I made a silly joke, and it made me think.

You may be aware that our own Christian Lawson-Perfect is running the Big Internet Math-Off here at the Aperiodical, a single-elimination tournament with sixteen competitors. I was knocked out in round one by the brilliant Alison Kiddle. I joked that if Alison went on to win, then I’d be joint second.

Much as I like and respect @ch_nira, I’ll be rooting for @ajk_44. If she goes on to win the #BigMathOff final and is crowned The World’s Most Interesting Mathematician, then I’m joint-second, right? https://t.co/8Jt37gHFif

— Peter Rowlett (@peterrowlett) July 10, 2018

I’ve been mulling this over and I felt there was something there in thinking about the placement of the non-winners in such a tournament, so I had a play.

I started by doing the only sensible thing: I used an online random name generator to generate sixteen names and arranged them into eight first-round matches. Lacking an actual competition and with the Big Internet Math-Off in recent memory, I decided each match by generating a random number $0$-$100$ for one opponent and taking $100$ minus this for the other. Think of it, if you like, as percentage of a public vote. Thankfully, there were no draws, so I didn’t have to invent a tiebreaker rule.

Here’s the outcome, written up using this lovely LaTeX tournament code example.

Clearly, Michal wins, by winning the final. But who came second? Norah lost the final, so clearly has a claim to being second. She won one fewer match than did the winner.

But what about Betsy? Michal beat Betsy, so if we choose to believe the fiction of the tournament, Betsy is worse (at being assigned a random number $>50$) than Michal. But is she better or worse that Norah? Think about it. I’d say we don’t know, because this hasn’t been tested. We know Michal is better than Betsy, and Michal is better than Norah, but whether Betsy is better or worse than Norah is untested. All we can do, therefore, is award them joint second place. They both lost one match, to Michal, and they didn’t play each other.

But there are a bunch of other people who also lost only one match to Michal and haven’t played each other. If they hadn’t been knocked out, perhaps they could have gone on to win, or at least prove themselves better than one of our second-place candidates. So perhaps they are all joint second. The people Michal knocked out are: Norah, Betsy, Camilla and Maia.

By the same logic, if you were knocked out by a second-place competitor, you might have a claim to being joint third. As Michal beat Norah and Norah beat Yasmin, there’s a clear ordering here that puts Yasmin third. But Siobhan, who was beaten by joint-second-place Betsy, hasn’t played Yasmin, so I don’t think we can reasonably say Yasmin is better (at being assigned a random number $>50$) than Siobhan. The people beaten by a second-place competitor are: Yasmin, Iris and Amirah (knocked out by Norah), Cathy and Hiba (knocked out by Betsy) and Abigail (knocked out by Camilla).

Who was knocked out by a third-place competitor? Lacie and Judy (knocked out by Yasmin), Niamh (knocked out by Iris) and Siobhan (knocked out by Cathy). By the same logic, these are joint fourth place.

There is now only one competitor not placed: Agnes, who was knocked out by Lacie. An argument could be constructed to put her as the only competitor in fifth place.

Here’s a diagram showing the tournament with these placements.

So in the **final**, we have two competitors: one in first place, one in second.

In the **semi-finals**, four: one first, two second and one third. That is, the two from the final are there in their original positions, plus they each knock out one competitor who is, therefore, placed one less than them.

In the **quarter-finals**: one first, three second, three third and one fourth.

In the **first round**: one first, four second, six third, four fourth and one fifth.

Round | 1st | 2nd | 3rd | 4th | 5th |
---|---|---|---|---|---|

Final | 1 | 1 | 0 | 0 | 0 |

Semi | 1 | 2 | 1 | 0 | 0 |

Quarter | 1 | 3 | 3 | 1 | 0 |

1st | 1 | 4 | 6 | 4 | 1 |

Do we recognise these numbers? Labelling from the final backwards, at each stage we keep the competitors from the stage that happens after it in their places and add as many again that they knocked out, each one place below them. So, e.g., the number of second place competitors in the semi-final is the sum of the number of first and second place competitors in the final. In general, the number in column $m$ in row $n$ is made from the number who were in column $m$ in row $n-1$ plus the number in column $m-1$ in row $n-1$. We are making Pascal’s triangle.

