Probability of at least one success

Does a 1% chance repeated 100 times guarantee a win? Predict the probability before revealing the calculation.

Formula and worked example

An average is not a guarantee

With a fixed 1% chance on each independent draw, the expected number of wins in 100 draws is one. But individual sets can contain zero, one or several wins. The expected count describes an average over many sets; it does not promise a win in yours.

Calculate the opposite event

A failed draw has probability 0.99. Failing every one of 100 independent draws has probability 0.99¹⁰⁰, about 36.6%. Subtract that from one: the chance of at least one success is about 63.4%. In general, P(at least one) = 1 − (1 − p)ⁿ, where p is the single-draw probability as a decimal.

Past failures do not create a debt

A fair coin that has landed heads five times still has a 50% chance of tails on its next independent flip. Predicting that tails is “due” confuses a past sequence with a future trial. Use the simulator repeatedly to see how small samples vary.

Assumptions and limits

This model assumes constant probability and independent draws. Guarantees, increasing odds, draws without replacement and changing conditions require another model. The simulation has no prizes, payments or real-world rewards. A higher probability is not a reason to spend more.

This is a learning exercise, not a validated assessment.