A riddle on uniform spacings!, namely when considering eight iid Uniform (0,1) variates as visiting times and three further iid Uniform (0,1) variates as server availability times, with unit service time, the question being the probability a server is available for a ninth visiting time, T⁹. Which can be decomposed into four cases:
- at least one server becomes available between the previous visiting time and T⁹
- at least two servers become available between the penultimate visiting time and the previous visiting time
- two servers become available between the antepenultimate visiting time and penultimate visiting time and one server becomes available between the penultimate visiting time and the previous visiting time
- three servers become available between the antepenultimate visiting time and the penultimate visiting time
with respective probabilities (using Devroye’s Theorem 2.1)
1/4, 8 9!3!/12!, 7 8!3!/12!, 7 8!3!/12!,
resulting in a total probability of 0.2934, compatible with a simulation assessment.