I don’t think about single-elimination tournaments very much (I didn’t know the term until very recently), probably because I have little-to-no interest in sport. Is this result obvious to everyone? I appreciate it’s not the most practical way to look at a tournament — I’m not sure that anyone would seriously buy my argument that someone who rocked up in round one and won nothing, but happened to be playing the eventual winner, deserves joint second-place with the person who won three matches to be knocked out in the final. But as a bit of fun with logic, I thought it was quite nice.

Sadly, Alison was knocked out in the second round by Nira Chamberlain, so, depending on Nira’s performance in later rounds, the best I can hope for is joint-third place, and I might yet be fifth.

]]>This claim was famously made by Carl Sagan in the seminal programme Cosmos.

The cosmos is rich beyond measure. The number of stars in the universe is larger than all the grains of sand on all the beaches of the planet Earth.

More or Less come to a fairly standard answer, that Sagan was correct. This sort of problem, which involves approximating unknowable numbers based on a series of estimates, is called a Fermi problem. I’ve written about Fermi problems here before. The More or Less approach to answering this raised a question from a reader of this blog.

But that's less than a factor of 3 difference! For Fermi estimates of numbers of that size, those two answers are essentially the same. It wouldn't take much of an error in either estimate to push sand ahead of stars…

— Paul Taylor (@aPaulTaylor) July 8, 2018

Alright, actually Paul is one of the writers of this blog, rather than a reader. Even so, are his concerns warranted?

The numbers in the More or Less piece are

- for stars: “1 followed by 22 zeros” (approx. 4:18 in the podcast audio), described as 10 sextillion;
- for sand: “3.65 with 21 zeros after it” (approx. 8:30), described as about 4 sextillion.

Fermi estimation involves making order-of-magnitude estimates based on informed guesses about the inputs. (Listen to the More or Less piece if you want to hear this process in action.) Then, given the level of estimation, you aren’t supposed to take too seriously differences within the same order of magnitude – 3 million and 5 million can’t be taken to be seriously different – but if one number comes out at least an order of magnitude bigger than the other, you take this to be your conclusion. In that sense, what More or Less did was correct – 10 sextillion is one order of magnitude bigger than 4 sextillion and so, we conclude, bigger.

There is an uncomfortable feeling which comes from how similar these numbers are, given the amount of guessing involved. Something about comparing so-many sextillions and some-other-number-of sextillions seems like there isn’t that much difference. Actually, the difference between $4 \times 10^{21}$ and $1 \times 10^{22}$ is $6 \times 10^{21}$, so we’d need More or Less to have missed out more than half the sand on Earth in estimating to have come to the wrong conclusion.

There are people who’ll tell you that the only number to pay attention to in a Fermi estimate is the $10^n$ term, because the other is just noise. But this leads to awkwardness. In this case, $4 \times 10^{21}$ is four times bigger than $1 \times 10^{21}$ but only two-and-a-half times smaller than $10^{22}$. Can we really say $1 \times 10^{21}$ and $9 \times 10^{21}$ are effectively the same, while $10 \times 10^{21}$ is significantly bigger?

It’d certainly be a clearer conclusion if the numbers came out several orders of magnitude different, but then obviously very differently-size numbers wouldn’t make for such an interesting piece of rhetoric, would they? The real question you have to ask, then, is whether any doubts you have about the way each number was estimated amounts to a big enough difference in either number.

I guess it is also possible to think about this estimation method too seriously — it’s only really designed to give a rough-and-ready, back-of-the-envelope guesstimate. There’s something about the way the More or Less piece suddenly comes to a conclusion and ends, with a definitive-sounding “Carl Sagan was right” and “he nailed it”. I’d say a few more caveats would have been appropriate. Some discussion of upper and lower bounds, or just an acknowledgement of how much error would cause the conclusion to flip, might have been nice. Paul points out, and I think he’s right, that the fact these numbers are pretty close is remarkable, and perhaps this could have been remarked upon. So, in conclusion, I don’t think the More or Less conclusion is necessarily wrong, but it could have been framed better.

]]>Yesterday morning, my son and I did something similar with our cat, Tabby. This is in response to Matt Parker’s latest initiative, Psychic Pets. Matt is hoping to get thousands of pet owners to make predictions, in order that the odds are good a pet can be found which predicted all prior results for both teams in the final. The good news is it’s fairly straightforward to take part.

Matt’s made a video which explains.

And, for readers who are teachers, there are teaching resources related to the project from Think Maths.

