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Geometric Distribution



Key Terms
The Probabilistic Experiment Behind geometric distribution
Notation Used
Parameters
Probability Mass Function (PMF) and Support (Range)
Cumulative Distribution Function (CDF)
Expected Value (Mean)
Variance and Standard Deviation
Mode and Median
Applications and Examples
Interactive Calculator
Geometric at a Glance



Geometric Distribution: Trials Until First Success


The geometric distribution is based on a sequence of independent Bernoulli trials that continues until the first success occurs. The number of trials is not fixed in advance; instead, the experiment stops as soon as success is observed. The random variable represents how long it takes to achieve the first success, making the timing of success — not the total count — the central focus.

Key Terms

Geometric Distribution— number of trials until first success
Bernoulli Experiment— the single trial repeated until success
Sequence of Bernoulli Trials— the repeated structure of the geometric
Negative Binomial Distribution— the geometric is the r=1r=1 special case
Expected Value— E[X]=1/pE[X] = 1/p
Variance— Var⁡(X)=(1−p)/p2\operatorname{Var}(X) = (1-p)/p^2

See All Probability Definitions →


The Probabilistic Experiment Behind geometric distribution


The geometric distribution serves to count the number of trials required until the first success occurs. Unlike the binomial distribution, the number of trials is not fixed in advance. Instead, trials continue indefinitely until success happens for the first time.

Each trial is a Bernoulli experiment: two outcomes, constant success probability, and independence between trials. The random variable counts when success occurs, not how many successes occur in total. This makes the geometric distribution fundamentally about waiting time rather than accumulation.

A defining characteristic of the geometric distribution is the memoryless property: the probability that success occurs in future trials does not depend on how many failures have already occurred. The process effectively "restarts" after each failure.



Example:

Flipping a coin repeatedly until the first heads appears. If X=4X=4, this means the first three flips were tails and the fourth flip was heads. The experiment stops as soon as success occurs.


Notation Used


X∼Geom(p)X \sim \text{Geom}(p) or X∼Geometric(p)X \sim \text{Geometric}(p) — distribution of the random variable.

Geom(p)\text{Geom}(p) — used to denote the distribution itself (not the random variable).

G(p)G(p) — less common shorthand in some texts or software contexts.

P(X=k)=(1−p)k−1p,for k=1,2,3,…P(X = k) = (1 - p)^{k - 1} p, \quad \text{for } k = 1, 2, 3, \ldots — probability mass function (PMF), where:

pp — probability of success on each trial

kk — number of trials until first success

Alternative PMF formulation:

P(X=k)=(1−p)kp,for k=0,1,2,…P(X = k) = (1 - p)^k p, \quad \text{for } k = 0, 1, 2, \ldots — number of failures before first success

Alternative notations:

q=1−pq = 1 - p — probability of failure, so PMF can be written as P(X=k)=qk−1pP(X = k) = q^{k-1} p

Key properties:

E[X]=1pE[X] = \frac{1}{p} — expected value (mean)

Var(X)=1−pp2\text{Var}(X) = \frac{1-p}{p^2} — variance

Memoryless property:

P(X>m+n∣X>m)=P(X>n)P(X > m + n \mid X > m) = P(X > n) — the distribution has no memory

See All Probability Symbols and Notations →

Parameters


𝑝𝑝: probability of success on a single trial, with 0<𝑝≤10<𝑝≤1

The geometric distribution models the number of trials needed to get the first success in a sequence of independent Bernoulli trials.

There's only one parameter — 𝑝𝑝, the chance of success each time — which completely determines the shape of the distribution.

The outcomes are positive integers: 1,2,3,…1,2,3,… where each value represents the trial number on which success first occurs.

Probability Mass Function (PMF) and Support (Range)


The probability mass function (PMF) of a geometric distribution is given by:

P(X=k)=(1−p)k−1p,k=1,2,3,…P(X = k) = (1-p)^{k-1} p, \quad k = 1, 2, 3, \ldots



* First Success: The geometric distribution models the number of trials needed to get the first success in a sequence of independent Bernoulli trials.

* Support (Range of the Random Variable):
• The random variable XX can take on values 1,2,3,…1, 2, 3, \ldots (all positive integers).
• X=kX = k means the first success occurs on the kk-th trial.
• The support is thus a countably infinite set.

