MathsJam November Inventory (Novemtory) Data

Age

We asked each MathsJam to provide an indication of the ages of the attendees – using whichever method they like, since we’re aware not everyone is willing to divulge this information. Some MathsJams, particularly those with lower attendance, were happy to give exact values, while others used clever methods to determine the average age without anyone having to give their own age.

As an example, the London MathsJam used a random seed to keep it all secret – the first person obtained a random number, and added it to their age. This total was passed quietly along the line of attendees, each of whom added their own age and passed on the new total. The final total was given back to the first person, who could subtract their random seed and divide the result by the number of people to obtain an average age.

• Average age of a MathsJam attendee: 32.37
• Estimated total age of all MathsJam attendees: 3398.4 years

Occupation

We’re interested to know what type of people attend a MathsJam – while it’s known that some teachers enjoy coming for the opportunity to share things they can use for classroom enrichment (and others like to get away from work, and remind themselves why they enjoy the subject!), we also wonder if the presence of a strong university-based contingent can affect the feel of the event. We also know some people who work in less mathematical careers might attend in order to indulge their hobby, while others who trained in maths but don’t get to use it any more will love being reminded of nice maths.

The results of this snapshot (admittedly only for one month’s attendance, and across the MathsJams which replied – 17 out of roughly 30 existing Jams), provides some surprising figures. We grouped the job titles given into categories, with anyone who’s a student or lecturer at a university in the same category, and IT/computing/developer jobs clumped together, while ‘science’ covers anyone in a technical or research role; this may contain people who should be in the ‘university’ category, but they weren’t specific enough.

Rough Demographic Breakdown of MathsJams:

Those Data Again As A Pie Chart, because Pie Chart:

Reminder Emails

Out of sheer curiosity, we investigated how well the MathsJam organisers are remembering to send their reminder emails before each event. In the previous sentence, read ‘sheer curiosity’ as ’wanting to know if we’re more or less organised than the average’. It would be interesting to investigate how promptness or otherwise of reminder emails impacts on attendance vs. average attendance, but we considered this dataset too small to be able to draw meaningful conclusions. Again, this is from the ones that replied.

• Average number of days before the MathsJam that a reminder email gets sent: 4.16
• MathsJam with highest average value: Nottingham; 6.33 days
• MathsJam with lowest average value: Edinburgh; 1 day (they literally always send it the day before, which is probably fine)

And the one you really want to know:

• MathsJam with highest standard deviation: Liverpool: 2.83 (tie for second place between Cardiff and Nottingham).

Attendance

Since we asked for the month each MathsJam started meeting, we could calculate the total number of monthly MathsJam meetings that have taken place ever, which we got as 272. This doesn’t include the MathsJams that didn’t reply, so the true total will be much higher, but there have been a few cancellations due to weather/organiser availability, which will bring it down a bit. We also calculated the average attendance – it turns out that November was pretty much exactly an average month, with 105 total attendees reported over all replying Jams, compared to a total average attendance of 104. For sheer stupidity, the estimated total number of occasions on which a person has attended a MathsJam (again, missing a lot of data).

• Mean monthly MathsJam attendance: 7.9 people
• Total number of Person-Jams: 2151

Gender

Even though mathematics as a subject is male-dominated, we’d hope that we’re a welcoming organisation and would maintain a relatively equitable gender split across MathsJammeurs and MathsJammeuses. We were pleased to note that we’re not totally overrun by gentlemen, and only two MathsJams (both small ones) reported a total absence of ladies.

• Mean gender split: 66% male (mean average)
• Lady-est MathsJam: Milton Keynes (although this was caused by a November attendance of 1 male and 2 females).

