German tank problem

During World War II, production of German tanks such as the Panther was accurately estimated by Allied intelligence using statistical methods.

In the statistical theory of estimation, the problem of estimating the maximum of a discrete uniform distribution from sampling without replacement is known in English as the German tank problem, due to its application in World War II to the estimation of the number of German tanks.

The analyses illustrate the difference between frequentist inference and Bayesian inference.

Estimating the population maximum based on a single sample yields divergent results, while the estimation based on multiple samples is an instructive practical estimation question whose answer is simple but not obvious.

Example

Suppose an intelligence officer has spotted k = 4 tanks with serial numbers, 2, 6, 7, and 14, with the maximum observed serial number, m = 14. The unknown total number of tanks is called N.

The formula for estimating the total number of tanks suggested by the frequentist approach outlined below is

$N\approx m-1+{\frac {m}{k}}=16.5$

Whereas, the Bayesian analysis below yields (primarily) a probability mass function for the number of tanks

$\Pr(N=n)={\begin{cases}0&{\text{if }}n<m\\{\frac {k-1}{k}}{\frac {\binom {m-1}{k-1}}{\binom {n}{k}}}&{\text{if }}n\geq m\end{cases}}$

from which we can estimate the number of tanks according to

${\begin{aligned}N&\approx \mu \pm \sigma =19.5\pm 10\\\mu &=(m-1){\frac {k-1}{k-2}}\\\sigma &={\sqrt {\frac {(k-1)(m-1)(m-k+1)}{(k-3)(k-2)^{2}}}}\end{aligned}}$

This distribution has positive skewness, related to the fact that there are at least 14 tanks.

Historical problem

Panther tanks are loaded for transport to frontline units, 1943.

During the course of the war the Western Allies made sustained efforts to determine the extent of German production, and approached this in two major ways: conventional intelligence gathering and statistical estimation. In many cases statistical analysis substantially improved on conventional intelligence. In some cases conventional intelligence was used in conjunction with statistical methods, as was the case in estimation of Panther tank production just prior to D-Day.

The allied command structure had thought the Panzer V (Panther) tanks seen in Italy, with their high velocity, long barreled 75 mm/L70 guns, were unusual heavy tanks, and would only be seen in northern France in small numbers, much the same way as the Tiger I was seen in Tunisia. The US Army was confident that the Sherman tank would perform well against the Panzer III and IV tanks that they expected to meet. Shortly before D-Day, rumors indicated that large numbers of Panzer V tanks were being used.

To ascertain if this were true the Allies attempted to estimate the number of tanks being produced. To do this they used the serial numbers on captured or destroyed tanks. The principal numbers used were gearbox numbers, as these fell in two unbroken sequences. Chassis and engine numbers were also used, though their use was more complicated. Various other components were used to cross-check the analysis. Similar analyses were done on tires, which were observed to be sequentially numbered (i.e., 1, 2, 3, ..., N).

The analysis of tank wheels yielded an estimate for the number of wheel molds that were in use. A discussion with British road wheel makers then estimated the number of wheels that could be produced from this many molds, which yielded the number of tanks that were being produced each month. Analysis of wheels from two tanks (48 wheels each, 96 wheels total) yielded an estimate of 270 produced in February 1944, substantially more than had previously been suspected.

German records after the war showed production for the month of February 1944 was 276. The statistical approach proved to be far more accurate than conventional intelligence methods, and the phrase German tank problem became accepted as a descriptor for this type of statistical analysis.

Estimating production was not the only use of this serial number analysis. It was also used to understand German production more generally, including number of factories, relative importance of factories, length of supply chain (based on lag between production and use), changes in production, and use of resources such as rubber.

Specific data

According to conventional Allied intelligence estimates the Germans were producing around 1,400 tanks a month between June 1940 and September 1942. Applying the formula below to the serial numbers of captured tanks, the number was calculated to be 246 a month. After the war, captured German production figures from the ministry of Albert Speer showed the actual number to be 245.

