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Pick-up sticks problems for exponential and Pareto distributions
Abstract
We deal with the following two pick-up sticks problems: given an integer n greater than or equal to 3, we choose n sticks with independent random lengths from an exponential distribution or Pareto type I distribution, and we are interested in the probability that no three of them can be the sides of a triangle. In case of exponential distribution, this problem has been solved by other authors: the probability in question is equal to n!/((F3−1)(F4−1)⋯(Fn+2−1)), where (Fi)i=1∞ is the sequence of Fibonacci numbers. We give a new, self-contained and detailed proof for this formula, which uses a different method compared to the earlier ones. In case of Pareto type I distribution, we derive new formulae for the corresponding probability.
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BibTeX
@misc{barczy2026pickup,
title = {{Pick-up sticks problems for exponential and Pareto distributions}},
author = {Matyas Barczy and Rezső Lovas},
howpublished = {arXiv},
year = {2026},
month = nov,
doi = {10.48550/arxiv.2611.04436},
eprint = {2611.04436},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2611.04436},
}
RIS
TY - UNPB
TI - Pick-up sticks problems for exponential and Pareto distributions
AU - Matyas Barczy
AU - Rezső Lovas
PY - 2026
DA - 2026/11/03
PB - arXiv
DO - 10.48550/arxiv.2611.04436
UR - https://arxiv.org/abs/2611.04436
AB - We deal with the following two pick-up sticks problems: given an integer n greater than or equal to 3, we choose n sticks with independent random lengths from an exponential distribution or Pareto type I distribution, and we are interested in the probability that no three of them can be the sides of a triangle. In case of exponential distribution, this problem has been solved by other authors: the probability in question is equal to n!/((F_3−1)(F_4−1)⋯(F_n+2−1)), where (F_i)_i=1^∞ is the sequence of Fibonacci numbers. We give a new, self-contained and detailed proof for this formula, which uses a different method compared to the earlier ones. In case of Pareto type I distribution, we derive new formulae for the corresponding probability.
ER -