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A capture-recapture model is a statistical model for repeated observations of identifiable individuals, including those missed on some occasions. A closed-population capture-recapture model estimates a fixed population cardinality; models allowing individuals to enter or leave can instead describe survival and recruitment. The sampling assumptions determine which likelihood function is appropriate.

For the two-sample closed-population capture-recapture model, condition on the sample sizes $M$ and $C$. If the second sample selects $C$ distinct individuals from $N$ by simple random sampling, then the overlap $R$ has the hypergeometric distribution:Here $r$ is an integer and the probability mass function is zero outside these bounds. The binomial coefficients count choices of $r$ marked and $C-r$ unmarked individuals, divided by all equally likely size-$C$ subsets. The hypergeometric distribution has expected valueEquating the observed overlap with this expected value gives the Lincoln-Petersen estimator. The underlying sampling without replacement, probability mass function, and expected value are developed in Raymond P. Guiteras, MIT 14.30 hypergeometric distribution handout, §§1–4 and 8.

For the toy counts $M=20$, $C=25$, and $R=5$, hold the observations fixed and vary the unknown integer $N$. The likelihood function isValues below forty cannot contain the observed set union. Exact comparison of consecutive likelihoods gives two tied maximum-likelihood estimates, $N=99$ and $N=100$. The Lincoln-Petersen estimator returns $100$; the Chapman estimator returns $90$ and is a different rule for estimation.

The vertical scale is $L(N)/{\max }_{n\ge 40}L(n)$. This relative likelihood compares how well candidate statistical parameters explain the same observations; it is not a posterior probability. The displayed range $40\le N\le 300$ is only a plotting window. The likelihood function remains positive for every integer $N\ge 40$ and tends to zero as $N$ increases, so the curve has a continuing right tail. The data do not justify truncating statistical inference at three hundred. The conditional hypergeometric distribution formulation also appears in Louis-Paul Rivest and Mamadou Yauck, Small Sample Estimators for Two-way Capture Recapture Experiments, §3, equation (5).

Zero recaptures require care. With $R=0$ and positive $M,C$, the Lincoln-Petersen estimator has no finite ratio, but the capture-recapture model still gives a valid likelihood function:This likelihood function increases toward one as $N$ grows, so there is no finite maximum-likelihood estimate. For the same sample sizes twenty and twenty-five with zero overlap, forty-five distinct individuals would be observed. Lack of overlap supplies a lower bound but very weak information about large population cardinalities.

In more elaborate capture-recapture models, a first-recapture array $m_{ij}$ counts individuals released at occasion $i$ and next observed at $j$. A probability of survival ${\varphi }_j$ and a probability of detection $P_{j+1}$ contribute a no-observation continuation factor ${\varphi }_j(1-P_{j+1})$. The resulting likelihood function depends on both survival and detection; these probabilities need not be separately identifiable at the last observation interval. Missing-count EM for capture-recapture describes one way to fit such a model, and identifiability remains a separate issue from computing an estimator output.

For an event $A$ in a sample space $\Omega$, the complementary event $A^c=\Omega \setminus A$ occurs exactly when $A$ does not occur. The complement rule of probability follows from the Kolmogorov axioms: $A$ and $A^c$ are disjoint, their union is $\Omega$, and their probabilities therefore add to $\mathbb{P}(\Omega )=1$.

This rule is useful when an event is easier to describe through its failure. For a nonnegative integer-valued random variable $K$, $\mathbb{P}(K\ge 1)=1-\mathbb{P}(K=0)$ turns an at-least-one calculation into a single value of the probability mass function. In sampling without replacement, the hypergeometric distribution gives the chance of missing all marked objects, so its complement gives the chance of finding at least one.

For a nonnegative integer $n$, the factorial $n!$ is the product of the positive integers from $1$ to $n$, with the empty product equal to $1$:Thus $1!=1$, $2!=2$, $3!=6$, $4!=24$ and $5!=120$. The factorial counts the permutations of $n$ distinct objects: arranging $A,B,C$ gives exactly $ABC,ACB,BAC,BCA,CAB,CBA$. The choice tree in permutation shows the three choices for the first position, two for the second, and one for the last.

