Gooble Gooble , fuck the Colts . <| sep |> It ' s more the way you responded to a genuine attempt at help with sarcasm . <| sep |> i think it has the highest breaking rate <| sep |> Just completed mine . Took my son out for a 3 . 33 mile walk . <| sep |> Perhaps I ' m misunderstanding the problem , but it looks like it will depend on the distribution of the table unavailability time ? If the random variable of the table unavailability time is constant , then this problem reduces to the classic problem of finding expectation of the min of 10 U 0 , 75 s , which evaluates to 7511 . But if the tables are taken for 1 minute 99 % of the time , and 7401 minutes for the other 1 %, then the average unavailability time is also 75 minutes , but the expected waiting time is clearly less than 1 minute .