Math Detours · Grade 8 · Connections · 3

Counting fish you cannot see

Tag 50 fish, release them, catch 40, find 4 tagged. Estimate the lake. Proportions, and the same trick for counting words a writer knew.

Name ____________________Date ____________

Tonight's idea

You cannot count the fish in a lake. But you can catch some.

Catch \(M\) fish, tag them, release them, and wait for them to mix in. Catch again: \(C\) fish this time, of which \(R\) carry a tag. In the whole lake the tagged fraction is \(M/N\), where \(N\) is the number you want. In the second catch it is \(R/C\). If the net is a fair sample of the lake, the two fractions should be about equal:

\(\dfrac{R}{C} \approx \dfrac{M}{N}\), so \(N \approx \dfrac{M \times C}{R}\).

second catch: 2 of 10 lake: 8 of 40 tagged
Tagged fish in orange. One fifth of the lake is tagged, and so is one fifth of the net.

This is the Lincoln–Petersen estimate, and it leans on assumptions: the tagged fish mix evenly, tags stay on, tagged fish are no easier or harder to catch, and the lake does not change between catches. Break one and the estimate drifts in a direction you can predict.

The same overlap trick counts the people a census missed, the bugs two testers both failed to find, and even the words an author knew but never wrote down: the known works tag the words, and a newly found text is the second catch. Statisticians have done exactly this for Shakespeare.

Watch one: 50 tagged, 40 caught, 4 tags

Here \(M = 50\), \(C = 40\), \(R = 4\). One tenth of the second catch is tagged, so about one tenth of the lake is tagged, and the lake holds about ten times 50:

\[N \approx \frac{50 \times 40}{4} = 500.\]

Notice how fragile that is. Had the net held 3 tagged fish instead of 4, the estimate would be \(2000/3 \approx 667\). Had it held 5, it would be 400. A single fish moves the answer by a hundred or more. The estimate is only as good as \(R\) is large, which is why real surveys tag thousands.

Your turn

  1. A ranger tags 120 deer in a forest. A month later a survey counts 90 deer, of which 15 are tagged. Estimate the number of deer.
  2. A lake is believed to hold about 2,000 fish. A biologist tags 80 of them and later nets 150. Roughly how many of the 150 should carry a tag? Write the proportion you used.
  3. Sixty fish are tagged, and sixty are caught in the second catch. (a) Give the estimate of \(N\) if \(R = 2, 3, 4\) and \(6\). (b) Some tagged fish lose their tags before the second catch. Does the estimate come out too high or too low? Explain using the formula.
  4. Two proofreaders read the same manuscript independently. Ana finds 30 errors, Ben finds 24, and 18 of the errors are on both lists. Treat Ana's errors as the tagged fish and Ben's reading as the second catch. Estimate the total number of errors in the manuscript, and how many neither of them found. Then suppose both are good at spotting spelling mistakes but neither notices wrong dates. Would your estimate be too high or too low?

Stretch: how many to tag

A lake might hold 10,000 fish. You tag 100 and catch 100. How many tagged fish do you expect in the net? Why is the estimate almost useless? If you tag and catch equal numbers, how many must each catch be for the expected \(R\) to be about 20?

Answers

  1. \(N \approx 120 \times 90 / 15 = 720\) deer.
  2. The tagged fraction of the lake is \(80/2000 = 1/25\), so about \(150/25 = 6\) of the net should be tagged. (The formula run backwards: \(R \approx MC/N\).)
  3. (a) \(3600/R\): 1800, 1200, 900 and 600. (b) Lost tags make \(R\) smaller than it should be. \(R\) is in the denominator, so the estimate comes out too high. The same happens if tagged fish learn to avoid the net.
  4. \(N \approx 30 \times 24 / 18 = 40\) errors. Between them they found \(30 + 24 - 18 = 36\), so about 4 errors escaped both readers. That assumes every error is equally easy for each reader to find, the proofreading version of a fair net. Errors that both readers miss, like the wrong dates, never show up in either list, so the estimate is too low. With 60 invisible dates on top of 40 findable errors, the lists could look exactly the same while 64 errors escaped.

Stretch. One per cent of the lake is tagged, so you expect \(100 \times 0.01 = 1\) tagged fish in the net. With \(R\) around 1, a single fish either way doubles or halves the estimate, and \(R = 0\) gives no estimate at all. You need \(M \times C / 10000 \approx 20\), so \(M \times C \approx 200{,}000\); with \(M = C\) that is \(\sqrt{200000} \approx 447\), so tag and catch about 450 each.

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