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