The Math Behind The Birthday Paradox

This post will explore the math behind the birthday paradox, which was covered in my previous post

This article is for those who are interested in knowing the math behind the birthday paradox, which I covered in my previous post

If you recall, from the birthday paradox post

p(atleast 2 people sharing bday) = 1 - p(Case 1)

Where

\(\textrm{p(Case 1)} = \frac{365}{365}* \frac{364}{365}*...*\frac{365-(n-1)}{365}\)

The above can be written concisely as:

\(\textrm{p(Case 1)} = \prod_{i=1}^{n}\frac{365-(i-1)}{365}=\prod_{i=1}^{n}[1-\frac{(i-1)}{365}]\)

We can make use of the below approximation for small values of x

\(e^{-x} \approx 1-x\)

Approximating, we get

\(1-\frac{(i-1)}{365} \approx e^{-\frac{(i-1)}{365}}\)

Substituting in the equation of p(Case 1)

\(\textrm{p(Case 1)} = \prod_{i=1}^{n}[1-\frac{(i-1)}{365}] \approx \prod_{i=1}^{n}e^{-\frac{(i-1)}{365}}=e^{-\sum_{i=1}^{n}\frac{i-1}{365}}\)

\(\sum_{i=1}^{n}(i-1) = \sum_{i=0}^{n-1}i = \sum_{i=1}^{n-1}i = \frac{n(n-1)}{2}\)

\(=> \textrm{p(Case 1)} \approx e^{-\sum_{i=1}^{n}\frac{(i-1)}{365}} = e^{-\frac{n(n-1)}{2*365}} = e^{-\frac{n(n-1)}{730}} \)

From this we obtain a compact expression

\(\textrm{p(atleast 2 people sharing bday)} \approx 1-e^{-\frac{n(n-1)}{730}}\)

By substituting the value for n (no. of people in the room), we can obtain the corresponding probability

Let us now try to verify the paradox: Is it actually true that the probability is 50% for 23 people

Substituting n=23 in the above equation, we get

\(\textrm{p(atleast 2 people sharing bday)} \approx 1-e^{-\frac{23*22}{730}} \approx 1-e^{-\frac{506}{730}} = 0.5 \)

This matches with the paradox and also the value obtained from the graph

Another approach to this calculation is given in the below website which I have referenced in the previous post also:

https://betterexplained.com/articles/understanding-the-birthday-paradox/

Thanks for reading!!!