And for fans of conditional probability:

• $P(\text{You’re a man} \;|\; \text{You attend MathsJam}) = 65.71\%$

Daft Questions

We ended the questionnaire with some questions to get the attendees thinking, including one about the number of ping-pong balls that could fit in the pub (answers varied from ‘one big one’ to Portsmouth’s probably overly specific 27216000.5). We also asked MathsJams to estimate the number of dominoes in this bag:

• Correctest answer to the ‘bag of dominoes’ question: 253 (Newcastle – answer was obtained by averaging the guesses of all attendees)
• Wrongest answer to the ‘bag of dominoes’ question: 850 (London)
• Actual number of dominoes that were in that bag: 254
• Amount that we are suspicious of Newcastle MathsJam: a lot

Time

We also asked for approximate start and end times for each MathsJam, for three reasons: 1. to check there’s some kind of consistency in start times; 2. to make sure what it says on the MathsJam website is correct; and 3. to see who is the most hardcore. In acknowledgement of the fact that having two people in the pub doesn’t lend itself to a long session like having 20 people does, we’ve weighted the below based on the number of people who attended. London MathsJam, being the oldest and most well-established, scored both highest attendance and longest session, clocking in at 4 hours 24 minutes with 21 people. Most MathsJams begin at 7pm, with a few starting at 7.30pm, and the average finish time is 10.09pm.

Average time spent jamming: 3 hours 25 minutes (weighted for number of attendees, assuming everyone’s there at the start).

• if the two squares above are the same colour, use that colour for the square below;
• if the two squares above are different colours, use the third (unused) colour for the square below.

Proceeding by these rules, it turns out you can predict the colour of the bottom square by applying the rule above to the two squares at the ends of the top row. This impressed our MathsJam attendees, some of whom had had as few as one and a half pints at this stage. This rule works for a row of ten squares as a starter, but will it work for other sizes of triangle?

Steve, in collaboration with Ehrhard Behrends, has written an article about this in The Mathematical Intelligencer, which describes their use of it as a street maths show, as well as their conjecture about which values of triangle size will work. I’ve since found the full text of the article, although some of our attendees were far too impatient for that and wrote a proof that this will always work for numbers of the form conjectured to work in the first part of the article. Meanwhile, the rest of us continued to have fun playing with square bits of coloured paper.

Another item briefly touched on was that of the Picross, or Nonogram (also known as Hanjie or Griddlers), which is a logic puzzle in which a blank grid is given along with a set of numbers for each row and column. The numbers describe the sizes of the groups of squares in that row or column, and you can use this information to reconstruct an image by colouring in the right squares. So for example, if a row is labelled 1,2,1, it means there’s a gap of indeterminate width, then a single square, then another gap, then two squares together, then another gap, then one square, and a gap at the end. There must be at least one square between clusters of squares, and it’s possible to use this along with the width of the whole row, and data from the columns, to make deductions about what lies where, (and crucially, which squares must be blank – like in minesweeper).

We had printed some easy, medium and hard puzzles for people to have a go at, and my plan was to see if we could come up with a method to check whether a given grid of black and white squares, when represented using this row and column data, can be reconstructed by solving such a puzzle. This won’t be true of all layouts (for example, a chessboard layout of squares has two valid solutions). It turned out that one of the attendees has already written a program to do this, which killed that one dead in the water, but at least I get to borrow it.

Since it was apparently World Book Night (not sure how that differs from World Book Day, 8th March), I also brought along some recommended mathematical reading material, which we all leafed through and chatted about, before we tried some of the puzzles from Ian Stewart’s Cabinet of Mathematical Curiosities, and discussion fell to how it is you know that of the two guards, one only ever lies and one tells the truth? Is there a sign which explains this? And if so, which guard made the sign?

We finished by playing a massive game of Perudo (also called Liar’s Dice), which we’ve played before, and since games take a long time, that was the rest of our MathsJam (other than occasionally chipping in with other MathsJam puzzles from Twitter, and cramming our faces with birthday cake and sharing bags of crisps). All in all, a successful night.

In order to get closer to a proper egg-shape, we then busted out the rulers and compasses, using a method I found using arcs to construct lovely egg shapes. We also had coloured pencils to colour them in, but everyone was distracted at this point by the arrival of 24 Cadbury’s Creme Eggs, brought by one of our regulars as a treat for everyone. This sparked off a hilarious conversation about Creme Eggs, inspired by the question: given the recommended daily maximum of 2500 calories, how many Creme Eggs could the average human consume in a lifetime?

