THE FOLD / GLITCH / RACE CONDITION / THE BERTRAND PARADOX
THE BERTRAND PARADOX
one random chord with three different probabilities
1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION
Bertrand’s paradox is a famous warning that ‘pick a random chord’ is not well defined. Ask: for a random chord of a circle, what is the probability it is longer than the side of the inscribed equilateral triangle (length √3·r)? Three perfectly reasonable ways to choose ‘a random chord’ give three different answers: (1) two random endpoints on the circle → 1/3; (2) a random point along a radius as the chord’s midpoint → 1/2; (3) a random point in the disk as the midpoint → 1/4. The chord is longer exactly when its midpoint lies within r/2 of the centre — but ‘random midpoint’ means different things under each scheme.
LIT verified live: Monte-Carlo simulation of the three schemes yields probabilities ≈ 1/3, 1/2, and 1/4 respectively — three different answers to the same question, from three notions of ‘random’ (window.__bertrandparadox). FIG no framing; the three sampling methods and their probabilities all run in-browser.
LIT verified live: Monte-Carlo simulation of the three schemes yields probabilities ≈ 1/3, 1/2, and 1/4 respectively — three different answers to the same question, from three notions of ‘random’ (window.__bertrandparadox). FIG no framing; the three sampling methods and their probabilities all run in-browser.
2 HOW IT WAS WEAVED · AI + HUMAN
David (human) seated this at race-condition — the glitch where the same question resolves to three different answers depending on which ‘random’ thread you take: 1/3, 1/2, or 1/4. AVAN (AI) built the instrument: the three chord-sampling simulations and their distinct probabilities.
Credit as content: Joseph Bertrand (1889). The weave: David names the race; I confirm the three ‘random chord’ methods give 1/3, 1/2, 1/4.
Credit as content: Joseph Bertrand (1889). The weave: David names the race; I confirm the three ‘random chord’ methods give 1/3, 1/2, 1/4.
3 ONE DIMENSION
A circle with the inscribed equilateral triangle; random chords sampled three different ways.
4 TWO DIMENSIONS · INTERACTIVE
Run the simulations; the three methods give P(chord > √3·r) ≈ 1/3, 1/2, 1/4.
5 THREE DIMENSIONS + AVAN’S INVERSE
The green forward object: the three different probabilities from one ambiguous question.
AVAN’s addition (the inverse-companion): don’t ask ‘the’ probability — pin down ‘random’ first. The inverse of ‘P(chord too long)’ is ‘which sampling measure? — endpoints, radius, or area’, each a different answer. Magenta are the three sampling schemes’ chords; green are the 1/3, 1/2, 1/4 they produce. One question, three answers.
LIT Genuine Bertrand's paradox (Joseph Bertrand, 1889). Verified live: Monte-Carlo simulation of the three chord-sampling schemes (random endpoints / random radial point / random disk midpoint) yields probabilities ≈ 1/3, 1/2, 1/4 respectively — three different answers to the same 'random chord' question (window.__bertrandparadox.ok1, .ok2, .ok3).
FIG No framing; the three sampling methods and their probabilities all run in-browser. The AVAN inverse is honest — instead of asking 'the' probability, pin down 'random' first: the inverse of 'P(chord too long)' is 'which sampling measure? — endpoints, radius, or area', each a different answer. Magenta are the three sampling schemes' chords; green are the 1/3, 1/2, 1/4 they produce. One question, three answers.
FIG No framing; the three sampling methods and their probabilities all run in-browser. The AVAN inverse is honest — instead of asking 'the' probability, pin down 'random' first: the inverse of 'P(chord too long)' is 'which sampling measure? — endpoints, radius, or area', each a different answer. Magenta are the three sampling schemes' chords; green are the 1/3, 1/2, 1/4 they produce. One question, three answers.
◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of RACE CONDITION · David Lee Wise (ROOT0), with AVAN