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Hypergeometric distribution

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From Wikipedia, the free encyclopedia

Discrete probability distribution

Not to be confused withGeometric distribution orHypergeometric function.

Hypergeometric
Probability mass function
Cumulative distribution function
Parameters	${\begin{aligned}N&\in \left\{0,1,2,\dots \right\}\\K&\in \left\{0,1,2,\dots ,N\right\}\\n&\in \left\{0,1,2,\dots ,N\right\}\end{aligned}}\,$
Support	$\scriptstyle {k\,\in \,\{\max {(0,\,n+K-N)},\,\dots ,\,\min {(n,\,K)}\}}\,$
PMF	${\frac {{\binom {K}{k}}{\binom {N-K}{n-k}}}{\binom {N}{n}}}$
CDF	$1-{{{n \choose {k+1}}{{N-n} \choose {K-k-1}}} \over {N \choose K}}\,_{3}F_{2}\!\!\left[{\begin{array}{c}1,\ k+1-K,\ k+1-n\\k+2,\ N+k+2-K-n\end{array}};1\right],$ where $\,_{p}F_{q}$ is thegeneralized hypergeometric function
Mean	$n{K \over N}$
Mode	$\left\lceil {\frac {(n+1)(K+1)}{N+2}}\right\rceil -1,\left\lfloor {\frac {(n+1)(K+1)}{N+2}}\right\rfloor$
Variance	$n{K \over N}{N-K \over N}{N-n \over N-1}$
Skewness	${\frac {(N-2K)(N-1)^{\frac {1}{2}}(N-2n)}{[nK(N-K)(N-n)]^{\frac {1}{2}}(N-2)}}$
Excess kurtosis	$\left.{\frac {1}{nK(N-K)(N-n)(N-2)(N-3)}}\cdot \right.$ ${\big [}(N-1)N^{2}{\big (}N(N+1)-6K(N-K)-6n(N-n){\big )}$ ${}+6nK(N-K)(N-n)(5N-6){\big ]}$
MGF	${\frac {{\binom {N-K}{n}}\,_{2}F_{1}(-n,-K\,;\,N-K-n+1\,;\,e^{t})}{\binom {N}{n}}}$
CF	${\frac {{\binom {N-K}{n}}\,_{2}F_{1}(-n,-K\,;\,N-K-n+1\,;\,e^{it})}{\binom {N}{n}}}$

Inprobability theory andstatistics, thehypergeometric distribution is adiscrete probability distribution that describes the probability of $k {\displaystyle k}$ successes (random draws for which the object drawn has a specified feature) in $n {\displaystyle n}$ draws,without replacement, from a finitepopulation of size $N {\displaystyle N}$ that contains exactly $K {\displaystyle K}$ objects with that feature, where in each draw is either a success or a failure. In contrast, thebinomial distribution describes the probability of $k {\displaystyle k}$ successes in $n {\displaystyle n}$ drawswith replacement.

Definitions

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Probability mass function

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The following conditions characterize the hypergeometric distribution:

The result of each draw (the elements of the population being sampled) can be classified into one oftwo mutually exclusive categories (e.g. Pass/Fail or Employed/Unemployed).
The probability of a success changes on each draw, as each draw decreases the population (sampling without replacement from a finite population).

Arandom variable $X {\displaystyle X}$ follows the hypergeometric distribution if itsprobability mass function (pmf) is given by^[1]

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

where

$N {\displaystyle N}$ is the population size,
$K {\displaystyle K}$ is the number of success states in the population,
$n {\displaystyle n}$ is the number of draws (i.e. quantity drawn in each trial),
$k {\displaystyle k}$ is the number of observed successes,
${\textstyle \textstyle {a \choose b}}$ is abinomial coefficient.

Thepmf is positive when $\max(0,n+K-N)\leq k\leq \min(K,n)$ .

A random variable distributed hypergeometrically with parameters $N {\displaystyle N}$ , $K {\displaystyle K}$ and $n {\displaystyle n}$ is written ${\textstyle X\sim \operatorname {Hypergeometric} (N,K,n)}$ and has probability mass function ${\textstyle p_{X}(k)}$ above.

