🎂 The Birthday Paradox

About this simulation

People arrive one at a time and each is assigned a random "bucket" — a day of the year, or, more generally, one of N equally likely slots. Every arrival flies to its bucket's pillar on the ring. The moment two people land on the same pillar, that pillar flashes red: a collision. The surprising part is how few arrivals it usually takes.

How it works

• With N buckets, the probability that n people include no collision is ∏(1 − i/N) for i = 0…n−1. Collision probability is one minus that product — it grows far faster than intuition predicts because it compares every pair, not just each new person against one fixed target.
• For N = 365, a 50% chance of a shared birthday needs only 23 people, and 99% needs just 70 — nowhere near 183.
• The expected number of draws before the first collision is close to √(πN/2) ≈ 1.25√N, plotted live against the empirical average of completed trials.
• The same math governs cryptographic hash collisions: a hash with an n-bit output behaves like N = 2ⁿ buckets, so a "birthday attack" can find two colliding digests in roughly √(2ⁿ) tries instead of the 2ⁿ tries brute force would need — why hash outputs must be twice as long as the desired security level.

Controls

• Interpretation — quick presets: birthdays, a small class, or hash digest spaces of a given bit width.
• Bucket count (N) — fine-tune the number of slots directly.
• Add speed — how fast new people/hashes arrive.
• ⏸ Pause / ↺ Reset run — freeze the animation or restart the current trial early. Drag to rotate, scroll to zoom.

Did you know?

MD5 (128-bit) and SHA-1 (160-bit) were both broken via birthday-style collision attacks, which is why modern protocols require at least 256-bit hash outputs to keep the birthday bound comfortably out of reach.