A secretary problem with two decision makers
A secretary problem with two decision makers
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两个决策者的秘书问题
DOI：
10.2307/3215019
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发表时间：
1997
影响因子：
1
通讯作者：
L. Shepp
中科院分区：
文献类型：
--
作者：
Robert W. Chen;B. Rosenberg;L. Shepp
n个具有类似资格的申请人在面试名单上，他们的工资要求来自已知的连续分布。两个经理，I和II，将一次采访一个。每次面试后，经理I总是有第一个机会决定是否雇用申请人，除非他已经雇用了一个。如果经理I决定不雇用当前的申请人，那么经理II可以决定是否雇用该申请人，除非他已经雇用了一个。如果两个经理都没有雇用当前的申请人，他们会面试下一个申请人，但都失去了雇用当前申请人的机会。如果其中一位经理确实雇用了现任经理，那么他们将继续面试，直到另一位经理也雇用了一位申请人。面试过程一直持续到两个经理各自雇用一名申请人。然而，在流程结束时，每个经理必须雇用一名申请人。在本文中，我们首先推导出他们的最优策略，使他所雇用的人要求的工资比另一个所雇用的人少的概率最大化。然后，我们推导出一个算法，计算经理II的获胜概率时，两个经理发挥最佳。最后，我们证明了经理II的获胜概率在n中严格增加，总是小于c，并且当n →∞时收敛于c，其中c = 0.3275624139 · ··是方程ln（2）+ x ln（x）= x的解。
n applicants of similar qualification are on an interview list and their salary demands are from a known and continuous distribution. Two managers, I and II, will interview them one at a time. Right after each interview, manager I always has the first opportunity to decide to hire the applicant or not unless he has hired one already. If manager I decides not to hire the current applicant, then manager II can decide to hire the applicant or not unless he has hired one already. If both managers fail to hire the current applicant, they interview the next applicant, but both lose the chance of hiring the current applicant. If one of the managers does hire the current one, then they proceed with interviews until the other manager also hires an applicant. The interview process continues until both managers hire an applicant each. However, at the end of the process, each manager must have hired an applicant. In this paper, we first derive the optimal strategies for them so that the probability that the one he hired demands less salary than the one hired by the other does is maximized. Then we derive an algorithm for computing manager II's winning probability when both managers play optimally. Finally, we show that manager II's winning probability is strictly increasing in n, is always less than c, and converges to c as n →∞, where c = 0.3275624139 · ·· is a solution of the equation ln(2) + x ln(x) = x.