Lazy employee finding a freed parking spot

An office has $10$ parking spaces, all taken by 9:00 AM. Three executives leave in the afternoon, each freeing their space at an independent uniform time between 12:00 and 3:00 PM. Nine lazy employees, Simon among them, each arrive at an independent uniform time in that same window.

When a lazy employee arrives, they take a free space if one exists and keep it until 5:00 PM; if none is free, they phone in sick and go back home. Find the chance that Simon ends up phoning in sick.

$\tfrac{1399}{1980}$. All $\binom{12}{3}=220$ interleavings of the three departures among the nine arrivals are equally likely. In each, an arrival parks if departures so far exceed earlier parkers; averaging gives $\tfrac{581}{220}$ parkers, so Simon parks with probability $\tfrac{581}{1980}$ and calls in sick otherwise.