The Park Census

Set how many animals are really out there, how many the park thinks it released, and whether the search stops once it reaches that number. Then look at what the sizes of the animals you did find are telling you.

Worked example — The Park Census

The search reports	12
What the sizes say	-2.6 breeding

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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A count that stops at the expected number can never report a surplus 🖖

Press The count that hides the surplus. Forty animals are out there, the park believes it released 12, and the search is told to stop once it reaches the expected total. It reports 12 of 12. The screen says the park is exactly right, and 28 animals are unaccounted for. Now lift the ceiling: the same park, the same sweeps, and the count comes back at 37. The reassuring number was produced by the instruction, not by the animals. This is not only a plot device. Any audit that stops when the books balance, any test suite that stops at the expected failures, any search that ends when it has found what it was looking for, has the same shape.

The distribution gives it away, but a glance will not 🖖

Malcolm reads breeding off a graph of the animals' heights: a controlled release comes in batches, so the sizes cluster; free breeding spreads ages out continuously. He is right, and the naive version of his argument does not work. Building this tool, counting the humps in the histogram got it backwards at realistic sample sizes — the noise in a continuous sample of 48 made it look lumpier than three genuine cohorts. What does work is comparing the two models directly, charging the cohort model for the extra parameters it spends. Switch between Released in batches and Breeding and watch the evidence figure: +68.3 one way, −18.4 the other, from the same 46 animals. The graph is what convinces you. The penalised likelihood ratio is what knows.

A ceiling does not only miscount, it starves the evidence 🖖

The animals you measure are the animals you caught, so the two failures compound. In the capped run there are 12 sizes to work with and the breeding evidence comes out at −2.6, pointing the right way and far too weakly to act on. Lift the ceiling and the same park yields 37 animals and −18.5, which is decisive. Detecting cohorts stays easy at any sample size, because clusters are loud. Proving the absence of cohorts is the hard direction and it needs numbers. So the instruction to stop counting removed the population evidence and the breeding evidence in one move, and left a screen that read normal.

Problems solved in full

Forty animals out there in a park that expects 12 7 steps
A park expects 12 animals. Forty are out there. Each sweep spots any given animal with probability 0.28, and the search runs 8 sweeps. What does the count say, with the ceiling on and off?
1. Start with coverage. An animal is missed by one sweep with probability 0.72, so it is missed by all eight with probability 0.72 to the eighth.
  
  $(1 - 0.28)^{8} = 0.7 2^{8} = 0.0722$
2. So the search sees about 93% of whatever is there. With 40 animals present that is 37.
  
  $1 - 0.0722 = 0.928 \Rightarrow 40 \times 0.928 = 37$
3. With the ceiling on, the search stops at 12. It never gets to 37, and it reports the expected number exactly.
  
  $min (37, 12) = 12$
4. The gap is the finding: 28 animals are out there and unrecorded, and the report shows a perfect match. The ceiling did not just cap the number, it removed the only signal that anything was wrong.
  
  $40 - 12 = 28$
5. Now the second instrument. Sizes are measured on the animals found, so the ceiling decides the sample size too: 12 animals with the ceiling, 37 without.
  
  $n_{capped} = 12 n_{free} = 37$
6. Comparing a three-cohort model against a continuous spread of ages gives a log-likelihood ratio of −2.6 on the twelve and −18.5 on the thirty-seven. Below about 5 in absolute value, a likelihood ratio is not telling you to act.
  
  $lo g \frac{L _{cohorts}}{L _{continuous}} = - 2.6 vs - 18.5$
7. Which is where the tool runs out, and where the novel is quietly right about something real. Both numbers point at breeding. Only one of them is strong enough to say so, and the difference between them was not caused by the animals or by the biology. It was caused by an instruction about when to stop looking.
  