Estimates for some specific months are given as:

Month	Statistical estimate	Intelligence estimate	German records
June 1940	169	1,000	122
June 1941	244	1,550	271
August 1942	327	1,550	342

Similar analyses

V-2 rocket production was accurately estimated by statistical methods.

Similar serial number analysis was used for other military equipment during World War II, most successfully for the V-2 rocket.

During World War II, German intelligence analyzed factory markings on Soviet military equipment, and during the Korean War factory markings on Soviet equipment were analyzed. The Soviets also estimated German tank production during World War II.

In the 1980s, some Americans were given access to the production line of Israel's Merkava tanks. The production numbers were classified, but the tanks had serial numbers, allowing estimation of production.

The formula has been used in non-military contexts, for example to estimate the number of Commodore 64 computers built, where the result (12.5 million) matches the official figures quite well.

Countermeasures

To prevent serial number analysis, serial numbers can be excluded, or usable auxiliary information reduced. Alternatively, serial numbers that resist cryptanalysis can be used, most effectively by randomly choosing numbers without replacement from a list that is much larger than the number of objects produced (compare the one-time pad), or produce random numbers and check them against the list of already assigned numbers; collisions are likely to occur unless the number of digits possible is more than twice the number of digits in the number of objects produced (where the serial number can be in any base); see birthday problem. For this, a cryptographically secure pseudorandom number generator may be used. All these methods require a lookup table (or breaking the cypher) to back out from serial number to production order, which complicates use of serial numbers: a range of serial numbers cannot be recalled, for instance, but each must be looked up individually, or a list generated.

Alternatively, sequential serial numbers can be encrypted, which allows easy decoding, but then there is a known-plaintext attack: even if starting from an arbitrary point, the plaintext has a pattern (namely, numbers are in sequence). One example is given in Ken Follett's novel "Code to Zero", where the encryption of the Jupiter C rocket serial numbers is described as:

H	U	N	T	S	V	I	L	E	X
1	2	3	4	5	6	7	8	9	0

The code word here is Huntsville (with repeated letters omitted) to get a 10-letter key. The rocket number 13 was therefore "HN", or the rocket number 24 was "UT".

Frequentist analysis

Minimum-variance unbiased estimator

For point estimation (estimating a single value for the total(${\hat {N}}$)), the minimum-variance unbiased estimator (MVUE, or UMVU estimator) is given by:

${\hat {N}}=m\left(1+k^{-1}\right)-1$

where m is the largest serial number observed (sample maximum) and k is the number of tanks observed (sample size). Note that once a serial number has been observed, it is no longer in the pool and will not be observed again.

This has a variance of

$\operatorname {var} ({\hat {N}})={\frac {1}{k}}{\frac {(N-k)(N+1)}{(k+2)}}\approx {\frac {N^{2}}{k^{2}}}{\text{ for small samples }}k\ll N$

so a standard deviation of approximately N/k, the (population) average size of a gap between samples; compare m/k above.

Intuition

The formula may be understood intuitively as:

"The sample maximum plus the average gap between observations in the sample",

the sample maximum being chosen as the initial estimator, due to being the maximum likelihood estimator, with the gap being added to compensate for the negative bias of the sample maximum as an estimator for the population maximum, and written as

${\hat {N}}=m+{\frac {m-k}{k}}=m+mk^{-1}-1=m\left(1+k^{-1}\right)-1$

This can be visualized by imagining that the samples are evenly spaced throughout the range, with additional samples just outside the range at 0 and N + 1. If starting with an initial gap between 0 and the lowest sample (sample minimum), the average gap between samples is $(m-k)/k$; the $-k$ being because the samples themselves are not counted in computing the gap between samples.

This philosophy is formalized and generalized in the method of maximum spacing estimation.

Derivation

The probability that the sample maximum equals m is ${\tbinom {m-1}{k-1}}{\big /}{\tbinom {N}{k}}$, where ${\tbinom {\cdot }{\cdot }}$ is the binomial coefficient.

