Birthday Coincidences and Poisson Approximations

This morning, Facebook was extremely keen to remind me via every available medium that four of my friends celebrate their birthday today. My first thought was that I hope they all enjoy their day, and my second thought was to ask what the chance of this was. I have about 200 Facebook friends, and so this struck me as an unlikely occurrence. But this problem has form, and it felt worthwhile to try some calculations to see if my intuition was well-founded.


Siméon Denis Poisson celebrated his 234th birthday on 21st June this year.

The classical birthday problem

The starting point is the question: how many friends do you have to have before you expect to start seeing anyone sharing a birthday? There are a ridiculous number of articles about this on the web already, so I will say little, except that I don’t want to call this the ‘birthday paradox’, because it’s not a paradox at all. At best it might be counter-intuitive, but then the moral should be to change our intuition for this type of problem.

Throughout, let’s discount February 29th, as this doesn’t add much. So then, to guarantee having a shared pair of birthdays, you need to have 366 friends. But if you have a mere 23 friends, then the probability of having some pair that share a birthday is slightly greater than a half. The disparity between these two numbers leads to the counter-intuition. Some people might find it helpful to think that instead of counting friends, we should instead be counting pairs of friends, but I don’t personally find this especially helpful.

For me, thinking about the calculation in very slightly more generality is helpful. Here, and throughout, let’s instead take N to be the number of days in a year, and K the number of friends, or kids in the class if you prefer. Then, as usual, it is easier to calculate the probability that no two share a birthday (that is, that all the birthdays are distinct) than the probability that some two share a birthday. We could think of the number of ways to pick the set of birthdays, or we could look at the kids one-at-a-time, and demand that their birthday is not one of those we’ve already seen. Naturally, we get the same answer, that is

\frac{^N P_K}{N^K} = 1\cdot \frac{N-1}{N}\cdot\ldots \frac{N-K+1}{N}.

We’ve assumed here that all birthdates are equally likely. We’ll come back to this assumption right at the end. For now, let’s assume that both N and K are large, and we’ll try to decide roughly how large K has to be in relation to N for this answer to be away from 0 and 1. If we pair opposite terms up, we might approximate this by

(\frac{N-\frac{K}{2}}{N})^K = (1-\frac{K}{2N})^K\approx e^{-K^2/2N}.

In fact, AM-GM says that this is an overestimate, and a bit more care can be used to show that this is a good-approximation to first order. So we see that if K=\Theta(\sqrt{N}) for large N, we get a non-trivial limit.

Challenges for four-way shared birthdays

So the original problem I posed is harder, because there isn’t (unless I’m missing something) a natural way to choose birthdays one-at-a-time, or describe the set of suitable birthday sets. There are two major obstacles in a calculation such as this. Firstly, the overlap of people, that is we might have five or more birthdays overlapping; secondly, the overlap of days, that is we might have several days with four (or more) birthdays. We’ll end up worrying more about the second situation.

We start by eliminating both problems, by asking for the probability that exactly four friends are born on January 1st. The general form of this probability is \frac{\binom{K}{4} }{N^4} \cdot (\frac{N-1}{N})^{K-4}. Now, if K\ll N, this final term should not be significant. Removing this is not exactly the same as specifying the probability that at least four birthdays are on January 1st. But in fact this removal turns a lower bound (because {exactly four}<{at least four}) into an upper (in fact a union) bound. So if the factor being removed is very close to one, we can use whichever expression is more convenient.

In the real life case of N=365, K=200, this term is not negligible. But accounting for this, we get that the probability of exactly four birthdays on 1st January is ~0.0021. Our upper bound on the probability of at least four is ~0.0036.

But now that we know the probability for a given day, can we calculate (1-0.0021)^{365} to estimate the probability that we never have four-overlap? When we did our previous iterative calculation, we were using independence of the different kids’ birthdays. But the event that we have four-overlap on January 1st is not quite independent of the event that we have four-overlap on January 2nd. Why? Well if we know at least four people were born on January 1st, there are fewer people left (potentially) to be born on January 2nd. But maybe this dependence is mild enough that we can ignore it?

We can, however, use some moment estimates. The expected number of days with four-overlap is 365\cdot 0.0021 \approx 0.77. So the probability that there is at least one day with four-overlap is at most ~0.77.

But we really want a lower bound. So, maybe we can borrow exactly the second-moment argument we tried (there for isolated vertices in the random graph) in the previous post? Here, the probability that both January 1st and January 2nd are four-overlapping is

\frac{\binom{K}{4}\binom{K-4}{4}}{N^8}\cdot (\frac{N-2}{N})^{K-8}\approx 4.3\times 10^{-6}.

