Is March 23rd an Exceptionally Good Day For Athletes To Be Born?

Here's another real-life-maths guest post by tutor, author and friend-of-the-blog Colin, of Flying Colours Maths. This time, it's Olympic.

Short answer: no.

An article in the Guardian suggests that because Mo Farah, Chris Hoy and Steve Redgrave share a birthday with Jason Kenny, Roger Bannister, Joe Calzaghe and Mike Atherton, March 23rd must be a Good Day to have Olympian kids.

It's unusual, right enough, for three gold medallists from these Games - Farah, Kenny and Hoy- to share March 23rd as a birthday. You can work out the odds against it, roughly, using a binomial distribution: The probability of any one athlete having March 23rd as a birthday is - give or take - 1/365. There were 44 British gold medallists at the games. So the probability of at least three gold medallists sharing March 23rd as a birthday works out to be

1 - 44C0 (1/365)⁰(364/365)⁴⁴ - 44C1 (1/365)¹(364/365)⁴³ - 44C2 (1/365)²(364/365)⁴²

(That is, the probability of there not being 0, 1 or 2 athletes with that birthday).

That works out to be 1 - 0.8863 - 0.1071 - 0.0063, which is about 1 in 4000. Tiny odds!

Trouble is, that's a deeply, deeply flawed experiment. We've started with the knowledge that three of the athletes share that particular birthday. The probability of three athletes sharing any birthday is much higher. Roughly speaking, it's 365 times as high, about one in 11, nowhere near enough to be statistically significant. Actually working it out properly at 8am is a little beyond me, but writing a little simulation isn't. Taking 44 random birthdays, repeatedly, the results look like this:

• No two people share a birthday about 7% of the time
• No more than two people share any birthday about 84% of the time
• Three people share a birthday about 8% of the time
• Four people share a birthday about 0.3% of the time
• Five people share a birthday about 0.005% of the time

(That 8% is pretty close to the one-in-11 I suggested earlier, which makes it look like I'm on the right track).

If, as the article does, you extend it to previous medallists and athletes from other sports, you're bound to be able to cherry-pick more people with the same birthday.

Really, you'd expect the Guardian to run these things past a mathematician before they go live.

Guest Post: Is It Worth Insuring My Mobile?

Here's another guest post from Colin!

I'm in the process of upgrading my phone, and was just offered insurance against losing or damaging my new shiny. "It'll cost you £299 to replace your phone," it says, and offers me peace of mind for just £6 per month over my two year contract.

How much?

Seems reasonable, on the face of it. But let's do some sums, and not think about the fact that my home insurance payment is only double that.

Over the 24 months, with the insurance, I'd end up paying a total of £24 x 6 = £144: nearly half of the value of the phone. So, is it worth it?




Well, that turns into a probability question. The value of the insurance to me is how much the phone would cost to replace multiplied by the probability of me claiming on it. For instance, if I had a 1% chance of losing the phone over the next two years, the insurance would be worth about 0.01 x £299 = £2.99.

So, if I call the value V, the probability p and the cost of the phone C, the equation would be: V = pC.

Should I?

That would be worthwhile if the value was greater than the cost of the insurance (I): V > I, and since we know V is the same thing as pC, I can replace that in the inequality:

pC > I

Now, I know two of those values: C = 299 and I = 144, so, the policy is good value if:

299 p > 144

Dividing both sides by 299 (inequalities work, for the most part, just like equals signs):

p > 144 / 299 ~ 0.4816.

That means, the policy is good value for me if I think I have a more than 48% chance of losing my phone over the next two years. And while I'm clumsy and forgetful, I've never lost or broken a mobile phone.

So, this policy is a really bad idea for me - I'd do much better to put £6 a month into a savings account and use it to buy something nice at the end of it: maybe a new phone!

How Can We Calculate the Angular Size of an Object in the Sky?

This post is written as a sister-post to this one over at Blogstronomy, and is intended to show more explicitly the mathematical process that is alluded to there.

Angular size, as I describe on the Blogstronomy post, is a way that astronomers use to talk about how big things appear in the sky. That post talks about how big Jupiter and its moons Io and Ganymede would appear in the sky above Europa. This post explains how to work it out.

In order to work it out we need to be fairly comfortable with trigonometry - that's that GCSE topic with SOH CAH TOA and right-angled triangles. We also need to know two things about the object we want to look at: its actual size, and its distance from us.

Here's the general maths that we need:

Drawn by me using Paper on the iPad

The grey lines are imaginary lines from our eyeball to the extents of the object in the sky. r stands for the radius of the object we're looking at, and the yellow line (d) represents the distance between us and the centre of the object. Luckily for us, it also cuts the (grey) isosceles triangle in half and gives us a right-angled triangle that we can get to work on with trig.