  $∣Λ∣ < 5 \Rightarrow \emptyset$
Answer

12 with the ceiling, 37 without, from a population of 40. The capped count reports a perfect match while hiding 28 animals, and the twelve it does return put the breeding evidence at −2.6 instead of −18.5. One instruction, two instruments blinded.
Forty-six animals counted, and the number the park actually wants 6 steps
Press Nobody is breeding. Fifty are out there, the search runs eight sweeps and comes back with 46. Run the tool backwards — recover the population from the count — then decide whether the park should sweep more often or measure its search better.
1. Coverage first, because everything here hangs off it. A sweep spots any given animal with probability 0.28, so it misses it with probability 0.72, and eight independent sweeps miss it with 0.72 to the eighth. Whatever is out there, the search finds 92.78% of it.
  
  $c = 1 - (1 - p)^{k} = 1 - 0.7 2^{8} = 1 - 0.0722 = 0.9278$
2. Which is where the 46 on the card comes from, and it is worth seeing the tool do this in the direction nobody ever has the data for: it was given the truth and asked for the count.
  
  $⌊ 50 \times 0.9278 ⌋ = 46$
3. A real survey is standing on the other side of that equation. It has 46 in a notebook and wants the number in the park, so divide instead of multiplying. Rounded to the nearest whole animal, 50.
  
  $\hat{N} = \frac{n}{c} = \frac{46}{0.9278} = 49.58 \Rightarrow 50$
4. That answer leans on a spotting rate the park guessed at, so the question is how hard it leans. Differentiate the logarithm: the fractional error in the estimate is this multiple of the fractional error in p.
  
  $\frac{d ln N ^}{d ln p} = \frac{k p ( 1 - p ) ^{k - 1}}{1 - ( 1 - p ) ^{k}}$
5. Work it out at one sweep and it is exactly 1 — every percent you are wrong about p is a percent you are wrong about the population, no better and no worse. At eight sweeps it is 0.242. The same misjudgement costs a quarter as much.
  
  $k = 1 : \frac{p}{p} = 1; k = 2 : 0.837; k = 8 : \frac{8 \times 0.28 \times 0.7 2 ^{7}}{0.9278} = 0.242$
6. Check that against the module rather than trusting the derivative. Let the true spotting rate be anywhere from 0.20 to 0.40 while the park keeps assuming 0.28: after eight sweeps the estimate wanders between 55 and 47, and a two-sweep search — which finds 24 animals, not 46 — wanders between 67 and 38.
  
  $p = 0.20 \to 0.40 : \hat{N}_{k = 8} = 55.3 \to 46.8, \hat{N}_{k = 2} = 66.7 \to 37.5$
Answer

Fifty animals, from 46 found and nothing else, and the eight-sweep design is what makes that number worth quoting. Sweeping repeatedly does two jobs at once: it raises coverage towards 1, and it drives the leverage of the spotting rate towards 0. A single sweep passes every error in p straight through; eight sweeps absorb three quarters of it.

So the answer to the park's question is sweep again, not calibrate. Measuring a detection probability well is expensive fieldwork — you need animals you already know about — and the returns are the ones above: the more sweeps you have run, the less the measurement is worth. It also says what the tool's own Sweeps slider is really for. It looks like a knob for finding more animals, and past about ten sweeps it has stopped doing that, because coverage is already 96%. What it is still doing is making the answer robust to the one number nobody in the park can check.

$k \to \infty lim \frac{k p ( 1 - p ) ^{k - 1}}{1 - ( 1 - p ) ^{k}} = 0, k = 1 \Rightarrow 1$

By the 42Math team · Updated 2026-08-25

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Nobody is breeding - Fifty animals, released in three batches, none of them breeding. The search finds 46 and the sizes fall into three tight clusters — the evidence for a controlled release is +68.3, which is not a close call.
The count that hides the surplus - Forty animals are out there and the park expects 12. The search stops at 12, reports a perfect match, and 28 animals go unrecorded. Worse: with only those 12 to measure, the breeding evidence comes out at −2.6, which is not enough to conclude anything.
Same park, no ceiling - The same park with the ceiling lifted. Now 37 animals turn up, 25 more than expected, and those 37 put the breeding evidence at −18.5. Nothing about the park changed. Only the instruction to stop counting.