. Then I found increasingly elegant and logical solutions. Then I shared it with some friends: I love it! Go read the whole thing.
Imagine I’m in a desert, and there are two wells where I can obtain water. If I want to go to the nearest well, which well do I visit? Clearly, it depends one where I am standing.
It’s possible to draw a line dividing the desert. To the ‘left’ of the line, it’s nearer to go to the well on the well on the ‘left’, to the ‘right’ of the line, it’s closer to go to
the well on the ‘right’…
So, I’ve not been well lately. And because a few days lying on my back with insufficient mental stimulation is a quick route to insanity for me, I’ve been trying to spend my
most-conscious moment doing things that keep my brain ticking over. And that’s how I ended up calculating pi.
When I say I’ve been unwell, that might be an understatement. But we’ll get to that another time.
Pi (or π) is, of course, the ratio of the circumference of a circle to its diameter, for every circle. You’ll probably have learned it in school as 3.14, 3.142, or 3.14159, unless you
were one of those creepy kids who tried to memorise a lot more digits. Over the years, we’ve been able to calculate it to increasing precision, and although there’s no practical or theoretical reason that we need to know it beyond the 32 digits worked out by
Ludolph van Ceulen in the 16th Century, it’s still a fascinating topic that attracts research and debate.
Our calculation of pi has rocketed since the development of the digital computer.
Most of the computer-based systems we use today are hard to explain, but there’s a really fun computer-based
experimental method that can be used to estimate the value of pi that I’m going to share with you. As I’ve been stuck in bed (and often asleep) for the last few days, I’ve not
been able to do much productive work, but I have found myself able to implement an example of how to calculate pi. Recovery like a nerd, am I right?
Pi goes on forever. Pie, sadly, comes to an end.
Remember in school, when you’ll have learned that the formula to describe a circle (of radius 1) on a cartesian coordinate system is x2 + y2 = 1? Well you can work
this backwards, too: if you have a point on a grid, (x,y), then you can tell whether it’s inside or outside that circle. If x2 + y2 < 1, it’s inside, and if
x2 + y2 > 1, it’s outside. Meanwhile, the difference between the area of a circle and the area of a square that exactly contains it is π/4.
Think back to your school days. Ever draw a circle like this? Do the words “Cartesian coordinates” ring any bells?
Take those two facts together and you can develop an experimental way to determine pi, called a Monte Carlo
method. Take a circle of radius 1 inside a square that exactly contains it. Then randomly choose points within the square. Statistically speaking, these random points have a
π/4 chance of occurring within the circle (rather than outside it). So if we take the number of points that lie within the circle, divide that by the total number of
points, and then multiply by 4, we should get something that approaches the value of pi. You could even do it by hand!
I wrote some software to do exactly that. Here’s what it looks like – the red points are inside the circle, and the black points are outside.
The software illustration I’ve written is raw JavaScript, HTML, and SVG, and should work in any modern web browser (though it can get a little slow once it’s drawn a few thousand
points!). Give it a go, here! When you go to that page, your browser will start drawing dots at random points, colouring them red if
the sum of the squares of their coordinates is less than 1, which is the radius of the circle (and the width of the square that encompasses it). As it goes along, it uses the formula I
described above to approximate the value of pi. You’ll probably get as far as 3.14 before you get bored, but there’s no reason that this method couldn’t be used to go as far as
you like: it’s not the best tool for the job, but it’s super-easy to understand and explain.
Oh, and it’s all completely open-source, so you’re welcome to take it and do with it what you wish. Turn off the graphical output
to make it run faster, and see if you can get an accurate approximation to 5 digits of pi! Or slow it down so you can see how the appearance of each and every point affects the
calculation. Or adapt it into a teaching tool and show your maths students one way that pi can be derived experimentally. It’s all yours: have fun.
And I’ll update you on my health at some other point.
When I learned to program, back when dinosaurs walked the earth and the internet had no cats, there was an idea: if you were good at math, you’d be good at programming. I was great at
math as a kid, but perhaps because I didn’t like it much, no one steered me towards programming. I…
Have you ever come across non-transitive dice? The classic set, that you can get in most magic shops, consists of three different-coloured six-sided dice:
A “Grime’s” style set of 3 non-transitive dice. Notice the unusual numbering.
There are several variants, but a common one, as discussed by
James Grime, involves one die with five “3” sides and one “6” side (described as red below), a second die with three “2” sides and three “5” sides (described as
green below), and a third die with one “1” side and five “four” sides (described as blue below).