MAN: Incidentally, our earlier Creme Egg musings were inspired by the fact that @diffractionman brought 24 for us. twitter.com/MathsJam/statu…

— Maths Jam (@MathsJam) March 19, 2013

We calculated an answer using estimates as around 380,000 Creme Eggs, which is roughly 187 times the average human bodyweight – however, we also decided that eating nothing but Creme Eggs daily from birth would probably mean you would neither live as long nor weigh an average amount. You would also probably feel sick all the time, not to mention the issues around Creme Eggs not being on sale all year round. Can you freeze them?

We also briefly considered this: If you count one Creme Egg as taking one hour off your life expectancy, and you ate them at 2500 calories-worth per day (about 14 Creme Eggs), would there come a point when you’d know the next Creme Egg you eat will kill you? It later transpired this isn’t hugely well-defined, since you’d need to know how long your life expectancy was to start with.

We finished our eggy maths chatter by discussing the ‘White/Yolk Theorem’, which it turns out isn’t called that but is in fact called the ‘Ham Sandwich Theorem‘. This also inspired brief discussion of the Pizza Theorem. What’s your favourite mathematical result named after a foodstuff?

Moving on from Easter-themed maths, we played games of SET and a new Gigamic game called Gobblet, which I found at a university Maths Arcade and bought in the closing-down sale of the IRL branch of an excellent board game shop in Greenwich. The game involves making a line of four in your own colour, but some pieces can be placed over the top of others and change their colour, and only big pieces can be placed over smaller pieces, making the game have quite an interesting strategy challenge.

One of our new attendees challenged us to determine the value of $\displaystyle\lim_{n \rightarrow \infty} n – \sqrt{n^2 + n}$. It’s not zero!

We looked at this probability puzzle, which was tweeted by @MathUpdate:

How many times should you draw a random number from [0,1] to have it sum over 1? sns.mx/aylty2

— Math Update (@MathUpdate) February 25, 2013

Having announced immediately what he thought the answer might be, Paul and others wrote a quick Python program and determined they were right. I then discovered that we had a copy of the book it’s taken from in the bottom of the MathsJam box. We also tried these domino puzzles, in which you have to fit all the dominoes into the grid using logic and a pen; and we looked at this matchstick puzzle from Leicester MathsJam, coming up with some slightly strained definitions of ‘moving’ and ‘equilateral’. I was quite impressed by the solution.

Other MathsJams went really well this month too, with Cardiff and Newcastle communicating via Crisp Tube Enigma, and new startup Winnipeg MathJam enjoyed being in the same timezone as Washington DC MathJam and had a nice chat via Twitter.

A MathsJam is an informal gathering of people with some kind of interest in maths, from teachers to professionals to students to authors to engineers to interested laypeople to passers-by who want to know what’s going on with the Pringles tubes. Puzzles and games are the order of the evening, and you definitely don’t need to know your $x$ from your $\times$ to win at Set. There’s also an annual conference in Cheshire in November.

You can find details of your local MathsJam at MathsJam.com, or by asking the mysterious @MathsJam twitter account. If there isn’t one near you… why not set one up? Email london@mathsjam.com to find out more.

More information

MathsJam.com

Newcastle and Manchester MathsJam recaps

Map of current and potential MathsJams

We started off with some quick mental arithmetic brainteasers: how many straight cuts do you need to make to slice a flat square cake into 196 equally sized square pieces? Several people got the answer quite quickly, while others tried to cheat by stacking cake pieces and moving them around between cuts. No cheating!

