0.08192
0.67232
0.32768
5
20
4.472136
4
1
0.08192
0.67232
0.32768
5
20
4.472136
4
1
The Geometric Distribution Calculator computes the probability mass function (PMF), cumulative distribution function (CDF), mean, variance, and standard deviation for the geometric distribution. The geometric distribution models the number of trials needed to achieve the first success in a sequence of independent Bernoulli trials, each with the same probability of success p.
While the binomial distribution asks "how many successes in n trials?", the geometric distribution asks the complementary question: "how many trials until the first success?" This distinction makes the geometric distribution the natural model for waiting-time problems in discrete settings. How many times must you roll a die before getting a six? How many customers do you pitch before making a sale? How many wells must you drill before striking oil? How many attempts before a free throw goes in?
The geometric distribution is the only memoryless discrete distribution, mirroring the exponential distribution's memoryless property in the continuous domain. If you have already failed k times, the probability of needing at least j more trials is the same as needing j trials from the start. This means past failures do not affect future prospects, a property that only holds when the success probability p remains constant across trials.
The distribution has a single parameter p (the probability of success on each trial), with support on k = 1, 2, 3, ... (the trial number of the first success). Note that there are two conventions: some sources define k as the number of failures before the first success (support k = 0, 1, 2, ...). This calculator uses the number of trials until first success convention (k starts at 1).
The mean 1/p gives the expected number of trials needed. With p = 0.1, you expect to need 10 trials on average. The variance (1-p)/p² quantifies the uncertainty around this average. The geometric distribution is heavily right-skewed: the most likely outcome is always k = 1 (success on the first try), but the tail extends indefinitely, allowing for very long waiting times. This calculator also provides the CDF, giving the probability that the first success occurs on or before trial k, which is useful for planning and resource allocation.
In reliability engineering, the geometric distribution models the number of test cycles until the first failure of a component undergoing repeated stress tests. In telecommunications, it describes the number of transmission attempts until a packet is successfully delivered over a noisy channel. In genetics, it models the number of offspring until the first child with a particular genotype. In sports analytics, it can model the number of at-bats until a baseball player's first hit, or the number of penalty kicks until the first goal. The geometric distribution's simplicity and analytical tractability make it an ideal starting point for understanding more complex waiting-time distributions.
The geometric PMF gives the probability that the first success occurs on trial k:
$$P(X = k) = (1 - p)^{k-1} \cdot p$$
This requires k-1 consecutive failures (each with probability 1-p) followed by one success (probability p). The CDF is:
$$P(X \leq k) = 1 - (1 - p)^k$$
which follows from the complement: P(X > k) = (1-p)^k is the probability of k consecutive failures. The key statistics are:
$$E[X] = \frac{1}{p}, \quad \text{Var}(X) = \frac{1 - p}{p^2}, \quad \sigma = \frac{\sqrt{1 - p}}{p}$$
The median is ⌈-1/log₂(1-p)⌉, which is often close to but not equal to the mean. The mode is always 1 (first trial), regardless of p.
The PMF P(X = k) is the probability that the first success occurs on exactly trial k. This value decreases geometrically with k (hence the name). The CDF P(X ≤ k) is the probability of achieving at least one success within k trials, useful for answering questions like 'what is the chance I succeed within 10 attempts?'. The mean 1/p is the average number of trials needed. For example, with p = 0.25, you need 4 trials on average. The variance increases rapidly as p decreases, indicating high unpredictability for rare events.
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The probability of rolling your first six on exactly the 6th roll is about 6.7%. The CDF shows a 66.5% chance of getting at least one six within 6 rolls. On average, you need about 6 rolls.
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With a 5% conversion rate, the probability of the first sale on exactly the 20th pitch is about 1.89%. There is a 64.2% chance of making at least one sale within 20 pitches. On average, 20 pitches are needed.
Convention 1 (used here): X = number of trials until first success, support {1, 2, 3, ...}, mean = 1/p. Convention 2: X = number of failures before first success, support {0, 1, 2, ...}, mean = (1-p)/p. The PMF in convention 2 is P(X=k) = (1-p)^k · p. Always check which convention your textbook uses.
P(X > s + t | X > s) = P(X > t) because (1-p)^(s+t) / (1-p)^s = (1-p)^t. In practical terms: if you have flipped 10 tails in a row, the probability of needing at least 5 more flips is the same as starting fresh. Each trial is independent, so past outcomes cannot affect future probabilities. This is the discrete analogue of the exponential distribution's memoryless property.
Use the CDF: P(X ≤ n) = 1 - (1-p)^n. For example, with p = 0.1, the probability of at least one success in 10 trials is 1 - 0.9^10 ≈ 0.6513 (65.13%). This is useful for planning: how many attempts do you need to achieve a desired confidence level of success?
The geometric distribution is a special case of the negative binomial distribution with r = 1. While the geometric models the number of trials until the first success, the negative binomial models the number of trials until the r-th success. If X₁, X₂, ..., Xᵣ are independent geometric(p) variables, then X₁ + X₂ + ... + Xᵣ follows a negative binomial distribution.
Yes, when p is small, the geometric distribution with parameter p is well-approximated by the exponential distribution with rate λ = p. Specifically, if X ~ Geometric(p) and p → 0, then pX converges in distribution to Exponential(1). This connection reinforces the geometric as the discrete counterpart of the exponential.
The PMF P(X = k) = (1-p)^(k-1) · p is a decreasing function of k for any 0 < p ≤ 1. The maximum always occurs at k = 1, where P(X = 1) = p. This makes intuitive sense: the most likely outcome is immediate success on the first try. However, if p is small, P(X = 1) is also small, meaning no single outcome is very likely.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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