I figure there are teams people Matt reaches are more likely to pick and some that are likely to be less well-picked, and so chose Peru to help Matt get good coverage and because their first match was later the same day. I wrote out sheets with team names and ‘draw’, and offered Tabby the choice of each combination of sheets with a cat biscuit on it. (I’m not completely on top of the details of the World Cup, but the Psychic Pets website has all the info.) Tabby is the sort of cat who will do almost anything for a cat biscuit, even take part in one of Matt’s silly projects. She made her basic three predictions, which I videoed and uploaded to the site.

Made a contribution for @standupmaths’ latest stupid project, @PsychicPets.

You can do it too! https://t.co/yGgW5rI6YB pic.twitter.com/mmHRrmFH9Z— Peter Rowlett (@peterrowlett) June 16, 2018

Well, it didn’t take long to learn that her first prediction, Peru for the win, wasn’t to be. Pat Parslow pointed out on Twitter that this doesn’t prove my cat isn’t psychic — she could be toying with me. But, either way, it doesn’t help advance Matt’s cause.

But, while it’s too late for Tabby, do you have a pet and could get involved? Get started at Psychic-Pets.com.

]]>Enrico Fermi apparently had a knack for making rough estimates with very little data. Fermi problems are problems which ask for estimations for which very little data is available. Some standard Fermi problems:

- How many piano tuners are there in New York City?
- How many hairs are there on a bear?
- How many miles does a person walk in a lifetime?
- How many people in the world are talking on their mobile phones right now?

Hopefully you get the idea. These are problems for which little data is available, but for which intelligent guesses can be made. I have used problems of this type with students as an exercise in estimation and making assumptions. Inspired by a tweet from Alison Kiddle, I have set these up as a comparison of which is bigger from two unknowable things. Are there more cats in Sheffield or train carriages passing through Sheffield station every day? That sort of thing.

The point of these is not to look up information or make wild guesses, but instead to come up with a back-of-the-envelope, ‘wrong, but useful‘, orders of magnitude estimate. Some ‘rules’, if you want to play with these the way I would:

- don’t look up information;
- don’t make precise calculations using calculator or computer;
- be imprecise — there are 400 days in a year, people are 2m tall, etc.;
- round numbers where possible and calculate in your head.

One approach is to estimate by bounding – come up with numbers that are definitely too small and too large, and then use an estimate that is an average of these. But which average?

Say I think some quantity is bigger than 2 but smaller than 400. The arithmetic mean would be $\mathrm{AM}(2,400)=\frac{2+400}{2}=201$. The geometric mean would be $ \mathrm{GM}(2,400)=\sqrt{2\times 400} = 28.28\!\ldots$.

Which is a better estimate? The arithmetic mean is half the upper bound, but 100 times the lower bound. On this basis, for an ‘order of magnitude’-type estimate, you might agree that the geometric mean is a better average to use here. Following my Maths Jam talk, Rob Low said that the geometric mean makes more sense for an order of magnitude estimate, since it corresponds to the arithmetic mean of logs. To see this, consider \[

\begin{align*}

\log(\mathrm{GM}(A, B)) &= \log(\sqrt{AB}) \\

&= \log((AB)^{\frac{1}{2}}) \\

&= \frac{1}{2}\log(AB) \\

&= \frac{1}{2}(\log(A) + \log(B)) = \mathrm{AM}(\log(A), \log(B)) \text{.}

\end{align*}

\]

So, geometric mean it is. However, taking a square root is not usually easy in your head, and we want to avoid making precise calculations by calculator or computer. Enter the approximate geometric mean.

For the approximate geometric mean, take $2=2 \times 10^0$ and $400=4 \times 10^2$, then the AGM of $2$ and $400$ is: \[ \begin{align*}

\frac{2+4}{2} \times 10^{\frac{0+2}{2}} &= 3 \times 10^1\\

&= 30 \approx 28.28\!\ldots = \sqrt{2\times 400} = \mathrm{GM}(2,400) \text{.}

\end{align*} \]

Why does this work? Let $A=a \times 10^x$ and $B=b \times 10^y$. Then \[

\begin{align*}

\mathrm{GM}(A,B)=\sqrt{AB}&=\sqrt{ab \times 10^{x+y}}\\

&=\sqrt{ab} \times 10^{\frac{x+y}{2}} \text{,}

\end{align*}

\]

and \[\mathrm{AGM}(A,B) = \frac{a+b}{2} \times 10^{\frac{x+y}{2}}\text{.}\]