The expected value of the sample maximum is

${\begin{aligned}\mu &=\sum _{m=k}^{N}m{\frac {\tbinom {m-1}{k-1}}{\tbinom {N}{k}}}={\frac {k(N+1)}{k+1}}\\\Rightarrow N&=\mu \left(1+k^{-1}\right)-1\end{aligned}}$

Then

${\begin{aligned}\mu \left(1+k^{-1}\right)-1&=E\left[m\left(1+k^{-1}\right)-1\right]\\\Rightarrow {\hat {N}}&=m\left(1+k^{-1}\right)-1\end{aligned}}$

is an unbiased estimator of N.

To show that this is the UMVU estimator:

• first show that the sample maximum is a sufficient statistic for the population maximum, using a method similar to that detailed at sufficiency: uniform distribution (but for the German tank problem we must exclude the outcomes in which a serial number occurs twice in the sample);
• Next, show that it is a complete statistic.
• Then the Lehmann–Scheffé theorem states that the sample maximum, corrected for bias as above to be unbiased, is the UMVU estimator.

Confidence intervals

Instead of, or in addition to, point estimation, interval estimation can be carried out, such as confidence intervals. These are easily computed, based on the observation that the probability that k samples will fall in an interval covering p of the range (0 ≤ p ≤ 1) is p^k (assuming in this section that draws are with replacement, to simplify computations; if draws are without replacement, this overstates the likelihood, and intervals will be overly conservative).

Thus the sampling distribution of the quantile of the sample maximum is the graph x^1/k from 0 to 1: the pth to qth quantile of the sample maximum m are the interval [p^1/kN, q^1/kN]. Inverting this yields the corresponding confidence interval for the population maximum of [m/q^1/k, m/p^1/k].

For example, taking the symmetric 95% interval p = 2.5% and q = 97.5% for k = 5 yields $\scriptstyle 0.025^{1/5}\;\approx \;0.48,\,$ $\scriptstyle 0.975^{1/5}\;\approx \;0.995$, so a confidence interval of approximately $\scriptstyle \left[1.005m,\,2.08m\right]$. The lower bound is very close to m, so more informative is the asymmetric confidence interval from p = 5% to 100%; for k = 5 this yields $\scriptstyle 0.05^{1/5}\;\approx \;0.55,$ so the interval [m, 1.82m].

More generally, the (downward biased) 95% confidence interval is $\scriptstyle \left[m,\,m/0.05^{1/k}\right]\;=\;\left[m,\,m\cdot 20^{1/k}\right]$. For a range of k, with the UMVU point estimator (plus 1 for legibility) for reference, this yields:

k	point estimate	confidence interval
1	$\scriptstyle 2m$	$\scriptstyle [m,\,20m]$
2	$\scriptstyle 1.5m$	$\scriptstyle [m,\,4.5m]$
5	$\scriptstyle 1.2m$	$\scriptstyle [m,\,1.82m]$
10	$\scriptstyle 1.1m$	$\scriptstyle [m,\,1.35m]$
20	$\scriptstyle 1.05m$	$\scriptstyle [m,\,1.16m]$

Immediate observations are:

• For small sample sizes, the confidence interval is very wide, reflecting great uncertainty in the estimate.
• The range shrinks rapidly, reflecting the exponentially decaying likelihood that all samples will be significantly below the maximum.
• The confidence interval exhibits positive skew, as N can never be below the sample maximum, but can potentially be arbitrarily high above it.

Note that m/k cannot be used naively (or rather (m + m/k − 1)/k) as an estimate of the standard error SE, as the standard error of an estimator is based on the population maximum (a parameter), and using an estimate to estimate the error in that very estimate is circular reasoning.

