Skip to main content

TL;DR

The probability that in a group of n people, at least two share a birthday. With 23 people, the probability exceeds 50%. The complement is \( \prod_{k=0}^{n-1}(1 - k/365) \).

By Valenke Exam Prep Team·Last updated 2026-06-03

Birthday Problem

The probability that in a group of n people, at least two share a birthday. With 23 people, the probability exceeds 50%. The complement is \( \prod_{k=0}^{n-1}(1 - k/365) \).

Why it matters for interviews

A classic probability problem that illustrates how human intuition about coincidences fails. It teaches the power of the complement method and appears frequently in interview warm-up questions.

Definition and Mathematical Foundation

The probability that in a group of n people, at least two share a birthday. With 23 people, the probability exceeds 50%. The complement is \( \prod_{k=0}^{n-1}(1 - k/365) \).

Application in Quantitative Finance

A classic probability problem that illustrates how human intuition about coincidences fails. It teaches the power of the complement method and appears frequently in interview warm-up questions.

Related Terms

Ready to practice for the Quant Trading Interview?

Adaptive practice powered by Item Response Theory targets your weak areas. Start with 3 free sessions.

Start free practice →

Frequently Asked Questions

Why is 23 the magic number?▼
With 23 people, there are \( \binom{23}{2} = 253 \) pairs. Each pair has a \( 1/365 \) chance of matching. The expected number of matches is \( 253/365 \approx 0.69 \). The exact probability (accounting for dependence) is 50.7%.
How does this relate to hash collisions?▼
In a hash table with m buckets, collisions become likely at \( \sqrt{m} \) entries. This birthday bound determines when hash functions need upgrading and affects collision-based attacks on cryptographic hash functions.
What is the generalized birthday problem?▼
Finding the expected number of draws to get k matches from n categories. For k=2 (standard case), the answer is \( \approx \sqrt{\pi n/2} \). For k=3, the scaling changes to \( \Theta(n^{2/3}) \).