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Expected Number of Trials Calculator

Enter the per-trial success probability to find the expected number of trials.

How it is calculated

Each trial is independent with the same success probability p. The tool reads p as a percentage by default (2 means 2%) and divides by 100; if the unit is set to a fraction, p is used as given (0.02). It then computes: expected trials 1 / p; standard deviation sqrt((1 - p) / p2); median ceil(ln 0.5 / ln(1 - p)); and trials for a 90% chance of at least one success ceil(ln 0.1 / ln(1 - p)). Medians and 90% counts are rounded up to whole trials. The probability must be above 0 and at most 100% (or 1); 0 or anything larger gives no result. At p = 100% every figure is 1 trial.

Worked example

A drop has a 2% chance per attempt, so p = 0.02. Expected trials = 1 / 0.02 = 50.00 trials. Standard deviation = sqrt(0.98 / 0.0004) = 49.50. Median = ln 0.5 / ln 0.98 = 34.3, rounded up to 35 trials. For 90% certainty: ln 0.1 / ln 0.98 = 113.97, rounded up to 114 trials. So the average is 50 attempts, but about half of players need 35 or fewer and you need 114 to be 90% sure.

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Common mistakes

  • Typing 0.02 while the unit is percent. That means 0.02%, giving an expected 5,000 trials instead of 50.
  • Typing 2 while the unit is fraction. A value above 1 is not a probability and returns no result.
  • Believing that after the expected number of trials success is guaranteed. At 50 expected trials there is still about a 36% chance of no success yet.
  • Using the mean as the typical experience. The distribution is skewed, so the median (35 here) is below the mean (50).
  • Applying it when the chance changes between attempts, such as pity timers or drawing without replacement. The tool assumes a fixed p.

About this calculator

When each independent trial has a fixed success probability p, the number of trials needed until the first success follows the geometric distribution. Its expected value is E[X] = 1/p, its median is the smallest integer m satisfying (1−p)^m ≤ 0.5, and the number of trials needed to reach 90% confidence of at least one success is the smallest n satisfying (1−p)^n ≤ 0.10.

Statisticians and quality control engineers use these figures to plan sampling or inspection processes, game designers use them to tune random item drop rates so players get a fair sense of how many attempts a rare drop should take, and anyone reasoning about repeated independent attempts — from A/B test iterations to games of chance — uses the geometric distribution to separate an average outcome from a reasonably confident upper bound.

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Comments (1)

  • Editorial team

    The average and the confident figure can be far apart, which matters for sampling plans. If a defect turns up in 0.1% of items, you expect to inspect 1,000 before the first one, but to be 90% sure of catching at least one you need 2,302. Plan around the second number when missing a defect is costly. What kind of repeated attempt are you modelling: inspections, drops, or something else?

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