What Is Birthday Paradox?
The birthday paradox is the surprising fact that a small group already has a good chance of a shared birthday. The probability that all n birthdays are different is the product (d/d) × ((d − 1)/d) × … × ((d − n + 1)/d) with d possible days, and the chance of at least one match is 1 minus that.
The calculator reports the probability for your group size, the number of pairs of people, the expected number of matching pairs, and the group size needed for 50 % and 99 %. It also shows a quick exponential approximation. Set d to a different number to model other situations, such as matching hash values.
Below you will find the formulas, a variables table, a worked example and answers to common questions such as why the answer feels counter-intuitive.
How to Use the Birthday Paradox Calculator
- Enter the number of people in the group.
- Leave the number of possible values at 365, or change it for other collision problems.
- Set a target probability to see the group size needed to reach it.
- Read the exact probability, the pairs count and the curve. Use the table for a quick lookup.
Birthday Paradox Formula
Probability that all n birthdays differ
Probability of at least one match
Approximation
Group size for 50% (rule of thumb)
Birthday Paradox Variables Explained
| Symbol | Meaning | Notes |
|---|---|---|
| n | Number of people | |
| d | Number of possible days (or values) | 365 by default |
| P(all differ) | No shared birthday | Product of falling fractions |
| P(match) | At least one shared birthday | 1 − P(all differ) |
| n(n − 1)/2 | Number of pairs |
Birthday Paradox Worked Example
With the example values loaded in the calculator above, the steps are:
- Probability that all 23 people have different birthdays = (365/365) × (364/365) × … × (343/365) = 0.49270277.
- So P(at least one shared birthday) = 1 − 0.49270277 = 0.50729723.
- Searching n upward, the smallest group with probability at least 90% is n = 41.
Result: Chance at least two of 23 share a birthday = 0.5072972; People needed for a 90% chance = 41.
Why the Birthday Answer Feels Wrong
With 23 people the chance that at least two share a birthday is 0.5073, just over a half. The surprise comes from counting pairs rather than people: 23 people form 253 pairs, and each pair has a 1-in-365 chance of matching. The expected number of matching pairs is 253 ÷ 365 = 0.693, and 1 − e^(−0.693) = 0.5, which is the quick approximation shown in the results.
The result assumes every day is equally likely and ignores 29 February. Real birthdays are not evenly spread, which makes a match slightly more likely, not less. The page also gives the group size for other confidence levels: 41 people for a 90% chance and 57 for 99%.
Why does it work with so few people?
The number of pairs grows with the square of the group. With n people there are n(n − 1)/2 pairs. Ten people have 45 pairs, 23 have 253, and 50 have 1,225. Each pair has a small chance of matching, but with hundreds of pairs those small chances add up quickly.
It is also easier to compute the chance that everyone has a different birthday and subtract from 1. The second person avoids the first person’s birthday with probability 364/365, the third avoids both with 363/365, and so on; multiplying these gives the chance of no match.
| People | Pairs | Probability of a match |
|---|---|---|
| 10 | 45 | 11.7% |
| 20 | 190 | 41.1% |
| 23 | 253 | 50.7% |
| 30 | 435 | 70.6% |
| 40 | 780 | 89.1% |
| 50 | 1,225 | 97.0% |
| 70 | 2,415 | 99.92% |
Assumptions and real-world differences
- Birthdays are uniformly spread over 365 days. In reality there are seasonal patterns, which make a match slightly more likely.
- February 29 is ignored. Using 366 changes the numbers only slightly.
- People are unrelated: twins or family members would increase the chance of a match.
- The question is about any pair sharing a birthday, not about sharing a particular person’s birthday. The chance someone matches your birthday needs about 253 people for 50%.
The same maths in computing and security
The birthday problem describes collisions: two items landing in the same bucket. In computing it explains why hash functions need many more bits than you might expect. A hash with d possible outputs starts producing collisions after about √d items, which is why a 128-bit hash gives only about 64 bits of collision resistance. Cryptographic “birthday attacks” exploit this. Change the “possible values” input to 2^32 = 4,294,967,296 to see that around 77,000 random 32-bit IDs already have a 50% chance of a collision.
Where Birthday Paradox Is Used
The classic puzzle
Find the chance that at least two of 23 people share a birthday, which is just over 50%.
Other sets of values
Change 365 to another number to model shared hash values, IDs or dice outcomes.
People needed
Find how many people give a 50%, 90% or 99% chance of a match.
Teaching probability
Show why the number of pairs grows much faster than the number of people.
Common Birthday Paradox Mistakes to Avoid
- To match a specific date, not any pair, use the “at least once” probability with p = 1/365 instead.
- For large d, the square-root rule n ≈ 1.18√d gives the 50% point.
- The curve is S-shaped: it moves slowly at first, then quickly, then flattens.
Birthday Paradox Terms and Definitions
- Collision
- Two items mapping to the same value.
- Pair
- A combination of two people; there are n(n − 1)/2 in a group of n.
Birthday Paradox FAQs
What is the birthday paradox?
The counter-intuitive fact that in a group of only 23 people there is about a 50.7% chance that two share a birthday.
How many people do you need for a 99% chance?
57 people give about 99% chance. With 70, the chance is over 99.9%.
Is it really a paradox?
No. It is only surprising because we compare with a single person’s birthday, while the question involves every pair in the group.
What is the chance someone shares my birthday?
That is a different problem. You would need about 253 people for a 50% chance, because only comparisons with your own date count.
How is it used in cryptography?
Birthday attacks look for hash collisions in about √d tries rather than d. That is why hash outputs need to be long.
Does leap day change the result?
Slightly. With 366 days the group size for 50% is still 23, and the probabilities differ only in the third decimal place.
Can I use a number of days other than 365?
Yes. Enter any number of possible values to model other collision problems.
How many people are needed for a 50% chance of a shared birthday?
23 people, which gives 50.7% with 365 days. For 90% you need 41 and for 99% you need 57.
Why is the birthday paradox surprising?
You are not comparing one person with the group but every pair. 23 people form 253 pairs, so matches pile up much faster than intuition expects.
How is the probability calculated?
Find the chance everyone is different, 365/365 × 364/365 × … for n people, and subtract from 1. The tool also shows the approximation 1 − e^(−n(n−1)/2d).
What assumptions does it make?
Birthdays are independent and equally likely, and leap days are ignored. Real birthdays are not perfectly uniform, which slightly raises the true chance of a match.