formula.expert

Expected Trials Until First Success

E[X]=1pE[X] = \frac{1}{p}

Worked example: Average wait of 4 attempts → p = 0.25 — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

Learning zone

Expected Trials Until First Success explained

pE[X]

If one attempt in p succeeds, you wait on average 1/p attempts for the first success. A fair die needs 6 rolls on average to show a six; a 1-in-1000 manufacturing fault surfaces every 1000 units on average. Enter p as a decimal from 0 to 1, or switch the unit to %.

The average is a poor summary of this distribution, and that is the trap. Waiting times here are heavily skewed: with p = 1/6 the mean is 6, yet the most likely single outcome is 1, and there is still a (5/6)¹² ≈ 11% chance of going twelve rolls with no six at all. Half of all waits finish by roll 4, while a long tail drags the average up. Reversed, the formula turns an observed waiting time into a rate: a machine that jams once every 40 cycles on average has p = 1/40 = 0.025 per cycle. In queueing and reliability work the continuous cousin of this result gives the exponential distribution's mean, 1/λ.

Expected Trials Until First Success formula

E[X]=1pE[X] = \frac{1}{p}
Where
  • E[X]E[X]= Expected number of trials
  • pp= Probability of success per trial

Missing one of these? Work it out first, then come back