Hypergeometric Distribution

Probability Distributions

Sampling Without Replacement — Finite Populations

The hypergeometric distribution models the number of successes when sampling without replacement from a finite population. Every draw changes the odds.

Card games — probability of drawing aces in a poker hand
Quality inspection — defective items in a batch sample
Ecology — capture-recapture population estimation
Legal sampling — random drug testing from a workforce

Unlike the binomial, the hypergeometric accounts for the fact that each draw depletes the population.

Core Concepts

The hypergeometric distribution models the number of successes when sampling without replacement from a finite population. It is the correct distribution for card games, quality inspection of small batches, and any scenario where the population is depleted as items are drawn.

DfHypergeometric Distribution

A random variable $X$ follows a hypergeometric distribution with parameters $(N, K, n)$ , written $X \sim \text{Hyp}(N, K, n)$ , if its PMF is:

P(X = k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}},

where $N$ is the population size, $K$ is the number of successes in the population, $n$ is the sample size, and $k$ satisfies $\max(0, n - (N-K)) \leq k \leq \min(n, K)$ .

Why This PMF Is Correct

The total number of ways to choose $n$ items from $N$ is $\binom{N}{n}$ . The number of ways to choose exactly $k$ successes from $K$ available and $n-k$ failures from $N-K$ available is $\binom{K}{k}\binom{N-K}{n-k}$ . The ratio gives the probability, since each subset of size $n$ is equally likely.

Mean and Variance

Hypergeometric Mean and Variance

E[X] = n\frac{K}{N}, \quad \text{Var}(X) = n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}

Here,

$N$ =Population size
$K$ =Successes in population
$n$ =Sample size
$\frac{N-n}{N-1}$ =Finite population correction factor

Derivation of the Mean

Let $X_i$ be the indicator that the $i$ -th draw is a success. By linearity of expectation:

E[X] = E\!\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n E[X_i] = n \cdot \frac{K}{N},

since by symmetry each draw has probability $K/N$ of being a success, regardless of the order. (Note: this holds even though the $X_i$ are not independent.)

Derivation of the Variance

Using $\text{Var}(X) = \sum_i \text{Var}(X_i) + 2\sum_{i<j}\text{Cov}(X_i, X_j)$ :

$\text{Var}(X_i) = \frac{K}{N}\left(1 - \frac{K}{N}\right) = \frac{K(N-K)}{N^2}$
For $i \neq j$ : $\text{Cov}(X_i, X_j) = -\frac{K(N-K)}{N^2(N-1)}$ (negative because drawing a success on draw $i$ makes a success on draw $j$ slightly less likely without replacement).

Therefore:

\text{Var}(X) = n\cdot\frac{K(N-K)}{N^2} + n(n-1)\cdot\left(-\frac{K(N-K)}{N^2(N-1)}\right) = n\frac{K(N-K)}{N^2}\left(1 - \frac{n-1}{N-1}\right) = n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}.

The Finite Population Correction Factor

ThFinite Population Correction

The ratio of hypergeometric to binomial variance is:

\frac{\text{Var}_{\text{hyp}}(X)}{\text{Var}_{\text{bin}}(X)} = \frac{N - n}{N - 1}.

This factor is always less than 1 for $n \geq 2$ , meaning sampling without replacement produces less variability than sampling with replacement.

Why Without Replacement Has Lower Variance

When you sample without replacement and observe a success, the remaining population has one fewer success, making the next draw slightly less likely to succeed. This negative correlation between draws reduces the overall variability of the count. The effect is negligible when $N \gg n$ .

Convergence to Binomial

ThBinomial Approximation

As $N \to \infty$ with $K/N \to p$ and $n$ fixed, the hypergeometric distribution converges to the binomial:

\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}} \longrightarrow \binom{n}{k}p^k(1-p)^{n-k}.

Proof Sketch

For fixed $k$ and $n$ :

\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}} = \binom{n}{k} \cdot \frac{K(K-1)\cdots(K-k+1)}{N(N-1)\cdots(N-k+1)} \cdot \frac{(N-K)(N-K-1)\cdots(N-K-n+k+1)}{N(N-1)\cdots(N-k+1)} \cdot \ldots

As $N \to \infty$ with $K/N \to p$ , each factor $\frac{K-i}{N-i} \to p$ and $\frac{N-K-i}{N-i} \to 1-p$ , yielding $\binom{n}{k}p^k(1-p)^{n-k}$ .

Symmetry Property

ThHypergeometric Symmetry

The hypergeometric distribution is symmetric under the transformation $k \to n-k$ when $K = N/2$ :

P(X = k) = P(X = n-k) \quad \text{when} \quad K = N/2.

More generally, $X \sim \text{Hyp}(N, K, n)$ has the same distribution as $n - X'$ where $X' \sim \text{Hyp}(N, N-K, n)$ : counting successes is equivalent to counting failures.

