A reader can know 95% of the running words in two passages while meeting the unfamiliar words in very different arrangements. Six unfamiliar occurrences might sit together in one stretch, or appear separately across the text. The overall percentage does not tell you the longest unfamiliar run or which sentence needs attention.
This guide develops that missing sequence information with original token records. It counts runs, derives bounds and checks local windows without turning any of those measures into a comprehension score. The records are artificial teaching constructions, not observations about named learners or a recommendation to rearrange real prose mechanically.
The existing vocabulary-load and lexical-coverage guide explains why centrality, phrases, context and reading purpose matter. Here the narrower question is mathematical: what does a fixed coverage total establish about where unfamiliar occurrences can be located?
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Distinguish token coverage from the order and meaning of unfamiliar occurrences. Follow the worked sequence or choose the route that locates the first uncertain connection.
ROUTE 1 · CHAPTERS 1–5
Define the reading record
Distinguish token coverage from the order and meaning of unfamiliar occurrences.
ROUTE 2 · CHAPTERS 6–10
Build and bound runs
Construct equal-coverage sequences, derive run bounds and count adjacent pairs.
ROUTE 3 · CHAPTERS 11–15
Examine local windows
Compare window counts, passage boundaries and a deliberately artificial shuffle model.
ROUTE 4 · CHAPTERS 16–21
Audit and transfer
Preserve knowledge conditions and solve a changed sequence without predicting comprehension.
ROUTE 5 · CHAPTERS 22–26
Practise and continue
Use the record for a focused teaching question, FAQs and connected reading.
Chapter contents
Define the reading record · Chapters 1–5
1. Define the running-word record
2. Separate tokens, types and meanings
3. Define a run without changing the coverage total
Build and bound runs · Chapters 6–10
6. Derive the maximum possible number of runs
7. Derive the smallest possible longest run
8. Use a known run count to sharpen the bound
9. Count adjacent unfamiliar pairs in a second way
10. Derive run count from starts rather than scanning lengths
Examine local windows · Chapters 11–15
11. Define a local window and its counting question
12. Compare windows in the two original arrangements
13. Preserve the edge effects in a window count
Audit and transfer · Chapters 16–21
16. Keep a shuffle reference separate from a language explanation
17. Keep the knowledge check attached to time and task
18. Leave uncertain marks visibly uncertain
19. Try a changed sequence without the earlier numbers
Practise and continue · Chapters 22–26
22. Repair a misleading coverage claim
23. Use a marked location to choose a language question
24. Choose a learner route from marks to meaning
Choose one text, one tokenisation rule and one reader-check condition. A token is a counted occurrence in that record. Repeated occurrences of the same written word can contribute several tokens.
For the first exercise, each of 120 token positions is marked familiar or unfamiliar under the supplied check. There are 114 familiar positions and six unfamiliar positions. Coverage is 114/120 = 95%.
The binary marks simplify a richer learning situation. Someone can recognise a form while missing its meaning in this sentence, understand one sense but not another, or know the word while struggling with a phrase. The check needs to specify what familiar means for this record.
Do not treat a frequency-list match as proof of the reader’s knowledge unless that is explicitly the proxy being studied. A list-based profile and a reader’s checked understanding are different sources of evidence.
Record the text version, counting rule and check condition before calculating. If the tokenisation or knowledge criterion changes, both the denominator and the pattern can change. A percentage is interpretable only while those definitions remain attached to it.
Six unfamiliar tokens could be six repetitions of one target word, three repetitions each of two targets, or six different targets. The token total alone does not identify the number of distinct word forms or concepts requiring attention.
The same written form can also have different senses. A learner might understand mean in everyday use but not in a mathematical question. Assigning a single permanent familiar flag to every occurrence of that spelling would hide the task-specific distinction.
The corpus linguistics guide introduces tokens, types, lemmas, families and profiling. Those counting units should remain separate from the within-sequence run calculation developed here.
For the artificial record, the six unfamiliar flags are stipulated at particular positions. We will not infer how many words the reader knows overall or how many teaching sessions they need.