From this, we can evaluate the expectation of the square of the number of days with four-overlap, and thus find that the variance is ~0.74. So we use Chebyshev, calling this number of days #D for now:

\mathbb{P}(\# D=0)\le \mathbb{P}(|\#D - \mathbb{E}\# D|^2 \ge (\mathbb{E}\# D)^2 ) \le \frac{\mathrm{Var} \# D}{(\mathbb{E} \#D)^2}.

In our case, this unfortunately gives us an upper bound greater than 1 on this probability, and thus a lower bound of zero on the probability that there is at least one day with four-overlap. Which isn’t especially interesting…

Fairly recently, I spoke about the Lovasz Local Lemma, which can be used to find lower bounds on the probabilities of intersections of events, many of which are independent (in a particular precise sense). Perhaps this might be useful here? The natural choice of ‘bad event’ is that particular 4-sets of people share a birthday. There are \binom{K}{4} such events, and each is independent of the collection of \binom{K-4}{4} disjoint events. Thus we can consider using LLL if e\cdot (\binom{K}{4}-\binom{K-4}{4})\cdot 0.0021 \le 1. Unfortunately, this difference of binomial coefficients is large in our example, and so in fact the LHS has order 10^3.

Random number of friends – coupling to a Poisson Process

All of these methods failed because without independence we had to use estimates which were really not tight at all. But we can re-introduce independence if we remove the constraints on the model. Suppose instead of demanding I have K friends, I instead demand that I have a random number of friends, with distribution Poisson(K). Now it is reasonable to assume that for each day, I have a Poisson(K/365) friends with that birthday, independently for each day.

If we end up having exactly K friends with this random choice, then the distribution of the number of 4-overlap days is exactly the same as in the original setup. However, crucially, if we end up having at most K friends with this random choice, the distribution of the number of 4-overlap days is stochastically dominated by the original distribution. So instead let’s assume we have Poisson(L) friends, where L<K, and see how well we can do. For definiteness, we’ll go back to N=365, K=200 now. Let’s say X is the distribution of birthdays in the original model, and \Xi for the distribution of birthdays in the model with a random number of friends


\mathbb{P}(\exists \ge 4\text{-overlap in }\Xi) = 1- \mathbb{P}(\mathrm{Po}(L/365)\le 3)^365. (*)

Now we can write the crucial domination relation as

\mathbb{P}(\exists \ge 4\text{-overlap in }X)\ge \mathbb{P}( \exists \ge 4\text{-overlap in }\Xi \,|\, |\Xi|\le 200),

and then use an inequality version of the law of total probability to bound further as

\ge \frac{ \mathbb{P}(\exists \ge 4\text{-overlap in }\Xi) - \mathbb{P}(|\Xi|>200)}{\mathbb{P}(|\Xi|\le 200)}.

This is a function of L, and in principle we could find its maximum, perhaps as N\rightarrow\infty. Here, though, let’s just take L=365/2 and see what happens. For (*) we get ~0.472.

To estimate \mathbb{P}(\mathrm{Po}(365/2)>200), observe that this event corresponds to 1.4 standard deviations above the mean, so we can approximate using quantiles of the normal distribution, via the CLT. (Obviously this isn’t completely precise, but it could be made precise if we really wanted.) I looked up a table, and this probability is, conveniently for calculations, roughly 0.1. Thus we obtain a lower bound of \frac{0.472-0.1}{0.9}. Allowing for the fairly weak estimates at various points, we still get a lower bound of around 0.4. Which is good, because it shows that my intuition wasn’t right, but that I was in the right ball-park for it being a ‘middle-probability event’.

Remarks and References

– The reason for doing the upper bound for the probability of exact 4-overlap is that the same argument for at-least-4-overlap would have given an upper bound of 1. However, this Poisson Process coupling is also a much better method for obtaining an upper bound on either event.

– Birthdays are not uniformly distributed through the year. The deviation is strong enough that even from the set of birth frequencies (rather than the sequence of birth frequencies), we can reject a null hypothesis of uniformity. Early September is pretty close to the maximum. Two comments: 1) this is the time of year where small variations in birth date have a big effect on education, especially in primary school; 2) we are 37 weeks into the year…

– It is known that 187 friends is the first time the probability of having at-least-4-overlap is greater than ½. You can find the full sequence on OEIS as A014088. I used to have about 650 Facebook friends, before I decided that I’d prefer instead the pleasant surprise of finding out what old acquaintances were up to when I next spoke to them. In this case, the median of the distribution of the largest number sharing a birthday would be seven.

– Eric Weisstein’s article on Mathworld is, in my opinion, the best resource for a mathematician on the first two pages of Google hits by at least two orders of magnitude. In the notation of this article, we were calculating P_4(n=200,d=365). There are also some good general asymptotics, or at least recipes for asymptotics, in equations (17) and (18).

– The paper Methods for Studying Coincidences by Diaconis and Mosteller is, as one might expect, extremely readable, and summarises many results and applications, including several generalisations.


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