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With 365 equally likely outcomes, independent draws with replacement, and 70 rounds, the chance of seeing no repeat is about 0.084%—so the chance of at least one repeat is about 99.916%. These are calculated birthday-problem probabilities, not results from a real draw system. The answer changes if the number of possible outcomes, their probabilities, or the draw rules change.
Why a repeat is so likely by round 70
The key is that the event is “any two rounds match,” not “a particular outcome appears again.” Across 70 rounds there are 2,415 possible pairs of rounds that could share an outcome. Each pair may have a small chance of matching, but the many possible pairs make at least one collision likely.
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That is the same counterintuitive effect behind the birthday problem. In Berkeley’s idealized 365-day model, the chance of a shared birthday exceeds 50% with 23 people and is over 99% with 60 people. A 70-draw calculation under the same uniform model yields an even higher repeat probability. Berkeley CS 70, Note 13 (Spring 2025)
How to calculate the chance of no repeat
Suppose there are N equally likely outcomes, and every draw is independent and made with replacement, so all N outcomes remain available each round. The first draw cannot repeat. The second must avoid the first outcome, the third must avoid the first two, and so on.
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For k draws:
P(no repeat) = (N/N) × ((N−1)/N) × ((N−2)/N) × … × ((N−k+1)/N) = ∏j=0k−1(1−j/N)
Then P(at least one repeat) = 1 − P(no repeat). This product-complement method is described in Berkeley’s notes and Pearson’s explanation of random sampling with replacement. Berkeley CS 70, Note 13 (Spring 2025); Pearson Random Sampling Calculator
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The 365-outcome, 70-round example
Putting N = 365 and k = 70 into the product gives a no-repeat probability of about 0.00084, or 0.084%. Its complement is about 0.99916, or 99.916%, for at least one repeated outcome. These percentages are arithmetic evaluations of the formula, not a separately published statistic.
What “random draw” must mean for this answer to apply
The 70-round result is not meaningful on its own. To use it for a real situation, identify all of the following:
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- Outcome count: How many distinct results are possible?
- Draw rule: Does each result remain eligible on every round, or is it removed after selection?
- Distribution: Are all results equally likely? If not, the uniform formula is not exact.
- Event: Are you asking whether any result repeats, whether one specified result returns, or how long it takes to see every result?
Under independent, uniform draws with replacement, each specific outcome remains just as likely on the next draw as it was before. The high chance of a repeat comes from the number of pairs available to match, not because an outcome becomes “due.” The model does not establish the behavior of any particular real-world draw system.
Independent reader supportYour contribution helps us test, update, and keep practical guides available for everyone.Replacement and collection are different problems
With replacement: a first repeat can happen
When every outcome stays eligible on every round, the same value can appear again. Use the product above to calculate the chance of at least one repeat after a chosen number of draws. Pearson Random Sampling Calculator
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Without replacement: a repeat is impossible
If each selected outcome is removed before the next draw, every result in the sample is distinct by construction. The independent-with-replacement repeat formula does not apply. Pearson Random Sampling Calculator
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Collecting every outcome: the coupon-collector problem
Waiting until all possible outcomes have appeared is not the same as waiting for the first duplicate. If there are N equally likely outcomes, after i−1 distinct outcomes have appeared, the next draw has probability 1−(i−1)/N of being new. The expected number of draws to collect all outcomes is N times the Nth harmonic number, approximately N ln N. This is the coupon-collector problem. Tufts Computer Science, The Coupon Collector Problem
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