At Least One Success Probability Calculator
Calculate the chance of at least one success in n independent trials, or find how many tries you need for a 90%, 95% or 99% target probability.
e.g. 0.01, 1% or 1/100
Tries needed for common confidence levels
| Confidence of at least one success | Tries needed |
|---|
Step-by-step solution
At Least One Success Formula
P(at least one) = 1 − (1 − p)ⁿ, where p is the same success probability on every try and n is the number of independent tries. The opposite event is zero successes: all n tries fail, with probability (1 − p)ⁿ.
This is the binomial event P(X ≥ 1) = 1 − P(X = 0). The binomial distribution calculator also handles exactly k, at most k and other count questions.
Probability of at least one success in n trials
With a 1% chance on each independent try, 100 tries give about 63.40%, not 100%. The table shows how the chance changes as you repeat the same experiment.
| Number of tries | Chance of at least one success | Try these inputs |
|---|---|---|
| 10 | 9.56% | Calculate 1% over 10 tries |
| 50 | 39.50% | Calculate 1% over 50 tries |
| 100 | 63.40% | Calculate 1% over 100 tries |
| 299 | 95.05% | Calculate 1% over 299 tries |
“At least one” includes one, two or more successes. If you need exactly one success, use the binomial calculator and choose Exactly k instead. All examples assume the same probability on every try and independent outcomes.
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.
Example 3: at least one head in 3 coin flips
Flip a fair coin 3 independent times. What is the probability of at least one head?
- Use p = 0.5 for a head and n = 3 flips.
- The complement is no heads: all 3 flips are tails, with probability 0.5³ = 0.125.
- P(at least one head) = 1 − 0.5³ = 0.875.
Answer: 87.5%. This includes one, two or three heads; it excludes only the all-tails outcome.
Example 4: at least one double six in 24 dice rolls
Roll two fair dice together 24 independent times. What is the probability of getting a double six at least once?
- One paired roll has 36 equally likely outcomes, so p = 1/36 and n = 24.
- The chance of no double six in one paired roll is 35/36.
- P(no double six in 24 rolls) = (35/36)²⁴ = 0.5085961239.
- P(at least one double six) = 1 − (35/36)²⁴ = 0.4914038761.
Answer: 49.14%. Even 24 paired rolls leave the chance of at least one double six slightly below one half.
Example 5: at least one defective item in 50
Each sampled item is independently defective with probability 0.02. In a sample of 50, what is the chance of finding at least one defective item?
- Set p = 0.02 and n = 50; a nondefective item has probability 0.98.
- P(no defective items) = 0.98⁵⁰ = 0.3641696801.
- P(at least one defective item) = 1 − 0.98⁵⁰ = 0.6358303199.
Answer: 63.58%. This assumes independent items with the same defect rate; sampling without replacement from a small finite batch can need a different model.
How Many Trials Do You Need?
For 0 < p < 1 and a target between 0 and 1, solve n ≥ ln(1 − target) / ln(1 − p) and round up. With p = 0.02 and a 90% target, the unrounded value is 113.974, so you need 114 trials. At 114 trials the probability is about 90.0052%; 113 trials fall below 90%.
The calculator already supports this reverse calculation through its target input. Try the 2% probability and 90% target. A zero success probability cannot reach a positive target; a certain success needs one trial. For 0 < p < 1, no finite number of trials guarantees a 100% outcome.
Choose what to calculate
At a constant 1% chance per independent try, 100 tries give about 63.4% probability of at least one success. A 95% target needs 299 tries. A “95% target” here is an event probability, not a statistical confidence interval.
Common Mistakes
- Adding the probabilities. 100 tries × 1% = 100% is wrong; the probabilities overlap, and the correct answer is 1 − 0.99¹⁰⁰ ≈ 63.4%.
- Using the formula for dependent trials. Drawing without replacement or changing odds after a failure does not satisfy the independent, fixed-p assumption.
- 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.
Past failures do not make the next independent try more likely to succeed. If a game has a pity counter, guaranteed drops, or changing odds, this fixed-p model does not describe it. For two or more successes, use the binomial distribution calculator.
FAQ
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 is the probability of at least one six in 4 dice rolls?
For a fair six-sided die, the chance of not rolling a six is 5/6 each time. Across 4 independent rolls, the chance of no six is (5/6)⁴, so the chance of at least one six is 1 − (5/6)⁴ = 671/1296, or about 51.77%.
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.
How do you calculate the probability of at least one success?
Use P(at least one) = 1 − (1 − p)ⁿ, where p is the probability of success on each try and n is the number of independent tries. The probability p must remain the same on every try.
Why do you use 1 minus the probability of zero successes?
Zero successes and at least one success are mutually exclusive and cover every outcome. Their probabilities add to 1. It is simpler to calculate the all-fail probability (1 − p)ⁿ and subtract it than to add the probabilities of one, two and more successes.
How many trials do I need for a 90% chance of at least one success?
It depends on p. For p = 0.02, solve n ≥ ln(0.10) / ln(0.98) ≈ 113.974 and round up to 114 independent trials. With 113 trials the chance is below 90%; with 114 it is just above 90%.
Can I add the probabilities of each trial together?
Not for an at-least-one event across independent trials: successes can occur on several trials, so direct addition double-counts overlapping outcomes. Four rolls of a fair die do not give 4 × 1/6 = 2/3 for at least one six; the complement calculation gives about 51.77%. Direct addition applies to mutually exclusive events.
Related calculators
More about this calculator, its assumptions and formulas
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”.
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.
Worked Probability Examples
What is the probability of getting exactly 3 heads when a fair coin is flipped 10 times? Read the probability of exactly 3 heads in 10 coin flips worked solution.
If a fair six-sided die is rolled 4 times, what is the probability that at least one roll is a six? Read the probability of at least one six in 4 dice rolls worked solution.
In a sample of 20 independent items, each with an 8% defect probability, what is the chance that exactly 2 are defective? Read the probability of exactly 2 defective items in 20 worked solution.