# How probable are 3 consecutive birthdays?

April 1, 2013Have 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.