Probability calculator

Use the complement for at least one

With a 10% chance on each of ten independent trials, the chance of no successes is 0.9 to the power of ten. Subtract that from one to get about 65.13% for at least one success.

All, none and exactly k differ

All successes uses p to the power of n. Exactly k successes also counts the different orders in which those successes can occur. The expected number, n times p, is an average and can be a non-integer.

Define the event before calculating

The formula assumes the same probability on independent trials. Sampling without replacement, correlated outcomes and a streak appearing anywhere in a longer session require other models. A past miss does not raise the next-trial chance under this model.

Drop chance and one-in probabilities

A one-in-N chance is 100/N percent: one in 10 is 10%; one in 5000 is 0.02%; one in 10000 is 0.01%. Enter the percent, then the attempts. For a one-in-10000 event repeated 10000 independent times, the chance of at least one success is about 63.21%, not 100%.

At least one success is 1 − (1−p)^n. This can represent a rare drop only when the item has a known fixed probability on independent attempts. A pity counter, changing drop table or correlated result needs its actual rules.

One percent of 10000 is 100, whereas one in 10000 is 0.01%. Also distinguish probability from payout odds: American +10000 corresponds to an implied chance of 1/101, about 0.99%, not one in 10000.

Choose a model that matches the event

Use lottery odds for draws without replacement, dice probabilities for sums, and bingo line probability for a specified line completed by a draw count. For other explicit models, compare safe picks on a mine grid, higher or lower card probabilities, and Plinko path probabilities. This calculator does not forecast playoffs or infer game-specific drop rates.

When more attempts still leave a substantial miss chance

A one-in-six success chance repeated six independent times gives 1−(5/6)⁶ ≈ 66.51% for at least one success. It is not 100%. The expected number of successes is one, but the actual count can be zero, one or more.

For four independent fair flips, exactly two heads has probability C(4,2) × 0.5⁴ = 37.5%. The example buttons load these two different event definitions. The tool assumes fixed independent trial probabilities; use without-replacement card probabilities when draws change the remaining pool.