Problem 901

Well Drilling

A driller drills for water. At each iteration the driller chooses a depth $d$ (a positive real number), drills to this depth and then checks if water was found. If so, the process terminates. Otherwise, a new depth is chosen and a new drilling starts from the ground level in a new location nearby.

Drilling to depth $d$ takes exactly $d$ hours. The groundwater depth is constant in the relevant area and its distribution is known to be an exponential random variable with expected value of $1$ . In other words, the probability that the groundwater is deeper than $d$ is $e^{- d}$ .

Assuming an optimal strategy, find the minimal expected drilling time in hours required to find water. Give your answer rounded to $9$ places after the decimal point.

钻井

一名钻井工人在钻探水源。每一轮，钻井工人选择一个（正实数）深度 $d$ ，一直钻到这个深度，然后检查是否找到水。如果找到了，就结束钻探。否则，工人会在附近找一个新位置，选择一个新的深度，从地面开始新的钻探。

钻到深度 $d$ 需要恰好 $d$ 小时。在这片区域内，地下水的深度是恒定的，且满足期望值为1的指数分布。换言之，地下水深度超过 $d$ 的概率是 $e^{- d}$ 。

假设钻井工人采用最优策略，求找到水所需的最小预期钻探时间（以小时为单位），并将你的答案四舍五入保留 $9$ 位小数。