* Logic Behind the Formula:
• (1−p)k−1(1-p)^{k-1}: The probability of getting k−1k-1 failures before the first success
• pp: The probability of success on the kk-th trial
• The total probability sums to 1:

∑k=1∞P(X=k)=∑k=1∞(1−p)k−1p=p∑k=1∞(1−p)k−1=p⋅11−(1−p)=p⋅1p=1\sum_{k=1}^{\infty} P(X = k) = \sum_{k=1}^{\infty} (1-p)^{k-1} p = p \sum_{k=1}^{\infty} (1-p)^{k-1} = p \cdot \frac{1}{1-(1-p)} = p \cdot \frac{1}{p} = 1

• This uses the geometric series formula: ∑k=0∞rk=11−r\sum_{k=0}^{\infty} r^k = \frac{1}{1-r} for ∣r∣<1|r| < 1

Geometric Distribution

Trials until first success (probability p)

Explanation

The geometric distribution models the number of trials needed to get the first success in repeated independent trials. The probability mass function is P(X=k)=(1−p)k−1pP(X = k) = (1-p)^{k-1} p. The expected value is E[X]=1pE[X] = \frac{1}{p} and the variance is Var(X)=1−pp2\text{Var}(X) = \frac{1-p}{p^2}. Common applications include counting coin flips until the first heads, attempts until the first sale is made, or trials until equipment failure occurs.


Cumulative Distribution Function (CDF)


The cumulative distribution function (CDF) of a geometric distribution is given by:

FX(k)=P(X≤k)=1−(1−p)kF_X(k) = P(X \leq k) = 1 - (1-p)^k


Where:
pp = probability of success on each trial
kk = number of trials until and including the first success (where k≥1k \geq 1)
(1−p)(1-p) = probability of failure on each trial

Intuition Behind the Formula


Definition: The CDF gives the probability that the first success occurs on or before trial kk.

Complement Approach:
Instead of summing probabilities from 1 to kk, it's easier to use the complement. The event "first success on or before trial kk" is the complement of "first success after trial kk", which means all of the first kk trials are failures:

P(X≤k)=1−P(all first k trials fail)=1−(1−p)kP(X \leq k) = 1 - P(\text{all first } k \text{ trials fail}) = 1 - (1-p)^k


Verification by Summation:
We can verify this by summing the PMF:

P(X≤k)=∑i=1k(1−p)i−1p=p∑i=0k−1(1−p)i=p⋅1−(1−p)k1−(1−p)=1−(1−p)kP(X \leq k) = \sum_{i=1}^{k} (1-p)^{i-1}p = p\sum_{i=0}^{k-1} (1-p)^i = p \cdot \frac{1-(1-p)^k}{1-(1-p)} = 1-(1-p)^k


Monotonic Property: As kk increases, (1−p)k(1-p)^k decreases toward 0, so FX(k)F_X(k) increases toward 1, which reflects the increasing certainty that success will eventually occur.

Geometric Distribution CDF

CDF for waiting time until first success

CDF Explanation

The geometric CDF has a closed form: F(k)=P(X≤k)=1−(1−p)kF(k) = P(X \leq k) = 1 - (1-p)^k for k≥1k \geq 1. This represents the probability that the first success occurs on or before trial kk. The CDF starts at F(1)=pF(1) = p (success on the first trial) and asymptotically approaches 1.0 as kk increases. The rate of increase depends on pp: larger values of pp lead to faster convergence to 1.0, while smaller values result in a more gradual increase, reflecting the longer expected waiting time.

Expected Value (Mean)


As explained in the general case for calculating expected value, the expected value of a discrete random variable is computed as a weighted sum where each possible value is multiplied by its probability:

E[X]=∑xx⋅P(X=x)E[X] = \sum_{x} x \cdot P(X = x)


For the geometric distribution, we apply this general formula to the specific probability mass function of this distribution.

Formula


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


Where:
pp = probability of success on each trial

Derivation and Intuition


Starting from the general definition and substituting the PMF P(X=k)=(1−p)k−1pP(X = k) = (1-p)^{k-1}p for k=1,2,3,…k = 1, 2, 3, \ldots:

E[X]=∑k=1∞k⋅(1−p)k−1p=p∑k=1∞k⋅(1−p)k−1E[X] = \sum_{k=1}^{\infty} k \cdot (1-p)^{k-1} p = p \sum_{k=1}^{\infty} k \cdot (1-p)^{k-1}


Using the formula for the derivative of a geometric series, we recognize that:

∑k=1∞k⋅rk−1=1(1−r)2\sum_{k=1}^{\infty} k \cdot r^{k-1} = \frac{1}{(1-r)^2}


Substituting r=1−pr = 1-p:

E[X]=p⋅1(1−(1−p))2=p⋅1p2=1pE[X] = p \cdot \frac{1}{(1-(1-p))^2} = p \cdot \frac{1}{p^2} = \frac{1}{p}


The result E[X]=1pE[X] = \frac{1}{p} captures an intuitive relationship: if the probability of success on each trial is pp, then on average you need 1p\frac{1}{p} trials to achieve the first success. The smaller the probability of success, the more trials you expect to need.

