Three bits of maths every parent needs to know

Searching for a useful formula, from arguments to tea to football stickers

How many arguments?

Parents all know the feeling. Two children are playing nicely, then another arrives and suddenly it’s argument central.

There’s actually a useful network result here, which tells us that if we go from 2 to 3 children in a group, the number of potential arguments triples. This is because we’ve moved from 1 potential route to an argument (child 1 arguing with child 2) to 3 routes (1 and 2 arguing, or 2 and 3, or 1 and 3).

We can generalise this into a mathematical rule. If n children are playing, the number of possible pairwise arguments is:

\(\text{Potential arguments} = \frac{n(n-1)}{2}\)

This is why gatherings often become disproportionately more argue-y when more children join in:

Don’t get mugged off by your mug

Drinking cold tea is a staple of being a parent. But what type of mug shape is best to mitigate against this?

Assuming you’ve got a fairly solid cylindrical mug, heat loss is proportional to the exposed top surface area. If the radius of the mug is r, then this is equal to the good old Archimedes formula:

\(\pi r^2\)

And if the mug height is equal to h, the cylinder volume is:

\(\pi r^2 h\)

Reducing heat loss means making the volume of liquid as big as possible relative to the exposed surface area. This means maximising the ratio of volume/area, or:

\(\frac{\pi r^2 h}{\pi r^2} = h\)

In other words, for a fixed volume, a tall narrow mug loses less heat than a short wide one because it has a smaller exposed surface. And a wide, short mug is just a cold tea machine.

This is a terrible mug.

Sticky sticker calculations

Back when I was at university, a friend of mine thought it would be nostalgically fun to buy some World Cup football stickers. But he soon realised a problem he’d overlooked in his solo effort: swapsies. Soon he had loads of duplicates and nobody to trade with.

If an album requires collecting N different stickers and you gather n stickers at random, each new sticker might match any sticker you already have. So when you’ve collected n stickers there are about n(n-1)/2 possible pairs (the same logic as our argument result above). Since each pair will have a probability 1/N of matching, the expected number of duplicates when n is small relative to N (so repeated stickers are mostly simple pairs rather than triples or larger clusters) is roughly:

\(\frac{n(n-1)}{2N}\)

But how long will it take to complete the set? Oliver Johnson recently had post about football sticker maths, which included a 1960 paper by Newman and Shepp with the answer. (As an aside: if you’re ever struggling with an everyday probability problem, there is generally a paper from the 1950s or 60s that will have solved it.)

In theory, if you’re extremely lucky, you’ll fill up a World Cup album with N spaces after buying N stickers. But that’s not how things work in practice. Newman showed that if you tackle the problem solo – like my struggling university friend – then on average you’ll have to buy around N log(N) stickers to fill up the album. But if you're part of a large group of friends who can easily swap duplicates, then the average number of stickers required per person falls dramatically, to around N log(log(N)).

Assuming you haven’t started arguing with them yet.

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