Counting it again

How many plants would it take sized the one experiment the frequency question needs: score each specimen Fibonacci or not by its counted pair, set the geometry’s share of 14.7 per cent against the ninety per cent a grown history gives, and four specimens separate them. The census wants a low count then carried the likeliest counting error through that sizing — a counter who closes the circle a few degrees early or late, multiplying both counts by the same factor — and found the cost falls on the high counts. With closing errors spread over 7.2°, the census needs six specimens counted at 13/21 and 449 at 55/89.

It also found half a remedy. At 34/55 almost every wrong reading shares a factor — 34/54, 33/54, 35/56 — and so announces itself. A census that sets such readings aside rather than scoring them as not Fibonacci needs fifteen kept specimens at 7.2° rather than fifty-four, at the price of the specimens set aside.

Setting a specimen aside is the obvious response and not the only one. A reading that announces itself says that this count went wrong. The head is still on the bench. The counter can count it again, from a fresh starting organ, and keep the new reading if it does not announce itself. This essay prices that, and answers the question the census essay ended on: whether counting at 34/55 with recounts ever becomes cheaper, in counts rather than in specimens, than counting at 13/21 once.

The policy, stated #

Count a specimen. If its reading shares a factor, count it again, independently, up to $K$ counts in all. Keep the first reading that does not announce itself; if all $K$ do, set the specimen aside. With one count allowed, this is the census essay’s setting aside exactly, and it reproduces that essay’s fifteen specimens to the last share.

Write $a$ for the chance one count of a specimen announces itself. The specimen is kept with chance $1 - a^K$, and the counts it costs, kept or not, are $1 + a + a^2 + \dots + a^{K-1}$, which is $(1 - a^K)/(1 - a)$. Those two lines are all the arithmetic a recount needs, apart from one thing: what the kept reading says.

A recount does not change what a kept reading says #

The census essay’s own statement of the question expected a recount to cut two ways. It recovers specimens, and it gives a silent error — a wrong reading that shares no factor and so passes as an ordinary report — a second chance to be drawn.

It does give the silent error a second chance. It gives a right reading exactly the same second chance, and the two cancel. Each fresh count is independent of the last, so the reading kept is the first of a run of independent readings that does not announce itself, and the first success in a run of independent trials is distributed as one trial conditioned on success. The kept reading is one reading conditioned on not announcing, whatever the cap.

For a grown plant counted at 34/55 with closing errors spread over 7.2°, a kept reading is a consecutive Fibonacci pair 70.2 per cent of the time and silently something else 29.8 per cent — with one count allowed, with three, with eight. What changes is whether there is a kept reading at all: half the time on one count, 74.9 per cent within two, 87.5 within three. The lowest line in the figure, the chance a specimen ends as a silent reading, rises with the cap, but only because the chance it ends as anything rises; the ratio between its two possible endings never moves.

That settles one worry cheaply. A recount cannot make the census’s readings worse, and it cannot make them better. It is a filter on specimens, not on readings, and whatever it does to the census, it does by changing which specimens are kept.

Which specimens setting aside keeps #

Setting aside keeps each specimen with chance $1 - a$, and $a$ is not the same for every specimen. A head counted high announces itself far more often than a head counted low, because more spirals cross a given sliver of it. The census’s two sides are made of different heads, so setting aside keeps different shares of each.

With one count allowed, a grown plant counted at 34/55 is kept 49.9 per cent of the time, while the census’s non-Fibonacci specimens — drawn from the geometry’s own pairs, which are small — are kept 58.5 per cent. So setting aside thins out exactly the plants the census is looking for, and the Fibonacci share it reads among grown plants is 62.1 per cent. Counting again puts them back: 65.2 per cent within three counts, 65.5 within eight.

The geometry’s side moves the other way, from 22.7 per cent to 21.8, because its non-whorled heads that announce a wrong count are restored as well and most of them are not Fibonacci. The gap between the two sides widens by about four points. That is the whole of what a recount does to the question — a selection effect removed — and it is enough to matter, because the census’s sample size falls steeply with the gap.

Specimens bought back #

At 34/55 the census needs fifteen kept specimens with one count allowed, twelve with two, and ten from three on. At 55/89 it needs twenty-eight, then twenty-two, then nineteen. At 13/21 the gain is small — nine to seven — because a low count rarely announces itself, and at 21/34 it is twelve to ten.