A useful learner question is: “What exactly does this six count?” A correct answer names unfamiliar occurrences in the included text under the stated criterion. If the answer becomes six unknown ideas, six new dictionary entries or six errors on an examination, the interpretation has moved beyond the supplied record.
CHAPTER 3 OF 26 · Define the reading record
3. Define a run without changing the coverage total
An unfamiliar-word run is a maximal consecutive stretch of unfamiliar token positions. Maximal means that the stretch cannot be extended left or right within the same sequence without meeting a familiar token or reaching the text boundary.
If unfamiliar positions are 61 through 66, they form one run of length six. If unfamiliar positions are 20, 40, 60, 80, 100 and 120, each is separated from the others, giving six runs of length one.
Both records have six unfamiliar tokens and 95% coverage. Their run counts and longest runs differ. These are properties of the supplied order, not different calculations of the same overall percentage.
A run count also needs a boundary rule. Two unfamiliar tokens on opposite sides of a paragraph break can be consecutive tokens in one continuous sequence, or belong to separate passage records if the analysis treats paragraphs independently. Neither choice should be left implicit.
Before counting, ask the learner to identify where each run starts and ends. This prevents a common mistake: counting every unfamiliar token as a separate run even when it immediately follows another unfamiliar occurrence.
CHAPTER 4 OF 26 · Define the reading record
4. Construct two equal-coverage arrangements
Call the record with unfamiliar positions 61–66 the concentrated arrangement. Call the record with positions 20, 40, 60, 80, 100 and 120 the separated arrangement. All other positions are familiar.
| Artificial record | Tokens | Unfamiliar tokens | Coverage | Runs | Longest run |
|---|---|---|---|---|---|
| Concentrated | 120 | 6 | 95% | 1 | 6 |
| Separated | 120 | 6 | 95% | 6 | 1 |
The table does not declare one text easier. The concentrated record leaves a long unfamiliar stretch in one location; the separated record places unfamiliar occurrences across more locations. Which arrangement matters to a reading task depends on the actual words, relationships and purpose.
The construction proves one limited result: coverage alone does not determine run count or longest run. Any claimed universal conversion from 95% to a particular run pattern fails on these two compatible records.
For independent checking, count unfamiliar positions directly, then calculate coverage. Next count run starts and identify the longest stretch. Keeping these operations separate makes it easier to see whether an error concerns the total, the order or the definition of a run.
CHAPTER 5 OF 26 · Define the reading record
5. Keep a descriptive pattern separate from comprehension
A six-token unfamiliar run can be worth inspecting, but its length does not establish a fixed comprehension loss. The unfamiliar occurrences might be names, redundant descriptions, a central technical relationship or something else. Familiar words can also form a difficult sentence.
The primary study by Schmitt, Jiang and Grabe, published in 2011, examined vocabulary coverage and comprehension using specified readers, texts and tasks. It did not establish a universal abrupt coverage boundary guaranteeing understanding. Its research context should not be replaced by a mechanical rule for an individual learner.
Our run calculations do not estimate a causal effect from moving words around. They describe what information an overall total omits. To investigate comprehension, retain meaningful text and use an appropriate task that checks the relationship the reader needs to understand.
A useful report says, “This marked record contains one unfamiliar run of six tokens, so inspect that location,” rather than “The reader cannot understand six words in a row, so the whole text is unsuitable.” The first statement identifies evidence and a next question. The second adds a conclusion the count has not established.
CHAPTER 6 OF 26 · Build and bound runs
6. Derive the maximum possible number of runs
Let U be the number of unfamiliar tokens and K the number of familiar tokens. Each unfamiliar run contains at least one unfamiliar token, so there cannot be more than U runs.
Separating two unfamiliar runs requires at least one familiar token between them. K familiar tokens can separate at most K + 1 unfamiliar runs, including possible runs before the first familiar token and after the last.
Therefore the maximum possible run count is min(U, K + 1). With U = 6 and K = 114, that maximum is six, reached by separating every unfamiliar occurrence.
