Calculators
Birthday Collision Calculator
Calculate the chance of at least one repeated outcome across independent uniform draws, including the classic birthday problem.
Calculate the chance of at least one repeated outcome in independent draws with replacement from equally likely outcomes. Use 0–100000 draws and 1–1000000000000 possible outcomes. For birthdays, 365 outcomes ignores leap days and seasonal variation. All processing stays in your browser.
How birthday collision probability works
For n independent draws with replacement from d equally likely outcomes, the chance of at least one repeated outcome is 1 − ∏(1 − i/d), for i from 0 to n−1. If n exceeds d, a collision is certain. With zero or one draw, a collision is impossible. For 23 people and 365 equally likely birthdays, the probability is about 50.7297234324%. The expected number of matching pairs is n(n−1)/(2d); multiple pairs can occur, so this count is not the collision probability. Birthday modeling ignores leap days and nonuniform birth frequencies. This calculator assumes independence and uniform outcomes; it does not model a supplied nonuniform distribution. Enter whole numbers: 0–100000 draws and 1–1000000000000 outcomes. Logarithms, log1p, expm1 and compensated summation preserve small probabilities without multiplying many tiny factors. Results use up to 12 significant digits and may round to 0% or 100%. All calculation runs in your browser.
Frequently Asked Questions
Everything you need to know about this tool, how it works, and privacy.
What is the classic birthday problem result?
With 23 people and 365 equally likely birthdays, at least two share a birthday with probability about 50.7297234324%.
What assumptions does this use?
Draws are independent and each possible outcome is equally likely. Birthday calculations ignore leap days and seasonal variation.
What if there are more draws than outcomes?
At least one repeated outcome is certain by the pigeonhole principle.
Are expected pairs a probability?
No. Expected matching pairs is n(n−1)/(2d). Several matching pairs can occur in a single group, so the expected count can exceed one.
Does my input leave my device?
No. All calculations run locally in your browser.