Why does 23 people give a 50% chance of a shared birthday?

The probability that NO two of n people share a birthday is the product (365/365)·(364/365)·…·((365−n+1)/365). At n=23 that product drops just below 0.5, so the chance of at least one match exceeds 50% (precisely 0.5073). The counterintuitive part is that you compare every pair, not match to one fixed date: 23 people form C(23,2)=253 pairs, each with a 1/365 chance, which is why the probability climbs so fast. By n=70 it reaches 99.9%.

Birthday paradox

23 strangers in a room, and the odds tip past 50%.

Drop people in one at a time. Each lands on a day of the year wheel. When two land on the same day, it flashes — and the probability curve on the right keeps climbing toward its landmarks at 23 and 70.

In the room

people

Theory

0.0%

P(≥1 shared)

Empirical

—

no rooms run

Collisions

shared dates

Year wheel · 365 days0 placed

empty taken collision

P(shared) vs. peoplen = 0

theoretical curve empirical (rooms) 50% · 99.9%

Add people to this room

Empirical test · many rooms

Rooms simulated0

Theory at this n—

Empirical at this n—

Each room redraws the current number of people and checks for a shared birthday. The fraction with a collision should converge on the theoretical curve.

How it works

Add people and watch them scatter across the year wheel — 365 ticks arranged clockwise from Jan 1. A repeat date flashes amber and counts as a collision. The right-hand curve plots the exact theoretical probability of at least one shared birthday against the number of people, with a dot marking where your room sits.

The empirical test runs whole rooms in bulk: it should land the amber line right on the purple theoretical curve.

Why so few people?

The trick is pairs, not individuals. With n people there are C(n,2) = n(n−1)/2 distinct pairs, each with a 1/365 chance of matching.

The probability that no pair shares is (365/365)·(364/365)·…·((365−n+1)/365). At n=23 this drops below 0.5, so the chance of at least one match exceeds 50% (precisely 0.5073). With 23 people that's C(23,2)=253 pairs — which is why it climbs so fast. By n=70 it's 99.9%.

Threshold table

5 people	2.7%
10	11.7%
20	41.1%
23	50.7%
30	70.6%
50	97.0%
70	99.9%
100	99.9999%

The birthday attack in cryptography

The same math underlies collision attacks on hash functions. For an n-bit hash you'd naively need ~2ⁿ attempts to hit a specific output — but only ~2^n/2 to find any collision among generated outputs.

That's why modern hashes (SHA-256) use 256-bit outputs: birthday-bound collision resistance needs 2¹²⁸ work, still infeasible.

Citations

Feller, W. (1968). An Introduction to Probability Theory and Its Applications, Vol. 1 (3rd ed.). Wiley. Ch. II — the birthday problem in its classical form.
von Mises, R. (1939). "Über Aufteilungs- und Besetzungswahrscheinlichkeiten." Revue de la Faculté des Sciences de l'Université d'Istanbul. The earliest known formal treatment of the problem.
McKinney, E.H. (1966). "Generalized Birthday Problem." American Mathematical Monthly 73(4): 385–387.
Yuval, G. (1979). "How to swindle Rabin." Cryptologia 3(3): 187–189. First explicit cryptographic birthday-attack construction.

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