The Birthday Paradox

19 April 24

By RW Bro Terry McCallum

Hazardous maths ahead!

Editor’s Note: For safety reasons – please ensure your head is securely screwed on before reading this piece.

You may recall an article from the Oct–Dec 2023 issue ’Three for Aquarius’ (Have Your Say), where three masons in a group of 60 realised that they all shared the same birthday. The piece ended with the Editor inviting any brethren to calculate the odds of this happening.

How did you go? Well, way back in the dawn of time I used to teach Electrical and Electronic Engineering at East Ham Technical College in London. The subjects were highly technology oriented and we got into some really heavy maths. I also worked as a quality control engineer for a while, and that whole field is based largely on the laws of probability.

So, with my historic feet in each of the maths and probability fields, here’s my offering.

The Concept – one step at a time

As you can imagine, the higher the number of people, the greater the probability that two of them will share a birthday. If you have 367 people in a group, you can GUARANTEE at least one birthday match.

So what is the answer to the question at the end of the last issue’s article? The odds are far greater than you might think. In fact, when you reach a group size of only 23 people, there will be about a 50% chance that two of them will share the same birthday. It’s a very common and popular question among mathematicians, and it is known as ‘The Birthday Paradox’. For simplicity – let’s forget 29 February and just pretend that there are 365 days in each and every year. If you want to complicate it with Leap Year Day then do it in your own time.

The theory is based on the assumption that all birthdays are spread evenly throughout the year. There’s a problem straight away, because a large portion of the population are born in the ninth month because... y’ know... Christmas and New Year celebrations and alcohol and merriment and... y’ know. So in month number nine, the fruits of your Christmas ‘labours’ are born (literally). The same applies to nine months after Anzac Day and Easter, but not so much.

Bang goes our uniform distribution of birthdays, but we’ll continue anyway.

The mathematics of it is based on the idea that in any group there are PAIRS of people that might share a birthday.

As mentioned above, the chance that any two people have the same birthday is roughly 1/365. Not that much. But in a room of 60 (as in this case) there are lots of pairs (30), ie 30 possible dates to produce a match.

Break it down

Let’s get hold of the concept first. For two people (A and B), there is a 1/365 chance of A having the same birthday as B. Let’s show that as A = B.

Now let’s add a third person (C) and look at the possible ‘same’ pairings. The possibilities now are:

A=B or A=C or B=C

We now have three possibilities of matching birthdays – whatever the date – just by adding one person.

Don’t stop there...

Now let’s add a fourth person (D) and again look at the possible ‘same’ pairings. We now have:

A=B or A=C or A=D (3 chances)

B=C or B=D or C=D (3 chances)

We again added just one person but doubled the odds of a birthday match. Six chances for only four people!

OK, one more – but I think you’re getting the picture...

Now let’s add a fifth person (E) and repeat the exercise. We now get:

A=B or A=C or A=D

or A=E (4 chances)

B=C or B=D or B=E (3 chances)

C=D or C=E or D=E (3 chances)

This time, adding just one person increased the possibilities by another four; 10 chances of a shared birthday! And we’ve only got five people!

This increasing paradox is fuelled by the fact that for each person added there is an extra date to choose from as well as more people in the group.

Go figure!

Before we expand our brains to calculate for three (as per Oct-Dec issue’s article) let’s first calculate for two people out of 60 sharing the same birthday.

Imagine our 60 people all lined up in a column. Go to the back end and count how many people are in front of the last person in the column (ie 59). 58 are ahead of that 59th person and so on. Now add all those numbers up

(ie 59 + 58 + 57 + (and so on down to) +1). It works out to 1,827 (and yes – if you can smell a little bit of reverse Fibonacci in there you are kind of correct, but not really).

What you have just done is calculate the possible pairings for that group of 60, like we did above for our groups of 2, 3, 4 and 5 above.

So now it’s pretty basic to see that if you repeat something with a 1/365 chance 1,827 times it’s pretty likely you’ll get a good few matches, because

you now have:

1,827/365, ie you are very likely to get five pairings in a group of 60.

I’ll let that settle because for three to match it starts to get serious. I’ll leave this as a cliffhanger. See what you can do with it, and I’ll come up with my version of an answer for three matches in the next issue.