Bayesian analysis

The Bayesian approach to the German tank problem is to consider the probability $\scriptstyle \Pr(N\;=\;n\mid M\;=\;m,\,K\;=\;k)$ that the number of enemy tanks $\scriptstyle N$ is equal to $\scriptstyle n$, when the number of observed tanks, $\scriptstyle K$ is equal to the number $\scriptstyle k$, and the largest of the serial numbers $\scriptstyle M$ is equal to $\scriptstyle m$.

For brevity $\scriptstyle \Pr(N\;=\;n\mid M\;=\;m,\,K\;=\;k)$ is written $\scriptstyle (n\mid m,\,k)$

The rule for conditional probability gives

$(n\mid m,k)=(m\mid n,k){\frac {(n\mid k)}{(m\mid k)}}$

The expression $\scriptstyle (m\mid n,\,k)\;=\;\Pr(M\;=\;m\mid N\;=\;n,\,K\;=\;k)$ is the conditional probability that the largest tank serial number observed is equal to $\scriptstyle m$, when the number of enemy tanks is known to be equal to $\scriptstyle n$, and $\scriptstyle k$ enemy tanks have been observed. It is

$(m\mid n,k)={\begin{cases}{\frac {\binom {m-1}{k-1}}{\binom {n}{k}}}&{\text{if }}k\leq m\leq n\\0&{\text{otherwise}}\end{cases}}$

where the binomial coefficient $\scriptstyle {\binom {n}{k}}$ is the number of $\scriptstyle k$-sized samples from an $\scriptstyle n$-sized population.

The expression $\scriptstyle (m|k)\;=\;\Pr(M\;=\;m\mid K\;=\;k)$ is the probability that the maximum serial number is equal to m once k tanks have been observed but before the serial numbers have actually been observed. $\scriptstyle (m\mid k)$ can be re-written in terms of the other quantities by marginalizing over all possible $\scriptstyle n$.

${\begin{aligned}(m|k)&=(m\mid k)\cdot 1\\&=(m\mid k){\sum _{n=0}^{\infty }(n\mid m,k)}\\&=(m\mid k){\sum _{n=0}^{\infty }(m\mid n,k){\frac {(n\mid k)}{(m\mid k)}}}\\&=\sum _{n=0}^{\infty }(m\mid n,k)(n\mid k)\end{aligned}}$

The expression $\scriptstyle (n|k)\;=\;\Pr(N\;=\;n|K\;=\;k)$ is the probability that the total number of tanks is equal to n when k tanks have been observed but before the serial numbers have actually been observed. Assume that it is some discrete uniform distribution

$(n|k)={\begin{cases}{\frac {1}{\Omega -k}}&{\text{if }}k\leq n<\Omega \\0&{\text{otherwise}}\end{cases}}$

The upper limit $\Omega$ must be finite, because the function

$f(n)=\lim _{\Omega \rightarrow \infty }{\begin{cases}{\frac {1}{\Omega -k}}&{\text{if }}k\leq n<\Omega \\0&{\text{otherwise}}\end{cases}}$

is

$f(n)=0$

which is not a probability mass function.

Then

$(n\mid m,k)={\begin{cases}{\frac {(m\mid n,k)}{\sum _{n=m}^{\Omega -1}(m\mid n,k)}}&{\text{if }}m\leq n<\Omega \\0&{\text{otherwise}}\end{cases}}$

If $\scriptstyle \sum _{n=m}^{\infty }(m\mid n,\,k)\;<\;\infty$, then the unwelcome variable $\scriptstyle \Omega$ disappears from the expression.

$(n|m,k)={\begin{cases}0&{\text{if }}n<m\\{\frac {(m\mid n,k)}{\sum _{n=m}^{\infty }(m\mid n,k)}}&{\text{if }}n\geq m\end{cases}}$

For k ≥ 1 the mode of the distribution of the number of enemy tanks is m.

