A Stopping Game on Zero-Sum Sequences

TL;DR

This paper introduces a stopping game involving permutations of zero-sum sequences, analyzes three algorithms for the binary case, and demonstrates that the simple algorithm achieves a worst-case optimal payoff proportional to the square root of sequence length.

Contribution

It presents three online algorithms for a zero-sum permutation game, analyzing their expected payoffs and establishing the optimality of the simplest algorithm in the general case.

Findings

Expected payoff of algorithms is Θ(√n) for binary sequences.

Algorithm 3 is simple and achieves worst-case optimal payoff.

The analysis extends to arbitrary zero-sum multisets, confirming Algorithm 3's optimality.

Abstract

We introduce and analyze a natural game formulated as follows. In this one-person game, the player is given a random permutation $A=(a_1,\dots, a_n)$ of a multiset $M$ of $n$ reals that sum up to $0$, where each of the $n!$ permutation sequences is equally likely. The player only knows the value of $n$ beforehand. The elements of the sequence are revealed one by one and the player can stop the game at any time. Once the process stops, say, after the $i$th element is revealed, the player collects the amount $\sum_{j=i+1}^{n} a_j$ as his/her payoff and the game is over (the payoff corresponds to the unrevealed part of the sequence). Three online algorithms are given for maximizing the expected payoff in the binary case when $M$ contains only $1$'s and $-1$'s. $\texttt{Algorithm 1}$ is slightly suboptimal, but is easier to analyze. Moreover, it can also be used when $n$ is only known with some approximation. $\texttt{Algorithm 2}$ is exactly optimal but not so easy to analyze on its own. $\texttt{Algorithm 3}$ is the simplest of all three. It turns out that the expected payoffs of the player are $\Theta(\sqrt{n})$ for all three algorithms. In the end, we address the general problem and deal with an arbitrary zero-sum multiset, for which we show that our $\texttt{Algorithm 3}$ returns a payoff proportional to $\sqrt{n}$, which is worst case-optimal.