Various people’s mathematical Christmas presents were present, including a tangram puzzle, a set of ‘Crazy Four’ in which you have to arrange four cubes in a line so that each face has the same colour all the way along (frustrating!), and a ridiculous Guinness World Record Domino Challenge set which was given to some of the Domino Computer volunteers as a hilarious sarcastic Christmas present, given that they’d already stacked enough dominoes to never want to see one again. The set claimed that the world record for placing horizontally flat dominoes in a vertical stack, one on top of the other, using one hand, was 19 dominoes in 30 seconds. We managed to beat this in a pub on a slightly wobbly table, so we’re not sure if this is a world record that will stand for long. It doesn’t appear to be listed on the Guinness website, although the Amazon listing for the product confirms in the description that the record is 19, as does the box. We concluded that this was just a ploy to sell Christmas gifts, especially as the record was apparently set in the Guinness World Records office by someone who presumably works there. It was also a good excuse to explain the Domino Computer to anyone who hadn’t heard about it, and relive former glories.

We discussed several problems from Futility Closet, an increasingly useful reference for MathsJam puzzles, including one about soldiers with different injuries (we agreed it needed restating in a less depressing way, possibly involving cake) and one about a belt around three pulleys, which took some people a while to realise there’s a nice shortcut.

We also tried to work out the trick in this probability teaser, in which the following game is played:

You’re about to play a game. A single person enters a room and two dice are rolled. If the result is double sixes, he is shot. Otherwise he leaves the room and nine new players enter. Again the dice are rolled, and if the result is double sixes, all nine are shot. If not, they leave and 90 new players enter.

And so on, the number of players increasing tenfold with each round. The game continues until double sixes are rolled and a group is executed, which is certain to happen eventually. The room is infinitely large, and there’s an infinite supply of players.

The puzzle asks, if you’re selected to enter the room, how worried should you be? Not particularly: Your chance of dying is only 1 in 36. Later your mother learns that you entered the room. How worried should she be? Extremely: About 90 percent of the people who played this game were shot.

It took us some time to work out what the difference was in the information your mother has, and the information you have, since each of the conclusions is valid from the perspective of the relevant person. It’s a nice illustration of how counterintuitive probability can be!

There was a good discussion of this train puzzle, which some people had seen before and others worked out slowly – which descended eventually into a discussion of countable and uncountable infinities, and why infinity squared (if you’re allowing it to be a thing) is the same as infinity. Minds were blown, cake pops were eaten.

We experimented with making Menger Sponges from cards (a Manchester MathsJam staple) although this time Katie had printed some cards up with a Sierpinski Carpet design, which meant the sponge pattern went down to extreme levels of detail. Since we were limited in time and ability to be bothered, we made single cubes, although the cards could easily be used to cover the outside of a sponge made of 20 cubes.

Ross, who’s been a recent addition to Manchester’s attendance, brought a puzzle he’d invented, involving six dice. You roll the dice, and then re-roll any subset of the dice which contains duplicates (so, for example, rolling 134466, you’d re-roll the 4s and 6s), and then repeat this process, looking at the whole set each time for duplicates. If you ever find you have 6 unique values (1 to 6), that means you win, and if you ever get to a point where you’d have to re-roll all 6 dice (e.g. if they’re in three pairs, or two triples, or all the same) then you lose. Are you more likely to win or lose?

Various on-paper calculations were done, several confident-sounding fractions were thrown around, and the puzzle was discussed at other MathsJams including the newly-restarted Washington DC MathsJam, which took place a few hours after we’d finished and gone to bed. I’m still not sure what the answer is, but I’ll work it out when I get a minute.

On a night when several other MathsJams were cancelled or curtailed due to the horrible weather, I’m glad we managed to get together and it was good to see both old and new faces! And cake.

I am 64¼ years old and I’ve been a maths teacher all my working life. In that time things have changed. Long gone are the days when gowned masters would sweep in, silence any murmur with half a raised eyebrow, and delight compliant uniformed schoolchildren with chalk-covered boards of mathematical exposition.

No, you’re right. That never happened outside the covers of Goodbye Mr Chips, even in my day.

The reality then. Schoolchildren have morphed into learners. Exam results rule. Quality (in the sense that Orwell might have used the word) is managed by quality managers. And so our working lives are driven by the pursuit of Ofsted targets, success rates, achievement rates, benchmarks, observation grades, results. And every joyless lesson has its own lesson plan, with aims, objectives, learning outcomes and action points. But above all, those damned results – and every year, year after year, they had to IMPROVE.

