At Least One Success Probability Calculator
Find the probability of at least one success in n independent tries, 1 − (1 − p)ⁿ, and the number of tries needed to reach 50%, 90%, 95% or 99% confidence.
e.g. 0.01, 1% or 1/100
Tries needed for common confidence levels
| Confidence of at least one success | Tries needed |
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Step-by-step solution
About this calculator
If something has a 1% chance of happening each time you try, does trying 100 times make it certain? It does not. This calculator answers the question exactly: given the probability p of success on one try and the number of tries n, it gives the probability of at least one success, the probability of none, and how many tries you need to reach a confidence level you choose.
It is useful for game drop rates, the chance that at least one component in a system fails, quality-control sampling, or any “will it happen at least once?” question where the attempts are independent.
How to use it
- Enter p, the chance of success on a single try, as a decimal (0.01), percentage (1%) or fraction (1/100).
- Enter n, the number of tries.
- Optionally change the target confidence (95% by default) to see how many tries you would need to reach it.
- The results and the table of common targets update as you type.
The formulas
Probability of no successes
P(none) = (1 − p)ⁿEach try fails with probability 1 − p. Because the tries are independent, the chance that all n fail is that number multiplied by itself n times.
Probability of at least one success
P(at least one) = 1 − (1 − p)ⁿ“At least one success” is everything except “no successes”, so it is the complement. This is much easier than adding up the chances of exactly 1, 2, 3, … successes.
Tries needed for a target confidence c
n = ⌈ ln(1 − c) / ln(1 − p) ⌉Solve 1 − (1 − p)ⁿ ≥ c for n by taking logarithms, then round up to the next whole number, because a fraction of a try is not possible.
Quick rule for small p
P(at least one) ≈ 1 − e^(−n·p) n₉₅ ≈ 3 / pWhen p is small, (1 − p)ⁿ is close to e^(−np). Setting that equal to 0.05 gives n ≈ ln(20) / p ≈ 3 / p, the “rule of three”.
Worked examples
Example 1: a 1% chance, 100 tries
An item has a 1% drop rate. You defeat the boss 100 times. What is the chance of getting the item at least once?
- P(no drop in one try) = 1 − 0.01 = 0.99.
- P(no drop in 100 tries) = 0.99¹⁰⁰ = 0.3660323413.
- P(at least one drop) = 1 − 0.3660323413 = 0.6339676587.
- Tries for 95% confidence: ln(0.05) / ln(0.99) = 298.07, rounded up to 299.
Answer: About 63.4% — not 100%. About 37% of players who try 100 times will still have nothing. Reaching 50% takes 69 tries, 95% takes 299 and 99% takes 459.
Example 2: rolling a six
What is the chance of at least one six in 4 rolls of a fair die, and how many rolls give a 95% chance? (This is half of a classic 17th-century gambling question associated with the Chevalier de Méré, which helped prompt Pascal and Fermat’s work on probability.)
- p = 1/6, so the chance of no six in one roll is 5/6.
- P(no six in 4 rolls) = (5/6)⁴ = 625/1296 = 0.4822530864.
- P(at least one six) = 1 − 0.4822530864 = 0.5177469136.
- Rolls for 95%: ln(0.05) / ln(5/6) = 16.431, rounded up to 17.
Answer: 51.77% for 4 rolls — just better than even, so an even-money bet on it wins slightly more often than it loses. You need 17 rolls for a 95% chance of at least one six.
When to use the at least one success probability calculator
- Each try either succeeds or fails.
- The chance of success p is the same on every try.
- Tries are independent: failing once does not make the next try more or less likely to succeed.
- You only care whether at least one success happens, not how many.
When it does not apply
- Systems with “pity” or guaranteed-drop mechanics, or anything that gets easier with practice: p changes between tries.
- Drawing without replacement, such as looking for an ace in a single deck: each failed draw changes the odds of the next.
- Different probabilities on each try: use 1 − (1 − p₁)(1 − p₂)…(1 − pₙ) instead.
Common mistakes
- Adding the probabilities. 100 tries × 1% = 100% is wrong; the probabilities overlap, and the correct answer is 1 − 0.99¹⁰⁰ ≈ 63.4%.
- Expecting a guaranteed success after 1/p tries. 1/p is the average number of tries until the first success, but about 37% of the time it takes longer (when p is small).
- Confusing “at least one” with “exactly one”. Exactly one is n·p·(1 − p)ⁿ⁻¹, which is smaller; use the binomial distribution calculator for that.
- Entering the percentage without a percent sign — 5 instead of 5% or 0.05.
Frequently asked questions
Why isn’t 100 tries at a 1% chance a guaranteed success?
Because each try can fail independently. The chance of failing all 100 is 0.99¹⁰⁰ ≈ 36.6%, so the chance of at least one success is about 63.4%. No finite number of tries makes it certain.
How many tries do I need for a 1-in-X chance?
For a small chance of 1 in X, you need about 0.69·X tries for a 50% chance, 2.3·X for 90%, 3·X for 95% and 4.6·X for 99%. The calculator gives the exact whole number for your p.
Is this the same as P(X ≥ 1) in a binomial distribution?
Yes. P(at least one success) is P(X ≥ 1) for a binomial distribution with n trials and probability p, which equals 1 − P(X = 0) = 1 − (1 − p)ⁿ.
What if the probability is different on each try?
Multiply the individual failure probabilities and subtract from 1: P(at least one) = 1 − (1 − p₁)(1 − p₂)…(1 − pₙ). The tries still need to be independent.
What is the expected number of tries until the first success?
On average 1/p tries (this is the mean of the geometric distribution). For a 1% chance that is 100 tries, even though 100 tries only gives about a 63% chance of success.