If we're trying to find the angle that's marked (that's half the angular size), we can label the radius as the 'opposite' side, the yellow line as the 'adjacent' side, and the grey line is the hypotenuse. Let's call the angle "x", to make things easy. We know the radius and distance, so from our school-days "SOH CAH TOA", we know that we have to use the Tangent function:

Tan(x) = Opposite / Adjacent

Or, using more sensible notation from the diagram:

Tan(x) = r / d

We want to find x, though, so we can rearrange:

x = Tan^-1(r / d)

But if we're being honest with ourselves, we want to find the entire angle between the two grey lines, so we need to double whatever result we get (calling the angular size "a"):

a = 2x = 2 x Tan^-1(r / d)

Now all we have to do is substitute the values we know for each of the objects. Here's the data you need:

• Io: r = 1830 km ; d = 249,300 km.
• Ganymede: r = 2631.2 km ; d = 399,300 km.
• Jupiter: r = 69,911 km ; d = 671,100 km.

All you have to do is whack those numbers in the right places into your favourite calculator, then we've found the angular size of these three bodies as viewed from Europa! Just make sure your calculator's set to 'degrees' and neither 'radians' nor 'stun'. The result for Io comes out at 0.841 degrees. If you want to find out what that means for our view and/or find out the answers for the Ganymede and Jupiter (but try them out yourself first, or it's no fun), head over to the post at Blogstronomy!

Guest Post: The Higgs Boson: How Sure is Pretty Sure?

You'll have heard, unless you particularly like the conditions under rocks, that the folks at CERN's LHC have found some pretty convincing evidence for the existence of what is known as the Higgs boson. At long last they've got something to work with beyond 'it should be there, and it should be a bit like this'. They're pretty sure that what they've found is the Higgs boson and not just something that happens to act a bit like one should act. But how sure is 'pretty sure'? Friend of the blog Colin Beveridge**** answers that question. Apologies for letting through a small instance of bad language. I thought it worked.

“They’re only pretty sure they’ve found the Higgs boson, aren’t they?”

If I had a billion dollars for every time I’d heard that question, I’d have enough to build a medium-sized hadron collider. Technically, that’s right: they’re only pretty sure. But, as with all things scientific, there’s ‘pretty sure’ and there’s ‘pretty sure.’

Luckily, science has a better way of talking about things than just saying ‘we’re pretty sure’, and it’s — surprise! — to put a number on things. You’ll quite often see something described as ‘a three-sigma event’ or — in the case of the Higgs — ‘a five-sigma discovery’. (There’s a business philosophy that, like most business philosophies, had a year or two in the spotlight and then vanished, called ‘six-sigma’ — it’s the same idea.) The higher the number, the more certain you are about the event⁠*.


So what is it this mysterious sigma represents?

Well, it’s to do with hypothesis testing, and it’s based on the mathematical idea of proof by contradiction. The idea is, if you want to show that something is true, you can prove it by showing that the opposite isn’t true. For instance, if you wanted to prove that there was no biggest number, you’d start by saying “let’s assume the opposite! Assume there is a biggest number.” Being a mathematician, you’d probably call it N, but you don’t have to. “Now, add one to that number — you’ve got a bigger number now, so that assumption about having a biggest number? Load of bollocks⁠**. Therefore, there can’t be a biggest number.”

Got it? To prove something is true, you can assume that it isn’t true, and show that reality is inconsistent with that assumption.

That’s sort of how hypothesis testing works, too, but it’s a little bit trickier. When you’re doing nice proofs about real numbers, all the properties are known and you can say for certain (most of the time) whether something is absolutely, definitely true or not. When you’re dealing with the real world, it’s not so simple. As Einstein said, “As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.” You really can’t show that something is definitely untrue in science — only that it’s extremely unlikely.

So that’s what you do! You assume the opposite of what you’re trying to prove (and call it the null hypothesis, if you’re that way inclined). You say what you’re trying to prove (the alternate hypothesis). You work out a threshold of how certain you want to be⁠*** — which is about the same thing as saying how unlikely the null hypothesis needs to be — and then you do the experiment.

If your observations would be absurdly unlikely under the null hypothesis, but quite plausible under the alternate hypothesis, you throw out the null hypothesis as a bad job, and accept the alternate hypothesis. If your observations aren’t unlikely enough, no matter how unlikely they are, you don’t have enough evidence for what you were trying to prove. Sorry.


So, where do the sigmas come in?

They come from the normal distribution, which looks to me like a boa constrictor that’s just eaten an elephant. It’s a shape that comes up repeatedly in science — most measurements are pretty close to the mean, in the middle of the graph, and the further away from the middle you get, the less likely the result. For instance, if you look at the heights of a large group of 25-year old adult males, you’d expect most of them to be about 175cm tall, most people within — I’d guess — 15cm of that, and possibly some extremely tall or extremely short people. The graph of those heights would looks something like a normal distribution.

