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。