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This page sets out the method for capture recapture in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Capture recapture topic page.
When you see two samples with some animals tagged in the first and some of those turning up in the second, set up the proportion.
When you see 'comment on the estimate' or 'what if tags fall off', think about how changes.
The tagged fraction in the sample estimates the tagged fraction in the population. This assumes the animals mix back in, none join or leave, and the second sample is random.
40 were tagged. A later sample of 50 contains 10 tagged. The estimate is 200.
If the population is known, the sample should contain the same fraction tagged.
10 tagged out of a sample of 50. The tagged fraction is 1/5.
If the estimate and the sample are known, you can recover how many were tagged first.
First tagged over population equals tagged in the sample over the sample size.
Use this when a first catch was marked and a later sample contains some marks. If 12 deer out of 48 are marked, about a quarter of the population is marked. With 36 marked first, the population estimate is 36 × 4 = 144.
Write 36/N = 12/48 before you multiply. The first catch is the marked part of the population. The recaptures are the marked part of the sample. Then N = (36 × 48)/12.
Method: 1. Write the marked fraction in the second sample. 2. Set that equal to the first catch over the population. 3. Rearrange to N = first catch times second sample, divided by recaptures. 4. Divide exactly and state the estimate.
Cancel before you multiply when a factor is shared. Here 12 divides 36 to leave 3, and 3 × 48 = 144. That matches 1728 ÷ 12. Do not stop at 48 ÷ 12 = 4. That 4 still has to be multiplied by the first catch.
Give the exact value of the formula. Show M × n ÷ m so the method is visible. Do not add the two sample sizes. Adding ignores the recaptures.
Method: 1. Substitute the three given numbers into N = M times n, divided by m. 2. Cancel a common factor of the first catch and the recaptures if you can. 3. Multiply by the second sample size. 4. Check that the marked fraction of N matches the marked fraction of the sample.
Four assumptions make the second sample a fair copy of the population. The population is closed: nobody joins or leaves. The marks hold. The marked animals mix back in. The second sample is random, so every animal has the same chance of being caught again.
If the question asks for an assumption, one clear statement is enough. Do not say every marked animal is caught again. The method expects only the same fraction. If an assumption fails, the formula still produces a number, but that number may be a poor estimate.
Method: 1. Assume the population does not change size between the samples. 2. Marks are not lost and are noticed in the second sample. 3. Marked animals mix fully back into the population. 4. Take the later sample at random.
If the population is given and the question asks how many were marked first, find M = (N × m)/n. For 240 fish, with 9 marked in a sample of 36, 9/36 = 1/4, so the first catch is 240 ÷ 4 = 60.
Do not put the population into the forward formula as if it were the first catch. Estimate the population means find N. How many were marked means find M. Both need the same assumptions, and both use exact division.
Method: 1. Decide whether the unknown is N or M. 2. For M, multiply the population by the marked count in the second sample. 3. Divide by the second sample size. 4. Simplify the second-sample fraction first when it cancels.
The marked fraction of the second sample is used as an estimate of the marked fraction of the whole population. Write that as an equation: M/N = m/n, where M is the number marked first, N is the population, n is the size of the second sample and m is the number of marked animals in it. Writing this line earns the method mark even if the arithmetic slips.
Then solve for N. Cross multiplying gives N = (M × n)/m. On a non-calculator paper the numbers are friendly, so cancel first. For M = 90, n = 120 and m = 20, divide 120 by 20 to get 6, then 90 × 6 = 540. Always say "about" or "an estimate", because the sample is only a guide to the real population and a different sample would give a slightly different answer.
Method: 1. Identify the four numbers: first catch M, second sample n, marked in the second sample m, and the unknown N. 2. Write the equation M/N = m/n with the numbers in place. 3. Rearrange to N = (M × n)/m and cancel before multiplying. 4. Write the answer as an estimate with the units, for example animals or fish.
Tip: The first catch and the number marked in it are the same thing. The recapture count, m, is the small number in the second sample.
Exams ask for an assumption, and then ask what happens if it is wrong. Keep the formula in mind: N = (M × n)/m. If the count m is too small, then N is too large, because m is the divisor. If m is too large, the estimate is too small. Use this to decide the direction.
If some marks rub off, fewer marked animals are found, so m is too small and the estimate is too large. If marked animals are easier to catch again, m is too large and the estimate is too small. If animals are born or arrive between the samples, the second sample contains more unmarked animals and the estimate changes. State the assumption, then the effect, in one sentence each.
Method: 1. State the assumption in a sentence, such as "no marks rub off between the two samples". 2. Decide whether the assumption being wrong makes m too small or too large. 3. Use N = (M × n)/m to decide whether the estimate is too large or too small. 4. Write the effect with a reason, for example "the estimate would be too large because fewer marked animals are counted".
Tip: Other assumptions that earn credit: the marked animals mix back in fully, the second sample is random, and the population does not change.
Some questions give a friend's estimate and ask you to explain why it cannot be right. A population must be at least as large as any single sample. If 64 animals are caught in one go, then an estimate of 60 cannot be correct. You can also say the marked fraction is high, so the population should be close to the number first marked.
The animals seen in both samples together must also be counted. At least M + (n - m) different animals have been seen, namely all of the marked animals plus the unmarked ones in the second sample. For M = 80, n = 100 and m = 25 this gives at least 80 + 75 = 155. An estimate below this would be impossible.
Method: 1. Compare the estimate with the size of each sample. It cannot be smaller. 2. Work out M + (n - m) for the smallest possible population from the data. 3. If the estimate is below either value, say it cannot be correct and give the number. 4. Check that m is not larger than M. You cannot recapture more marked animals than were marked.
The same equation can be solved for any of the four letters. Given an estimated population, the number to expect in the second sample is m = (M × n)/N. Given the second sample, the number first marked is M = (N × m)/n. Read the question carefully so you put each number in its right place, and say in a few words which quantity you are finding.
Questions often link to percentages or algebra. You may be asked to find an estimate and then reduce the population by 30%, then add new animals. Write each stage on its own line. In algebra versions the formula is left in terms of one letter, such as N = 2700/m, and you substitute values to find N or m. Keep the order of the numbers the same on every line so nothing is swapped by accident.
Method: 1. Write the proportion equation with the unknown letter in place. 2. Rearrange to isolate the unknown before substituting numbers. 3. For linked parts, finish the estimate first, then apply the percentage change or addition. 4. Check by putting your answer back into the proportion.
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