Collecting Coupons with Random Initial Stake

TL;DR

This paper precisely analyzes the coupon collector problem with a random initial set, revealing it is roughly half a round faster than starting with half the coupons, with implications for randomized local search algorithms.

Contribution

The paper provides a detailed analysis of the coupon collector problem with a random initial set, deriving exact expected times and implications for heuristic optimization.

Findings

Expected rounds: $nH_{n/2} - 1/2 ext{ (plus small error)}$

Random initial stake speeds up collection by about half a round

Implication for local search: $nH_{n/2} - 1/2$ iterations to find optima

Abstract

Motivated by a problem in the theory of randomized search heuristics, we give a very precise analysis for the coupon collector problem where the collector starts with a random set of coupons (chosen uniformly from all sets). We show that the expected number of rounds until we have a coupon of each type is $nH_{n/2} - 1/2 \pm o(1)$, where $H_{n/2}$ denotes the $(n/2)$th harmonic number when $n$ is even, and $H_{n/2}:= (1/2) H_{\lfloor n/2 \rfloor} + (1/2) H_{\lceil n/2 \rceil}$ when $n$ is odd. Consequently, the coupon collector with random initial stake is by half a round faster than the one starting with exactly $n/2$ coupons (apart from additive $o(1)$ terms). This result implies that classic simple heuristic called \emph{randomized local search} needs an expected number of $nH_{n/2} - 1/2 \pm o(1)$ iterations to find the optimum of any monotonic function defined on bit-strings of length $n$.