Not sure what you are looking for , but you may want to try my version of the CP efficiency calculator original by uvlfph . <| sep |> Thanks for all the comments , the person who asked me the question used integrals to solve it and claimed it can be solved using basic combinatorics but has indeed reached the same result you have mentioned of 7510 . I ' ve reached a different answer for some reason , and I ' d really appreciate an explanation on why it ' s wrong , he mentioned something about the median and mean and needing to prove it ' s symmetrical or something along those lines , but I didn ' t really get it 7475 The probability of a table remaining unavailable during a minute 747510 The probability of 10 tables remaining unavailable during a minute 747510 k The probability of 10 tables remaining unavailable during k minutes 747510 x 0 . 5 x is the number of minutes required for the probability of 10 tables remaining unavailable to be 50 %, which means it ' s also the number of minutes required for the probability of at least one table becoming available . x 5 tel 986076415 If there ' s a 50 % chance of at least one table becoming available , and for any longer timespan it ' ll be a higher rate , and for any shorter timespan it would be a lower rate , does that not mean that the average waiting time would be X ?