For k ≥ 2, the probability that the number of enemy tanks is equal to $n$, is

$\Pr(N=n\mid M=m\geq k,K=k\geq 2)={\begin{cases}0&{\text{if }}n<m\\{\frac {k-1}{k}}{\frac {\binom {m-1}{k-1}}{\binom {n}{k}}}&{\text{if }}n\geq m\end{cases}}$

and the probability that the number of enemy tanks, $\scriptstyle N$, is greater than $\scriptstyle n$, is

$\Pr(N>n\mid M=m\geq k,K=k\geq 2)={\begin{cases}1&{\text{if }}n<m\\{\frac {\binom {m-1}{k-1}}{\binom {n}{k-1}}}&{\text{if }}n\geq m\end{cases}}$

For k ≥ 3, $N$ has the finite mean value:

${\frac {(m-1)(k-1)}{k-2}}$

For k ≥ 4, $\scriptstyle N$ has the finite standard deviation:

${\sqrt {\frac {(m-1)(k-1)(m+1-k)}{(k-2)^{2}(k-3)}}}$

These formulas are derived below.

Summation formula

The following important formula 14 from Binomial coefficient#Identities involving binomial coefficients is used below for simplifying series relating to the German Tank Problem.

$\sum _{n=m}^{\infty }{\frac {1}{\binom {n}{k}}}={\frac {k}{k-1}}{\frac {1}{\binom {m-1}{k-1}}}$

This sum formula is somewhat analogous to the integral formula

$\int _{n=m}^{\infty }{\frac {dn}{n^{k}}}={\frac {1}{k-1}}{\frac {1}{m^{k-1}}}$

These formulas apply for k > 1.

One tank

Observing one tank randomly out of a population of n tanks gives the serial number m with probability 1/n for m ≤ n, and zero probability for m > n. Using Iverson bracket notation this is written

$\Pr(M=m\mid N=n,K=1)=(m|n)=[m\leq n]{\frac {1}{n}}$

This is the conditional probability mass function of $\scriptstyle m$.

When considered a function of n for fixed m this is a likelihood function.

${\mathcal {L}}(n)=[n\geq m]{\frac {1}{n}}$

The maximum likelihood estimate for the total number of tanks is N₀ = m.

The total likelihood is infinite, being a tail of the harmonic series.

$\sum _{n}{\mathcal {L}}(n)=\sum _{n=m}^{\infty }{\frac {1}{n}}=\infty$

but

${\begin{aligned}\sum _{n}{\mathcal {L}}(n)[n<\Omega ]&=\sum _{n=m}^{\Omega -1}{\frac {1}{n}}\\&=H_{\Omega -1}-H_{m-1}\end{aligned}}$

where $H_{n}$ is the harmonic number.

The probability mass function depends explicitly on the prior limit $\scriptstyle \Omega$:

${\begin{aligned}&\Pr(N=n\mid M=m,K=1)\\={}&(n\mid m)={\frac {[m\leq n]}{n}}\cdot {\frac {[n<\Omega ]}{H_{\Omega -1}-H_{m-1}}}\end{aligned}}$

The mean value of $\scriptstyle N$ is

${\begin{aligned}\sum _{n}n\cdot (n\mid m)&=\sum _{n=m}^{\Omega -1}{\frac {1}{H_{\Omega -1}-H_{m-1}}}\\&={\frac {\Omega -m}{H_{\Omega -1}-H_{m-1}}}\\&\approx {\frac {\Omega -m}{\log \left({\frac {\Omega -1}{m-1}}\right)}}\end{aligned}}$

Two tanks

If two tanks rather than one are observed, then the probability that the larger of the observed two serial numbers is equal to m, is