There are two important bits of information there: the 175cm, which is the mean of the distribution — how tall you expect a random person picked from the crowd to be; and the 15cm, which is the standard deviation. The standard deviation is a bit harder to explain — in very loose terms, it’s the ‘give or take’: if you say “a person from this group will be between 160cm and 190cm tall”, you should be right about 70% of the time. That’s the nature of the normal distribution.

Sigma is just the number of standard deviations an observation is away from the mean. So, if you had a person in the group who was exactly 175cm tall, that would be a zero-sigma event. Someone 160cm tall is 15cm away from the mean, or one standard deviations, making him a 1-sigma event. Someone 205cm tall — a big lad — would be 30cm, or two standard deviations from the mean, making him a 2-sigma event.

The thing about the normal distribution is, it tails off very quickly. A one-sigma event in a given direction shows up about 14% of the time (around one in seven). A two-sigma event, only about 2% (one in 44), and a three-sigma event, barely 0.1% (1 in 740). By the time you get to 4 sigma, you’re talking about one in 32,000, and 5 sigma is one in three and a half million — about the same as winning the lottery if you buy four tickets.


So, the answer is ‘yes, they’re only pretty sure’, but in this context, it means that the alternative is preposterously unlikely. Not impossible, of course, they could just have got lucky and found something Higgs-shaped exactly where they happened to be looking for it — and if that’s the case, we’ll find out more about it in the next few months.

* I’ll get into the details of what the sigma means later.
** Probably not the technical language you’d use in a formal proof.
*** I’m getting to the sigmas, promise!
**** Colin's a mathematician with his finger under many belts, if you'll excuse the mixed metaphor: he's a maths tutor, a maths blogger, a maths writer, and a folk guitarist too.

RL Maths: The Great Pizza Ripoff

I went to a well-known pizza restaurant chain for dinner last night. I won't tell you which one, but I'll refer to it as 'Pizza Shed' and you can read from that what you will. When we were ordering, my girlfriend and I briefly wondered whether we should get a regular pizza each, or a large pizza with 'half-and-half' toppings. We went for the latter, not really thinking too much about it (we wanted a stuffed crust anyway).

However, when the bill came we were surprised to note that our half-and-half large pizza had been charged as two regular ones. This got us thinking: is one large pizza really worth the same as two regular ones? The good thing is that we don't even need to know the price to work this out. We do need to know the sizes, though: a regular pizza of the type we ordered is 11" across, and a large is 14".

At first glance, you might say "the large pizza isn't twice the size of the regular one, so no, it's not worth it!" but be careful: you're making a mistake that loads of people make when considering the mathematics of pizza ordering*.

Consider the hastily produced graphic to the left**. It shows two small pizzas, side by side, plonked on top of a large pizza. The important bit is that the two small pizzas together have the same diameter as the large pizza, but when plonked on top there's still a large expanse of pizza that isn't hidden. This means that twice the diameter means much more than twice the pizza.

So we've got to be a bit more cleverer about this. When talking about how much pizza you get, it makes more sense to talk about the area of a pizza than its diameter (which is the standard quoted measurement for pizzas worldwide). If you've got a maths GCSE (or O-Level) you'll have at least a dim recollection that the area of a circle and its radius (which is half the diameter) are linked. Specifically, the formula is:

A = π x r²

Where A stands for 'area' and r stands for 'radius'. And π, far from being something scary, is just a symbol that means "a bit more than 3"***.

So, for our pizzas, we can find the regular one's area by working out π x 5.5², and finding the answer to π x 7² will give us the area of a large pizza.

A good idea in maths is to estimate things before working them out exactly. This gives you a feeling for what kind of number you're expecting the actual answer to be, so you can tell if you've messed something up (or hit the wrong button on the calculator if you've got fat fingers like me). In maths 'estimating' is not just guessing, but working out a simpler version of the problem. I did this, last night, in a minute or so whilst driving home (that's how easy it is****).

So the regular pizza has an estimated area of "about 3 multiplied by about 30*****", which is about 90 square inches, and the large pizza has an estimated area of "about 3 multiplied by about 50******", which is about 150 square inches. It looks like two regular pizzas gives you more total pizza than one large one!

What's the solution, exactly?

• π x 5.5²  = 95.03 square inches (to 2dp).
• π x 7² = 153.94 square inches (to 2dp).

My estimates were both slightly low, but not by a lot. The same result stands: two regular pizzas would have given us more pizza than one large one, yet they charged us the same amount!

The scoundrels!

* O.k, the number of people who actually stop to consider the mathematics of pizza ordering might be quite small.
** I am available for freelance graphic design jobs for a modest fee.
*** π is, of course, a specific number (3.14159... etc), but to write it down accurately would take more time and energy than is available to us, even if we enlist our children's children's children, so we're lazy and use a symbol instead.
**** If you'd like some tips for being more accurate with your mental maths for not much extra effort, you could do worse than check out my pal Colin's 'Mathematical Ninja' series.
***** 5² is 25, and 6² is 36, so 5.5² is about half way between, which is about 30.
****** 7² is 49, which is about 50.