Constrained-Order Prophet Inequalities

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Authors Makis Arsenis, Odysseas Drosis, Robert Kleinberg arXiv ID 2010.09705 Category cs.DS: Data Structures & Algorithms Citations 15 Venue ACM-SIAM Symposium on Discrete Algorithms Last Checked 3 months ago
Abstract
Free order prophet inequalities bound the ratio between the expected value obtained by two parties each selecting a value from a set of independent random variables: a "prophet" who knows the value of each variable and may select the maximum one, and a "gambler" who is free to choose the order in which to observe the values but must select one of them immediately after observing it, without knowing what values will be sampled for the unobserved variables. It is known that the gambler can always ensure an expected payoff at least $0.669\dots$ times as great as that of the prophet. In fact, there exists a threshold stopping rule which guarantees a gambler-to-prophet ratio of at least $1-\frac1e=0.632\dots$. In contrast, if the gambler must observe the values in a predetermined order, the tight bound for the gambler-to-prophet ratio is $1/2$. In this work we investigate a model that interpolates between these two extremes. We assume there is a predefined set of permutations, and the gambler is free to choose the order of observation to be any one of these predefined permutations. Surprisingly, we show that even when only two orderings are allowed---namely, the forward and reverse orderings---the gambler-to-prophet ratio improves to $\varphi^{-1}=0.618\dots$, the inverse of the golden ratio. As the number of allowed permutations grows beyond 2, a striking "double plateau" phenomenon emerges: after increasing from $0.5$ to $\varphi^{-1}$, the gambler-to-prophet ratio achievable by threshold stopping rules does not exceed $\varphi^{-1}+o(1)$ until the number of allowed permutations grows to $O(\log n)$. The ratio reaches $1-\frac1e-\varepsilon$ for a suitably chosen set of $O(\text{poly}(\varepsilon^{-1})\cdot\log n)$ permutations and does not exceed $1-\frac1e$ even when the full set of $n!$ permutations is allowed.
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