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With 365 equally likely outcomes and independent draws made with replacement, the chance of seeing no repeat after 70 draws is about 0.084%. That means the chance of at least one repeated outcome is about 99.916%. These figures are calculated from the standard collision-probability formula; round 70 has no universal significance unless the number of possible outcomes and the drawing rules are specified.
Why a repeat is so likely by draw 70
The key question is whether any two draws match—not whether a particular outcome appears again. After 70 draws, there are 70 × 69 ÷ 2 = 2,415 possible pairs of draws that could match. Each individual pair may seem unlikely to coincide, but thousands of possible pairings give a repeat many opportunities to occur. This is the collision effect behind the birthday problem.
For a familiar illustration, Berkeley’s CS 70 Spring 2025 notes show that in an idealized model with 365 equally likely birthdays, the probability of a shared birthday exceeds 50% among 23 people and exceeds 99% among 60 people. These are model-based birthday examples, not claims about actual birthday patterns. Berkeley CS 70, Note 13 (Spring 2025)
How the 70-draw probability is calculated
Let N be the number of equally likely outcomes, and let k be the number of independent draws with replacement. The first draw cannot be a repeat. For every later draw to be new, it must avoid all outcomes already drawn:
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P(no repeat after k draws) = (N/N) × ((N−1)/N) × ((N−2)/N) × … × ((N−k+1)/N) = ∏j=0k−1(1−j/N).
The probability of at least one repeat is one minus that product. With N = 365 and k = 70, the product is approximately 0.00084, or 0.084%; its complement is approximately 0.99916, or 99.916%. These are arithmetic evaluations of the formula, not separately published statistics. Berkeley’s notes and Pearson’s sampling reference describe the product-and-complement method. Berkeley CS 70, Note 13 (Spring 2025); Pearson Random Sampling Calculator
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What changes the answer
The number of possible outcomes
The calculation depends on N. A 70-draw process with thousands of possible outcomes does not have the same collision probability as one with 365. “No repeat through 70” is incomplete unless the outcome-space size is known.
Whether outcomes can be drawn again
The formula assumes replacement: after each draw, every outcome remains available. If each selected outcome is removed, the process is sampling without replacement and a repeat cannot occur within that sample. Pearson distinguishes these two sampling rules. Pearson Random Sampling Calculator
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Whether outcomes are equally likely
The 365-outcome calculation assumes every outcome has the same probability and that draws are independent. If some outcomes are more common than others, the uniform formula is not exact; the distribution and dependence between draws must be accounted for.
Which event you mean
“Any outcome repeats” is different from “a particular outcome returns.” Under the uniform independent model, each specific next outcome remains equally likely on each draw; an outcome does not become due because it has not appeared recently.
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A first repeat is not the same as collecting everything
Waiting for any duplicate is the birthday or collision problem. Waiting until every possible outcome has appeared is the coupon-collector problem, a different stopping rule. In the uniform model, if i−1 distinct outcomes have appeared, the probability that the next draw is new is 1−(i−1)/N. The expected number of draws needed to collect all N outcomes is N times the Nth harmonic number, approximately N ln N. Tufts University Computer Science: The Coupon Collector Problem
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