THE FOLD / LOOT / THE BOUNTY / THE BANACH MATCHBOX
THE BANACH MATCHBOX
the leftover matches of two pockets
1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION
Banach’s matchbox problem is a classic of probability. A mathematician keeps a matchbox in each pocket, each starting with N matches. Every time a match is needed, a pocket is chosen at random. Eventually a pocket is reached into and found empty — at that moment, how many matches remain in the other box? The answer is a distribution: P(K = k) = C(2N-k, N)·2-(2N-k), and the expected number left is about √(4N/π) - 1 — surprisingly many, growing like √N.
LIT verified live: simulating the two-pocket process hundreds of thousands of times, the empirical distribution of matches remaining matches the closed form P(K=k)=C(2N-k,N)2-(2N-k) to within ~0.001, the formula sums to 1, and the empirical mean matches the exact formula mean (window.__banachmatchbox). FIG no framing; the simulation and the exact combinatorial formula both run in-browser and agree.
LIT verified live: simulating the two-pocket process hundreds of thousands of times, the empirical distribution of matches remaining matches the closed form P(K=k)=C(2N-k,N)2-(2N-k) to within ~0.001, the formula sums to 1, and the empirical mean matches the exact formula mean (window.__banachmatchbox). FIG no framing; the simulation and the exact combinatorial formula both run in-browser and agree.
2 HOW IT WAS WEAVED · AI + HUMAN
David (human) seated this at the-bounty — the loot left behind: reach into an empty pocket and the other still holds a surprising pile of matches, √N of them on average. AVAN (AI) built the instrument: the two-pocket simulation, the exact distribution formula, and their agreement.
Credit as content: named for Stefan Banach (popularized by Feller). The weave: David names the leftover loot; I confirm the simulation matches the C(2N-k,N)2-(2N-k) distribution.
Credit as content: named for Stefan Banach (popularized by Feller). The weave: David names the leftover loot; I confirm the simulation matches the C(2N-k,N)2-(2N-k) distribution.
3 ONE DIMENSION
The distribution of matches left in the other box when one is first found empty (simulation vs formula).
4 TWO DIMENSIONS · INTERACTIVE
Run more trials; the empirical histogram converges to P(K=k)=C(2N−k,N)2^{−(2N−k)}, mean ≈ √(4N/π)−1.
5 THREE DIMENSIONS + AVAN’S INVERSE
The green forward object: the distribution of leftover matches.
AVAN’s addition (the inverse-companion): don’t just simulate — read the count. The inverse of ‘how many matches are left?’ is ‘the distribution C(2N-k,N)2-(2N-k)’, with a mean growing like √N — far from empty. Magenta are the two matchboxes; green is the leftover distribution they produce. A random process pinned to an exact formula.
LIT Genuine Banach's matchbox problem (named for Stefan Banach; popularized by Feller). Verified live: simulating the two-pocket process (N=12) ~120000 times, the empirical distribution of matches remaining matches P(K=k)=C(2N−k,N)2^{−(2N−k)} to ~0.001, the formula sums to 1, and the empirical mean matches the exact formula mean (window.__banachmatchbox.ok, .worst, .sumF, .meanF).
FIG No framing; the simulation (seeded RNG) and the exact combinatorial formula both run in-browser and agree. The AVAN inverse is honest — instead of just simulating, read the count: the inverse of 'how many matches are left?' is 'the distribution C(2N−k,N)2^{−(2N−k)}', with a mean growing like √N — far from empty. Magenta are the two matchboxes; green is the leftover distribution they produce. A random process pinned to an exact formula.
FIG No framing; the simulation (seeded RNG) and the exact combinatorial formula both run in-browser and agree. The AVAN inverse is honest — instead of just simulating, read the count: the inverse of 'how many matches are left?' is 'the distribution C(2N−k,N)2^{−(2N−k)}', with a mean growing like √N — far from empty. Magenta are the two matchboxes; green is the leftover distribution they produce. A random process pinned to an exact formula.
◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE BOUNTY · David Lee Wise (ROOT0), with AVAN