A treasure is placed in one of M boxes according to a known distribution and k searchers are searching for it in parallel during T rounds. How can one incentivize selfish players so that the probability that at least one player finds the treasure is maximized? We focus on congestion policies $C(\ell )$ specifying the reward a player receives being one of the ℓ players that (simultaneously) find the treasure first. We prove that the exclusive policy, in which $C(1)=1$ and $C(\ell )=0$ for $\ell >1$, yields a price of anarchy of ${(1-{(1-1/k)}^k)}^{-1}$, which is the best among all symmetric reward policies. We advocate the use of symmetric equilibria, and show that besides being fair, they are highly robust to crashes of players. Indeed, in many cases, if some small fraction of players crash, symmetric equilibria remain efficient in terms of their group performance while also serving as approximate equilibria.

根据已知分布，将一个宝藏放置在M个盒子中的一个中，k个搜索者正在T轮中并行搜索它。如何激励自私的玩家，以使至少一名玩家找到宝藏的概率最大化？我们专注于拥堵政策 $C（\ell ）$指定玩家（同时）首先找到宝藏的ℓ玩家之一获得的奖励。我们证明了排他性政策，其中$C（\mathrm{1个}）=\mathrm{1个}$ 和 $C（\ell ）=0$ 对于 $\ell >\mathrm{1个}$，产生无政府状态的价格${（\mathrm{1个}-{（\mathrm{1个}-\mathrm{1个}/ķ）}^{ķ}）}^{-\mathrm{1个}}$，这是所有对称奖励政策中最好的。我们提倡使用对称均衡，并表明除了公平之外，它们还对玩家的崩溃非常有力。确实，在许多情况下，如果一小部分参与者崩溃，对称均衡在团队表现方面仍然保持高效，同时还可以充当近似均衡。