The last of three servers

Idea. Split the waiting time into the gaps between successive finishes. Thanks to the memoryless property, each gap is a fresh minimum of exponentials, whose mean is easy.

Step 1: the first gap. With $3$ jobs running, the first finish happens at the minimum of $3$ independent $\text{Exp}(1)$ variables, which is $\text{Exp}(3)$, with mean $\frac{1}{3}$.

Step 2: the next gaps. At that moment, the other two jobs have been running for a while, but by memorylessness their remaining durations are again independent $\text{Exp}(1)$. So the next gap is $\text{Exp}(2)$, with mean $\frac{1}{2}$, and the last one is $\text{Exp}(1)$, with mean $1$.

Step 3: add. $$\mathbb{E}[\max ]=\frac{1}{3}+\frac{1}{2}+1=\frac{11}{6}\approx 1.83\text{ hours}.$$

In general. The maximum of $n$ independent $\text{Exp}(1)$ variables has mean $H_n=1+\frac{1}{2}+\cdots +\frac{1}{n}$. Waiting for the slowest of $n$ machines costs about $\ln n$ times the mean job length.