Combinatorial identities

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As required, we have

\sum _{0\leq k\leq {\textrm {min}}(n,K)}{{K \choose k}{N-K \choose n-k} \over {N \choose n}}=1,

which essentially follows fromVandermonde's identity fromcombinatorics.

Also note that

{{K \choose k}{N-K \choose n-k} \over {N \choose n}}={{{n \choose k}{{N-n} \choose {K-k}}} \over {N \choose K}};

This identity can be shown by expressing the binomial coefficients in terms of factorials and rearranging the latter. Additionally, itfollows from the symmetry of the problem, described in two different but interchangeable ways.

For example, consider two rounds of drawing without replacement. In the first round, $K {\displaystyle K}$ out of $N {\displaystyle N}$ neutral marbles are drawn from an urn without replacement and coloured green. Then the colored marbles are put back. In the second round, $n {\displaystyle n}$ marbles are drawn without replacement and colored red. Then, the number of marbles with both colors on them (that is, the number of marbles that have been drawn twice) has the hypergeometric distribution. The symmetry in $K {\displaystyle K}$ and $n {\displaystyle n}$ stems from the fact that the two rounds are independent, and one could have started by drawing $n {\displaystyle n}$ balls and colouring them red first.

Note that we are interested in the probability of $k {\displaystyle k}$ successes in $n {\displaystyle n}$ drawswithout replacement, since the probability of success on each trial is not the same, as the size of the remaining population changes as we remove each marble. Keep in mind not to confuse with thebinomial distribution, which describes the probability of $k {\displaystyle k}$ successes in $n {\displaystyle n}$ drawswith replacement.

Properties

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Working example

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The classical application of the hypergeometric distribution issampling without replacement. Think of anurn with two colors ofmarbles, red and green. Define drawing a green marble as a success and drawing a red marble as a failure. LetN describe the number ofall marbles in the urn (see contingency table below) andK describe the number ofgreen marbles, thenN − K corresponds to the number ofred marbles. Now, standing next to the urn, you close your eyes and draw n marbles without replacement. DefineX as arandom variable whose outcome isk, the number of green marbles drawn in the experiment. This situation is illustrated by the followingcontingency table:

	drawn	not drawn	total
green marbles	k	K −k	K
red marbles	n −k	N + k − n − K	N − K
total	n	N − n	N

Indeed, we are interested in calculating the probability of drawing k green marbles in n draws, given that there are K green marbles out of a total of N marbles. For this example, assume that there are5 green and45 red marbles in the urn. Standing next to the urn, you close your eyes and draw10 marbles without replacement. What is the probability that exactly4 of the10 are green?

This problem is summarized by the following contingency table:

	drawn	not drawn	total
green marbles	k =4	K −k =1	K =5
red marbles	n −k =6	N + k − n − K =39	N − K =45
total	n =10	N − n =40	N =50

To find the probability ofdrawing k green marbles in exactly n draws out of N total draws, we identify X as a hyper-geometric random variable to use the formula

$P(X=k)=f(k;N,K,n)={{{K \choose k}{{N-K} \choose {n-k}}} \over {N \choose n}}.$

To intuitively explain the given formula, consider the two symmetric problems represented by the identity

${{K \choose k}{N-K \choose n-k} \over {N \choose n}}={{{n \choose k}{{N-n} \choose {K-k}}} \over {N \choose K}}$

left-hand side - drawing a total of only n marbles out of the urn. We want to find the probability of the outcome ofdrawing k green marbles out of K total green marbles, and drawing n-k red marbles out of N-K red marbles, in these n rounds.
right hand side - alternatively, drawing all N marbles out of the urn. We want to find the probability of the outcome of drawingk green marbles in n draws out of the total N draws, and K-k green marbles in the rest N-n draws.

Back to the calculations, we use the formula above to calculate the probability of drawing exactlyk green marbles

P(X=4)=f(4;50,5,10)={{{5 \choose 4}{{45} \choose {6}}} \over {50 \choose 10}}={5\cdot 8145060 \over 10272278170}=0.003964583\dots .

