Birthday paradox
Drag the room size and watch the probability climb. The curve crosses 50% earlier than almost anyone guesses, and tops 99% before you fill a single bus.
P(at least one shared birthday)
50.73%
With 23 people in the room. Expected matching pairs: 0.69.
You only need 23 people for a 50% chance, and 57 for 99%.
Same formula at common headcounts. The 50% crossover sits between 22 and 23.
| People (N) | P(shared) | Expected pairs |
|---|---|---|
| 5 | 2.71% | 0.03 |
| 10 | 11.69% | 0.12 |
| 15 | 25.29% | 0.29 |
| 20 | 41.14% | 0.52 |
| 23 | 50.73% | 0.69 |
| 30 | 70.63% | 1.19 |
| 40 | 89.12% | 2.14 |
| 50 | 97.04% | 3.36 |
| 70 | >99.9% | 6.62 |
| 100 | >99.9% | 13.56 |
The birthday paradox is not a paradox in the logical sense. It is counterintuitive because we instinctively count people rather than pairs. With 23 people you do not have 23 chances at a match. You have 253. That is C(23, 2) = 253 distinct pairs, and each pair has a 1 in 365 chance of sharing a date. The probability that none of those 253 pairs match drops below half almost exactly at N = 23.
The clean way to compute it is to ask the opposite question: what is the chance that everyone has a different birthday? The first person picks any day (365/365). The second avoids it (364/365). The third avoids two days (363/365). Multiply the fractions together for N people and subtract from 1. The product shrinks fast because each extra person multiplies in a smaller fraction. By N = 50 the chance of a match is already 97%, and by N = 70 it is over 99.9%.
You can sanity check this against the real world. An NFL roster holds 53 active players, so any given team has roughly a 98% chance that two players share a birthday. NBA teams carry 15 players and land around 25%. A typical classroom of 30 students sits near 70%. Once you have a feel for the pair count C(n,2) growing as n squared, the surprise disappears. The math is honest about how much faster pairs grow than people.