Well I was no good at any of this stuff – and consequently I always got on very badly with my managers. Until one year…

AS Maths results

It had been my job to produce an annual report on the progress or otherwise of the maths department at my college, and to give a presentation on this to management. In 2007 I prepared my stats for AS “high grades” (that is, grades A or B). The table below shows the percentages achieving high grades in 2007, and 2006 for comparison.

20072006

Quite clearly I’m in trouble. The results have failed to IMPROVE. And I know it will not help my defence if I appeal to sporadic fluctuation or regression to the mean. Or even tell the truth – that is, that in 2007 we had a not so good group of students (sorry, learners).

Then I had a great idea. I would split the group into those who had taken a Mechanics module, and those who’d taken a Statistics module as part of their AS Maths. What happened was absolutely amazing!

Here is the revised version.

20072006

Both the split groups IMPROVE.

I’m happy. Management’s happy. Everyone’s a winner.

This was my eureka moment. But it set me thinking.

What if splitting the group in this way hadn’t worked to my advantage? Could I have got the favourable outcome I wanted by choosing different “splitting criteria”. For example – gender, those who wear glasses and those who don’t, those who watch Strictly Come Dancing and… (you get the idea!).

Batting averages

Four years later I came upon another weird thing. I was trying to decide which of two Leicestershire county cricketers, Jigar Naik or Tom New, was the better batsman. (Why? Don’t ask!)

Here are their batting averages for 2010 and 2011.

	2010			2011
	Runs	Outs	Batting Average	Runs	Outs	Batting Average
Jigar Naik	301	9	33.4	545	25	21.8
Tom New	746	23	32.4	412	19	21.7

Clearly it’s close, but Jigar Naik consistently seems to have the edge both seasons. Based on this evidence, the better batsman is undeniably Jigar Naik. No one would disagree with that.

No one – except possibly Tom New. He might argue that it would have been fairer to combine the data for the two years. And that when you do this, the apparently obvious conclusion is turned completely on its head.

2010 and 2011 combined
	Runs	Outs	Batting Average
Jigar Naik	846	34	24.9
Tom New	1158	42	27.6

Simpson’s Paradox

Both these wondrous anomalies are manifestations of Simpson’s Paradox. Surely this has to be the greatest gift ever to practitioners of the dark art of statistical conjuring.

How it works is like this.

All we do is find integers $a, b, c, d, p, q, r, s$ such that

$\displaystyle{\frac{a}{b} \gt \frac{p}{q}}$ and $\displaystyle{\frac{c}{d} \gt \frac{r}{s}}$ but $\displaystyle{\frac{a+c}{b+d} \lt \frac{p+r}{q+s}}$.

To go back to the two examples, you can see that Simpson’s Paradox “works” in both cases if you convert the percentages back into fractions.

AS results:

$\displaystyle{ \frac{18}{57} \lt \frac{12}{34} }$ but $\displaystyle{ \frac{5}{9} \gt \frac{6}{11} }$ and $\displaystyle{ \frac{13}{48} \gt \frac{6}{23} }$.

Batting averages:

$\displaystyle{ \frac{301}{9} \gt \frac{746}{23} }$ and $\displaystyle{ \frac{545}{25} \gt \frac{412}{19} }$ but $\displaystyle{ \frac{846}{34} \lt \frac{1158}{42} }$.

A simple example, represented graphically

The simplest example is

$\displaystyle{ \frac{1}{2} \gt \frac{3}{7} }$ and $\displaystyle{ \frac{1}{5} \gt \frac{1}{6} }$ but $\displaystyle{ \frac{2}{7} \lt \frac{4}{13} }$,

and I’ll use this example to illustrate Simpson’s Paradox in a nice graphical way.