Setting aside the $10^{\frac{x+y}{2}}$ term, which appears in both averages, is it obvious that, for single digit numbers $>0$, \[\mathrm{GM}(a,b)=\sqrt{ab} \approx \frac{a+b}{2}=\mathrm{AM}(a,b) \text{?} \]

There is a standard result that says \[ \begin{align*}

0 \le (x-y)^2 &= x^2 – 2xy + y^2\\

&= x^2 + 2xy + y^2 – 4xy\\

&= (x+y)^2 – 4xy \text{.}

\end{align*} \]

Hence \[ \begin{align*}

4xy &\le (x+y)^2\\

\sqrt{xy} &\le \frac{x+y}{2} \text{,}

\end{align*} \]

with equality iff $x=y$. So $\mathrm{GM}(a,b)\le\mathrm{AM}(a,b)$, but are they necessarily close?

By exhaustion, it is straightforward to show (for single-digit integers, given the rule to round numbers where possible) that the largest error occurs when $a=1$ and $b=9$. Then \[ \sqrt{1 \times 9} = 3 \ne 5 = \frac{1+9}{2} \] and the error is $2$ which, relative to the biggest number $9$ might be seen as quite significant.

I’d say you are not likely to use this method if the numbers are of the same order of magnitude, because the idea is to come up with fairly wild approximations and if they were quite close it might be sensible to think of them as not really different. Then the error is going to be at least one order of magnitude smaller than the upper bound, i.e. $10^\frac{x+y}{2} \ll 10^y$. For example, if your numbers were $1$ and $900$ (as a pretty bad case), then: \[ \mathrm{GM}(1,900)=\sqrt{900}=30 \ne 50=\mathrm{AGM}(1,900) \] and a difference of $20$ on a top value of $900$ is not as significant as a difference of $2$ was on a top value of $9$.

So I suppose I would argue that this makes the error relatively insignificant. However, this thinking left me somewhat unsatisfied. I felt there ought to be a nicer way to demonstrate why the approximate geometric mean works as an approximation for the geometric mean. Following my talk at Maths Jam, Philipp Reinhard has been thinking about this, and he will share his thoughts in a post here in a few days (the post is now online).

I didn’t have time to fit into my talk what I would recommend if the two numbers differed by an odd number of orders of magnitude. For example, $\mathrm{AGM}(1,1000)$ generates another square root in $1 \times 10^{\frac{3}{2}}$ – precisely what we were trying to avoid! What I have recommended to students is to simply rewrite one of the numbers so that the difference in exponents is even. For example, writing $1=1 \times 10^0$ and $1000 = 10 \times 10^2$ gives \[\mathrm{AGM}(1,1000)=5.5 \times 10^{1} \text{.}\]

Following Maths Jam, the esteemed Colin Beveridge made the sensible suggestion of just treating $10^{\frac{1}{2}}$ as $3$, making \[

\begin{align*}

&\mathrm{AGM}(1,1000)\\

&= 1 \times 10^{\frac{3}{2}}\\

&\approx 1 \times 3^3 = 27\text{.}

\end{align*}

\]

This increases our problems, though, because we have the potential to deal with larger differences (hence larger errors) than when dealing with single-digit numbers. Actually, it was wondering why this increased error happens that got me thinking seriously on this topic in the first place. I’ll stop now to let Philipp share what he has been thinking on this.

]]>I just noticed that last Wednesday was ten years since that lecture. It was basic maths for forensic science students. I was given a booklet of notes and told to either use it or write my own (I used it), had a short chat about how the module might work with another lecturer, and there I was in front of the students. That was spring in the academic year 2007/8 and this is the 21st teaching semester since then. This one is the 15th semester during which I have taught — the last 12 in a row, during which I got a full-time contract and ended ten years of part-time working.

I have this awful feeling this might lead people to imagine I’m one of the people who knows what they are doing.

P.S. The other thing that I started when I started working for the IMA was blogging – yesterday marks ten years since my first post. So this post represents the start of my second ten years of blogging.

]]>The first will reward a well-made, delicious item; the second will reward the item which has been decorated the most beautifully and looks most like what it’s supposed to be; and the third will reward the most ingenious mathematical theming.

You can view the entries from this year on the MathsJam website.

