The birthday problem, worked out

It was recently my wife’s birthday. As such she was sat doing her yearly facebook browse, and noticed that two of her ex-colleagues shared the same birthday.

“What’s the chances of that?” she asked.

Silly question to ask a statistics nerd.

The wrong answer

Superficially, this seems remarkably improbable. What are the chances that any three individuals share the same birthday? Well, if we assume:

1. Nobody is born on on a leap year
2. Birthdays are randomly and uniformly distributed throughout the year (they are not)
3. The three people’s birthdays are independent of each other

then this becomes a simple problem:

$$P(wife)+P(colleagu{e}_1)+P(colleagu{e}_2)=1\ast \frac{1}{365}\ast \frac{1}{365}=\frac{1}{133225}$$

So the chances of that happening were about 0.00075%? Wow, very rare.

Except it isn’t.

How likely is it that any one colleague shares her birthday?

The previous calculation was for the very specific case of those particular two colleagues sharing her birthday. What we actually want is the the probability that at least one of her colleagues shares her birthday, which is a different - and far more likely - scenario known as “the birthday paradox”.

If we assume my wife has $n$ colleagues, which can calcuate firstly the probability of that any one specific colleague does not share her birthday. The probability that a specific colleague shares that birthday is $\frac{1}{365}$, therefore the probability that a specific colleague does not share that birthday is $1-\frac{1}{365}=\frac{364}{365}$.

Assuming each colleague’s birthday is independent, we can then work out the probability of all $n$ colleagues not sharing her birthday. If the probability that the first colleague does not share her birthday is 364/365, then the probability that the second colleague does not share her birthday is also 364/365… and so on for all $n$ colleagues. Therefore the probability that none of the $n$ colleagues share her birthday is:

$$P(\text{none share})=\frac{364}{365}\ast \frac{364}{365}\ast ...\ast \frac{364}{365}\text{(n times)}$$ $$P(\text{none share})={\frac{364}{365}}^n$$

It follows from this that the probability of at least one colleague sharing her birthday is is the complement of the event “no colleagues share her birthday, therefore the probability of at least one colleague sharing her birthday is:

$$\begin{gathered} P(\text{at least one shares})=1-P(noneshare) \\ P(\text{at least one shares})=1-{\frac{364}{365}}^n \end{gathered}$$

So as you can see, this depends on the number of colleagues she has (and had). The more people on her Facebook, the more chance that at least one will share a birthday.

In fact, if there’s any more than 253 then there’s more than a 50% chance she’ll share a birthday with at least one! Which for an ICU nurse is an almost certainty (but we’ll come to that).

What about two colleagues?

But we haven’t yet answered the question. She had two colleagues with the same birthday, so lets address that.

This is a binomial probability problem. So if we define terms as:

• $n$: The total number of colleagues
• $k$: The number of colleagues who share the birthday (in this case, k = 2)
• $p$: The probability that any one specific colleague shares her birthday, subject to the same assumptions above ($)
• $q$: The probability that any one specific colleague does not share her birthday (i.e. $q=1-p=1-\frac{1}{365}=\frac{364}{365}$)

The formula for binomial probability is:

$$P(X=k)=C(n,k)\cdot p^k\cdot q^{(n-k)}$$ where $C(n,k)$ is the number of combinations of choosing $k$ items from a set of $n$ items, also known as “$n$ choose $k$”):

$$C(n,k)=\frac{n!}{k!(n-k)!}$$ So for exactly two colleagues ($k=2$) sharing the birthday:

$$P(\text{exactly 2 share})=C(n,2)\cdot {\frac{1}{365}}^2\cdot {\frac{364}{365}}^{(n-2)}$$ $$P(\text{exactly 2 share})=\frac{n!}{2!\cdot (n-2)!}\cdot {\frac{1}{365}}^2\cdot {\frac{364}{365}}^{(n-2)}$$ $$P(\text{exactly 2 share})=\frac{n(n-1)}{2}\cdot {\frac{1}{365}}^2\cdot {\frac{364}{365}}^{(n-2)}$$

So this is kind of interesting. As the number of potential colleagues rises, so does the chance of exactly two colleagues sharing her birthday - until it begins to fall again. Well this makes sense - the ultimately the potential pool of birthday sharers grows so much that it’s far more likely that 3 or more collagues will share her birthday, making exatly two increasingly improbable.

Probability of at least two colleagues

However, what I suspect she was actually asking (or indeed not asking at all, because who’d intentionally set me off like this?) if what what’s the chance that at least two of her $n$ colleagues share her birthday? Given that it’s unlikely she’s still in touch with all of them, and that all of them have their accurate birthday on facebook.

Calculating $P(X\ge 2)$ (the probability of 2 or more successes) directly would involve summing the probabilities for $X=2,X=3,...,X=n$, which would be tedious. A much easier way is to use the complement rule.

