If you're watching a race with 20 runners and the time it takes each of the runners to finish is an independent exponential random variable with mean 60 minutes. What is the expected time that you leave, if you leave after 3 runners finish?
Since all runners start at the same time wouldn't the answer technically be 60? I feel like I'm missing something here?
You are missing something, which a simple example brings out. Assume every runner takes $1,2,$ or $3$ minutes to finish the race, with each outcome being equally likely. If there are $99$ runners, then the expected number of runners who finish in $1$ minute is $33>3$ so you the expected time you leave at should be $1$ minute, not $2$ minutes.
The key here is that although the expected value of the mean is certainly $60$, you're not interested in the expected value of the mean. You're interested in the expected value of the third smallest value. In the above example, I exploited the discreteness of the output to simplify the calculation - a similar argument for an exponential distribution doesn't work. To get started on the actual problem, I would start off with finding the expected value of the smallest value, then try to figure out the expected value of the second smallest value, and finally figure out the expected value of the third smallest value.