Capture-Recapture
Learn the capture-recapture method for NSW Year 12 Mathematics Standard 2. It estimates the size of an animal population that is too large to count by tagging \(M\) animals, releasing them, then taking a second sample of \(n\) and counting the \(r\) that are tagged.
By equating the tagged fractions \(\dfrac{r}{n}=\dfrac{M}{N}\), you rearrange to \(N=\dfrac{M\times n}{r}\) to estimate the population, and learn the assumptions the estimate relies on — a core ratios skill for Standard 2 that is used in fisheries management and wildlife conservation.
Theory
Capture-recapture is a ratio technique in Year 12 Standard 2 (NSW) for estimating an animal population you cannot count directly. You tag \(M\) animals, release them, then take a second sample of \(n\) with \(r\) tagged, and equate the tagged fractions \(\dfrac{r}{n}=\dfrac{M}{N}\) to estimate \(N=\dfrac{M\times n}{r}\). This guide covers the formula, the assumptions and four worked examples.
Capture-recapture (also called mark-recapture) is a method for estimating the size of an animal population \(N\) that is too large to count one by one. You capture a known number of animals, mark them, release them, and later take a second sample to see what fraction of it is marked.
Let \(M\) be the number tagged in the first capture, \(n\) the size of the second sample, and \(r\) the number of tagged animals recaptured in that sample. If the tagged animals mix evenly, the tagged fraction of the sample equals the tagged fraction of the whole population: \(\dfrac{r}{n}=\dfrac{M}{N}\).
Rearranging gives \(N=\dfrac{M\times n}{r}\). In this Year 12 Standard 2 (NSW) topic the answer is always an estimate — it relies on the population being well mixed and closed (no births, deaths or migration between the two samples).
With \(M\) tagged in the first capture, a second sample of size \(n\) and \(r\) tagged recaptured, the tagged fractions are equal:
Make the unknown population \(N\) the subject:
How to estimate a population by capture-recapture
- Identify \(M\) (number tagged and released), \(n\) (size of the second sample) and \(r\) (number of tagged animals recaptured).
- Write the proportion \(\dfrac{r}{n}=\dfrac{M}{N}\).
- Rearrange to \(N=\dfrac{M\times n}{r}\).
- Substitute and evaluate, then round to the nearest whole number and interpret the estimate in context.
Equate the tagged fractions with \(M=60,\ n=48,\ r=8\), then solve for \(N\).
| \(\dfrac{r}{n}\) | \(=\) | \(\dfrac{M}{N}\) |
| \(\dfrac{8}{48}\) | \(=\) | \(\dfrac{60}{N}\) |
| \(N\) | \(=\) | \(\dfrac{60\times 48}{8}\) |
| \(N\) | \(=\) | \(360\) |
There are about \(360\) bass in the reservoir.
Substitute \(M=80,\ n=50,\ r=10\) into \(N=\dfrac{M\times n}{r}\).
| \(\dfrac{10}{50}\) | \(=\) | \(\dfrac{80}{N}\) |
| \(N\) | \(=\) | \(\dfrac{80\times 50}{10}\) |
| \(N\) | \(=\) | \(400\) |
The estimated population is \(400\) possums.
Estimate \(N\) first, then the untagged count is \(N-M\) and the percentage is \(\dfrac{M}{N}\times100\%\).
| \(\dfrac{18}{90}\) | \(=\) | \(\dfrac{150}{N}\) |
| \(N\) | \(=\) | \(\dfrac{150\times 90}{18} = 750\) |
| \(\text{untagged}\) | \(=\) | \(750-150 = 600\) |
| \(\text{tagged}\) | \(=\) | \(\dfrac{150}{750}\times100\% = 20\%\) |
(i) \(750\) kangaroos, (ii) \(600\) untagged, (iii) \(20\%\) tagged.
Use \(M=300,\ n=64,\ r=16\); then recall what must stay true between samples.
| \(\dfrac{16}{64}\) | \(=\) | \(\dfrac{300}{N}\) |
| \(N\) | \(=\) | \(\dfrac{300\times 64}{16}\) |
| \(N\) | \(=\) | \(1200\) |
(i) About \(1200\) lorikeets. (ii) The estimate assumes the banded birds mix evenly and that no births, deaths or migration change the population between the two samples (a closed population).
Common pitfalls
Frequently asked questions
What is the capture-recapture method?
Capture-recapture is a way to estimate how many animals are in a population without counting them all. You catch, tag and release a number of animals, then take a second sample later and see what fraction of it is tagged. That fraction is used to estimate the whole population.
What is the formula for capture-recapture?
The population estimate is N equals M times n divided by r, where M is the number tagged in the first capture, n is the size of the second sample, and r is the number of tagged animals recaptured. It comes from equating the tagged fractions, r over n equals M over N.
What do M, n and r stand for?
M is the number of animals tagged and released in the first capture. n is the total number of animals caught in the second sample. r is how many of that second sample are tagged, meaning they were recaptured.
What assumptions does capture-recapture rely on?
It assumes the tagged animals mix evenly back through the population, the population is closed with no births, deaths or migration between the two samples, and that tagging does not fall off or change an animal's chance of being caught again. If these do not hold, the estimate is less reliable.
Why is capture-recapture only an estimate?
Because it is based on a sample, not a full count, and on assumptions that are never perfectly true in the wild. Different samples give slightly different values of r, so the method gives an approximate population size rather than an exact one.
Where is capture-recapture used in real life?
Ecologists and fisheries managers use it to estimate fish stocks in dams and lakes and the number of possums, kangaroos, birds or insects in a reserve. It guides decisions such as fishing limits and conservation programs.