## A harder birthday problem

It is a well-known non-intuitive result that in a group of $n=23$ people– conveniently the size of a classroom of students– the probability is at least 1/2 that $k=2$ or more of them share a birthday.  This is a nice problem for several reasons:

1. Its solution involves looking at the problem in a non-obvious way, in this case by considering the complementary event that all birthdays are distinct.
2. Once the approach in (1) is understood, computing the answer is relatively easy: the probability is $1-(365)_{n}/365^n$, where $(x)_{n}$ is the falling factorial.
3. The answer is surprising.  When I ask students, “How many people are needed for the probability of a shared birthday to exceed 1/2?”, guesses as high as 180 are common.
4. Picking on the not-quite-realistic assumption of all $d=365$ birthdays being equally likely actually helps; that is, with a non-uniform distribution of birthdays, the probability of coincidence is higher.

But what about larger $k$?  For example, suppose that while surveying your students’ birthdays in preparation for this problem, you find that three of them share a birthday?  What is the probability of this happening?

The Wikipedia page on the birthday problem doesn’t address this generalization at all.  There are several blog and forum posts addressing the $k=3$ case specifically, by grouping the prohibited cases according to the number of pairs of people with shared birthdays.  This is generalized to arbitrary $k \geq 2$ on the Wolfram MathWorld page; the resulting recurrence relation is pretty complex, but it’s a nice exercise to prove that it works.

Probability that at least (2,3,4,5) people share a birthday, vs. group size.

The motivation for this post is to describe what I think is a relatively simpler solution, for arbitrary $k$, including Python source code to perform the calculation.  Let’s fix the number of equally likely possible birthdays $d=365$, and the desired number $k$ of people sharing a birthday, and define the function

$G(x)=(1+x+\frac{x^2}{2} \ldots + \frac{x^{k-1}}{(k-1)!})^d$

Then $G(x)$ is the exponential generating function for the number of “prohibited” assignments of birthdays to $n$ people where no more than $k-1$ share a birthday.  That is, the number of such prohibited assignments is $n!$ times the coefficient of $x^n$ in $G(x)$.

(When working through why this works, it’s interesting how often it can be helpful to transform the problem into a different context.  For example, in this case, we are also counting the number of length-$n$ strings over an alphabet of $d$ characters, where no character appears more than $k-1$ times.)

The rest of the calculation follows in the usual manner: divide by the total number of possible assignments $d^n$ to get the complementary probability, then subtract from 1.  The following Python code performs this calculation, either exactly– using the fractions module, which can take a while– or in double precision, which is much faster.

import numpy as np
from numpy.polynomial import polynomial as P
import fractions
import operator

def p(k, n, d=365, exact=False):
f = fractions.Fraction if exact else operator.truediv
q = 0 if n > d * (k - 1) else P.polypow(np.array(
[f(1, np.math.factorial(j)) for j in range(k)], dtype=object),
d)[n]
for j in range(1, n + 1):
q = q * f(j, d)
return 1 - q

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### 2 Responses to A harder birthday problem

1. Nice python modules. I don’t know this python modules fractions and operator.
Anyway you have a error:

>>> p(1,100)
Traceback (most recent call last):
File “”, line 1, in
File “”, line 6, in p
IndexError: index 100 is out of bounds for axis 0 with size 1
>>> p(2,100)
0.9999996927510721

• Note that k=1 doesn’t really yield a useful calculation, since we’re asking for the probability that “1 or more” of n people have the same birthday. In that case, the generating function described in the post is just the constant G(x)=1.

More generally, we could add the edge case check for n>(k-1)*d, in which case p(k,n)=1 by the pigeonhole principle, but the simple array indexing in the example code won’t work to extract the corresponding generating function coefficient (which is zero).

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