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Equation 6 · Cognitive Bias Was the Label; Attention Sink Is the Suspect

What does this equation mean?

σ=p0(1−p0)/n≈0.0086\sigma = \sqrt{p_0(1-p_0)/n} \approx 0.0086

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Inputs and operationssqrtp_0(1-p_0)/n ≈ 0.0086
Result or conditionσ
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This equation gives an approximation: it relates the quantities while allowing an approximation. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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σ\sigma

Symbol σ

σ is part of the quantity the equation computes from the expression on the right.

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p0p_0

Symbol p_0

p0p_0 is one of the signed contributions combined to compute the quantity on the left.

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nn

Symbol n

n is one of the signed contributions combined to compute the quantity on the left.

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=

=

The expressions on both sides represent the same quantity under the stated assumptions.

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√

√

Take a square root.

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≈

≈

Approximately equal to; the equality is not exact.

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subscript

subscript

The lower label selects a particular version, component, or indexed member of the quantity. For example, x₀ and xₜ can be values at different positions.

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How to interpret it

Its accuracy depends on the assumptions and range of use described in the article. Read it with the definitions, units, and assumptions supplied by the article.

What the article says around this equation

A skeptical reader’s first move should be to ask whether that headline number could be sampling noise dressed up by a generous threshold. It is worth actually running that check rather than asserting an answer to it. Under a null model in which a classifier’s selections are entirely independent of label position — no cognitive bias, no architectural quirk, nothing but content — the expected share of selections landing in any one third of an evenly split label list is p0p_0 = 1/3 , and with n = 3000 independent trials per dataset the standard deviation of that observed share is σ\sigma = p0(1−p0)/n\sqrt{p_0(1-p_0)/n} ≈\approx 0.0086 , well under one percentage point. The paper’s own forty-percent threshold…
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A skeptical reader’s first move should be to ask whether that headline number could be sampling noise dressed up by a generous threshold. It is worth actually running that check rather than asserting an answer to it. Under a null model in which a classifier’s selections are entirely independent of label position — no cognitive bias, no architectural quirk, nothing but content — the expected share of selections landing in any one third of an evenly split label list is p0p_0 = 1/3 , and with n = 3000 independent trials per dataset the standard deviation of that observed share is σ\sigma = p0(1−p0)/n\sqrt{p_0(1-p_0)/n} ≈\approx 0.0086 , well under one percentage point. The paper’s own forty-percent threshold sits 0.40 - 0.3333 ≈\approx 0.0667 above that null mean, which is z ≈\approx 7.75 standard deviations out — a one-sided tail probability, by the normal approximation to the binomial (valid here since np0(1−p0)p_0(1-p_0) ≈\approx 667 , far above the usual rule-of-thumb minimum), of roughly 4.7 ×\times 10^{-15} . At three thousand trials per dataset, in other words, a classifier with no position sensitivity of any kind would essentially never cross this threshold by chance, in either direction. Whatever explains the measured pattern, it is not noise inflated by a lenient bar. This computation does not touch the paper’s causal claim in either direction — it only confirms that the thing being explained is real, which sharpens rather than weakens the actual dispute: the seventy-three-configuration majority is not the part anyone should contest. What follows contests only which sentence gets to explain it.

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