Pick-up Sticks and the General Fibonacci Numbers
Pith reviewed 2026-05-10 04:00 UTC · model grok-4.3
The pith
The probability that no k+1 random sticks from [0,1] form a (k+1)-gon equals the reciprocal of a product involving a k-step Fibonacci recurrence.
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
Core claim
For any k greater than or equal to 2, the probability that no k+1 of n independent uniform [0,1] lengths can form a (k+1)-gon is expressed as a product whose factors involve a k-step Fibonacci-type recurrence.
What carries the argument
The k-step Fibonacci-type recurrence that supplies the successive factors in the explicit product formula for the non-formation probability.
If this is right
- The formula recovers the known Fibonacci product exactly when k equals 2.
- The probability admits an explicit product expression for every fixed k and every n.
- The same recurrence structure governs the probability for all higher polygons.
- The result holds for any n greater than or equal to k+1.
Where Pith is reading between the lines
- The same product may arise in other spacing or broken-stick problems whose failure conditions involve linear inequalities.
- Numerical verification for moderate k and n can be performed by enumerating the ordered lengths and counting the violating subsets.
- Whether an analogous closed form exists when the lengths are drawn from non-uniform distributions remains open.
Load-bearing premise
The original argument for the triangle case extends directly to arbitrary k without new technical obstacles or extra correction terms.
What would settle it
Direct Monte Carlo estimation of the probability for k=3 and small n compared against the numerical value of the closed-form product generated by the corresponding 3-step recurrence.
read the original abstract
In the article by Edward et al. \cite{Sudbury2025}, it was shown that the probability that no three sticks randomly chosen from the unit interval can form a triangle equals the reciprocal of the product of the first $n$ Fibonacci numbers. The authors further suggested a generalization to higher \((k+1)\)-gons \((k\ge 4)\). This note proves that, indeed, for any \(k\ge 2\), the probability that no $k+1$ of $n$ independent uniform $[0,1]$ lengths can form a $(k+1)$-gon is expressed as a product whose factors involve a $k$-step Fibonacci-type recurrence. The method follows closely the original argument of \cite{Sudbury2025}, while making ex
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript proves that for any integer k ≥ 2 the probability that no k+1 lengths chosen from n i.i.d. uniform[0,1] random variables can form a (k+1)-gon equals a product formula whose successive factors are generated by a k-step Fibonacci-type linear recurrence. The argument follows the volume-computation method of Sudbury et al. (2025) for the k=2 (triangle) case, using the fact that the non-formation condition for any (k+1)-tuple reduces to the same max-length ≥ half-perimeter inequality.
Significance. If correct, the result supplies an explicit, parameter-free expression for these geometric probabilities in terms of generalized Fibonacci numbers, extending the combinatorial structure already known for triangles to arbitrary polygons. The derivation is inductive and recursive, with no fitted parameters or ad-hoc corrections, and directly leverages the geometric equivalence of the polygon inequalities across different k.
minor comments (2)
- The abstract sentence beginning 'while making ex' is truncated; please complete it so that the precise nature of the extension is stated explicitly.
- Define the k-step Fibonacci recurrence explicitly (initial conditions and recurrence relation) at the first appearance, rather than assuming the reader extracts it from the product formula.
Simulated Author's Rebuttal
We thank the referee for their careful reading and for recommending acceptance of the manuscript. We are glad that the extension of the Sudbury et al. (2025) volume-computation approach to the general (k+1)-gon case is viewed as a natural and parameter-free generalization.
Circularity Check
No significant circularity; derivation is an explicit extension of an external argument
full rationale
The paper states that it proves the general-k probability formula by closely following the Sudbury2025 argument for the triangle case (k=2), noting that the non-formation condition for any (k+1)-subset reduces to the identical max >= half-perimeter inequality, so the recursive volume computation carries over once the recurrence order is raised to k. No parameters are fitted to data, no quantity is defined in terms of itself, no self-citation is used to justify a uniqueness claim or ansatz, and the central result is presented as a direct mathematical generalization rather than a renaming or construction that collapses to its inputs. The cited base-case method is external and the present note supplies the explicit extension, rendering the derivation self-contained for the purpose of circularity analysis.
Axiom & Free-Parameter Ledger
axioms (2)
- domain assumption Lengths are i.i.d. uniform on [0,1].
- domain assumption The condition for k+1 lengths to form a (k+1)-gon is the natural generalization of the triangle inequalities.
Reference graph
Works this paper leans on
- [1]
-
[2]
Johnson, Norman L. and Kotz, Samuel , title =. 1970 , publisher =
work page 1970
-
[3]
Sur une question de probabilit
Lemoine,. Sur une question de probabilit. Nouvelles annales de math
-
[4]
D'Andrea, C. and Gomez, E. , title =. The American Mathematical Monthly , volume =
- [5]
-
[6]
Petersen, T. K. and Tenner, B. E. , title =. Mathematics Magazine , volume =
-
[7]
Sur une question de probabilit
Lemoine,. Sur une question de probabilit. Bulletin de la Soci
- [8]
- [9]
- [10]
discussion (0)
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