The value $0!=1$ has a counting explanation: the empty set has exactly one ordering, the empty list. Equivalently, it has exactly one bijection to itself, the empty function. There are no choices to make, and one completed outcome. The factorial therefore obeys the recurrence relationThe useful distinction is how many positions are filled. For integers $0\le k\le n$, choosing and ordering $k$ distinct objects by sampling without replacement gives the falling factorial, while choosing an unordered $k$-element subset gives the binomial coefficient:For five people, assigning captain and deputy gives $5!/3!=20$ outcomes; choosing a two-person team gives $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. Every team has exactly $2!$ assignments of the two roles, explaining the division. When $k=0$, both counts are one. The factorial is useful here because one count of full orderings yields both formulas by cancellation and by forgetting order. ^[ref]

Distinctness matters. If the two copies of $A$ in $AAB$ are indistinguishable, the visible arrangements are just $AAB,ABA,BAA$, numbering $3!/2!=3$. Temporarily labelling them $A_1,A_2$ gives six permutations, but exchanging the two labels leaves each visible word unchanged, so every word was counted $2!$ times.

Proof outline. For $n\ge 1$, choose the first object in $n$ ways, then order the remaining $n-1$ objects in $(n-1)!$ ways. The rule of product gives the recurrence relation; repeated application gives the displayed product, starting from the one empty ordering. This explains why the factorial grows quickly: every additional distinct object introduces another multiplicative factor.

The hypergeometric distribution counts marked objects in a uniformly chosen sample of fixed size from a finite population, using sampling without replacement. Let $n\ge 1$ be the population size, let $0\le c\le n$ objects be marked, and sample $0\le r\le n$ distinct objects. If $K$ counts the marked objects selected, its probability mass function isThe bounds describe the support of a probability distribution and $k$ must be an integer; the mass is zero elsewhere. The denominator counts all size-$r$ subsets, while the numerator counts choices of $k$ marked and $r-k$ unmarked objects using binomial coefficients. This counting argument requires every size-$r$ sample to have the same probability.^[ref]

For a constructed example, a box contains ten labelled objects, four red and six blue. Select five without replacement. There are $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$ possible samples. The probabilities for $K=0,1,2,3,4$ areIn particular, $\mathbb{P}(K=2)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{3}\right)/252=120/252=10/21$. Five red objects are impossible because the box only contains four. The expected value is $r(c/n)=2$, and the variance is $2/3$.

The count law also answers a detection question: what is the chance that five draws find at least one red object? By the complement rule of probability, it is $1-\mathbb{P}(K=0)=41/42$, approximately $0.9762$. This can model inspecting five distinct objects from a known finite collection, with a mark indicating the property we want to detect. If repeated inspections can select the same object again, the replacement model instead gives $1-0.6^5=0.92224$. Equal expected values therefore do not imply equal chances of detecting a marked object.

The corresponding binomial distribution counts red draws when each selected object is put back before the next independent draw: $K_{\mathrm{rep}}\sim \mathrm{Bin}(5,0.4)$. It has the same expected value $2$, but variance $5(0.4)(0.6)=6/5$, and it permits five red draws. Without replacement, a red first draw leaves red probability $3/9$ for the next draw; a blue first draw leaves $4/9$. This statistical dependence pulls the total toward the fixed composition of the population, producing the finite population correction.

The useful modelling question is whether previously selected objects remain available. A fixed-size sample of distinct items calls for a hypergeometric distribution; fresh independent trials with constant success probability call for a binomial distribution. When the population grows with $c/n\to p$ while $r$ stays fixed, the hypergeometric masses tend to the corresponding binomial distribution masses: removing finitely many objects has an increasingly small effect on the next draw's probability.

Mark and recapture estimates the cardinality of a finite set of individuals when counting everybody is impractical. The useful information is the overlap between observations: with fixed sample sizes, finding many of the same individuals again suggests a smaller population, while finding few suggests a larger one. More elaborate capture-recapture models also estimate survival and changing probabilities of detection.

