Birthday Paradox Calculator
The birthday paradox: with just 23 people, the probability that at least two share a birthday already exceeds 50% — far fewer than most people guess. This tool computes the probability exactly and shows the minimum group size needed to reach 50%, 90% and 99%.
Group size must be a whole number between 1 and 3660; days must be an integer ≥ 2.
Probability of at least one shared birthday
50.73%
p ≈ 0.50730
Approximation: 50.00%
Probability bar
Group size needed to reach a given probability
Formula
P(at least one shared birthday) = 1 − ∏ᵢ₌₀ⁿ⁻¹ (1 − i/D), where D is the number of days in a year
- · Assumes birthdays are uniformly distributed (real distributions are slightly skewed but only shift the third decimal place)
- · n = 23 → p ≈ 50.73%; n = 41 → 90%; n = 57 → 99%; n = 70 → 99.92%
- · When n ≥ D + 1, the pigeonhole principle forces probability = 1
- · The approximation 1 − exp(−n(n−1)/(2D)) is shown alongside the exact value (~1% error)
- · Change the "days in a year" field to use any bucket count (12 zodiac signs, 52 cards, etc.) for analogous collision problems
- · Source: Wikipedia — Birthday problem
Frequently asked
Only 23 people for 50%? Surely that is a miscalculation?
Intuition usually confuses "someone in the group shares MY birthday" (which is 23/365 ≈ 6.3%) with "any two people in the group share a birthday." The latter has C(23, 2) = 253 possible pairs, each colliding with probability 1/365 — the independence approximation gives 1 − (364/365)²⁵³ ≈ 0.4995. The exact value via ∏(1 − i/365) is 0.5073. The 50% threshold is real; only the intuition is wrong.
How many people guarantee a shared birthday with certainty?
By the pigeonhole principle: with 366 possible birthdays (including Feb 29), a group of 367 people forces a duplicate. But practically, n = 100 already gives p ≈ 99.99997% — indistinguishable from certainty in most settings.
Can I use this tool for other collision problems?
Yes. Change the "days in a year" field to any bucket count: 12 zodiac signs → 4 friends already give ~43.75% collision; 52-card deck → 9 draws give ~50% chance of a repeat. The same formula underpins hash-collision analysis — the "square-root rule" predicts collisions at about √(D × ln 2) draws, which gives ~22.5 for D = 365, matching the exact answer of 23.
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