Same filter
Comparing Two Lists Honestly
The method, in four steps, and the one assumption it rests on. It works even for a reader who rejects the coin convention entirely.
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Two lists, two filters, one question: which admits more?
The method is short, and it has a property worth stating first — it survives disagreement about the record's conventions. A reader who thinks counting subsets is a poor way to model anything can still use it, for reasons the last section of this page sets out.
The method
One. For each list, count the compulsory items. Call them e. That part contributes a factor of one over two to the e.
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Two. For each list, count the optional items and find the threshold. Call them n and m. That part contributes the sum of the binomial coefficients from m to n, over two to the n.
Three. Multiply the two factors. That is the intake, as an exact fraction.
Four. Compare the fractions.
A worked pair
List A: four compulsory items, no optional ones. Intake one in sixteen.
List B: one compulsory item, eight optional, five required. The compulsory part halves; the optional part admits ninety-three in two hundred and fifty-six. Intake ninety-three in five hundred and twelve, which is a little under one in five and a half.
| Items | Intake | |
|---|---|---|
| List A | 4 | 1/16 |
| List B | 9 | 93/512 |
List B is roughly three times wider, despite having more than twice as many items. The comparison by eye would have gone the other way and gone wrong by a factor of three.
The assumption it rests on
Only one: that both lists are counted the same way.
That assumption is free, because the counting is the record's own and identical on both sides. It does not require believing that the coin convention describes any population, that items are independent, or that the fractions correspond to anybody's chances.
Two readings from the same ruler can be compared even if the ruler is calibrated in units nobody asked for. What cannot be compared is two readings from different rulers, which is what comparing by impression amounts to.
Why this works for a sceptic
Somebody who rejects the coin convention entirely is rejecting the claim that one in sixteen means anything about applicants. Fine — the comparison does not need that claim.
What it needs is that if List A admits one in sixteen and List B admits roughly one in five and a half under the same counting, then List B is the wider filter. That is a statement about the two documents, not about the world, and it holds regardless of what the fractions are taken to represent.
Where the method fails
When a rule is missing. If either list has no stated threshold, step two cannot be completed, and the comparison is between one known filter and an unknown one.
In that case the honest answer is that the comparison cannot be made, and the useful response is to ask for the rule rather than to guess at it. The section on the unwritten rule covers what guessing produces.
A second pair, where the answer reverses
Two lists, both of which look like the kind of thing anybody might write.
The first: four compulsory items and four optional, two of the four needed. Eight items, two hundred and fifty-six profiles, eleven of the sixteen optional configurations meet two or more, so the intake is eleven in two hundred and fifty-six.
Now the other document. It names only two things as required and puts six under a rule of five. Of the sixty-four ways to meet and miss those six, seven reach five or more — six plus one — and two halvings over that leave seven in two hundred and fifty-six.
| List | Shape | Admits |
|---|---|---|
| first | 4 compulsory, 4 optional, 2 needed | 11/256 |
| second | 2 compulsory, 6 optional, 5 needed | 7/256 |
Same length, same number of items, and the one with fewer requirements is the narrower filter by half again. Swapping a single threshold would reverse the ranking, which is the whole argument for computing instead of reading.
Keeping a record of the comparison
Write down the three numbers, not the conclusion.
A conclusion — "the second one is stricter" — is unusable six months later, because nobody will remember which version of either document it referred to. The shape, written as three numbers and a fraction, survives: anybody can check it against the current document and see whether the list has moved.
That is the entire record-keeping burden this method imposes. Three numbers per list, set down next to it, and every later comparison becomes a matter of arithmetic rather than of memory.
The thirty-second version
Count the absolutes. Count the optional items and the threshold. Multiply. Compare.
Anybody who does that for two documents knows something real about both. Anybody who compares them by length knows nothing, as the previous page spent itself establishing.
Near this one
- Twenty-One Lists, One Filter
Twenty-one shapes in this record admit exactly one profile in two hundred and fifty-six. They run from eight i - The Ladder Of Classes
Forty-six values here are shared by more than one shape, and the largest classes form a descending ladder: twe - Eight Items And Sixteen, The Same Filter
The extreme case: a list of eight and a list of sixteen that admit identically. What the extra eight lines are - Why They Agree, And Why It Is Not A Coincidence
Shapes that admit identically do so for a reason that can be stated in one line, and the reason explains the w