How probable are 3 consecutive birthdays?

April 1, 2013

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.