Worked Example: Lottery / Card Draw

Example: Drawing Aces from a Deck

A standard deck has $N = 52$ cards, $K = 4$ aces. We draw $n = 5$ cards without replacement. Let $X$ = number of aces drawn.

PMF:

P(X = k) = \frac{\binom{4}{k}\binom{48}{5-k}}{\binom{52}{5}}.

Compute each value:

$P(X=0) = \frac{\binom{4}{0}\binom{48}{5}}{\binom{52}{5}} = \frac{1 \cdot 1{,}712{,}304}{2{,}598{,}960} \approx 0.6588$
$P(X=1) = \frac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}} = \frac{4 \cdot 194{,}580}{2{,}598{,}960} \approx 0.2995$
$P(X=2) = \frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}} = \frac{6 \cdot 17{,}296}{2{,}598{,}960} \approx 0.0399$
$P(X=3) = \frac{\binom{4}{3}\binom{48}{2}}{\binom{52}{5}} = \frac{4 \cdot 1{,}128}{2{,}598{,}960} \approx 0.0017$
$P(X=4) = \frac{\binom{4}{4}\binom{48}{1}}{\binom{52}{5}} = \frac{1 \cdot 48}{2{,}598{,}960} \approx 0.0000185$

Mean: $E[X] = 5 \cdot \frac{4}{52} = \frac{20}{52} \approx 0.3846$ .

Variance: $\text{Var}(X) = 5 \cdot \frac{4}{52} \cdot \frac{48}{52} \cdot \frac{47}{51} \approx 0.3205$ .

Compare with binomial approximation: $\text{Binomial}(5, 4/52)$ gives $\text{Var} = 5 \cdot \frac{4}{52} \cdot \frac{48}{52} \approx 0.3550$ .

The finite population correction factor $47/51 \approx 0.922$ reduces the variance by about $8\%$ .

Python Implementation

import numpy as np
from scipy import stats

np.random.seed(42)

# Hypergeometric parameters
N, K, n = 52, 4, 5  # deck, aces, draws

# Theoretical mean and variance
mean_theory = n * K / N
var_theory = n * (K/N) * (1 - K/N) * (N - n) / (N - 1)
print(f"Hyp({N}, {K}, {n}):")
print(f"  Theoretical mean: {mean_theory:.4f}")
print(f"  Theoretical variance: {var_theory:.4f}")

# Simulate
n_sims = 100000
samples = np.random.hypergeometric(K, N - K, n, size=n_sims)
print(f"  Empirical mean:   {np.mean(samples):.4f}")
print(f"  Empirical variance: {np.var(samples, ddof=0):.4f}")

# Compare with binomial (with replacement)
binom_var = n * (K/N) * (1 - K/N)
fpc = (N - n) / (N - 1)
print(f"\n  Binomial variance (with replacement): {binom_var:.4f}")
print(f"  Finite population correction: {fpc:.4f}")
print(f"  Hyp var / Bin var = {var_theory / binom_var:.4f}  (should be {fpc:.4f})")

Python Implementation: Batch Quality Inspection

import numpy as np
from scipy import stats

np.random.seed(42)

# Quality control: batch of 100 items, 8 defective, sample 10
N, K, n = 100, 8, 10
samples = np.random.hypergeometric(K, N - K, n, size=50000)

print(f"Quality Inspection: N={N}, K={K} defective, sample n={n}")
print(f"PMF values:")
for k in range(min(n, K) + 1):
    pmf = stats.hypergeom.pmf(k, N, K, n)
    print(f"  P(X={k}) = {pmf:.4f}")

print(f"\nSimulated mean: {np.mean(samples):.4f} (theoretical: {n*K/N:.4f})")

# Probability of finding at least 2 defectives
prob_ge_2 = 1 - stats.hypergeom.cdf(1, N, K, n)
print(f"P(X >= 2 defectives in sample) = {prob_ge_2:.4f}")

Key Takeaways

Summary: Hypergeometric Distribution

Models sampling without replacement from a finite population of size $N$ with $K$ successes
PMF: $P(X=k) = \binom{K}{k}\binom{N-K}{n-k}/\binom{N}{n}$
Mean: $E[X] = nK/N$ (same as binomial — expectation doesn't need independence)
Variance: $n\frac{K}{N}(1-\frac{K}{N})\frac{N-n}{N-1}$ — includes finite population correction $(N-n)/(N-1) < 1$
Negative covariance between draws: $\text{Cov}(X_i, X_j) = -\frac{K(N-K)}{N^2(N-1)}$
Converges to Binomial $(n, K/N)$ as $N \to \infty$ with $K/N$ fixed
Variance is always less than the binomial counterpart (less variability without replacement)

Hypergeometric Distribution — Sampling Without Replacement