Case 1
- Success probability (p)
- 0.2
- Trial of first success (k)
- 3
0.128000
Open with these valuesGeometric Distribution Calculator
0.128000
Result: 0.128000This asks how long you wait, not how many you get: the chance that the first success falls exactly on trial k. With a 20 % chance per trial, the third trial is the first success 12.8 % of the time, and the expected wait is 1 ÷ p = 5 trials.
Worked examples
Case 2
- Success probability (p)
- 0.5
- Trial of first success (k)
- 3
0.125000
Open with these valuesCase 3
- Success probability (p)
- 0.1
- Trial of first success (k)
- 4
0.072900
Open with these valuesHow it's calculated
P(X = k) = (1 − p)ᵏ⁻¹ × p
- StepEnter the chance that a single trial succeeds, as a decimal.
- StepEnter the trial number on which the first success is to land.
- ResultRead the probability of exactly that wait.
What this number means
Counts trials, not the failures before them
Here k is the trial the first success lands on, so k runs 1, 2, 3, … and P(X = 1) is simply p. The other convention counts the failures ahead of the success and shifts every k by one.
Exactly on trial k, not by trial k
The result is the chance that the first success falls on that trial and on no earlier one: 0.128000 for p = 0.2 at k = 3. Succeeding somewhere within three trials is a different, larger number.
The typical wait is 1 ÷ p
With p = 0.2 the first success arrives on the fifth trial on average, and the variance is (1 − p) ÷ p². The smaller p is, the longer the typical wait.
p must be the same on every trial
The formula assumes independent trials that each succeed with the same probability p. If p drifts from trial to trial, this is the wrong model.
Commonly misread
With p = 0.2 the first trial gives 0.8 × 0.2 = 0.16.
That is the other convention, in which k counts failures. Here k = 1 is the trial of the first success itself, so P(X = 1) = p = 0.2.
P(X = 3) = 0.128 is the chance of a success within the first three trials.
It is the chance that the first success lands exactly on the third trial, the first two having failed. A success anywhere in three trials is a larger number.
The geometric distribution counts successes in a fixed number of trials.
That is the binomial distribution. The geometric fixes the success at the first one and lets the number of trials vary.
Reference table
| p, k | Wait | P(X = k) |
|---|---|---|
| 0.2, 1 | first try | 0.200000 |
| 0.2, 3 | 20 % per trial, third try | 0.128000 |
| 0.25, 2 | second try at 25 % | 0.187500 |
| 0.5, 3 | fair coin, third flip | 0.125000 |
| 0.1, 4 | 10 % per trial, fourth try | 0.072900 |
Questions
How do I calculate a geometric probability?
Use P(X = k) = (1 − p)ᵏ⁻¹ × p, where p is the per-trial success probability and k is the trial on which the first success occurs. With p = 0.2 and k = 3 that is 0.8² × 0.2 = 0.128, about a 12.8 % chance.
What is the geometric distribution?
It is the distribution of how many independent trials it takes to get the first success, when every trial succeeds with the same probability p. It answers questions such as how likely the first heads is on the fourth flip, or the first defective part the seventh one inspected.
What is the mean of the geometric distribution?
The expected trial of the first success is 1 ÷ p, so with p = 0.2 the first success arrives on the fifth trial on average. The variance is (1 − p) ÷ p². The smaller p is, the longer the typical wait.
Does it count trials or failures?
There are two conventions, and this calculator uses the trial of the first success, with k = 1, 2, 3, … and P(X = k) = (1 − p)ᵏ⁻¹ × p. Some textbooks count the failures before the first success instead, which shifts every k by one — check which version your source uses.
How is this different from the binomial distribution?
The binomial counts how many successes occur in a fixed number of trials. The geometric counts how many trials it takes to reach the first success, so the number of trials is what varies. Both assume independent trials with a constant p.
Sources and last check
Information, not professional advice.