This is from exercise 25, chapter 5, in Ross' Introduction to Probability Models, 11th ed. It goes as follows:
Customers can be served by any of three servers, where the service times of server $i$ are exponentially distributed with rate $μ_i$, $i = 1,2,3$. Whenever a server becomes free, the customer who has been waiting the longest begins service with that server.
(a) If you arrive to find all three servers busy and no one waiting, find the expected time until you depart the system.
Here's my solution, which agrees with other solutions I've seen:
Let $T$ denote your time in the system. Want $E[T]$. This is equal to $E[\min\{X_1,X_2,X_3\} + S]$, where $X_i$ is service time of server $i$, and $S$ your service time. We get
$E[\min\{X_1,X_2,X_3\} + S] = E[\min\{X_1,X_2,X_3\}] + E[S] = \frac{1}{\mu_1 + \mu_2 + \mu_3} + E[S]$.
To find $E[S]$, condition on which service time is shortest:
\begin{align*} E[S] = E[E[S|X_i \ \text{smallest}]] = \sum_i E[S|X_i \ \text{smallest}] P(X_i \ \text{smallest}) = E[S|X_1 \ \text{smallest}]P(X_1 \ \text{smallest}) + E[S|X_2 \ \text{smallest}]P(X_2 \ \text{smallest}) + E[S|X_3 \ \text{smallest}]P(X_3 \ \text{smallest}) = \frac{1}{\mu_1} \frac{\mu_1}{\mu_1+\mu_2+\mu_3} + \frac{1}{\mu_2} \frac{\mu_2}{\mu_1+\mu_2+\mu_3} + \frac{1}{\mu_3} \frac{\mu_3}{\mu_1 + \mu_2 + \mu_3} = \frac{3}{\mu_1 + \mu_2 + \mu_3} \end{align*}
I was just wondering, why doesn't $E[T]$ simply work out to be just: $$2\cdot E[\min\{X_1,X_2,X_3\}] = \frac{2}{\mu_1 + \mu_2 + \mu_3}\text{ ?}$$ I reason that when you enter the system, expected time to wait for a free server will be $E[\min\{X_1,X_2,X_3\}]$, and once you're being served, expected time to leave will be $E[\min\{X_1,X_2,X_3\}]$, since you're being served by the same server -- the server with the shortest service time. This would imply that $E[S] = E[\min\{X_1,X_2,X_3\}]$, so that $E[T] = 2 E[\min\{X_1,X_2,X_3\}]$.