They’re all fair dice, and – like a normal six-sided dice – they all have an average score of 3.5. But they’ve got an interesting property, which you can use for all kinds of magic
tricks and gambling games. Typically: the red die will beat the green die, the green die will beat the blue die, and the blue die will beat the red die! (think Rock, Paper,
Scissors…)
Seemingly paradoxically, the dice will generally beat one another in a circular pattern.
If you want to beat your opponent, have them pick a die first. If they pick green, you take red. If they take red, you take blue. If they take blue, you take green. You now have about a
60% chance of getting the highest roll (normally you’d have about a 33% chance of winning, and a 17% chance of a draw, so a 60% chance is significantly better). To make sure that you’ve
got the best odds, play “best of 10” or similar: the more times you play, the less-likely you are to be caught out by an unfortunate unlucky streak.
But if that doesn’t bake your noodle enough, try grabbing two sets of nontransitive dice and try again. Now you’ll see that the pattern reverses: the green pair tends to beat the red
pair, the red pair tends to beat the blue pair, and the blue pair tends to beat the green pair! (this makes for a great second act to your efforts to fleece somebody of their money in a
gambling game: once they’ve worked out how you keep winning, give them the chance to go “double or nothing”, using two dice, and you’ll even offer to choose first!)
When you pair up the dice, the cycle reverses! While red beats green, double-green beats double-red!
The properties of these dice – and of the more-exotic forms, like Oskar van Deventer’s seven-dice set (suitable for playing a game with three players and beating both of your
opponents) and like the polyhedral
varieties discussed on Wikipedia – intrigue the game theorist and board games designer in me. Could there be the potential for this mechanic to exist in a board game? I’m thinking
something with Risk-like combat (dice ‘knock out’ one another from highest to lowest) but with a “dice acquisition” mechanic (so players perform actions, perhaps in
an auction format, to acquire dice of particular colours – each with their own strengths and weaknesses among other dice – to support their “hand” of dice). There’s a discussion going on in /r/tabletopgamedesign…
I’ve even written a program (which you’re welcome to download, adapt, and use) to calulate the
odds of any combination of any variety of non-transitive dice against one another, or even to help you develop your own non-transitive dice sets.
Penney’s game
Heads or tails? Image courtesy David M. Diaz.
Here’s another non-transitive game for you, but this time: I’ve made it into a real, playable game that you can try out right now. In this game, you and I will each, in turn, predict
three consecutive flips of a fair coin – so you might predict “tails, heads, heads”. Then we’ll start flipping a coin, again and again, until one of our sequences comes up. And more
often than not, I’ll win.
If you win 10 times (or you lose 20 times, which is more likely!), then I’ll explain how the game works, so you know how I “cheated”. I’ll remind you: the coin flips are fair, and it’s
nothing to do with a computer – if we played this game face-to-face, with a real coin, I’d still win. Now go play!
The other Three Ringers and I are working hard to wrap up Milestone:
Jethrik, the latest version of the software. I was optimising some of the older volunteer availability-management code when, by coincidence, I noticed this new bug:
Well, at least she’s being rational about it.
I suppose it’s true: Lucy (who’s an imaginary piece of test data) will celebrate her birthday in 13/1 days. Or 13.0 days, if you prefer. But most humans seem to be happier
with their periods of time not expressed as top-heavy fractions, for some reason, so I suppose we’d better fix that one.
They’re busy days for Three Rings, right now, as we’re also making arrangements for our 10th
Birthday Conference, next month. Between my Three Rings work, a busy stretch at my day job, voluntary work at Oxford Friend, yet-more-executor-stuff, and three different courses, I don’t have much time for anything else!
But I’m still alive, and I’m sure I’ll have more to say about all of the things I’ve been getting up to sometime. Maybe at half term. Or Christmas!
Like the TARDIS, your time machine has a fault. The fault isn’t a failure of its chameleon circuit, but a quirk in its ability to jump to particular dates. Picture courtesy
aussiegall (Flickr), licensed Creative Commons.
You own a time machine with an unusual property: it can only travel to 29th February. It can jump to any 29th February, anywhere at all, in any year (even back
before we invented the Gregorian Calendar, and far into the future after we’ve stopped using it), but it can only
finish its journey on a 29th of February, in a Gregorian leap year (for this reason, it can only jump to years which are leap years).
One day, you decide to take it for a spin. So you get into your time machine and press the “random” button. Moments later, you have arrived: it is now 29th February in a
random year!
Without knowing what year it is: what is the probability that it is a Monday? (hint: the answer is not1/7 – half of your challenge is to work
out why!).