In a closed-population capture-recapture model, let $N$ be the unknown population cardinality. Capture $M$ distinct individuals, mark them, release them all, and allow them to mix with the rest. Next take a second sample of $C$ distinct individuals using sampling without replacement, and count the $R$ individuals carrying a mark. The observed marked fraction $R/C$ estimates the population's marked fraction $M/N$, giving the Lincoln-Petersen estimatorRelease happens between the two samples; sampling without replacement means that an individual is counted at most once within the second sample. See Richard J. Barker, Open Population Mark-Recapture Models, §1.1.1.

For a constructed teaching example, mark $M=20$ individuals and later capture $C=25$, including $R=5$ marked individuals. One fifth of the second sample is marked, suggesting that the twenty marked individuals are one fifth of the population:Let $A$ and $B$ be the sets of individuals in the first and second samples. Their intersection has cardinality five. By the inclusion-exclusion principle, the set union containsThus $N\ge 40$ is an observed lower bound, whereas $100$ is the output of an estimator based on the capture-recapture model. These are toy counts, not measurements from a medical or ecological study.

The diagram shows only observed individuals. It does not draw an assumed population of one hundred. The capture-recapture model uses the overlap to infer how many individuals may have escaped both samples; its likelihood function shows why one estimator output cannot describe all that uncertainty.

Sampling without replacement selects distinct labelled units: after a unit is selected it is unavailable for the remaining draws. Distinct labels can carry equal numerical values, so repeated values do not necessarily imply a repeated unit. The rule alone does not specify selection probabilities; a procedure favoring particular units is still sampling without replacement but need not be simple random sampling.

In uniform sequential sampling without replacement, the next label is chosen uniformly among those remaining. For a population of $N$ labels, any ordered sequence of $n$ distinct labels has probability $1/[N(N-1)\cdots (N-n+1)]$. Each size-$n$ subset has $n!$ possible orders, so its probability is ${\left(\genfrac{}{}{0ex}{}{N}{n}\right)}^{-1}$. This gives the fixed-size simple random sampling design. Equivalently, take the first $n$ labels of a uniformly random permutation. Restricting a uniformly random permutation to a specified label set preserves uniformity; biased modulo mappings should not replace uniform selection.

The draw positions have identical marginal probability distributions, but exhibit statistical dependence. If the first label is $i$, the conditional probability that the second label is $i$ is zero, whereas its marginal probability is $1/N$. Thus these draws are not independent and identically distributed when $N>1$ and $n>1$. For a marked subset of labels, counting size-$n$ subsets with binomial coefficients gives the hypergeometric distribution. The resulting negative covariance explains the finite population correction for both counts and sample means.

Here simple random sampling means a random unordered subset $A$ of exactly $n$ distinct labels from $\{1,\dots ,N\}$, with every size-$n$ subset equally likely. The sample size is fixed in advance, $1\le n\le N$, and sampling without replacement prevents duplicate units. This is SRSWOR; some texts also use "simple random sampling" for independent uniform draws with replacement, which is a different design.^[ref]

Every label has inclusion probability $n/N$, but equal inclusion probabilities alone are insufficient. For example, from six labels choose either $\{1,2,3\}$ or $\{4,5,6\}$ with probability $1/2$. Each label is included with probability $1/2$, yet eighteen of the twenty size-three subsets are impossible: this is a two-group cluster sampling design. Unequal probability sampling deliberately changes inclusion probabilities, often to improve precision for a particular estimand, and generally requires an appropriately weighted estimator rather than the plain sample mean.

Consider the original mathematical population with labelled values $(2,4,6,8,10,12)$, and let $n=2$. The fifteen equiprobable subsets, followed by their sample means, areTheir sample means sum to $105$, giving expected value $105/15=7$. Their squared deviations from $7$ sum to $70$, giving variance $70/15=14/3$. This is an enumeration over hypothetical selections from six stipulated numbers, not measured survey data.

With independent uniform sampling with replacement, the $6^2=36$ ordered outcomes each have probability $1/36$. The two orders $(2,4)$ and $(4,2)$ are distinct outcomes, and $(2,2)$ is now possible. Forgetting order does not make the resulting unordered multisets equiprobable: $\{2,4\}$ has probability $2/36$ but $\{2,2\}$ has probability $1/36$. The thirty-six sample means still have expected value $7$, but their squared deviations sum to $210$, giving variance $210/36=35/6$. Avoiding duplicate labels improves precision for this fixed population.