Probability of At Least One Success

"At least one" is the complement of "none at all", and "none at all" is a simple product: all n attempts must fail, which happens with probability (1 − p)^n. At least one six in four rolls of a die is 1 − (5/6)⁴ = 1 − 625/1296 ≈ 0.518, a favourable bet. The Chevalier de Méré made money on exactly that wager, then reasoned that 24 throws of two dice should be equally good since 24/36 matches 4/6 — but 1 − (35/36)²⁴ ≈ 0.491, just under even, and his losses prompted the 1654 Pascal–Fermat correspondence that founded the subject.

Enter p as a decimal from 0 to 1, or switch the unit to %. Solving for n answers the planning question: to reach a 99% chance of at least one success when each attempt works 10% of the time, n = ln(0.01)/ln(0.9) ≈ 43.7, so 44 attempts. The trap is adding probabilities instead — ten 10% attempts do not give 100%, they give 1 − 0.9¹⁰ ≈ 0.651. Note the useful rule of thumb that n = 1/p attempts lands near 1 − e⁻¹ ≈ 63%, never certainty.