Key takeaways
- With 23 people and 365 equally likely birthdays, the chance that at least two share one is about 50.7%.
- Compute it through the complement: the chance everyone differs is (365/365)(364/365)⋯(343/365).
- The surprise comes from counting pairs: 23 people form 253 pairs, each a chance for a match.
- Matching a specific birthday, such as yours, needs about 253 people for a 50% chance.
- The same mathematics sets how many items can be hashed before a collision becomes likely: roughly the square root of the number of possible values.
How many people do you need in a room for it to be more likely than not that two share a birthday? Half of 365, about 180, can sound like a reasonable guess. The answer is 23. That gap between intuition and arithmetic is why this is sometimes called the birthday paradox, although nothing about it is contradictory once the counting is done carefully.
The assumptions
The standard version assumes 365 possible birthdays (ignoring 29 February), each equally likely, with different people's birthdays independent. MathWorld sets the problem up the same way: each birthday uniformly distributed over the possible days. Real birthdays are not perfectly uniform, and the effect of that is discussed below.
The calculation, through the complement
"At least two share" covers many cases: one pair, two pairs, three people on one day, and so on. Its complement, "all birthdays are different", is a single case and much easier.
Line people up one at a time.
- The first person can have any birthday: 365/365.
- The second must avoid the first: 364/365.
- The third must avoid both: 363/365.
- The nth must avoid n − 1 days: (366 − n)/365.
These multiply, by the multiplication rule for probabilities covered in OpenStax's Contemporary Mathematics, because each step is conditional on the earlier ones being distinct:
P(all different) = (365 × 364 × ⋯ × (366 − n))/365ⁿ,
and P(at least one shared birthday) = 1 − P(all different). In factorial form, the numerator is 365!/(365 − n)!, the number of ways to give n people distinct birthdays in order, a permutation count from permutations versus combinations.
The numbers
| People | Chance of a shared birthday |
|---|---|
| 10 | 11.7% |
| 20 | 41.1% |
| 23 | 50.7% |
| 30 | 70.6% |
| 40 | 89.1% |
| 50 | 97.0% |
| 57 | 99.0% |
| 70 | 99.9% |
At 23 people the chance first passes one half. A class of 30 has a better than two-in-three chance. By 70 a match is close to certain, even though the pigeonhole principle only guarantees one at 366.
MathWorld also gives the answer for larger coincidences: 88 people are needed for an even chance that three share a birthday, and 187 for four.
Why intuition gets it wrong
The intuitive guess compares people with days. The right comparison is pairs with days. A match can happen between any two people, and 23 people form
C(23, 2) = 253 pairs.
Each pair matches with probability 1/365, so there are 253 chances at a 1-in-365 event. The expected number of matching pairs is 253/365 ≈ 0.69, as the expected value guide shows using linearity, and a probability of about one half follows naturally.
A handy approximation makes this precise. For small probabilities, each pair avoids a match with probability about e^(−1/365), and treating the pairs as roughly independent gives
P(no match) ≈ e^(−n(n − 1)/730).
For n = 23 that is e^(−0.693) ≈ 0.500, close to the exact value of about 0.493. Setting the exponent equal to ln 2 shows the 50% point grows like the square root of the number of days: about 1.18√365 ≈ 22.5.
Matching your own birthday is different
If the question is "will anyone here share my birthday?", only the pairs that include you count, n − 1 of them rather than C(n, 2). Each other person misses your birthday with probability 364/365, so
P(nobody shares mine) = (364/365)ⁿ,
which first drops below one half at n = 253. Confusing "some pair matches" with "someone matches me" is exactly the error behind the intuitive guess.
Real birthdays
Births are not spread perfectly evenly across the year; MathWorld notes that in the United States September runs more than 6% above the average. Unevenness can only increase the chance of a match. For a single pair this is easy to check: two people match with probability p₁² + p₂² + ⋯ + p₃₆₅², where pₖ is the chance of being born on day k, and that sum is smallest when every pₖ equals 1/365. The same holds for groups of any size, so the uniform model gives the smallest possible probability, and the table above is, if anything, an underestimate for real groups.
Hash collisions and the birthday attack
The same mathematics governs hash collisions in computing. A hash function maps items into N possible values, like people into days. A collision, two items with the same hash, becomes likely after only about √N items, not N/2. For a 32-bit hash with about 4.3 billion values, an even chance of a collision arrives after roughly 77,000 items.
The same square-root rule applies to an attacker searching for any two inputs with the same hash, a "birthday attack": the effort is on the order of the square root of the number of possible outputs, not the number itself. Hashing itself is built on remainders, the arithmetic covered in modular arithmetic.
Try it
Check the table for your own group size, or compute it step by step: multiply 365/365 × 364/365 × … for n terms and subtract from 1. For a deeper look at combining probabilities this way, see conditional probability and Bayes' theorem.
Probability questions like this one belong to the probability unit of Discrete Math: Study & Quiz. It is part of our education app collection, and the discrete math hub links the counting guides that the calculation relies on.
Good to know
Frequently asked questions
What is the birthday problem?
How do you calculate the probability of a shared birthday?
How many people do you need for someone to share my birthday?
Is the birthday problem affected by birthdays not being evenly spread?
Sources
- Birthday Problem (opens in a new tab)Wolfram MathWorld — accessed
- 7.9 Conditional Probability and the Multiplication Rule (Contemporary Mathematics) (opens in a new tab)OpenStax — accessed
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