Take each fraction $\frac{a}{b}$ to be the gradient of the line from the origin to the point $(b,a)$. Then the $\frac{2}{7}$ line is a vector with components $\frac{1}{5}$ and $\frac{1}{2}$, and similarly the $\frac{4}{13}$ vector has components $\frac{3}{7}$ and $\frac{1}{6}$.

So the sketch below shows that the ordering of the gradients of the resultant vectors is reversed by the ordering of the gradients of the corresponding component vectors.

Can you make Simpson’s Paradox happen? And if so, how?

Thinking about the AS Maths Results example, a couple of questions come to mind.

“Can I always find a way of splitting the teaching group into smaller groups such that the original conclusion is reversed?” (The criterion for assigning the smaller groups can of course be decided afterwards!)

and

“If I can’t always do this, then under what conditions can I? and how?”

I would love to know the answers to these questions.

Case studies

A wonderfully entertaining case study is to be found on Alan Crowe’s blog.

I’m only sad that it isn’t true!

But there are many well-documented instances of Simpson’s Paradox that have found their way into academic publications.

1975 – Sex Bias in Graduate Admissions: Data from Berkeley, P.J. Bickel, E.A. Hammel, J.W. O’Connell.

1986 – Success rates of two treatments for kidney stones, C.R. Charig, D.R. Webb, S.R. Payne, J.E. Wickham.

2004 – Baseball batting averages, Ken Ross.

2006 – Low birth rate for mothers who smoke, Allen J. Wilcox.

A bit of background

Edward H. Simpson had been a codebreaker at Bletchley Park during the Second World War. The phenomenon that bears his name has been widely known since the publication in 1951 of the paper The Interpretation of Interaction in Contingency Tables. But Karl Pearson (of χ-squared and correlation coefficient fame) had written in 1899 about the paradox in a paper entitled Genetic (reproductive) selection: Inheritance of fertility in men.

Two questions

But at the end of it all, what I really want to know is…

“Who was the better batsman?”

and

“Did my AS Maths results improve?”

Sources

Impossible? Surprising solutions to counterintuitive conundrums by Julian Havil (Princeton University Press)

Wikipedia

Here’s a collection of clips we recorded while people were digesting both their dinners and the first day’s talks.

And here are Pat Ashforth and Steve Plummer of Woolly Thoughts, showing me their fantastic illusion knit.

Thanks to Katie for drawing the short straw and editing these clips together.

The formula for the surface area of a sphere, $A=4\pi r^{2}$, is the derivative of the formula for the volume of a sphere: $V=\frac{4}{3}\pi r^{3}$.

This result does not hold for a cube with side length $a$ if the surface area and volume are written in terms of $a$. However, if the surface area and volume are written in terms of half the side length, $r=\frac{a}{2}$, you get the surface area $A=24 r^{2}$, which is the derivative of the volume, $V=8 r^{3}$.

If you apply a similar technique to a tetrahedron with side length $a$, you need to write the surface area and volume formulae in terms of $r=\frac{\sqrt{6}}{12}a$.  This gives $A=24\sqrt{3} r^{2}$ which is the derivative of the volume, $V=8\sqrt{3} r^{3}$.

$\frac{a}{2}$ is the inradius (radius of the insphere) of a cube with side length $a$, and $\frac{\sqrt{6}}{12}a$ is the inradius of a tetrahedron with side length $a$. This leads to a marvellous result:

The formula for the surface area is the derivative of the formula for the volume if they are both written in terms of the inradius.

The result holds for all the platonic solids. For example an octahedron with side length $a$ has inradius $r=\frac{\sqrt{6}}{6}a$. This gives surface area $A=12\sqrt{3} r^{2}$, which is the derivative of the volume, $V=4\sqrt{3} r^{3}$.

I think it also works for any polyhedron whose faces are all tangent to a single sphere.

The equivalent result holds in 2D for regular polygons: the formula for the perimeter is the derivative of the formula for the area if they are both written in terms of the radius of the incircle. I think this works for non-regular polygons too, provided that all the sides are tangents to the same circle.

The original version of this was presented at the annual MathsJam conference, November 2012.  David Fontaine’s table of properties of platonic solids was very useful.