The bound is attainable for any nonnegative integer counts: use familiar tokens as separators until either every unfamiliar token is isolated or all K + 1 gaps are occupied. Put any remaining unfamiliar tokens into existing runs.
This is a slot argument. Picture the familiar tokens as dividers creating gaps. The gaps can hold unfamiliar tokens without changing how many familiar tokens exist. The argument counts possible sequence arrangements; it says nothing about whether a particular arrangement forms natural language or supports a particular reading purpose.
The K familiar tokens create at most K + 1 gaps in which unfamiliar tokens can be placed. To make the longest unfamiliar run as short as possible, distribute U tokens as evenly as possible among those gaps.
The smallest achievable longest run is ceiling(U/(K + 1)), with the value zero when U is zero. The ceiling appears because a run contains a whole number of tokens.
For six unfamiliar and 114 familiar tokens, the result is one. For eleven unfamiliar and three familiar tokens, there are four gaps, so the longest run must be at least ceiling(11/4) = 3. A distribution of lengths 3, 3, 3 and 2 attains that bound.
Check the construction rather than stopping at the inequality. Three familiar separators can indeed keep those four unfamiliar groups apart. Their total length is eleven.
The largest possible longest run is U, obtained by putting every unfamiliar token together. Thus coverage and text length provide a range for the longest run, but generally not its actual value. Sequence information is needed to locate the real record within that range.
CHAPTER 8 OF 26 · Build and bound runs
8. Use a known run count to sharpen the bound
Suppose U unfamiliar tokens are already known to form exactly R nonempty runs. Their run lengths are positive integers adding to U.
At least one run must have length at least ceiling(U/R). Otherwise, R shorter runs could not hold all U tokens. The longest run cannot exceed U − R + 1, because the other R − 1 runs each require at least one token.
For U = 6 and R = 3, the longest run can range from two to four. Lengths 2, 2, 2 attain the lower endpoint. Lengths 4, 1, 1 attain the upper endpoint. Both need enough familiar tokens to separate three runs.
The feasibility condition is R ≤ min(U, K + 1) when U is positive. A claimed run count outside that range is inconsistent with the supplied totals.
This extra information narrows the account without pretending to recover the full sequence. Even knowing that the longest run is two does not tell you whether it appears in an essential instruction or an optional descriptive sentence. Mathematical precision about one feature should not erase the remaining language question.
CHAPTER 9 OF 26 · Build and bound runs
9. Count adjacent unfamiliar pairs in a second way
A run of length l contains l − 1 adjacent unfamiliar pairs. For example, a length-six run contains five neighbouring pairs, while a length-one run contains none.
If the run lengths sum to U and there are R runs, the total adjacent unfamiliar-pair count A is U − R. This identity gives a second route to checking a marked sequence.
For the concentrated 120-token record, U = 6 and R = 1, so A = 5. For the separated record, R = 6 and A = 0. The same unfamiliar-token total produces different adjacency totals.
The pairs overlap: a three-token run has a pair across positions one and two and another across positions two and three. Counting those pairs does not mean there are four unfamiliar tokens. The middle token belongs to both pairs.
A useful checking routine records U, run-start count R and adjacent-pair count A independently, then verifies U = R + A. If that equality fails, inspect the marking or boundary rule. It is an arithmetic consistency check, not a complete assessment of vocabulary knowledge.
CHAPTER 10 OF 26 · Build and bound runs
10. Derive run count from starts rather than scanning lengths
A run begins at the first token if that token is unfamiliar. At every later position, a run begins when the current token is unfamiliar and the previous token is familiar.
Counting those starts gives R directly. Each unfamiliar token is either a start or follows another unfamiliar token. Those two categories are disjoint and exhaustive, which explains the identity U = R + A from another direction.
For a small invented record with unfamiliar positions 2, 3, 7, 10 and 11, run starts occur at 2, 7 and 10. Thus U = 5, R = 3 and A = 2. The run lengths are 2, 1 and 2.
This representation is useful for a spreadsheet or a short program, but the counting rule should remain understandable without code. Test the first position separately so a run at the beginning is not lost.
