What is the birthday paradox?

The birthday paradox is the counterintuitive result that in a group of just 23 people, there is a greater than 50% probability that at least two people share the same birthday. People expect the threshold to be much higher — around 183 (half of 365) — but the maths shows otherwise, because the number of possible pairs grows quadratically with group size.

What is the exact formula used here?

The probability that all n people have distinct birthdays in a d-day year is P(all distinct) = (d/d) x ((d-1)/d) x ((d-2)/d) x ... x ((d-n+1)/d). The probability of at least one shared birthday is P(shared) = 1 - P(all distinct). For large n the calculator uses log-space arithmetic — summing logarithms rather than multiplying fractions — to avoid floating-point underflow.

Why does 23 people give roughly 50% probability?

With n = 23 people there are C(23,2) = 253 distinct pairs. Each pair has a 1/365 chance of matching, and these pairs are nearly independent for small collision counts, so intuitively the expected number of matching pairs is 253/365 ≈ 0.69, which corresponds to a Poisson-approximation probability of 1 - e^(-0.69) ≈ 50%. The exact combinatorial formula gives 50.7%.

How many people are needed for a 99% probability?

In a standard 365-day year, 57 people are enough to push the probability of a shared birthday above 99%. The milestone table in the calculator shows this figure alongside thresholds for 25%, 50%, 70%, and 99.99% so you can see the full picture at a glance.

Does the calculator assume birthdays are equally likely on all days?

Yes. The standard birthday-paradox formula assumes a uniform distribution across all d days of the year. In reality, births are slightly more common in late summer and early autumn in the Northern Hemisphere, which makes real-world collision probability marginally higher than the uniform model predicts. For most practical purposes the difference is negligible.

Can I use this for other types of collision problems?

Absolutely. The formula is general: replace 365 with any pool size d, and replace "people" with any objects drawn from that pool. Common uses include hash-function collision analysis (d = 2^k for a k-bit hash), coupon-collector problems, randomised password uniqueness checks, and cryptographic birthday attacks. Just enter the appropriate d in the year-length field.

Birthday Paradox Calculator

The birthday paradox is one of probability theory’s most famous surprises: you need far fewer people than you would intuitively guess for a shared birthday to become more likely than not. This calculator computes the exact probability using the standard combinatorial formula, shows the full probability curve from n = 1 to n = 100, and includes a milestone table so you can instantly see the minimum group size required to hit any target probability.

How it works

Suppose there are d equally likely birthdays (d = 365 for a standard year). The probability that the first person has a unique birthday is d/d = 1. The second person must avoid the first person’s birthday, so their probability of being unique is (d−1)/d. The third person must avoid two occupied days, giving (d−2)/d, and so on. Multiplying these together gives the probability that all n people have distinct birthdays:

P(all distinct) = d/d × (d−1)/d × (d−2)/d × … × (d−n+1)/d

The probability of at least one shared birthday is simply the complement:

P(shared) = 1 − P(all distinct)

For large groups this product would underflow to zero in floating-point arithmetic, so the calculator sums logarithms instead of multiplying fractions, then exponentiates once at the end — preserving full double-precision accuracy for any group size up to d + 1.

Worked example

Consider a standard year with d = 365 days and a group of n = 23 people:

n	P(all distinct)	P(shared birthday)
10	88.31%	11.69%
20	58.86%	41.14%
23	49.27%	50.73%
30	29.37%	70.63%
40	10.87%	89.13%
57	less than 1%	more than 99%

At just 23 people the probability crosses 50% — a result that surprises nearly everyone when they first encounter it. By 57 people the probability is above 99%, and by 70 people it is effectively certain (99.9%).

Why the intuition fails

Most people anchor on the idea that one specific person needs someone else to share their birthday. But the question is whether any pair shares a birthday, and the number of pairs grows as n(n−1)/2. With 23 people there are 253 pairs. Each pair has a 1/365 ≈ 0.27% chance of matching, and 253 × 0.27% ≈ 69% expected matches — more than enough to push the probability over 50%. The quadratic growth of pairs is what makes the threshold so much lower than linear intuition suggests.

Beyond birthdays

The same formula underpins the birthday attack in cryptography: finding two inputs that hash to the same value. For a k-bit hash function the “collision space” has d = 2^k entries; a collision becomes probable after roughly √(2^k) × √(ln 2) ≈ 1.18 × 2^(k/2) attempts. This is why a 64-bit hash is considered insecure (collision after ~4 billion tries) while 256-bit hashes remain safe even for nation-state attackers.

Enter any pool size in the “Days in a year” field to run the same analysis for hash collisions, random identifier uniqueness, coupon-collector problems, or any other combinatorial collision scenario.

Formula note

All calculations run entirely in your browser using JavaScript’s 64-bit IEEE 754 doubles. Log-space computation (summing Math.log(d − k) − Math.log(d) for k = 0 … n−1, then exponentiating once) gives results accurate to approximately 15 significant figures for any group size up to d + 1. No data is ever sent to a server.