Hash Collision Birthday Paradox Calculator

The birthday paradox states that in a group of just 23 people, there is a 50% chance two share a birthday — far fewer than the 183 most people intuit. The same math applies to hash functions.

Why collisions matter. A collision occurs when two different inputs produce the same hash output. In hash tables, collisions degrade performance. In cryptography, they can completely break a scheme. The birthday paradox tells us how quickly collisions appear even with large hash spaces.

The probability of at least one collision:

P(collision) ≈ 1 - e^(-n(n-1) / (2 × 2^b))

For small n relative to 2^b, this simplifies to approximately n² / 2^(b+1).

where n is the number of values and b is the hash size in bits.

The 50% collision threshold:

n₅₀ ≈ 1.177 × 2^(b/2)

For a 32-bit hash: n₅₀ ≈ 77,000 values. For a 64-bit hash: n₅₀ ≈ 5.1 billion values. For a 128-bit hash: n₅₀ ≈ 2^64 values — essentially impossible in practice. For SHA-256 (256 bits): n₅₀ ≈ 2^128 values.

This is why MD5 (128-bit output) is broken for collision resistance — attackers can find collisions in roughly 2^64 operations, which is feasible. SHA-256 requires 2^128 operations for a collision, which is currently computationally impossible.

The practical lesson: a b-bit hash only provides b/2 bits of collision resistance, not b bits.