Updated 12 September 2026

Birthday problem

New to the notation? Start with the connected foundation for the underlying definitions and a worked example.

Under independent, uniformly distributed birthdays on 365 calendar days, 23 people are enough for the probability of at least one shared birthday to exceed one half. Leap days, dependence, and unequal day probabilities change the model.

Why so few

The trick is counting pairs, not people. With n people there are $n(n-1)/2$ pairs. At n = 23 that’s 253 pairs — 253 chances for a match. Each individual pair has only a 1/365 chance of a collision, but many chances add up fast, although their dependencies mean we cannot multiply 253 independent non-match probabilities.

The formula above just tracks that directly: for each of the n people, the probability their birthday is new given all previous births is $(365-k)/365$. Multiply those together for P(no collision); subtract from 1.

What to notice

• The curve is non-linear — it rises slowly then accelerates. 50% arrives earlier than intuition suggests, 99% by just n = 57.
• The empirical fraction fluctuates across seeded repetitions; its variance is P(1−P)/trials under the simulation model. Raising the trial count reduces Monte Carlo uncertainty.
• This is why hash collisions become likely much sooner than hash-output size implies. For uniform independent 128-bit outputs, the generic collision scale is roughly 2^64 samples. This is a collision-search statement, not a statement that finding a preimage for a specified output takes only 2^64 attempts.

Derive the complement and its approximation

The first person can have any birthday. Conditional on all k previous birthdays being distinct, person k+1 avoids them with probability 1−k/365. Multiplying these conditional probabilities gives the chance that every birthday is distinct. The factors are conditional; they do not assert independence of pair-match events.

For group sizes small relative to 365, using log(1−x) ≈ −x gives:

At n=23, the exact probability is 0.507297 and the approximation is about 0.500002. The expected matching-pair count is 253/365 ≈ 0.693151, which is neither of those probabilities. Groups with three equal birthdays contain three matching pairs but still count as only one group with a collision.

Match anyone, one person, or one day

Among 23 people, the chance that a designated person’s birthday matches any of the other 22 is $1-(364/365)^{22}\approx0.058571$, far below 0.507297. The chance that at least one of all 23 people was born on a preselected calendar day is $1-(364/365)^{23}\approx0.061151$. The demo reports these separately.

Unequal birthdays

The weight control creates a synthetic population in which each of days 1–90 is more probable than each remaining day. The weights are normalized, and birthdays remain independent. The exact no-collision probability is n! times the sum of the products of probabilities over every n-day subset. A recurrence evaluates this without enumerating all subsets. The first-group table exposes actual sampled days and multiplicities.

For probabilities p₁,…,p₃₆₅, the expected pair count is $\binom n2\sum_d p_d^2$. A Poisson-style rare-pair approximation uses that count in 1−exp(−count), but it is not an exact formula for dependent pair indicators. The exact curve is retained to show the approximation’s error.

Reference

Random Services: the birthday problem develops the finite-sampling complement formula. The weighted recurrence, designated-person calculation, and sample table state the specific extensions used here. They do not substitute simulated seasonality for measured birthday data.