The complement of “at least two successes” is “fewer than two successes”, meaning either:

• Exactly zero successes ($X=0$)
• Exactly one success ($X=1$)

So, the logic is:

$$P(\text{at least two share})=1-P(\text{fewer than two share})$$ $$P(X\ge 2)=1-[P(X=0)+P(X=1)]$$

We already know how to calculate $P(X=k)$ using the binomial probability formula. $P(X=0))$ means all $n$ colleagues do not share her birthday, so usingthe binomial formula $P(X=0)=C(n,0)\cdot p^0\cdot q^{(n-0)}$ where $C(n,0)=n!/(0!\cdot (n-0)!)=n!/(1\cdot n!)=1$ as there’s only one way for zero successes to occur (everybody “fails”) and $p^0=1$ by definition.

So this rattles out as:

$$P(X=0)=1\cdot 1\cdot q^n=q^n$$

This intuitively means the probability that all $n$ colleagues do not share the birthday is $(364/365{)}^n$ - i.e. increasingly unlikely as the numbers get bigger.

We derived $P(X=1)$ earlier:

$$P(X=1)=C(n,1)\cdot p^1\cdot q^{(n-1)}=n\cdot p\cdot q^{(n-1)}$$

Now we can substitute $P(X=0)$ and $P(X=1)$ into the complement rule formula:

$$P(X\ge 2)=1-[q^n+(n\cdot p\cdot q^{(n-1)})]$$

But how many colleagues are there?

Ok, so that’s lots of fun probability. But to put an actual number on her not-a-question-please-stop-talking-about-maths-now question, we need to estimate her number of potential colleagues.

We can estimate the Whole Time Equivalent (WTE) number of nurses needed for an e.g. 30-bedded intensive care unit based on GPICS and general NHS staffing principles.

ICU patients are categorised by the level of nursing care they require. Level 3 patients, or those with the highest dependency, typically require a 1:1 nurse-to-patient ratio while level 2 patients (“high dependency”) typically require a 1:2 nurse-to-patient ratio.

Both GPICS and BACCN standards emphasise these ratios for safe and effective care. So let’s make some assumptions about her 30-bedded unit to work out how many nurses are needed per shift, and scale from there.

If we assume an average requirement reflecting a high-acuity unit, perhaps averaging around 0.75 to 0.8 nurses per patient on the floor at any given time when averaged across all beds, for 30 beds: 30 beds * 0.8 nurses/bed = 24 nurses per shift for direct care.

Beyond direct patient care per shift, the WTE calculation needs to account for 24/7 cover, annual leave, sickness, and professional development. A common multiplier for 24/7 cover is approximately 4.2 to 4.8 WTEs to cover one nurse being present at all times, but recently the NHS has recommended figures closer to 5.6 WTEs per post to truly account for all leave and ensure consistent staffing while avoiding burnout.

GPICS also recommends a supernumerary shift clinical coordinator 24/7 for each unit, meaning at least one extra experienced nurse per shift not in the direct patient ratio count, as well as a ratio of 1:75 or ideally 1:50 WTE for clinical educators.

So let’s take a conservative estimate of 25 nurses per shift, meaning we need around 25 * 5.6 ≈ 140 nurses to cover the unit.

She’s also been there for around a decade, and during that time some of that nursing staff will have turned over.

Across the whole of the NHS the Nuffield Trust reports a leaver rate of 11.5% during 2021-2022, although recently this has fallen slightly to ~10.1%.

However, while UK-wide, long-term ICU-specific turnover statistics are less commonly published in headline figures, ICUs are known to be high-pressure environments. International studies often indicate that ICU turnover can be significantly higher than general nursing, with 24% of ICU nurses in France leaving every year. The Royal College of Nursing similarly highlighted increasing numbers of nurses leaving the NMC register early in their careers.

If we’re conservative and assume that the average annual turnover rate for experienced ICU nursing staff in a specific unit might be somewhere in the range of 10% to 15% per year over this 10-year period, we can estimating her total number of colleagues over 10 years based on:

• $N_0$: Estimated number of WTE nurses/colleagues at any one time = 140.
• Annual Turnover Rate ($T$): 0.12 (or 12%).
• Period: 10 years.

Number of nurses leaving (and presumably being replaced) each year $N_i=N_0\ast T=16.8$ new nurses per year. Over 10 years, the number of new faces due to turnover = 168 nurses new nurses. So the total distinct nursing colleagues $=140+168=308$.

Admittedly this is a gross oversimplification, as it doesn’t account for:

• People leaving and posts remaining vacant for periods.
• The same person leaving, then returning.
• Expansion or contraction of the unit’s WTE over the 10 years.
• People working less than full time

But we can finally answer the question - what is (approximately) the chance that at least two of my wife’s colleagues (from a nursing background, ignoring all the medical, admin, and AHP staff she’s also worked with) share her birthday:

$$P(X\ge 2)=1-[q^n+(n\cdot p\cdot q^{(n-1)})]$$ $$P(X\ge 2)=1-[{\frac{364}{365}}^{308}+(308\cdot \frac{1}{365}\cdot {\frac{364}{365}}^{(308-1)})]$$ $$P(X\ge 2)=0.206966$$

So there’s a 21% chance (or approximately 1 in 5) of it happening. Which is still slightly impressive… but by this point she’d lost interest in what I was wittering about and (understandably) gone off to celebrate with people far less boring than I.