In each case the gain stops after two or three counts, and the reason is the thinning. By the third count 87.5 per cent of grown plants at 34/55 are kept, so there is little selection left to remove, and the kept readings were never going to improve. The sample size is set by the gap between the two Fibonacci shares, and the gap is capped by what a single reading conditioned on not announcing can say: 70.2 per cent Fibonacci for a grown plant at 34/55, against 90 per cent if every count were right. No number of recounts closes that, because the readings a recount cannot rescue are the silent ones, and it keeps them.

That distinction is the one the one-mark essay drew between a wrong reading that announces itself and one that does not, now priced in specimens. A census can buy back everything the announcing errors cost it. The silent errors are the floor.

Where the counts go #

A kept specimen costs more than one count, and the cost differs between the two sides of the question — because the geometry’s census contains something no recount can rescue.

About a third of the geometry’s divergence angles give a pair that shares a factor: the census’s whorled bucket, full of genuinely jugate heads, $k$ and $2k$. A head like that announces itself on every count, because it really is whorled. Under a recount policy it is counted to the cap and set aside, every time.

With one count allowed, the geometry’s census costs 1.55 counts per kept specimen, and 0.55 of that goes on specimens set aside. With three counts, 2.56, of which 1.52 is set aside; with eight, 4.86, of which 3.66 is set aside. The share of all counts spent on whorled heads rises from 35.4 per cent to 61.6 at three counts and 72.4 at five. On the grown side the pattern reverses: counts per kept specimen barely move, 1.97 to 2.26, and the part spent on specimens set aside falls, from 0.97 to 0.34, because a grown plant that announces itself once is usually kept on the next count.

So a recount’s cost depends on which answer is true. If the plants are grown, recounts are cheap and restore specimens. If the plants are the geometry’s, recounts are spent on the whorled heads, whose announcement is not an error at all. A counter cannot tell in advance which case holds — that is what the census is for — so the cost to budget for is the worse of the two.

The whorled bucket stops misleading #

The census essay noticed a hazard in setting aside. The geometry’s whorled bucket holds about a third of its divergence angles, and a survey of grown plants counted high, with one count each, fills its own bucket to about the same level — not with jugate heads but with closing errors. A whorled share near a third, found in a survey, then says nothing about which kind of plant was surveyed.

A recount changes that, and the arithmetic of the last section is why. A grown plant is set aside only if every one of its counts announces itself, which gets rapidly less likely with each count allowed. A whorled head is set aside whatever the cap.

With one count at 7.2°, grown plants counted at 34/55 are set aside 49.2 per cent of the time — more often than the geometry’s heads, at 35.4. A survey counted that way would find more “whorled” plants among grown ones than the geometry predicts, which is the census essay’s hazard at its sharpest. With two counts the grown share falls to 26.6 per cent, with three to 15.2, with five to 6.7, while the geometry’s stays near a third, 32.7 at five counts, since only its non-whorled heads are rescued.

So a recounted survey’s set-aside share becomes evidence in its own right. On that share alone — ignoring what the kept readings say — a census counted at 34/55 with five counts a specimen separates grown plants from the geometry’s with nineteen kept specimens, and with eight counts, sixteen. That is weaker than the Fibonacci share, which needs ten, but it reads a different number from the same counts: a head that announces itself on every one of five counts is very probably whorled in fact, and a survey that records how many counts each head took has recorded that for free.

What it costs, in counts #

The census’s cost in counts is the kept specimens it needs times the counts each kept specimen costs, on each side.

At 34/55 and 7.2°, setting aside after one count spends 29.5 counts if the plants are grown and 23.2 if they are the geometry’s. Two counts a specimen spend 24.4 and 24.7. Three spend 20.7 and 25.6; eight spend 22.6 and 48.6. The grown side’s cost falls with the specimens bought back and then creeps up; the geometry’s side climbs throughout, paying for whorled heads. The cheapest cap on the worse side is two counts, at about twenty-five counts.

Counting 13/21 once, scoring every reading, spends six. That is the answer to the census essay’s question at this spread: a recounted census counted at 34/55 costs about four times the counts of a census counted low once. At 55/89 the cheapest cap is three counts, spending 51.7 and 48.6 — eight or nine times as many.

Heads rather than counts #

Counts are not always the scarce thing. A survey of living plants in a field can count a head as often as it likes; a survey of fossil cones, herbarium sheets or a photograph archive has the heads it has. For that kind of survey the number to minimise is heads examined, which is kept specimens divided by the share kept, and there the recount is at its strongest.