$\Pr(M=m\mid N=n,K=2)=(m|n)=[m\leq n]{\frac {m-1}{\binom {n}{2}}}$

When considered a function of n for fixed m this is a likelihood function

${\mathcal {L}}(n)=[n\geq m]{\frac {m-1}{\binom {n}{2}}}$

The total likelihood is

${\begin{aligned}\sum _{n}{\mathcal {L}}(n)&={\frac {m-1}{1}}\sum _{n=m}^{\infty }{\frac {1}{\binom {n}{2}}}\\&={\frac {m-1}{1}}\cdot {\frac {2}{2-1}}\cdot {\frac {1}{\binom {m-1}{2-1}}}\\&=2\end{aligned}}$

and the probability mass function is

${\begin{aligned}&\Pr(N=n\mid M=m,K=2)\\={}&(n\mid m)\\={}&{\frac {{\mathcal {L}}(n)}{\sum _{n}{\mathcal {L}}(n)}}\\={}&[n\geq m]{\frac {m-1}{n(n-1)}}\end{aligned}}$

The median $\scriptstyle {\tilde {N}}$ satisfies

$\sum _{n}[n\geq {\tilde {N}}](n\mid m)={\frac {1}{2}}$

so

${\frac {m-1}{{\tilde {N}}-1}}={\frac {1}{2}}$

and so the median is

${\tilde {N}}=2m-1$

but the mean value of N is infinite

$\mu =\sum _{n}n\cdot (n|m)={\frac {m-1}{1}}\sum _{n=m}^{\infty }{\frac {1}{n-1}}=\infty$

Many tanks

Probability mass function

The conditional probability that the largest of k observations taken from the serial numbers {1,...,n}, is equal to m, is

${\begin{aligned}&\Pr(M=m\mid N=n,K=k\geq 2)\\={}&(m\mid n,k)\\={}&[m\leq n]{\frac {\binom {m-1}{k-1}}{\binom {n}{k}}}\end{aligned}}$

The likelihood function of n is the same expression

${\mathcal {L}}(n)=[n\geq m]{\frac {\binom {m-1}{k-1}}{\binom {n}{k}}}$

The total likelihood is finite for k ≥ 2:

${\begin{aligned}\sum _{n}{\mathcal {L}}(n)&={\frac {\binom {m-1}{k-1}}{1}}\sum _{n=m}^{\infty }{1 \over {\binom {n}{k}}}\\&={\frac {\binom {m-1}{k-1}}{1}}\cdot {\frac {k}{k-1}}\cdot {\frac {1}{\binom {m-1}{k-1}}}\\&={\frac {k}{k-1}}\end{aligned}}$

The probability mass function is

${\begin{aligned}&\Pr(N=n\mid M=m,K=k\geq 2)=(n\mid m,k)\\={}&{\frac {{\mathcal {L}}(n)}{\sum _{n}{\mathcal {L}}(n)}}\\={}&[n\geq m]{\frac {k-1}{k}}\cdot {\frac {\binom {m-1}{k-1}}{\binom {n}{k}}}\\={}&[n\geq m]{\frac {m-1}{n}}\cdot {\frac {\binom {m-2}{k-2}}{\binom {n-1}{k-1}}}\\={}&[n\geq m]{\frac {m-1}{n}}\cdot {\frac {m-2}{n-1}}\cdot {\frac {k-1}{k-2}}\cdot {\frac {\binom {m-3}{k-3}}{\binom {n-2}{k-2}}}\end{aligned}}$

The complementary cumulative distribution function is the probability that N > x

${\begin{aligned}&\Pr(N>x\mid M=m,K=k)\\={}&{\begin{cases}1&{\text{if }}x<m\\\sum _{n=x+1}^{\infty }(n\mid m,k)&{\text{if }}x\geq m\end{cases}}\\={}&[x<m]+[x\geq m]\sum _{n=x+1}^{\infty }{\frac {k-1}{k}}{\frac {\binom {m-1}{k-1}}{\binom {N}{k}}}\\={}&[x<m]+[x\geq m]{\frac {k-1}{k}}\cdot {\frac {\binom {m-1}{k-1}}{1}}\sum _{n=x+1}^{\infty }{\frac {1}{\binom {n}{k}}}\\={}&[x<m]+[x\geq m]{\frac {k-1}{k}}\cdot {\frac {\binom {m-1}{k-1}}{1}}\cdot {\frac {k}{k-1}}\cdot {\frac {1}{\binom {x}{k-1}}}\\={}&[x<m]+[x\geq m]{\frac {\binom {m-1}{k-1}}{\binom {x}{k-1}}}\end{aligned}}$