Then test an all-familiar record and an all-unfamiliar record. The first has zero starts; a nonempty all-unfamiliar record has one. These simple boundary cases can reveal an implementation that mistakenly counts every unfamiliar token as a new run.
CHAPTER 11 OF 26 · Examine local windows
11. Define a local window and its counting question
A window of length w contains w consecutive token positions. In a sequence of N tokens, there are N − w + 1 full windows when 1 ≤ w ≤ N, using a one-token sliding step.
For N = 120 and w = 20, there are 101 windows. The first covers positions 1–20; the last covers 101–120. Each window can be assigned a local unfamiliar count.
Several summaries are possible: the largest unfamiliar count in any window, the number of windows containing at least one unfamiliar token, or the distribution of local counts. They are different measurements.
A twenty-token window is an artificial analysis unit, not automatically a sentence, a line of print or a unit of working memory. A meaningful teaching review should also examine actual sentence and paragraph boundaries.
State the step size. Nonoverlapping blocks of twenty tokens would give six blocks, while sliding one-token windows give 101 overlapping windows. Their counts cannot be compared as though they were the same set of observations. The chosen window method belongs in the definition, before an apparently precise local-density result is reported.
CHAPTER 12 OF 26 · Examine local windows
12. Compare windows in the two original arrangements
In the concentrated record, unfamiliar positions 61–66 are all inside some twenty-token windows. A window contains at least one of them when its start lies from 42 through 66, giving 25 affected windows.
The largest local unfamiliar count is six. Those windows show a concentrated patch, while many other windows contain none.
In the separated record, unfamiliar positions occur every twenty tokens, including position 120. Every full twenty-token window contains exactly one unfamiliar occurrence. All 101 windows are affected, while the largest local unfamiliar count is only one.
The separated record therefore has a smaller peak local count but more windows containing any unfamiliar token. Calling it better on every measure would be incorrect even within this purely descriptive model.
This contrast is useful because it prevents one preferred summary from becoming an unexamined score. A question about a long concentrated stretch differs from a question about how widely unfamiliar occurrences are encountered. Neither summary, by itself, establishes reading difficulty. Return to the actual words and task after identifying the sequence feature worth inspecting.
CHAPTER 13 OF 26 · Examine local windows
13. Preserve the edge effects in a window count
A token in the middle of a long sequence can belong to w different length-w sliding windows. A token at the first or last position belongs to only one full window.
For N = 120 and w = 20, position 60 belongs to windows starting at 41 through 60. Position 120 belongs only to the final window, starting at 101. Therefore summing window loads does not always equal U multiplied by w.
The exact contribution of position i is the number of allowed window starts between max(1, i − w + 1) and min(i, N − w + 1), inclusive. Count it only when the lower endpoint does not exceed the upper.
This formula derives from containment: a window starts no later than i and ends no earlier than i. The valid-start range must also stay inside the sequence.
For a learner check, use a ten-token record and windows of length four. Place the only unfamiliar token at position one, then at position five. Coverage remains 90%, but the total number of affected windows changes from one to four. The boundary effect follows from the measurement design, without any claim about a reader’s comprehension.
CHAPTER 14 OF 26 · Examine local windows
14. Preserve passage boundaries when combining records
Suppose one passage ends with an unfamiliar run of length two and the next begins with an unfamiliar run of length three. If the passages are concatenated with no boundary marker, those stretches become one run of length five.
The combined unfamiliar-token total is unchanged, but the combined run count is one less than the sum of the two separate run counts. If either boundary token is familiar, that particular merger does not occur.
The longer run can therefore be an artefact of joining records rather than a feature of either original passage. A corpus file boundary, paragraph boundary and continuous reading sequence need not mean the same thing.
Adding an artificial familiar separator prevents the merger, but also changes the token denominator if it is counted as a token. Another option is to preserve an explicit break rule and calculate runs within each record. State which method is used.
For a reproducibility check, reconstruct the separate passage summaries after combining the data. The original boundaries should still be available. A processing convenience should not silently turn the end of one text and the start of another into an observed stretch that a learner actually read continuously.