Intuitively we would expect it to be even more unlikely that all 5 green marbles will be among the 10 drawn.

P(X=5)=f(5;50,5,10)={{{5 \choose 5}{{45} \choose {5}}} \over {50 \choose 10}}={1\cdot 1221759 \over 10272278170}=0.0001189375\dots ,

As expected, the probability of drawing 5 green marbles is roughly 35 times less likely than that of drawing 4.

Symmetries

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Swapping the roles of green and red marbles:

f(k;N,K,n)=f(n-k;N,N-K,n)

Swapping the roles of drawn and not drawn marbles:

f(k;N,K,n)=f(K-k;N,K,N-n)

Swapping the roles of green and drawn marbles:

f(k;N,K,n)=f(k;N,n,K)

These symmetries generate thedihedral group $D_{4}$ .

Order of draws

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The probability of drawing any set of green and red marbles (the hypergeometric distribution) depends only on the numbers of green and red marbles, not on the order in which they appear; i.e., it is anexchangeable distribution. As a result, the probability of drawing a green marble in the $i^{\text{th}}$ draw is^[2]

P(G_{i})={\frac {K}{N}}.

This is anex ante probability—that is, it is based on not knowing the results of the previous draws.

Tail bounds

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Let $X\sim \operatorname {Hypergeometric} (N,K,n)$ and $p=K/N$ . Then for $0<t<K/N$ we can derive the following bounds:^[3]

{\begin{aligned}\Pr[X\leq (p-t)n]&\leq e^{-n{\text{D}}(p-t\parallel p)}\leq e^{-2t^{2}n}\\\Pr[X\geq (p+t)n]&\leq e^{-n{\text{D}}(p+t\parallel p)}\leq e^{-2t^{2}n}\\\end{aligned}}\!

where

D(a\parallel b)=a\log {\frac {a}{b}}+(1-a)\log {\frac {1-a}{1-b}}

is theKullback–Leibler divergence and it is used that $D(a\parallel b)\geq 2(a-b)^{2}$ .^[4]

Note: In order to derive the previous bounds, one has to start by observing that $X={\frac {\sum _{i=1}^{n}Y_{i}}{n}}$ where $Y_{i}$ aredependent random variables with a specific distribution $D {\displaystyle D}$ . Because most of the theorems about bounds in sum of random variables are concerned withindependent sequences of them, one has to first create a sequence $Z_{i}$ ofindependent random variables with the same distribution $D {\displaystyle D}$ and apply the theorems on $X'={\frac {\sum _{i=1}^{n}Z_{i}}{n}}$ . Then, it is proved from Hoeffding^[3] that the results and bounds obtained via this process hold for $X {\displaystyle X}$ as well.

Ifn is larger thanN/2, it can be useful to apply symmetry to "invert" the bounds, which give you the following:^[4]^[5]

{\begin{aligned}\Pr[X\leq (p-t)n]&\leq e^{-(N-n){\text{D}}(p+{\tfrac {tn}{N-n}}||p)}\leq e^{-2t^{2}n{\tfrac {n}{N-n}}}\\\\\Pr[X\geq (p+t)n]&\leq e^{-(N-n){\text{D}}(p-{\tfrac {tn}{N-n}}||p)}\leq e^{-2t^{2}n{\tfrac {n}{N-n}}}\\\end{aligned}}\!

Statistical Inference

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Hypergeometric test

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Thehypergeometric test uses the hypergeometric distribution to measure the statistical significance of having drawn a sample consisting of a specific number of $k {\displaystyle k}$ successes (out of $n {\displaystyle n}$ total draws) from a population of size $N {\displaystyle N}$ containing $K {\displaystyle K}$ successes. In a test for over-representation of successes in the sample, the hypergeometric p-value is calculated as the probability of randomly drawing $k {\displaystyle k}$ or more successes from the population in $n {\displaystyle n}$ total draws. In a test for under-representation, the p-value is the probability of randomly drawing $k {\displaystyle k}$ or fewer successes.