The cumulative distribution function is the probability that N ≤ x

${\begin{aligned}&\Pr(N\leq x\mid M=m,K=k)\\={}&1-\Pr(N>x\mid M=m,K=k)\\={}&[x\geq m]\left(1-{\frac {\binom {m-1}{k-1}}{\binom {x}{k-1}}}\right)\end{aligned}}$

Expected value and standard deviation

The expected number of enemy tanks is

${\begin{aligned}\mu &=\sum _{n}n\cdot \Pr(N=n\mid M=m,K=k)\\&=\sum _{n}n\cdot [n\geq m]{\frac {m-1}{n}}\cdot {\frac {\binom {m-2}{k-2}}{\binom {n-1}{k-1}}}\\&={\frac {m-1}{1}}\cdot {\frac {\binom {m-2}{k-2}}{1}}\sum _{n=m}^{\infty }{\frac {1}{\binom {n-1}{k-1}}}\\&={\frac {m-1}{1}}\cdot {\frac {\binom {m-2}{k-2}}{1}}\cdot {\frac {k-1}{k-2}}\cdot {\frac {1}{\binom {m-2}{k-2}}}\\&={\frac {m-1}{1}}\cdot {\frac {k-1}{k-2}}\end{aligned}}$

The standard deviation σ satisfies the equation

$\sigma ^{2}+\mu ^{2}=\sum _{n}n^{2}\cdot \Pr(N=n\mid M=m,K=k)$

So

${\begin{aligned}\sigma ^{2}+\mu ^{2}-\mu &=\sum _{n}n(n-1)\cdot \Pr(N=n\mid M=m,K=k)\\&=\sum _{n=m}^{\infty }n(n-1){\frac {m-1}{n}}\cdot {\frac {m-2}{n-1}}\cdot {\frac {k-1}{k-2}}\cdot {\frac {\binom {m-3}{k-3}}{\binom {n-2}{k-2}}}\\&={\frac {m-1}{1}}\cdot {\frac {m-2}{1}}\cdot {\frac {k-1}{k-2}}\cdot {\frac {\binom {m-3}{k-3}}{1}}\sum _{n=m}^{\infty }{\frac {1}{\binom {n-2}{k-2}}}\\&={\frac {m-1}{1}}\cdot {\frac {m-2}{1}}\cdot {\frac {k-1}{k-2}}\cdot {\frac {\binom {m-3}{k-3}}{1}}\cdot {\frac {k-2}{k-3}}\cdot {\frac {1}{\binom {m-3}{k-3}}}\\&={\frac {m-1}{1}}\cdot {\frac {m-2}{1}}\cdot {\frac {k-1}{k-3}}\\&\end{aligned}}$

and the standard deviation is

${\begin{aligned}\sigma &={\sqrt {{\frac {m-1}{1}}\cdot {\frac {m-2}{1}}\cdot {\frac {k-1}{k-3}}+\mu -\mu ^{2}}}\\&={\sqrt {\frac {(k-1)(m-1)(m-k+1)}{(k-3)(k-2)^{2}}}}\\&\end{aligned}}$

The variance-to-mean ratio is simply

${\frac {\sigma ^{2}}{\mu }}={\frac {m-k+1}{(k-3)(k-2)}}$

See also

• Capture-recapture, other method of estimating population size
• Maximum spacing estimation, which generalizes the intuition of "assume uniformly distributed"

Other discussions of the estimation

• Maximum likelihood#Bias
• Bias of an estimator#Maximum of a discrete uniform distribution
• Likelihood function#Example 2

References

Notes

Citations

Bibliography