How to Find the Probability of at Least One Success

This finds the chance that something happens at least once over repeated attempts. Counting up exactly one, exactly two and so on is laborious; taking the opposite needs a single multiplication.

$p$ is the chance each time and $n$ the number of attempts. $(1-p)^n$ is the chance of it never happening, and everything else counts as at least once.

Example

Thirty draws at 3% each. The chance of never winning is $0.97^{30} = 0.401$, or 40.1%, so the chance of winning at least once is 59.9%. The expected number of wins is $30 \times 0.03 = 0.9$.

An expectation near one is no guarantee

Even with 0.9 wins expected, four times in ten there is no win at all. The expected count and the chance of at least one are different quantities. For small $p$, the number of attempts needed to pass a 50% chance is roughly $0.69 \div p$, which here is 23.

Taking the opposite

Anything phrased as at least one is usually quicker through its opposite, none at all. The chance of rolling at least one six in four throws of a die falls out at once as $1 - (5/6)^4 = 51.8$%.

Notes

This assumes the chance is the same every time and unaffected by earlier results. It does not apply to drawing without replacement.