The test based on the hypergeometric distribution (hypergeometric test) is identical to the corresponding one-tailed version ofFisher's exact test.^[6] Reciprocally, the p-value of a two-sided Fisher's exact test can be calculated as the sum of two appropriate hypergeometric tests (for more information see^[7]).

The test is often used to identify which sub-populations are over- or under-represented in a sample. This test has a wide range of applications. For example, a marketing group could use the test to understand their customer base by testing a set of known customers for over-representation of various demographic subgroups (e.g., women, people under 30).

Related distributions

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Let $X\sim \operatorname {Hypergeometric} (N,K,n)$ and $p=K/N$ .

If $n=1$ then $X {\displaystyle X}$ has aBernoulli distribution with parameter $p {\displaystyle p}$ .
Let $Y {\displaystyle Y}$ have abinomial distribution with parameters $n {\displaystyle n}$ and $p {\displaystyle p}$ ; this models the number of successes in the analogous sampling problemwith replacement. If $N {\displaystyle N}$ and $K {\displaystyle K}$ are large compared to $n {\displaystyle n}$ , and $p {\displaystyle p}$ is not close to 0 or 1, then $X {\displaystyle X}$ and $Y {\displaystyle Y}$ have similar distributions, i.e., $P(X\leq k)\approx P(Y\leq k)$ .
If $n {\displaystyle n}$ is large, $N {\displaystyle N}$ and $K {\displaystyle K}$ are large compared to $n {\displaystyle n}$ , and $p {\displaystyle p}$ is not close to 0 or 1, then

P(X\leq k)\approx \Phi \left({\frac {k-np}{\sqrt {np(1-p)}}}\right)

where $\Phi$ is thestandard normal distribution function

If the probabilities of drawing a green or red marble are not equal (e.g. because green marbles are bigger/easier to grasp than red marbles) then $X {\displaystyle X}$ has anoncentral hypergeometric distribution
Thebeta-binomial distribution is aconjugate prior for the hypergeometric distribution.

The following table describes four distributions related to the number of successes in a sequence of draws:

	With replacements	No replacements
Given number of draws	binomial distribution	hypergeometric distribution
Given number of failures	negative binomial distribution	negative hypergeometric distribution

Multivariate hypergeometric distribution

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Multivariate hypergeometric distribution
Parameters	$c\in \mathbb {N} _{+}=\lbrace 1,2,\ldots \rbrace$ $(K_{1},\ldots ,K_{c})\in \mathbb {N} ^{c}$ $N=\sum _{i=1}^{c}K_{i}$ $n\in \lbrace 0,\ldots ,N\rbrace$
Support	$\left\{\mathbf {k} \in \left(\mathbb {Z} _{0+}\right)^{c}\,:\,\forall i\ k_{i}\leq K_{i},\sum _{i=1}^{c}k_{i}=n\right\}$
PMF	${\frac {\prod \limits _{i=1}^{c}{\binom {K_{i}}{k_{i}}}}{\binom {N}{n}}}$
Mean	$\operatorname {E} (k_{i})=n{\frac {K_{i}}{N}}$
Variance	$\operatorname {Var} (k_{i})=n{\frac {N-n}{N-1}}\;{\frac {K_{i}}{N}}\left(1-{\frac {K_{i}}{N}}\right)$ $\operatorname {Cov} (k_{i},k_{j})=-n{\frac {N-n}{N-1}}\;{\frac {K_{i}}{N}}{\frac {K_{j}}{N}},i\neq j$ $\operatorname {Corr} (k_{i},k_{j})=-{\sqrt {\frac {K_{i}K_{j}}{\left(N-K_{i}\right)\left(N-K_{j}\right)}}}$

The model of anurn with green and red marbles can be extended to the case where there are more than two colors of marbles. If there areK_i marbles of colori in the urn and you taken marbles at random without replacement, then the number of marbles of each color in the sample (k₁,k₂,...,k_c) has the multivariate hypergeometric distribution:

\Pr(X_{1}=k_{1},\ldots ,X_{c}=k_{c})={\frac {\prod \limits _{i=1}^{c}{\binom {K_{i}}{k_{i}}}}{\binom {N}{n}}}

This has the same relationship to themultinomial distribution that the hypergeometric distribution has to the binomial distribution—the multinomial distribution is the "with-replacement" distribution and the multivariate hypergeometric is the "without-replacement" distribution.