CHAPTER 15 OF 26 · Examine local windows
15. Explore a deliberately artificial shuffle model
Imagine uniformly shuffling the familiar and unfamiliar marks while keeping N and U fixed. This is a mathematical reference model for marks, not a model of how natural text or a reader’s knowledge is generated.
For N greater than one, the probability that a specified adjacent pair is unfamiliar in both positions is U/N multiplied by (U − 1)/(N − 1). There are N − 1 adjacent pairs, so their expected total is U(U − 1)/N.
For N = 120 and U = 6, the expected adjacent-pair count is 6 × 5/120 = 0.25. Because R = U − A, the expected run count is 6 − 0.25 = 5.75.
A particular shuffled sequence still has a whole-number run count. The fractional expectation summarises repetitions of the artificial random procedure.
The two pair positions are not independent under fixed totals: after an unfamiliar mark occupies the first, only U − 1 remain among N − 1 positions. Using (U/N) squared would replace the stipulated shuffle with another model. Preserve the counting constraint when deriving the reference expectation.
CHAPTER 16 OF 26 · Audit and transfer
16. Keep a shuffle reference separate from a language explanation
An observed concentrated pattern can differ sharply from the shuffle expectation. That does not establish that the concentration causes difficulty or that the shuffle would produce better prose.
Real language is structured. Related terms can appear together because a paragraph develops one concept, repeated words can maintain reference, and grammatical relationships constrain order. Shuffling actual words would usually destroy those relationships.
The artificial reference model answers a narrow conditional question: what adjacency count would this mark total produce under uniform rearrangement? It does not describe the distribution of genuine text arrangements or supply a universal significance test for a learner’s reading record.
If a statistical comparison is intended, specify how texts or learners were sampled, what the null process means and which outcome is being tested. A convenient shuffle is not automatically the appropriate null model.
For teaching, the simpler use is often enough. Compare the actual marked location with a differently arranged paper record, then return to the meaningful sentence. Ask what knowledge or relationship is missing there. The mathematical contrast identifies information lost by the overall percentage; the language task establishes why that information matters in this particular passage.
CHAPTER 17 OF 26 · Audit and transfer
17. Keep the knowledge check attached to time and task
A mark can change after explanation, rereading or a different prompt. That change may be useful evidence, but it should not be hidden inside a comparison presented as though the checking conditions were identical.
For example, a learner might recognise a target when shown a translation but fail to explain its meaning in the original sentence without support. A later record with the translation visible measures a different condition from the earlier unsupported record.
The existing guide to vocabulary scores when cues change develops that assessment distinction. The run calculation assumes the marks have already been defined; it cannot make unlike knowledge checks comparable.
A useful note records the check date, prompts, available references and criterion. If the aim is independent use, include a later changed-context task without the original model visible. Do not infer durable knowledge merely because a highlighted word was replaced by a familiar mark immediately after explanation.
This preserves a humane interpretation too. A marked passage describes a condition under which a learner worked. It should guide a specific next step, rather than become a permanent label on the learner’s ability.
Suppose three token positions are confirmed unfamiliar and two more are unresolved. The overall unfamiliar total lies between three and five, but the possible longest runs depend on where those unresolved positions sit.
If the confirmed unfamiliar positions are 4, 5 and 8, with unresolved positions 6 and 7, the longest run could be two when both unresolved positions are familiar, or five when both are unfamiliar. The positions matter, not just the number of uncertain marks.
To bound a feature, evaluate compatible complete marking arrangements. A minimum or maximum should be supported by an arrangement reaching it, while preserving the confirmed marks.
An empty text is another boundary case. Coverage has no defined percentage when the token denominator is zero. A nonempty all-familiar record has zero unfamiliar runs, while a nonempty all-unfamiliar record has one run covering the whole sequence.
Keep missing observations separate from familiar ones. A position not checked is not established as understood. The useful response is to describe the incomplete record and identify a small check that would resolve the important uncertainty, rather than fill every blank with the most flattering classification.