The other regular competition is the Competition Competition. This invites attendees to submit a competition, which other attendees can enter. There are some rules, including minimum font size, paper size and maximum value of prize. To be clear, the rules say “any type of competition is permitted as long as it can be judged by the setter (or a winner randomly chosen from the correct entries)”.

Prizes are awarded for best competition, popular vote winner (the competition with the most entries) and “best attempt at circumvention of the rules while still strictly sticking to the rules”. Seeing people attempt the latter is quite delightful.

Chatting to people at MathsJam this year, I was reminded of my entry into the Competition Competition when it first ran in 2014. I invited entrants to write down an integer between 0 and 100, then I said that I would run a Shapiro-Wilk test of normality on the numbers people had written down. This tests the null hypothesis that the data come from a normally distributed population. The competitive element of the competition asked people to guess the p-value obtained from that test.

While we were chatting about this, The Aperiodical’s own Paul Taylor asked me what p-value came out as. I couldn’t remember, but I’ve looked it up. The numbers entered were as follows:

Number entered |
Number of people entering it |
---|---|

2 | 1 |

6 | 1 |

12 | 1 |

16 | 1 |

50 | 1 |

71 | 1 |

72 | 2 |

73 | 2 |

86 | 2 |

97 | 2 |

99 | 7 |

In a sample of 21 integers from 1-99 where only four numbers are below 50 and one third are precisely 99, it may not surprise you to learn that the statistical test gave strong evidence to reject the hypothesis that these data were from a normally distributed population. The p-value (from R) was 0.0002211865 and the winner was Francis Hunt, who guessed 0.0001.

You can find out about the MathsJam competitions and other side activities that take place on the MathsJam Extras page.

]]>The MTaP carnival traditionally starts with a puzzle, game or piece of trivia. As this edition marks the start of a new year, I thought I’d share a new year puzzle I played with my students in the last session before the Christmas break.

Good news everyone – the 2017 game works (though it's easier than previous years, I think). pic.twitter.com/VRyfQ1bWym

— Peter Rowlett (@peterrowlett) December 12, 2016

In fact, this bring me straight onto the submissions this month. Let’s start with some puzzles and games. Denise Gaskins (@letsplaymath) submitted a post about the 2017 Mathematics Game, providing a set of rules similar to the game above that open it up to 1-100. Denise points out that “the goal is adjustable: Young children can start with looking for 1-10, middle grades with 1-25.”

Manan Shah (@shahlock) offers Alphanumeric Puzzle #1, saying “I thought readers might get a kick out of this”. It’s a type of puzzle known as a cryptarithmetic puhezzle.

Megan Schmidt (@veganmathbeagle) submitted Math Game: Draw 10 by Annie Forest. She says “Annie always has thought provoking ideas to share and this one is no different. She brings the math and the excitement for those days before Christmas when engagement is tough to come by.”

Next up, Measurement Games and Activities, a list of measurement/estimation games submitted by Crystal (@Tri_Learning), who says “The ability to estimate and measure are important skills for every day life and many professions. Find measurement games and activities for all ages.”

And if you still haven’t had enough games, Denise Gaskins (@letsplaymath) has it covered in My Favorite Math Games. Denise says “I like to use games as a warm-up with my co-op math circle. Here’s a collection of my favorites.”

Next, there are several posts offering stories and mathematical investigations.

This entertaining post by James Propp tells a tale centred around the (Prouhet-)Thue-Morse Sequence which covers religious law, fussy eaters, fractals, music, poetry, chess, rowing and more: Avoiding chazakah with the Prouhet-Thue-Morse sequence.

1226221 Is A Fascinating Number was submitted by Manan Shah (@shahlock), who says “What’s so fascinating about 1226221? It’s a palindrome, but that’s not all!”

I saw this fun post by Ben Orlin which asks the question: how many apparently different looking but actually the same sudoku puzzles can be generated from a single sudoku: 1.2 Trillion Ways to Play the Same Sudoku.

This being a mathematics teacher blog carnival, it may not come as a surprise that most submissions were tips and approaches to bringing a little fun to the classroom.

First, Tips for Teaching Students How to Identify Functions was submitted by Mrs. E (@mrseteachesmath), who says “This is a fun lesson idea for teaching students how to identify functions. It keeps kids engaged and a song helps them remember the definition of a function.”