The properties of this distribution are given in the adjacent table,^[8] wherec is the number of different colors and $N=\sum _{i=1}^{c}K_{i}$ is the total number of marbles in the urn.

Example

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Suppose there are 5 black, 10 white, and 15 red marbles in an urn. If six marbles are chosen without replacement, the probability that exactly two of each color are chosen is

P(2{\text{ black}},2{\text{ white}},2{\text{ red}})={{{5 \choose 2}{10 \choose 2}{15 \choose 2}} \over {30 \choose 6}}=0.079575596816976

Occurrence and applications

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Application to auditing elections

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Samples used for election audits and resulting chance of missing a problem

Election audits typically test a sample of machine-counted precincts to see if recounts by hand or machine match the original counts. Mismatches result in either a report or a larger recount. The sampling rates are usually defined by law, not statistical design, so for a legally defined sample sizen, what is the probability of missing a problem which is present inK precincts, such as a hack or bug? This is the probability thatk = 0 . Bugs are often obscure, and a hacker can minimize detection by affecting only a few precincts, which will still affect close elections, so a plausible scenario is forK to be on the order of 5% ofN. Audits typically cover 1% to 10% of precincts (often 3%),^[9]^[10]^[11] so they have a high chance of missing a problem. For example, if a problem is present in 5 of 100 precincts, a 3% sample has 86% probability thatk = 0 so the problem would not be noticed, and only 14% probability of the problem appearing in the sample (positive k):

{\begin{aligned}\operatorname {\boldsymbol {\mathcal {P}}} \{\ X=0\ \}&={\frac {\ \left[\ {\binom {\text{Hack}}{0}}{\binom {N\ -\ {\text{Hack}}}{n\ -\ 0}}\ \right]\ }{\left[\ {\binom {N}{n}}\ \right]}}={\frac {\ \left[\ {\binom {N\ -\ {\text{Hack}}}{n}}\ \right]}{\ \left[\ {\binom {N}{n}}\ \right]\ }}={\frac {\ \left[\ {\frac {\ (N\ -\ {\text{Hack}})!\ }{n!(N\ -\ {\text{Hack}}-n)!}}\ \right]\ }{\left[\ {\frac {N!}{n!(N\ -\ n)!}}\ \right]}}={\frac {\ \left[\ {\frac {(N-{\text{Hack}})!}{(N\ -\ {\text{Hack}}\ -\ n)!}}\ \right]\ }{\left[\ {\frac {N!}{(N\ -\ n)!}}\ \right]}}\\[8pt]&={\frac {\ \left[\ {\binom {100-5}{3}}\ \right]\ }{\ \left[\ {\binom {100}{3}}\ \right]\ }}={\frac {\ \left[\ {\frac {(100-5)!}{(100-5-3)!}}\ \right]\ }{\left[\ {\frac {100!}{(100-3)!}}\ \right]}}={\frac {\ \left[\ {\frac {95!}{92!}}\ \right]\ }{\ \left[\ {\frac {100!}{97!}}\ \right]\ }}={\frac {\ 95\times 94\times 93\ }{100\times 99\times 98}}=86\%\end{aligned}}

The sample would need 45 precincts in order to have probability under 5% thatk = 0 in the sample, and thus have probability over 95% of finding the problem:

\operatorname {\boldsymbol {\mathcal {P}}} \{\ X=0\ \}={\frac {\ \left[\ {\binom {100-5}{45}}\ \right]\ }{\left[\ {\binom {100}{45}}\ \right]}}={\frac {\ \left[\ {\frac {95!}{50!}}\ \right]\ }{\left[\ {\frac {100!}{55!}}\ \right]}}={\frac {\ 95\times 94\times \cdots \times 51\ }{\ 100\times 99\times \cdots \times 56\ }}={\frac {\ 55\times 54\times 53\times 52\times 51\ }{\ 100\times 99\times 98\times 97\times 96\ }}=4.6\%~.

