The Binomial Limit of Hypergeometric Sampling

Why a Binomial Limit Appears

When $n$ individuals are sampled without replacement from a finite population, their success indicators are dependent: observing a success changes the success fraction among the individuals left to draw. A binomial model instead describes $n$ independent trials with the same success probability. As the population becomes large while $n$ stays fixed, each draw has a diminishing effect on the next draw. This is why the hypergeometric count approaches a binomial count.

The limiting statement has a specific setup. Let the population size be $N$, let $K_N$ individuals be successes, and let $X_N$ count the successes in a simple random sample of size $n$, drawn without replacement. The population success fraction is $p_N=K_N/N$. We let $N$ grow while holding $n$ fixed and assuming $p_N$ converges to $p$, where $0\le p\le 1$.

Definition: If $X_N$ is the number of successes in a sample of size $n$ drawn without replacement from $N$ individuals, of whom $K_N$ are successes, then $X_N$ has a hypergeometric distribution. Its probability mass function is $$ P(X_N=j)=\frac{\binom{K_N}{j}\binom{N-K_N}{n-j}}{\binom{N}{n}}, $$ for feasible counts $j$, and is zero otherwise. Here $n$ is fixed as $N$ grows.

The candidate limiting distribution is $\operatorname{Bin}(n,p)$: the number of successes in $n$ independent Bernoulli trials, each with success probability $p$. Its probability at $j$ is $\binom{n}{j}p^j(1-p)^{n-j}$, for $j=0,\ldots,n$. The limiting result says the two probability mass functions get close at every possible count.

The Binomial Limit

Theorem: Suppose $n$ is a fixed nonnegative integer, $N\to\infty$, $0\le K_N\le N$, and $K_N/N\to p$ for some $p\in[0,1]$. If $X_N$ is hypergeometric with population size $N$, $K_N$ successes, and sample size $n$, then $$ P(X_N=j)\longrightarrow \binom{n}{j}p^j(1-p)^{n-j},\qquad j=0,\ldots,n. $$ Consequently, the distributions converge in total variation to $\operatorname{Bin}(n,p)$. At $p=0$ or $p=1$, the binomial distribution is interpreted as concentrated at $0$ or $n$, respectively.

To prove the theorem, write the hypergeometric probability using falling factorials. For an integer $a\ge 0$, define $(a)_r=a(a-1)\cdots(a-r+1)$ for $r\ge 1$, and $(a)_0=1$. For $j=0,\ldots,n$, the probability can be written as $$ P(X_N=j)=\binom{n}{j}\frac{(K_N)_j(N-K_N)_{n-j}}{(N)_n}. $$ This expression gives zero automatically if there are too few successes or failures for the specified count.

Separate the population-size factors.
Divide the falling-factorial numerator and denominator by $N^n$. The probability becomes $$ \binom{n}{j} \frac{\displaystyle\prod_{r=0}^{j-1}\left(\frac{K_N}{N}-\frac{r}{N}\right) \displaystyle\prod_{s=0}^{n-j-1}\left(1-\frac{K_N}{N}-\frac{s}{N}\right)} {\displaystyle\prod_{t=0}^{n-1}\left(1-\frac{t}{N}\right)}. $$ An empty product is defined to equal $1$.

Take the limit.
There are only finitely many factors because $n$ is fixed. Each success factor tends to $p$, each failure factor to $1-p$, and each denominator factor to $1$. Thus the expression tends to $\binom{n}{j}p^j(1-p)^{n-j}$. This reasoning also covers $p=0$ and $p=1$.

Extend the conclusion to the whole distribution.
Both distributions assign probability only to the finite set $\{0,\ldots,n\}$. Since the probability at each of these $n+1$ counts converges, the sum of their absolute differences converges to zero. In particular, their total variation distance, one half of that sum, tends to zero.

This proof works because the sample size is fixed: only finitely many factors are involved, and the possible counts do not expand as $N$ grows. It establishes more than convergence for a single event. Any event determined by the count has probability converging to its binomial-model probability, since it is a sum over a subset of the same finite set.

Worked Example: Comparing the Probability Mass Functions

A population has $N=20$ individuals, $K_N=6$ successes, and a sample of size $n=3$. Compare its hypergeometric probabilities with those of $\operatorname{Bin}(3,0.3)$. Then compare with a population of $N=200$ having $K_N=60$, again with $n=3$.

For $N=20$, the denominator is $\binom{20}{3}=1140$. The four hypergeometric probabilities, in order for $j=0,1,2,3$, are $$ \frac{\binom{14}{3}}{1140}=\frac{364}{1140}\approx0.3193,\quad \frac{6\binom{14}{2}}{1140}=\frac{546}{1140}\approx0.4789, $$ $$ \frac{\binom{6}{2}14}{1140}=\frac{210}{1140}\approx0.1842,\quad \frac{\binom{6}{3}}{1140}=\frac{20}{1140}\approx0.0175. $$ They sum to $1140/1140=1$. The binomial probabilities are $0.7^3=0.343$, $3(0.3)(0.7)^2=0.441$, $3(0.3)^2(0.7)=0.189$, and $(0.3)^3=0.027$.

