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Equation 56 · A Chatbot Confessed to Being Built by a Company That Never Trained It

What does this equation mean?

s=ln⁡(0.55/0.450.25/0.75)≈1.30s=\ln\left(\frac{0.55/0.45}{0.25/0.75}\right)\approx1.30

Read the formula alongside the article passage below. Each part has a deeper page with its role in the equation, the supporting passage and nearby citations.

Start with0.55/0.45
Divide by0.25/0.75
This relates tos
How to read the two sides of this formula. Follow the article passage for the meaning of each quantity.

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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ss

Symbol s

s 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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fraction

fraction

Divide the expression above the line by the one below it.

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0.55/0.450.55/0.45

Numerator: 0.55/0.45

The complete quantity above the fraction bar.

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0.25/0.750.25/0.75

Denominator: 0.25/0.75

The complete quantity below the fraction bar; it must be nonzero for this division.

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How to interpret it

With a fixed numerator, increasing a nonzero denominator reduces the fraction. Read it with the definitions, units, and assumptions supplied by the article.

What the article says around this equation

Suppose further that among the trait-positive horizontally exposed models, 70 percent produce a response diverging from the seeded canonical form beyond the declared threshold, μH\mu_H=0.70 , against 15 percent for the vertically exposed group, μV\mu_V=0.15 : horizontal transmission would be lossier, consistent with an indirect channel and with Mesoudi and Whiten’s finding that transmission-chain experiments on human cultural information typically show cumulative drift across successive links rather than the stable replication a simple copying model would predict [ 14 ] . And suppose a curation pass applied before the horizontally exposed artifact reached its downstream trainer moved the trait’s…
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Suppose further that among the trait-positive horizontally exposed models, 70 percent produce a response diverging from the seeded canonical form beyond the declared threshold, μH\mu_H=0.70 , against 15 percent for the vertically exposed group, μV\mu_V=0.15 : horizontal transmission would be lossier, consistent with an indirect channel and with Mesoudi and Whiten’s finding that transmission-chain experiments on human cultural information typically show cumulative drift across successive links rather than the stable replication a simple copying model would predict [ 14 ] . And suppose a curation pass applied before the horizontally exposed artifact reached its downstream trainer moved the trait’s prevalence from pprep_{\mathrm{pre}}=0.25 to ppostp_{\mathrm{post}}=0.55 , curators keeping the more verbose, apparently more careful outputs that happened to carry the trait, giving s=ln⁡\ln(0.55/0.450.25/0.75)\left(\frac{0.55/0.45}{0.25/0.75}\right)≈\approx1.30 : a positive selection coefficient, meaning the pipeline itself, not only the source model, pushed the trait toward survival. None of these six numbers has been measured. They exist to give the sealed-lineage trial proposed below something concrete to confirm or embarrass.

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