Calculate expected trials, target chance, or per-trial probability from success odds, observed results, and desired confidence levels.
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Expected Number of Trials Formula
For an event with a constant success chance per trial, the expected number of trials until the first success follows the geometric distribution.
E = 1 / p
- E = expected number of trials until the first success (includes the successful trial)
- p = probability of success on a single trial (0 < p ≤ 1)
To find how many trials you need to reach a target chance of at least one success:
n = ln(1 − T) / ln(1 − p)
- n = trials needed, rounded up
- T = target probability of at least one success (e.g. 0.95)
To estimate p from observations, use p = successes / trials. Assumptions: trials are independent, p does not change between trials, and each trial has only two outcomes (success or failure).
Reference Tables
Use these to sanity check the calculator output without re-running it.
| Success chance (p) | Expected trials (1/p) | Trials for 90% | Trials for 95% | Trials for 99% |
|---|---|---|---|---|
| 50% | 2 | 4 | 5 | 7 |
| 25% | 4 | 9 | 11 | 17 |
| 10% | 10 | 22 | 29 | 44 |
| 5% | 20 | 45 | 59 | 90 |
| 1% | 100 | 230 | 299 | 459 |
| 0.1% | 1,000 | 2,302 | 2,995 | 4,603 |
| Result | What it means |
|---|---|
| E = 1/p | Long-run average. Half of all attempts finish before this point, half after. |
| Median ≈ ln(0.5)/ln(1−p) | The 50/50 mark, always lower than the mean for a geometric distribution. |
| Std. deviation = √(1−p)/p | For low p, this is close to E itself, so spread is wide. |
| Trials at 95% target | Roughly 3× the expected count for small p. |
Worked Example
A drop has a 5% chance per kill. The expected number of kills to get one drop is 1 / 0.05 = 20.
That does not mean you are guaranteed a drop by kill 20. The actual chance of at least one drop in 20 kills is 1 − (0.95)20 ≈ 64.2%. To hit 95% confidence, solve n = ln(0.05) / ln(0.95) ≈ 58.4, which rounds up to 59 kills.
FAQ
Does the expected count include the successful trial? Yes. E = 1/p counts the success itself. Expected failures before the first success is (1 − p) / p.
Why does my actual run take so much longer than expected? Because the geometric distribution is skewed. The mean is pulled up by long tails, so more than half of all runs finish faster than the mean, and a smaller share take much longer.
Can I use this if p changes between trials? No. The formula assumes p is constant. If the rate changes, the geometric model does not apply.
What if I want a specific number of successes, not just one? The expected trials for k successes is k / p. This calculator handles the first-success case.