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Equation 4 · Cognitive Bias Was the Label; Attention Sink Is the Suspect

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p0=1/3p_0 = 1/3

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Inputs and operations1/3
Result or conditionp_0
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p0p_0

Symbol p_0

p0p_0 is part of the quantity the equation computes from the expression on the right.

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=

=

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

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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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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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