CHAPTER 19 OF 26 · Audit and transfer
19. Try a changed sequence without the earlier numbers
A new artificial record contains twenty tokens. Eight are unfamiliar and twelve familiar. Calculate coverage, the maximum possible number of unfamiliar runs, and the smallest possible longest run.
Next, suppose the eight unfamiliar tokens form exactly three runs. Find the smallest and largest possible longest run, then construct run lengths reaching both bounds. Check whether twelve familiar tokens are enough to keep the three runs separate.
For a complete observed pattern, take unfamiliar positions 2, 3, 4, 9, 13, 14, 15 and 16. Calculate the run count, longest run and adjacent unfamiliar-pair count. Verify the identity connecting those quantities.
For an extension, use five-token sliding windows. Find the greatest local unfamiliar count in the observed pattern. State how many full windows exist before examining their contents.
Do the construction before the arithmetic summary. A learner can obtain a plausible bound from a formula while overlooking that a supplied run count is impossible under the familiar-token total. The next chapter checks both the numerical answers and the underlying arrangement.
CHAPTER 20 OF 26 · Audit and transfer
20. Check the changed sequence through independent accounts
Coverage is 12/20 = 60%. The maximum run count is min(8, 13) = 8, and the smallest achievable longest run is ceiling(8/13) = 1.
With exactly three runs, the longest lies between ceiling(8/3) = 3 and 8 − 3 + 1 = 6. Lengths 3, 3, 2 attain the lower endpoint; 6, 1, 1 attain the upper. Two familiar separators suffice to keep three runs distinct, so the twelve available familiar tokens do not obstruct either construction.
In the observed pattern, the runs are positions 2–4, position 9, and positions 13–16. Their lengths are 3, 1 and 4. Thus R = 3, the longest run is four and A = (3 − 1) + (1 − 1) + (4 − 1) = 5. The identity U = R + A gives 8 = 3 + 5.
There are 20 − 5 + 1 = 16 full five-token windows. The maximum local unfamiliar count is four, attained by a window containing positions 13–16. Those results describe the artificial sequence only; they do not predict a 60% comprehension score.
A useful record identifies the text version, token rule, reader-check condition and included positions. Preserve the unfamiliar marks in order, plus any uncertain marks and passage boundaries.
Report the global and local quantities separately. A compact summary can include N, U, coverage, run count, longest run, adjacent-pair count and a specifically defined window summary. It need not turn them into one combined score.
Show a checkable representation. For a short record, list unfamiliar positions and run lengths. For a longer one, retain the position-level data so a summary can be reconstructed. A screenshot of a percentage alone does not preserve the sequence.
State which decisions the account can inform. It might identify a stretch for closer reading, a repeated target to examine or a discrepancy between two marking rules. It cannot establish why comprehension failed without further evidence about language, concepts and the task.
If the text is revised or teaching changes the reader’s knowledge, keep the old and new conditions distinct. A change in run length may reflect an edited passage, different tokenisation, a new prompt or new learning. A clear record lets those explanations be examined instead of assuming that every numerical improvement means the same thing.
Consider the invented sentence: “The learner knows 95% of this text, so there will be one unfamiliar word every twenty words.” The percentage gives a ratio, not a regular spacing rule.
In the original 120-token constructions, six unfamiliar occurrences can be spaced every twenty tokens, but they can also form one length-six run. Both satisfy 95% coverage.
A repaired sentence might say: “The marked record contains six unfamiliar occurrences among 120 tokens. Their positions must be inspected to establish the run pattern or local concentrations. The percentage alone does not determine spacing or comprehension.”
The same repair applies to statements such as “only six unfamiliar words means at most a short difficulty” or “one long run makes the whole passage inappropriate.” Those conclusions require the actual words, their role and evidence about the reading task.
For practice, ask the learner to draw a second arrangement with the same totals but a longest run of three. Then ask for its run count. A valid answer might use two length-three runs or several runs with a different length pattern. The exercise tests whether the learner understands what the coverage total leaves free.
CHAPTER 23 OF 26 · Practise and continue
23. Use a marked location to choose a language question
Suppose a fictional science sentence contains an unfamiliar verb linking two familiar objects. The unfamiliar occurrence might be central because it states the relationship the question asks about. A different sentence might contain several unfamiliar descriptive names whose exact meanings are less important for the immediate task.
