The entropy of lottery drawing machines Suppose we have a fair lottery drawing machine where you have a container of numbered balls that is rotated many times such that interaction of the balls with themselves and the container produces a random distribution of the balls. We can say that this distribution of balls has a certain entropy, call it $E$. Now we have some kind (the details are unimportant for the question) of extracting mechanism that extracts balls from the container, there is an entropy associated with this extractor , call it $E-prime$. Would it be correct to say that the resulting lottery picks are the result of the superposition of entropies $E$ and $E-prime$ or the product of both entropies?
 A: 
Would it be correct to say that the resulting lottery picks are the result of the superposition of entropies E and E-prime or the product of both entropies?

TLDR: the entropy of the total system is $E+E’$
Full answer: the entropy of a system is related to the number of micro states that a system could be in for a given macro state. Now, it is a little odd to discuss macroscopic machines in terms of macrostates as though we cannot observe the relevant details of the machine. But nevertheless we can discuss macro and micro states in general. 
So for any system the entropy is related to the number of microstates by $E=k \ln(\Omega)$.  Assuming that the two systems are independent and that each microstate is equally likely then the number of possible microstates in the combined system is the product of the number of microstates in the two sub systems. I.e. for every microstate of one subsystem the other could be in any of its possible microstates. 
Then, the entropy of the combined system can thus be directly calculated as $k \ln(\Omega \; \Omega’) = k \ln(\Omega) + k \ln(\Omega’) = E + E’$