Application to Texas hold'em poker

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Inhold'em poker players make the best hand they can combining the two cards in their hand with the 5 cards (community cards) eventually turned up on the table. The deck has 52 and there are 13 of each suit.For this example assume a player has 2 clubs in the hand and there are 3 cards showing on the table, 2 of which are also clubs. The player would like to know the probability of one of the next 2 cards to be shown being a club to complete theflush.
(Note that the probability calculated in this example assumes no information is known about the cards in the other players' hands; however, experienced poker players may consider how the other players place their bets (check, call, raise, or fold) in considering the probability for each scenario. Strictly speaking, the approach to calculating success probabilities outlined here is accurate in a scenario where there is just one player at the table; in a multiplayer game this probability might be adjusted somewhat based on the betting play of the opponents.)

There are 4 clubs showing so there are 9 clubs still unseen. There are 5 cards showing (2 in the hand and 3 on the table) so there are $52-5=47$ still unseen.

The probability that one of the next two cards turned is a club can be calculated using hypergeometric with $k=1,n=2,K=9$ and $N=47$ . (about 31.64%)

The probability that both of the next two cards turned are clubs can be calculated using hypergeometric with $k=2,n=2,K=9$ and $N=47$ . (about 3.33%)

The probability that neither of the next two cards turned are clubs can be calculated using hypergeometric with $k=0,n=2,K=9$ and $N=47$ . (about 65.03%)

Application to Keno

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The hypergeometric distribution is indispensable for calculatingKeno odds. In Keno, 20 balls are randomly drawn from a collection of 80 numbered balls in a container, rather likeAmerican Bingo. Prior to each draw, a player selects a certain number ofspots by marking a paper form supplied for this purpose. For example, a player mightplay a 6-spot by marking 6 numbers, each from a range of 1 through 80 inclusive. Then (after all players have taken their forms to a cashier and been given a duplicate of their marked form, and paid their wager) 20 balls are drawn. Some of the balls drawn may match some or all of the balls selected by the player. Generally speaking, the morehits (balls drawn that match player numbers selected) the greater the payoff.

For example, if a customer bets ("plays") $1 for a 6-spot (not an uncommon example) and hits 4 out of the 6, the casino would pay out $4. Payouts can vary from one casino to the next, but $4 is a typical value here. The probability of this event is:

P(X=4)=f(4;80,6,20)={{{6 \choose 4}{{80-6} \choose {20-4}}} \over {80 \choose 20}}\approx 0.02853791

Similarly, the chance for hitting 5 spots out of 6 selected is ${{{6 \choose 5}{{74} \choose {15}}} \over {80 \choose 20}}\approx 0.003095639$ while a typical payout might be $88. The payout for hitting all 6 would be around $1500 (probability ≈ 0.000128985 or 7752-to-1). The only other nonzero payout might be $1 for hitting 3 numbers (i.e., you get your bet back), which has a probability near 0.129819548.

Taking the sum of products of payouts times corresponding probabilities we get an expected return of 0.70986492 or roughly 71% for a 6-spot, for a house advantage of 29%. Other spots-played have a similar expected return. This very poor return (for the player) is usually explained by the large overhead (floor space, equipment, personnel) required for the game.

References

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Citations

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^Rice, John A. (2007).Mathematical Statistics and Data Analysis (Third ed.). Duxbury Press. p. 42.
^Pollard, David (Spring 2010)."Symmetry"(PDF).Stat 330/600 course handouts. Yale University. Retrieved2025-01-19.
^^a ^bHoeffding, Wassily (1963)."Probability inequalities for sums of bounded random variables"(PDF).Journal of the American Statistical Association.58 (301):13–30.doi:10.2307/2282952.JSTOR 2282952..
^^a ^b"Another Tail of the Hypergeometric Distribution".wordpress.com. 8 December 2015. Retrieved19 March 2018.
^Serfling, Robert (1974). "Probability inequalities for the sum in sampling without replacement".The Annals of Statistics.2 (1):39–48.doi:10.1214/aos/1176342611..
^Rivals, I.; Personnaz, L.; Taing, L.; Potier, M.-C (2007)."Enrichment or depletion of a GO category within a class of genes: which test?".Bioinformatics.23 (4):401–407.doi:10.1093/bioinformatics/btl633.PMID 17182697.
^K. Preacher and N. Briggs."Calculation for Fisher's Exact Test: An interactive calculation tool for Fisher's exact probability test for 2 x 2 tables (interactive page)".
^Duan, X. G. (2021). "Better understanding of the multivariate hypergeometric distribution with implications in design-based survey sampling".arXiv:2101.00548 [math.ST].
^Glazer, Amanda; Spertus, Jacob (10 February 2020). "Start spreading the news: New York's post-election audit has major flaws".SSRN 3536011.
^"State audit laws".Verified Voting. 10 February 2017. Archived fromthe original on 4 January 2020. Retrieved2 April 2018.
^"Post-election audits".ncsl.org. National Conference of State Legislatures. Retrieved2 April 2018.

