Takács has shown that, in the M/G/1 queue, the probability P(k | i) that the maximum number of customers present simultaneously during a busy period that begins with i customers present is P(k | i) = Qk –i /Qk, where the Q's are easily calculated by recurrence in terms of an arbitrary Q 0 ≠ 0. We augment Takács's theorem by showing that P(k | i) = bk –i /bk, where bn is the mean busy period in the M/G/1 queue with finite waiting room of size n; that is, if we take Q 0 equal to the mean service time, then Qn =bn.