The sequence count does not assign that centrality. Read the sentence, identify the intended claim and ask which missing meaning blocks it. A glossary entry can help with a word while leaving a causal or grammatical relationship unresolved.
A useful follow-up has two parts: explain the meaning in the original sentence, then use or recognise the relationship in a changed context. The second task checks whether the explanation became available beyond the original highlighted line.
Keep the scope proportionate. The purpose of locating a cluster is to focus attention, not to require teaching every nearby word before any reading can continue. The existing vocabulary-load guide offers the broader diagnostic route.
This chapter connects the mathematics to practice without claiming an experimentally established benefit from a particular run length. A descriptive feature becomes useful when it leads to a precise language question whose answer can actually be checked.
CHAPTER 24 OF 26 · Practise and continue
24. Choose a learner route from marks to meaning
For a first route, use twenty token cards with a fixed number of unfamiliar marks. Keep the count unchanged while arranging the marks in one group, several groups and isolated positions. Count starts and longest stretches aloud.
For a secondary mathematics route, derive the gap bound using familiar cards as separators. Check the adjacent-pair identity and compare nonoverlapping blocks with sliding windows. Include a boundary example so the learner cannot rely only on a central cluster.
For an advanced route, enumerate all short binary arrangements with fixed totals and verify the attainable run bounds. Explore a uniform-shuffle expectation as a stated artificial model. Then explain why the model is not a theory of natural text or reading comprehension.
Repair means locating a confusion between a total and an arrangement, or between an unfamiliar occurrence and an unfamiliar word type. Stabilisation means obtaining consistent summaries across changed sequences. Extension means adding real sentence meaning and an appropriately checked learning task.
The successful learner can explain both the calculation and its limit. They can identify a location worth examining while resisting the temptation to turn a neat sequence statistic into a complete diagnosis of a reader.
CHAPTER 25 OF 26 · Practise and continue
25. Questions about coverage and unfamiliar-word runs
Does 95% coverage mean one unfamiliar word every twenty tokens?
It means that five percent of the counted tokens meet the unfamiliar criterion. Their spacing can vary. Regular spacing is one compatible arrangement, not a consequence of the percentage.
Is a shorter longest run always better?
This guide does not establish that. A shorter maximum can coexist with unfamiliar occurrences spread across more windows. The actual words, relationships and reading purpose still matter.
Are runs the same as corpus dispersion?
They are related descriptive ideas with different definitions. This guide counts consecutive marked occurrences within a sequence. The existing dispersion article compares target occurrence shares across corpus parts of potentially unequal length.
Can a repeated word be unfamiliar more than once?
Yes, if the check marks each occurrence that way. Alternatively, learning or a changed sense can alter the mark. State the rule and do not confuse token occurrences with distinct target forms.
Does a shuffle calculation predict comprehension?
No. It describes a stipulated random rearrangement of marks with fixed totals. Reading requires meaningful language and evidence about the relevant task; the shuffle does not provide that evidence.
CHAPTER 26 OF 26 · Practise and continue
26. Continue through vocabulary, sequence and research
Return to the Vocabulary subject guide for meaning, sentence patterns, retrieval and transfer. Use vocabulary load and lexical coverage when the main question concerns centrality, phrases, concepts or reading purpose.
For a different distribution question, read word dispersion across unequal text lengths. For assessment conditions, read test scores when cues change. The corpus linguistics guide linked earlier supplies the broader counting vocabulary.
Primary research context is The Percentage of Words Known in a Text and Reading Comprehension, by Schmitt, Jiang and Grabe, published in 2011. Its stated texts, readers and assessment conditions matter. The original sequences, run bounds, adjacency identities, window calculations and shuffle exercises here are mathematical teaching constructions, not findings attributed to that study. Sources checked 23 October 2026.
Connect the work to How X Works and The Gold Standard of Education: learning transfers into life. The transferable skill is to preserve arrangement information when a total percentage cannot answer a local question.