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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.

For uniform sampling without replacement from a population of $n>1$ objects, the finite population correction accounts for the reduced spread caused by drawing distinct objects. For the marked-object count $K$ in a hypergeometric distribution, put $p=c/n$. ThenRelative to the corresponding binomial distribution, the variance is multiplied by $f^2=(n-r)/(n-1)$ and the standard deviation by $f$. The square-root convention for the correction is used in Philip Stark's discussion of sample sums.^[ref]

The factor has a direct interpretation. For $r=1$, replacement makes no difference and $f=1$. For $r=n$, the entire population is observed and the marked count is exactly $c$, so $f=0$. In the ten-object, five-draw example, $f^2=5/9$ and $(6/5)(5/9)=2/3$. The exceptional population size $n=1$ is deterministic and should be handled directly rather than inserting it into a zero denominator.

A short explanation uses an indicator random variable $I_j$ for a marked object at draw $j$. Each has expected value $p$ and variance $p(1-p)$. For distinct draw positions $i,j$,Since $K={\sum }_{j=1}^r{I}_j$, the variance of a sum includes $r$ individual terms and $r(r-1)$ ordered covariance terms, giving the displayed correction. Linearity of expectation still gives $rp$ despite this dependence.

For arbitrary fixed real population values, use $N>1$ for population size and $1\le n\le N$ for sample size. Write the finite population mean $\bar{Y}=N^{-1}{\sum }_i{y}_i$, the finite population variance $S^2=(N-1{)}^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$ and ${\sigma }_N^2=N^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$. Under simple random sampling without replacement, the sample mean satisfiesThus $1-n/N$ is the variance factor relative to $S^2/n$, while $(N-n)/(N-1)$ is the exact variance ratio to independent uniform sampling with replacement, whose sample mean has variance ${\sigma }_N^2/n$. The standard error uses the square root of the corresponding factor. These formulas agree because ${\sigma }_N^2=(N-1)S^2/N$; the denominator convention must be specified. The sample mean under simple random sampling gives a derivation and an exact six-value example.

When $m\ge 1$ items among $N$ are marked, inspect a uniformly random permutation until the first marked item. For its first position to be $k$, the other $m-1$ marked positions must lie among the last $N-k$ slots. This yields the displayed law with support $1\le k\le N-m+1$. It is the first-success form of sampling without replacement. For one marked item the position is uniform, while its conditional success chance after $r$ failures increases to $1/(N-r)$.

Draw balls sequentially without replacement from $g$ green and $r$ red balls. Let $T$ be the first positive draw count at which the numbers of drawn green and red balls agree. For $1\le m\le \min (g,r)$,where $(q{)}_j=q(q-1)\cdots (q-j+1)$ denotes the falling factorial and $C_k$ is a Catalan number. Odd first return times are impossible because equality requires an even total number of draws.

To count the successful colour orders, encode green by a step $+1$ and red by a step $-1$. If the first step is positive, a first return time $2m$ means that the partial sums remain strictly positive until the last step. Delete the first positive step and the last negative step, and shift the remaining heights down by one. The result is a Dyck path of length $2m-2$, of which there are $C_{m-1}$. Interchanging the colours doubles this count. Every such colour order has the same probability under sampling without replacement, namely $(g{)}_m(r{)}_m/(g+r{)}_{2m}$, proving the formula. In particular, the numbers of successful colour orders at lengths two, four and six are two, two and four; balanced orders with an earlier return must not be counted.

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.

For the positive-integer geometric distribution with success probability $0<p\le 1$, the displayed identity holds for nonnegative integers $m,r$ whenever the conditioning event has positive probability. The first $m$ attempts have failed; future independent attempts use the same success probability. Equivalently, divide the tail $(1-p{)}^{m+r}$ by $(1-p{)}^m$. For a zero-based failures count $F=T-1$, use $\mathbb{P}(F\ge m+r\mid F\ge m)=\mathbb{P}(F\ge r)$. Constant success probability and independence are essential: accumulated wear or learning would require a different model.

The full remaining-count law is unchanged, not only one tail probability. With $0<p<1$, after $m$ failures set $R=T-m$ on the event $T>m$. For $r=1,2,\dots$, its probability mass function under this conditional probability isHere independence makes the future trials unaffected by the observed failures, and the unchanged success probability makes their law identical to the original one. Consequently $\mathbb{E}[T-m\mid T>m]=1/p$, while $\mathbb{E}[T\mid T>m]=m+1/p$. For the constructed $p=0.3$ mechanism, five failures leave a mean of $10/3$ more attempts, making the conditional mean total $25/3$; the five already spent are still counted.

The same mechanism has constant discrete hazard:Thus each reached attempt has the same chance of success. This differs from the decreasing unconditional probability $\mathbb{P}(T=k)$: later success requires surviving earlier attempts. At $p=1$, only conditioning on an event of positive probability is allowed: $T>m$ permits only $m=0$, and $T\ge k$ only $k=1$.

