Geometric & Negative Binomial Expected Trials Calculator
In a sequence of independent Bernoulli trials with success probability p, the Negative Binomial / Geometric waiting-time distribution requires an average of E[X] = k / p total trials to observe k successes.
Average number of attempts until k successes occur
1 for first success (Geometric), or k ≥ 1 for Negative Binomial
Probability of success per attempt (0.0001 to 1.0)
Step-by-Step Calculation
In a sequence of independent Bernoulli trials with success probability p, the Negative Binomial / Geometric waiting-time distribution requires an average of E[X] = k / p total trials to observe k successes.
Real-World Reference Values
Expected attempts until the very first success
200 attempts expected for k = 10 wins
50% cumulative probability occurs at ~0.693 / p trials
20-Row Comparison Table
Expected Total Trials Required (E[X]) · Target Number of Successes (1 = Geometric) (k)
| # | Target Number of Successes (1 = Geometric) (k) | Expected Total Trials Required (E[X]) | Δ % |
|---|---|---|---|
| 01 | 1 Current | 20 | 0.00% |
| 02 | 73.3612 | 1,467.22 | +7236.12% |
| 03 | 145.7224 | 2,914.45 | +14472.24% |
| 04 | 218.0836 | 4,361.67 | +21708.36% |
| 05 | 290.4447 | 5,808.89 | +28944.47% |
| 06 | 362.8059 | 7,256.12 | +36180.59% |
| 07 | 435.1671 | 8,703.34 | +43416.71% |
| 08 | 507.5283 | 10,150.57 | +50652.83% |
| 09 | 579.8895 | 11,597.79 | +57888.95% |
| 10 | 652.2507 | 13,045.01 | +65125.07% |
| 11 | 724.6118 | 14,492.24 | +72361.18% |
| 12 | 796.973 | 15,939.46 | +79597.30% |
| 13 | 869.3342 | 17,386.68 | +86833.42% |
| 14 | 941.6954 | 18,833.91 | +94069.54% |
| 15 | 1,014.06 | 20,281.13 | +101305.66% |
| 16 | 1,086.42 | 21,728.36 | +108541.78% |
| 17 | 1,158.78 | 23,175.58 | +115777.89% |
| 18 | 1,231.14 | 24,622.80 | +123014.01% |
| 19 | 1,303.50 | 26,070.03 | +130250.13% |
| 20 | 1,375.86 | 27,517.25 | +137486.25% |
Frequently Asked Questions
Why is the median number of attempts for first success less than the mean 1/p?+
Because the Geometric distribution is right-skewed by occasional long losing streaks, 50% of first successes occur by ~0.693 / p attempts, below the mean 1 / p.
What is the Memoryless Property of the Geometric distribution?+
If you have already failed 10 times in a row, the expected number of additional future trials to get your first success is still 1 / p.
How do I use this for multiple sales quota closes?+
Set k to your quota target (e.g., k = 12 deals) and p to your close rate to see total pipeline prospects E[X] required.