Thanks for the nice point, Thomas. Generalising, if the impact is 0 for productivity P_0, and P_av is the productivity of random employees, an employee N times as productive as random employees would be (N*P_av—P_0)/(P_av—P_0) as impactful as random employees. Assuming the cost of employing someone is proportional to their productivity, the cost-effectiveness as a fraction of that of random employees would be (P_av—P_0/N)/(P_av—P_0). So the cost-effectiveness of an infinitely productive employee as a fraction of that of random employees would be P_av/(P_av—P_0) = 1/(1 - P_0/P_av). For your parameters (and cost proportional to productivity), employing an infinitely productive employee would be 3 (= 1/(1 − 100⁄150)) times as cost-effective as employing random employees.
Thanks for the nice point, Thomas. Generalising, if the impact is 0 for productivity P_0, and P_av is the productivity of random employees, an employee N times as productive as random employees would be (N*P_av—P_0)/(P_av—P_0) as impactful as random employees. Assuming the cost of employing someone is proportional to their productivity, the cost-effectiveness as a fraction of that of random employees would be (P_av—P_0/N)/(P_av—P_0). So the cost-effectiveness of an infinitely productive employee as a fraction of that of random employees would be P_av/(P_av—P_0) = 1/(1 - P_0/P_av). For your parameters (and cost proportional to productivity), employing an infinitely productive employee would be 3 (= 1/(1 − 100⁄150)) times as cost-effective as employing random employees.