Among proper laws on the positive integers, memorylessness of the geometric distribution also characterizes the geometric distribution. Write $S(n)=\mathbb{P}(T>n)$. The identity gives $S(m+r)=S(m)S(r)$; if $S(m)=0$, both sides vanish anyway. Since $S(0)=1$, induction yields $S(n)=S(1{)}^n$. Properness excludes $S(1)=1$, so setting $p=1-S(1)>0$ recovers the geometric distribution, including $p=1$.^[ref]

The retry example stipulates independent Bernoulli trials with a known constant $p$; it is not evidence that a real retry system behaves this way. Shared changing conditions or adapting attempts can alter the next success probability. Even constant individual success probabilities alone do not imply independence.

For an original contrasting model, uniformly shuffle one success card and three failure cards, then draw by sampling without replacement. Its success position $U$ has the discrete uniform distribution on $\{1,2,3,4\}$. Initially the success probability is $1/4$; after one failure it is $1/3$. In particular,This shows concretely why sampling without replacement need not be independent and why the remaining wait can change. Every draw still has unconditional success probability $1/4$ when all four positions are considered; the relevant discrete hazard changes after failures.

Suppose a finite population is split into $m$ categories with fixed sizes $c_1,\dots ,c_m$ summing to $n$. A uniform size-$r$ sample drawn by sampling without replacement has category counts $K_1,\dots ,K_m$ with the displayed joint probability mass function, for integers $0\le k_i\le c_i$ summing to $r$. It extends the two-category hypergeometric distribution; replacing the sampled objects instead gives a multinomial distribution.

For a box with category sizes $(5,3,2)$ and a sample of four distinct objects, the count pattern $(2,1,1)$ has probabilityPut $p_i=c_i/n$. When $n>1$, its covariance matrix is the corresponding multinomial distribution covariance matrix multiplied by the squared finite population correction:This connects the fixed-total dependence between categories with the further reduction from sampling without replacement. X. G. Duan gives a symmetry-based explanation of this shared covariance structure.^[ref]

The sampling without replacement makes all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs equally likely. The count of red-red pairs is $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$, while the count of mixed pairs is $3(n-3)$. Thus their probabilities satisfyand cancellation gives $n-3=3$. Hence $\boxed{n=6}$, with three black socks. The count of black-black pairs is also $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$, soThe mixed, red-red and black-black probabilities are respectively $3/5,1/5,1/5$, which add to one and verify the given ratio.

All size-six subsets are equally likely under uniform sampling without replacement. There are $\left(\genfrac{}{}{0ex}{}{12}{6}\right)$ subsets in total, and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)\left(\genfrac{}{}{0ex}{}{6}{3}\right)$ contain the required three members from each group. Thus the hypergeometric distribution gives

Count unordered sets of distinct labelled fish: there are $\left(\genfrac{}{}{0ex}{}{N}{M}\right)$ possible hauls. To obtain exactly $j$ trout, choose $j$ labels from the $K$ trout and $M-j$ from the $N-K$ carp, giving $\left(\genfrac{}{}{0ex}{}{K}{j}\right)\left(\genfrac{}{}{0ex}{}{N-K}{M-j}\right)$ possibilities. Uniform sampling without replacement therefore gives the hypergeometric distributionOutside this integer range the probability is zero. If ordered catches were counted instead, every haul would have $M!$ orders and that factor would cancel from the probability ratio.

Here equal catchability means equal selection probability for each individual remaining fish. It does not mean that the two species have probability $1/2$ on each draw regardless of their populations. This interpretation is the one that yields the stated hypergeometric law.

The three preceding probabilities satisfyTheir expected value and second moment areSubtracting the squared mean givesThe value $\mu =2(2/10)$ also follows from linearity of expectation. The variance includes the finite-population reduction caused by sampling without replacement.

When only a tiny fraction of either species is removed, the trout fraction among the remaining fish stays nearly constant at $p=K/N$. Successive catches are then approximately independent trials with that fixed success probability. The count therefore approaches the binomial distributionThe approximation comes from negligible depletion, not from an exact independence property of sampling without replacement.

Uniform sampling without replacement is equivalent to revealing a uniformly random permutation of seven labelled balls. Each draw position is equally likely to contain any label, so its colour has the original proportions. ThusFor example the law of total probability gives $\mathbb{P}(B_2)=(4/7)(3/6)+(3/7)(4/6)=4/7$. Equal marginal probabilities express exchangeability, rather than independence between successive draws.

For sampling without replacement, sum the two outcomes with a blue second draw:The marginal blue probability equals the initial blue fraction even though the two draws are dependent.

Random sampling without replacement assigns each of the ten distinguishable pieces to each lunch with the same probability. Five pieces of each type therefore giveOne can check the second probability by conditioning on the first lunch:The identical marginal probabilities do not imply independence of the draws.

Let $I_j$ be the indicator random variable of an apple in lunch $j$. The number remaining is $R=5-I_1-I_2-I_3$. Each expected value $E[I_j]=P(A_j)=1/2$. The linearity of expectation therefore givesAlthough the indicator random variables are dependent under sampling without replacement, the linearity of expectation needs no independence assumption.

With $N/2$ white balls, uniform sampling without replacement gives the probabilityRearranging, $4p_w(N-1)=N-2$, soA finite admissible population has even $N\ge 2$ and $0\le p_w<1/4$, so the denominator is nonzero.