At 34/55 and 7.2°, setting aside after one count examines 29.5 grown heads to keep fifteen. Three counts a specimen examine 11.8 to keep ten — 15.1 if the heads are the geometry’s — and eight counts, 10.4. So for a fixed collection a recount buys back two heads in every three that setting aside would have spent, and it does so without touching what a kept reading says.

At the low count it does the reverse. At 13/21 and 7.2°, scoring every reading needs six heads, and setting aside or recounting at any cap needs more than seven. The reason is on the other side of the question: setting a reading aside also sets aside the geometry’s whorled heads, which lifts the geometry’s Fibonacci share from 14.7 per cent to 22.7, narrowing the gap the census is trying to see. A low count rarely announces a closing error, so there is little to gain, and the whorled heads are a real loss. Only at the widest spread read, 10.8°, does a second count help even 13/21: fourteen heads scored once, 10.7 grown heads with two counts.

At every spread #

A spread of 7.2° might be unkind to high counts in a way a steadier counter escapes. It is not, and the same comparison at every spread read shows why.

At 1.8°, 13/21 counted once costs four counts and the cheapest recount at 34/55 or 55/89 costs 6.2; at 3.6°, four against 6.3 and 9.3; at 5.4°, five against 12.5 and 24.8; at 10.8°, fourteen against 76.7 and 306.8. At no spread and no cap from one count to eight does a recounted census counted at 34/55 or 55/89 spend fewer counts than 13/21 counted once, on either side of the question.

What the recount does do is make a high count usable at all. Scored once, a census counted at 55/89 at 7.2° spends 449 counts, and at 10.8° it cannot be sized; recounted, it spends 52 and 307. Even at 3.6°, setting aside after one count at 34/55 spends about 6.3 counts against ten for scoring every reading. So of the two ways to count high, recounting is by far the better. It is still a way to count high, and the census essay’s conclusion survives it: the count that pins the angle is the one a census should avoid.

Why the low count wins #

The reason is not subtle once the pieces are laid out. A census needs Fibonacci readings that are Fibonacci because the plant is, and non-Fibonacci readings that are non-Fibonacci because the plant is. Every error pushes readings toward the middle. Announced errors can be removed, by setting aside or by recounting, but silent errors cannot, and a high count has more of them: at 7.2° a kept reading of a grown plant counted at 34/55 is silently wrong 29.8 per cent of the time.

A low count has few errors of either kind. At 13/21, 94.6 per cent of grown plants yield a kept reading on the first count, and so there is little for a recount to do. The census is cheap because the readings were never much wrong, not because anything rescued them. What a count is worth valued a high count for the angle it pins, and that value is real; it is simply a different question from whether a plant is Fibonacci, which a low count answers almost as surely and with far fewer errors in the answer.

What the recount model assumes #

It assumes each recount is independent of the last. A counter who closed the circle late at one organ may well close it late again at the same organ, and two marks chosen by one eye found that a counter’s closing errors are likely to be correlated. A correlated recount would tend to repeat the announced reading, so it would buy back fewer specimens and cost more counts than the independent recount here, and the comparison with 13/21 would only widen.

It keeps the census essay’s other assumptions: closing errors normal about zero, the geometry’s census at a rise of 0.008 as the null, and grown plants ninety per cent Fibonacci. A count carries no error asked that a published count say how it could have been wrong; the recount adds that a published census should say how many times each specimen was counted and how many were set aside.

What would undo it #

A spread of closing error, within 1.8° to 10.8°, at which a recounted census at 34/55 spends fewer counts than 13/21 counted once. A cap at which the Fibonacci share of a kept reading differs from its share at one count. A census of grown plants whose kept share falls as the cap rises. Each is a statement about arithmetic that the model either satisfies or does not, and none depends on a sample.

Still open: a recount that remembers the last one #

Every recount here starts from a fresh organ with a fresh closing error. A counter who knows the last reading announced itself has information: the closing mark was probably misjudged in the direction that made the larger count move. A recount that deliberately closes a little earlier or later, or at an organ on the other side of the head, is a different policy and possibly a much better one. The next measurement is a recount whose mark is chosen from the announced reading — how often a single informed recount turns 34/54 into 34/55 rather than into a silent 33/53, and whether an informed recount at 34/55 finally undercuts 13/21 counted once.