This article includes a list ofgeneral references, butit lacks sufficient correspondinginline citations. Please help toimprove this article byintroducing more precise citations.(August 2011) (Learn how and when to remove this message)

Sources

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Berkopec, Aleš (2007)."HyperQuick algorithm for discrete hypergeometric distribution".Journal of Discrete Algorithms.5 (2):341–347.doi:10.1016/j.jda.2006.01.001.
Skala, M. (2011). "Hypergeometric tail inequalities: ending the insanity".arXiv:1311.5939 [math.PR]. unpublished note

External links

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Probability distributions (list)

Discrete
univariate

with finite support	Benford Bernoulli Beta-binomial Binomial Categorical Hypergeometric Negative Poisson binomial Rademacher Soliton Discrete uniform Zipf Zipf–Mandelbrot
with infinite support	Beta negative binomial Borel Conway–Maxwell–Poisson Discrete phase-type Delaporte Extended negative binomial Flory–Schulz Gauss–Kuzmin Geometric Logarithmic Mixed Poisson Negative binomial Panjer Parabolic fractal Poisson Skellam Yule–Simon Zeta

Continuous
univariate

supported on a bounded interval	Arcsine ARGUS Balding–Nichols Bates Beta Generalized Beta rectangular Continuous Bernoulli Irwin–Hall Kumaraswamy Logit-normal Noncentral beta PERT Power function Raised cosine Reciprocal Triangular U-quadratic Uniform Wigner semicircle
supported on a semi-infinite interval	Benini Benktander 1st kind Benktander 2nd kind Beta prime Burr Chi Chi-squared Noncentral Inverse Scaled Dagum Davis Erlang Hyper Exponential Hyperexponential Hypoexponential Logarithmic F Noncentral Folded normal Fréchet Gamma Generalized Inverse gamma/Gompertz Gompertz Shifted Half-logistic Half-normal Hotelling'sT-squared Hartman–Watson Inverse Gaussian Generalized Kolmogorov Lévy Log-Cauchy Log-Laplace Log-logistic Log-normal Log-t Lomax Matrix-exponential Maxwell–Boltzmann Maxwell–Jüttner Mittag-Leffler Nakagami Pareto Phase-type Poly-Weibull Rayleigh Relativistic Breit–Wigner Rice Truncated normal type-2 Gumbel Weibull Discrete Wilks's lambda
supported on the whole real line	Cauchy Exponential power Fisher'sz Kaniadakis κ-Gaussian Gaussianq Generalized hyperbolic Generalized logistic (logistic-beta) Generalized normal Geometric stable Gumbel Holtsmark Hyperbolic secant Johnson'sS_U Landau Laplace Asymmetric Logistic Noncentralt Normal (Gaussian) Normal-inverse Gaussian Skew normal Slash Stable Student'st Tracy–Widom Variance-gamma Voigt
with support whose type varies	Generalized chi-squared Generalized extreme value Generalized Pareto Marchenko–Pastur Kaniadakisκ-exponential Kaniadakisκ-Gamma Kaniadakisκ-Weibull Kaniadakisκ-Logistic Kaniadakisκ-Erlang q-exponential q-Gaussian q-Weibull Shifted log-logistic Tukey lambda