Example


Consider rolling a die until you get a 6, where p=16p = \frac{1}{6}:

E[X]=11/6=6E[X] = \frac{1}{1/6} = 6


On average, you expect to roll the die 6 times before seeing the first 6. This makes intuitive sense: with a 1-in-6 chance per roll, the average wait is 6 rolls.

Variance and Standard Deviation


The variance of a discrete random variable measures how spread out the values are around the expected value. It is computed as:

Var(X)=E[(X−μ)2]=∑x(x−μ)2P(X=x)\mathrm{Var}(X) = \mathbb{E}[(X - \mu)^2] = \sum_{x} (x - \mu)^2 P(X = x)


Or using the shortcut formula:

Var(X)=E[X2]−μ2\mathrm{Var}(X) = \mathbb{E}[X^2] - \mu^2


For the geometric distribution, we apply this formula to derive the variance.

Formula


Var(X)=1−pp2\mathrm{Var}(X) = \frac{1-p}{p^2}


Where:
pp = probability of success on each trial
(1−p)=q(1-p) = q = probability of failure on each trial

Derivation and Intuition


Starting with the shortcut formula, we need to calculate E[X2]\mathbb{E}[X^2].

We know from the expected value section that μ=1p\mu = \frac{1}{p}.

Using the PMF P(X=k)=(1−p)k−1pP(X = k) = (1-p)^{k-1}p and applying summation techniques involving derivatives of geometric series:

E[X2]=∑k=1∞k2(1−p)k−1p=2−pp2\mathbb{E}[X^2] = \sum_{k=1}^{\infty} k^2 (1-p)^{k-1} p = \frac{2-p}{p^2}


Applying the shortcut formula:

Var(X)=2−pp2−(1p)2=2−pp2−1p2=1−pp2\mathrm{Var}(X) = \frac{2-p}{p^2} - \left(\frac{1}{p}\right)^2 = \frac{2-p}{p^2} - \frac{1}{p^2} = \frac{1-p}{p^2}


The result Var(X)=1−pp2\mathrm{Var}(X) = \frac{1-p}{p^2} shows that variance increases as pp decreases. When success is rare (small pp), the waiting time becomes highly variable—sometimes you succeed quickly, sometimes you wait a very long time. The quadratic relationship with pp in the denominator means variance grows rapidly as pp approaches zero.

Standard Deviation


σ=1−pp2=1−pp\sigma = \sqrt{\frac{1-p}{p^2}} = \frac{\sqrt{1-p}}{p}


Example


Consider rolling a die until you get a 6, where p=16p = \frac{1}{6}:

Var(X)=1−16(16)2=56136=56×36=30\mathrm{Var}(X) = \frac{1 - \frac{1}{6}}{(\frac{1}{6})^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = \frac{5}{6} \times 36 = 30


σ=30≈5.477\sigma = \sqrt{30} \approx 5.477


The variance of 30 and standard deviation of about 5.5 indicate substantial variability around the expected wait time of 6 rolls. You might get lucky and succeed on roll 2, or unlucky and wait 15+ rolls.

Mode and Median

Mode


The mode is the value of kk (number of trials until first success) with the highest probability—the peak of the PMF.

For the geometric distribution, the mode is always:

Mode=1\text{Mode} = 1


Intuition: The geometric PMF is pX(k)=(1−p)k−1pp_X(k) = (1-p)^{k-1}p, which decreases monotonically. The probability is highest at the first trial and decreases exponentially with each subsequent trial. Since each additional trial multiplies the probability by (1−p)<1(1-p) < 1, the maximum probability always occurs at k=1k = 1.

Example:
For p=0.3p = 0.3:
• P(X=1)=0.3P(X = 1) = 0.3
• P(X=2)=0.7×0.3=0.21P(X = 2) = 0.7 \times 0.3 = 0.21
• P(X=3)=0.49×0.3=0.147P(X = 3) = 0.49 \times 0.3 = 0.147

The mode is 1, representing that getting success on the first trial is always the most likely single outcome.

This holds regardless of the value of pp—whether success is rare (pp near 0) or common (pp near 1), the first trial is always most probable.

Median


The median is the value mm such that P(X≤m)≥0.5P(X \leq m) \geq 0.5 and P(X≥m)≥0.5P(X \geq m) \geq 0.5.