For $N=200$, $K_N=60$, and $n=3$, the denominator is $\binom{200}{3}=1{,}313{,}400$. The four probabilities are $$ \frac{447{,}580}{1{,}313{,}400}\approx0.3408,\quad \frac{583{,}800}{1{,}313{,}400}\approx0.4445,\quad \frac{247{,}800}{1{,}313{,}400}\approx0.1887,\quad \frac{34{,}220}{1{,}313{,}400}\approx0.0261. $$ These are closer to the binomial probabilities. In both populations the success fraction is $0.3$; increasing the population while keeping $n=3$ fixed brings the hypergeometric distribution toward $\operatorname{Bin}(3,0.3)$.

Worked Example: The Probability of Two Successes

Suppose $n=2$ and the population success fraction is $0.4$. The binomial limit assigns probability $0.4^2=0.16$ to two successes. For a finite population, the exact probability is $$ P(X_N=2)=\frac{K_N(K_N-1)}{N(N-1)}. $$

With $N=10$ and $K_N=4$, this probability is $4\cdot3/(10\cdot9)=12/90=2/15\approx0.1333$. The second success is less likely after the first success has been drawn, because only three successes remain among nine individuals.

With $N=100$ and $K_N=40$, the probability is $40\cdot39/(100\cdot99)=1560/9900=26/165\approx0.1576$. As the population grows with $K_N/N=0.4$, the probability tends to $0.16$, as the theorem predicts.

What Changes, and What Does Not

In a hypergeometric sample, the draws are not independent. The binomial limit does not claim that they become exactly independent for any finite population; it says that the distribution of the fixed-size count approaches the binomial distribution under the stated limiting conditions. The population fraction must converge, and the sample size must remain fixed.

The two models also illustrate the effect of sampling without replacement on variability. For a hypergeometric count with $N>1$, its mean and variance are $$ E[X_N]=np_N,\qquad \operatorname{Var}(X_N)=np_N(1-p_N)\frac{N-n}{N-1}. $$ The factor $(N-n)/(N-1)$ approaches $1$ when $n$ is fixed and $N\to\infty$. The limiting mean and variance are therefore $np$ and $np(1-p)$, those of $\operatorname{Bin}(n,p)$. One way to see the variance factor is to write the count as a sum of $n$ success indicators: each has variance $p_N(1-p_N)$, and each distinct pair has covariance $-p_N(1-p_N)/(N-1)$. Summing the $n$ variances and $n(n-1)$ covariances gives the stated formula.

Worked Example: Checking the Variance Change

Consider populations with success fraction $0.4$, sampled with $n=4$. The binomial variance is $4(0.4)(0.6)=0.96$, and the mean is $4(0.4)=1.6$.

For $N=50$ and $K_N=20$, the hypergeometric mean is also $1.6$. Its variance is $$ 4(0.4)(0.6)\frac{50-4}{50-1} =0.96\frac{46}{49}\approx0.9012. $$ For $N=500$ and $K_N=200$, the mean remains $1.6$, while the variance is $$ 0.96\frac{500-4}{500-1} =0.96\frac{496}{499}\approx0.9542. $$ The variance moves toward the binomial value as the population grows. This is consistent with the distributional limit; it is not, by itself, a proof of that limit.

Common Mistake: Treating “Large” as a Complete Condition

A tempting argument is: “The population is large, so the hypergeometric count is binomial.” This leaves out the sample size and the success fraction. The result proved here requires a fixed sample size $n$ and a convergent success fraction $K_N/N$. If $n$ grows with $N$, this theorem does not establish a binomial limit. For example, if the whole population is sampled, then $X_N=K_N$ with certainty, while a binomial variable with parameters $N$ and $K_N/N$ is generally random.

The correction is to identify the sampling design and asymptotic conditions before choosing an approximation. Sampling without replacement leads to a hypergeometric count; its fixed-$n$, large-population limit is binomial when the population success fraction converges. This conclusion concerns the hypergeometric-to-binomial limit, not every possible approximation involving count distributions.

Key takeaway: If the sample size $n$ stays fixed and $K_N/N\to p$ as $N\to\infty$, then a hypergeometric count converges to $\operatorname{Bin}(n,p)$. The convergence holds for each count and, because the support is finite, for the entire distribution.

Check Your Understanding

Use the fixed-sample-size condition carefully in each question.

A population has size $N$, $K_N$ successes, and a fixed sample size $n$. What condition on $K_N/N$ gives a binomial limit?
Why does the proof consider only the probabilities at $j=0,\ldots,n$, even as $N$ grows?
If $n=2$ and $K_N/N\to p$, what is the limiting probability of observing two successes?
Why does sampling without replacement generally produce a smaller variance than the corresponding binomial model for a finite population?
Does the theorem apply when the sample size grows at the same time as the population? Explain what condition in the theorem would fail to apply.

The Binomial Limit of Hypergeometric Sampling

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