Capture and Recapture

Counting what cannot be listed

The fish in a pond cannot be lined up and counted, and a list of them cannot be made. Capture and recapture estimates how many there are from two catches.

Tag and release

Catch 20 fish, put a small tag on each one, and release them back into the pond. Give them time to swim off and mix with the other fish.

Once they have mixed, the 20 tagged fish are spread through the pond. Call the number of fish in the pond N. Then 20 out of the N fish carry a tag: the tagged fraction of the pond is $\frac{20}{N}$20/N.

The pond’s fish, with the tagged ones as one part of the whole. How large that part is, as a fraction of the pond, is what the second catch measures.

The second catch

Later, catch fish again. This time the catch is 30 fish, and 6 of them carry a tag. The tagged fraction of the catch is $\frac{6}{30}=\frac{1}{5}$6/30 = 1/5: one fish in five.

The second catch: 6 tagged fish and 24 untagged ones, 30 in all.

Matching the fractions

If the second catch is representative of the pond, the tagged fraction of the catch is the same as the tagged fraction of the pond: $\frac{6}{30}=\frac{20}{N}$6/30 = 20/N.

Solve for N. Multiply both sides by 30N: 6N = 20 × 30 = 600. Divide both sides by 6: $N=600/6=100$N = 600 / 6 = 100. So there are about 100 fish in the pond. Check: 20 tagged fish out of 100 is $\frac{1}{5}$1/5, the same fraction as in the catch.

In general, if m fish are tagged, and k of a second catch of n fish carry tags, then $\frac{k}{n}=\frac{m}{N}$k/n = m/N, so $N\approx$N ≈ mn/k. The guide’s example tags 60 fish and finds 12 tagged in a catch of 80: $N\approx 60\times 80/12=4$N ≈ 60 × 80 / 12 = 4,800 ÷ 12 = 400.

It is an estimate

The second catch is a sample, and a different catch of 30 would have held a different number of tagged fish. With 5 tagged, the estimate is 20 × 30 ÷ 5 = 120 fish. With 7, it is $600÷7\approx 85.7$600 ÷ 7 ≈ 85.7, about 86 fish. So 100 is a best estimate, not a count, and larger catches give steadier estimates.

This pond holds 200 fish. Tag m = 20 and catch n = 30, and 3 of the catch carry tags, so the estimate is 20 × 30 ÷ 3 = 200. Drag m and n: tagging 30 and catching 30 finds 8 tagged and estimates 113, while tagging 80 and catching 100 finds 41 tagged and estimates 195.

What the estimate assumes

The method works only if the second catch is representative of the pond, and that rests on several assumptions.

The tagged fish mix fully through the pond. If the second catch is taken where the tagged fish were released, before they have spread out, the catch holds too many of them. The tagged fraction of the catch is then too large, and the estimate of N is too small.

Every tag stays on. A fish that loses its tag is counted as untagged, so the tagged fraction of the catch is too small, and the estimate of N is too large.

Tagged and untagged fish are equally likely to be caught, and the population does not change between the two catches, with no fish born, dying, arriving or leaving in large numbers.

The usual mistakes

Multiplying without dividing. 20 × 30 = 600 is not the estimate; it still has to be divided by the 6 tagged fish in the catch.

Writing the equation upside down on one side, $\frac{6}{30}=\frac{N}{20}$6/30 = N/20. Both sides are a tagged count over a total: tags over the catch, and tags over the pond.

Adding the two catches. 20 + 30 = 50 counts fish that were caught, not the fish in the pond, and some of the 30 had already been caught once.

Deciding that lost tags make the estimate too small. They make the tagged share of the catch smaller, and a smaller share gives a larger estimate.