April 1, 2013

# How probable are 3 consecutive birthdays?

Have you ever wondered what the likelihood is of having three consecutive birthdays in a row? It’s a fun excursion from the classical statisical brain teaser on the birthday problem.

To begin with we’ll consider the probability of a given three people - let’s call them Alex, Brian and Catherine - having consecutive birthdays. For those familiar with sets, we’ll represent them with the set $P = { Alex, Brian, Catherine }$.

## Making assumptions

To keep the Math reasonably simple, we’ll make a few assumptions:

• Births aren’t evenly distributed throughout the year, but we’ll assume they are.
• We’ll ignore leapyears and take a year as 365 days.
• We’ll assume these three people are picked at random.

## Applying Combinatorics

Recall that we defined the set of the 3 people as $P = { Alex, Brian, Catherine }$.

P can be permuted (ordered) in several different ways, for instance $( Brian, Alex, Catherine )$ or $( Catherine, Alex, Brian )$. How many different permutations are there? Time for a little Combinatorics.

 If you’ve done this thing before, recall there are $m!$ ways to rearrange $m$ elements. If you’ve no idea what I just said, take a look at this explanation. There are $\mathrm{P} = 3$ birthdays, so there are $3! = 3 \times 2 \times 1 = \mathbf{6}$ permutations.

## Applying Probability

Thanks to the addition rule we just need to find the probability of a single permutation. We’ll consider $( Alex, Brian, Catherine )$:

• Alex’s birthday can happen on any day of the year, with probability $\frac{365}{365} = 1$.
• Brian’s birthday must happen on the day after that of Alex, with probability $\frac{1}{365}$.
• Catherine’s must happen on the day after that of Brian, again with probability $\frac{1}{365}$.

These events are all independent, so we can just use the multiplication rule to find the chance of them all occurring:

$\frac{365}{365} \times \frac{1}{365} \times \frac{1}{365} = \frac{1}{365^2} = 0.000007506\ldots$

That’s roughly a $0.00075\%$ chance. Now let’s account for the 6 permutations:

$6 \times \frac{1}{365^2} = \frac{6}{365^2} = 0.00004503\ldots$

So the probability of three people having consecutive birthdays is roughly $0.0045\%$.

## Applying this result to groups of people

The original Birthday problem ultimately concerns how many people you need in a group for two or more of them to share birthdays.

To do this for our similar problem the key is to use how many different groups of 3 people can arise from a group of $x$ people. Where $^{n}C_{k}$ represents these k-combinations, plot the function:

$x \in \mathbb{Z}, x \ge 3,\ y =\ ^{x}C_{3} \times \frac{6}{365^2}$

Then find the first value of $x$ where $y > 0.5$, and you’ve got the smallest number of people where it is probable for there to be three consecutive birthdays.

I’ll leave this final step as an exercise for the reader.

## Footnote

You might get confused if trying to derive this. There are 4 different ways to make 3 days of birthdays as $( Alex, Brian, Catherine )$, depending on whether the second two birthdays are the day before or after. However since these 4 different ways are each only of $\frac{1}{4}$ probability it all cancels out.

Many thanks to the wonderful people that reviewed drafts of this post: the statistical Samuel Littley, the inventive Nemil Dalal, the organised Andy Bursh, the clarifying Isabell Long & the syntactical David Kendal.

Any statistical shortcomings are the fault of the author, Michael Mokrysz.

UPDATE 1: I noticed I’d missed out the 4 different ways the birthdays can arise, so corrected for that.

UPDATE 2: Samuel Littley pointed out that whilst there are indeed 4 ways to make each triple, the probability of any specific one happening means that we divide by 4 again. Post updated with this reasoning.