This second post submitted by Mrs. E is about using paper folding to help students discover and think about geometry constructions: Altitudes and Angle Bisectors Paper Folding Activity.

And this third from Mrs. E is about using a puzzle/pattern-spotting approach to introducing logarithms: How I Teach Intro to Logs.

Crystal (@Tri_Learning) submitted Our Favorite Math Manipulatives, about using physical objects to help students learn, for students at various levels.

Next, the submitter didn’t leave their name, but we have Tell me everything you know about… by Jen McAleer, which offers some ideas and in-class activities to improve engagement, first by asking a loosely-defined question and second by holding a quick competition.

Amber Thomas submitted Playing with Our Food, Nutrition, and Fraction Line Plots, saying:

I found this fun nutrition website that I fondly refer to as “Smash My Food.” The kids see how much oil, salt, and sugar is in food (by watching it get squeezed out). It’s good, gross fun! Where the math comes in, is I take those amounts and create fractions on a line plot. You know, that pesky common core standard that math textbooks don’t cover at all (CCSS Math 4.MD.B.4)? Yeah, that one. Kids find the information relevant and relate-able, and I’ve gotten good feedback from others who have tried teaching fraction line plots this way.

Rupesh Gesota submitted Understanding v/s Answer-getting, a post about trying to teach understanding, even if it doesn’t use the standard set of rules. Rupesh says “This article shows some interesting (non-standard) ways of approaching and solving couple of mathematical problems. These are the ways developed by the students themselves based on understanding rather than procedures/ rules instructed by the teacher…”

Finally, in one last burst of teaching-related fun, Jo Morgan (@mathsjem) submitted the results of her World Cup of Maths, a Twitter competition to find UK maths teachers’ favourite topic to teach.

To round off this carnival, although I should disclaim that I am mentioned in it, this post by mathsbyagirl offers a guide to getting started listening to mathematical podcasts in 5 Maths Podcasts.

If you enjoyed this and the MTaP carnival in general, consider also reading the Carnival of Mathematics blog carnival which is coordinated by Katie Steckles here at The Aperiodical. Both Carnivals are always open for submissions and always happy to hear from potential future hosts. Submit a post you’ve enjoyed (whether you wrote it or not), and consider hosting a carnival at your blog!

]]>I’m so far from understanding the mind of a mathematical genius that it’s simply inconceivable that you could tell a person an apparently random number and he could intuit or deduce the kind of fact that he deduced about that taxi license number. I mean, I can’t run a four-minute mile, but I once ran a five-minute mile, and I can extrapolate from my own experience, in a way understand how someone might just be a lot better than me at something that, in an inferior way, I can also do. But Ramanujan isn’t like that. It’s as though this man were a different species, not just a superior example of the same species. Can you learn to do this kind of thing? Could I, if I had applied myself? Or is it that goddess again, is it really just genius?

Answers on a postcard!

]]>The Math Teachers at Play (MTaP) blog carnival is a monthly collection of tips, tidbits, games, and activities for students and teachers of preschool through pre-college mathematics. We welcome entries from parents, students, teachers, homeschoolers, and just plain folks. If you like to learn new things and play around with ideas, you are sure to find something of interest.

I’ll be hosting the January 2017 edition of MTaP here at Travels in a Mathematical World. Of course, a blog carnival is only as good as its submissions, so if you join me in aspiring to the claim “you are sure to find something of interest” then please keep your eyes open for interesting blog posts and submit them to MTaP. Please submit posts you’ve enjoyed by others or yourself. Posts you wrote that are appropriate to the theme are strongly encouraged. Submit through the MTaP submission form, leave a comment here or tweet me. Thank you!

Submissions are open now, and anything received by Friday 20th January 2017 will be considered for the edition hosted here.

]]>This space was renovated for mathematics a little before I arrived. It was designed to enhance student engagement and to create this sense of community, to allow collaborative learning and encourage inter-year interactions.

Over the last year, we conducted a study of use of the space. This included observations of use of the space as well as questionnaires and interviews with students about their use of the space, including students who had studied in the department in the old and new locations.

The results have just been published as ‘The role of informal learning spaces in enhancing student engagement with mathematical sciences‘ by Jeff Waldock, Peter Rowlett, Claire Cornock, Mike Robinson & Hannah Bartholomew, which is online now and will appear in a future issue of *International Journal of Mathematical Education in Science and Technology* (doi:10.1080/0020739X.2016.1262470).