For the geometric distribution, the median can be found by solving:

∑k=1m(1−p)k−1p≥0.5\sum_{k=1}^{m} (1-p)^{k-1}p \geq 0.5


Using the CDF F(k)=1−(1−p)kF(k) = 1 - (1-p)^k, we solve:

1−(1−p)m≥0.51 - (1-p)^m \geq 0.5


(1−p)m≤0.5(1-p)^m \leq 0.5


m≥ln⁡(0.5)ln⁡(1−p)=−ln⁡(2)ln⁡(1−p)m \geq \frac{\ln(0.5)}{\ln(1-p)} = \frac{-\ln(2)}{\ln(1-p)}


The median is:

Median=⌈−ln⁡(2)ln⁡(1−p)⌉\text{Median} = \left\lceil \frac{-\ln(2)}{\ln(1-p)} \right\rceil


Example:
For p=0.5p = 0.5:

Median=⌈−ln⁡(2)ln⁡(0.5)⌉=⌈0.6930.693⌉=1\text{Median} = \left\lceil \frac{-\ln(2)}{\ln(0.5)} \right\rceil = \left\lceil \frac{0.693}{0.693} \right\rceil = 1


Example:
For p=0.3p = 0.3:

Median=⌈−ln⁡(2)ln⁡(0.7)⌉=⌈0.6930.357⌉=⌈1.94⌉=2\text{Median} = \left\lceil \frac{-\ln(2)}{\ln(0.7)} \right\rceil = \left\lceil \frac{0.693}{0.357} \right\rceil = \lceil 1.94 \rceil = 2


Properties:
• The median is always close to 1p\frac{1}{p} (the expected value)
• For high pp, median = mode = 1
• For low pp, median > mode, reflecting right skew
• The geometric distribution is always right-skewed (median ≥ mode)


Applications and Examples


Practical Example


Suppose you're rolling a fair six-sided die until you get a 6. The probability of rolling a 6 is p=16p = \frac{1}{6}. The probability that you need exactly k=4k = 4 rolls to get your first 6 is:

P(X=4)=(56)4−1⋅16=(56)3⋅16≈0.096P(X = 4) = \left(\frac{5}{6}\right)^{4-1} \cdot \frac{1}{6} = \left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6} \approx 0.096

This means there's about a 9.6% chance that you'll need exactly 4 rolls to get your first 6.

Interactive Calculator


This interactive calculator computes probabilities for the geometric distribution, which models the number of trials needed until the first success occurs. Enter your success probability (pp) and choose whether you want the full distribution or specific probabilities like P(X=k)P(X=k), P(X<k)P(X<k), or P(X≥k)P(X≥k) to find how likely you are to achieve your first success on a particular trial. Perfect for modeling scenarios like sales conversions, quality testing, or any situation where you're waiting for the first success.

1. Select probability type: 'All values' shows full distribution, or choose a specific query like P(X=k) or P(X≤k)
2. Enter p (success probability) - must be between 0 and 1, cannot be 0
3. For specific queries, enter k (trial number) - must be a positive integer
4. Click Calculate to see results including mean, variance, and probability distribution

Geometric Distribution Calculator

Calculate probabilities and distribution properties

Probability of success on each trial


Geometric at a Glance

The table below collects the full anatomy of the geometric distribution into a single reference card — its single parameter, support, PMF and CDF, the mean and variance formulas, mode and median behavior, the defining memoryless property, the relationship to the negative binomial, and a canonical die-rolling example.
Aspect Formula / statement Note / example
Parameter p ∈ (0, 1] — success probability per trial independent Bernoulli trials with constant p
Support k ∈ {1, 2, 3, ...} k is the trial on which the first success occurs
PMF P(X = k) = (1 − p)k − 1 · p (k − 1) failures, then a success — strictly decreasing in k
CDF F(k) = 1 − (1 − p)k complement of "all first k trials fail"
Expected value E[X] = 1 / p rolling for a 6: p = 1/6, so E[X] = 6 trials on average
Variance Var(X) = (1 − p) / p2; σ = √(1 − p) / p grows rapidly as p → 0 — rare successes are highly variable
Mode and median mode = 1 always; median = ⌈−ln 2 / ln(1 − p)⌉ PMF is strictly decreasing, so trial 1 is always most likely
Memoryless property P(X > m + n | X > m) = P(X > n) the only discrete distribution with this property
Relation Geom(p) = NegBin(1, p) geometric is the r = 1 special case of the negative binomial
Canonical example rolling a die until the first 6: p = 1/6 E[X] = 6, Var(X) = 30, σ ≈ 5.48