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Apologies, it has been a terrible while since I've last had to use probabilities, and I can not find the proper terminology for my question

I'll be using the marble bag analogy since it's the easiest for me. I want to maximize the probability of "A" and "C" of being selected.

Consider 3 lists or bags:
L1 = [A, B, B, B]
L2 = [C, C, D, D]
L3 = [E]

Consider 2 types of events:
Z1 is a pick from L1 and L3
Z2 is a pick from all 3 lists

With a limited amount X of event Z1 and limited amount Y of event Z2, in what order to use Z1 & Z2 to maximise the amount of A and C picked - and how to generalize this ? Will the order be different for different amounts of X & Y ?
How is this kind of problem named and where could I learn more? "multiple sets k-permutations" ?

Precision: The usecase here is without replacement: an element once selected is removed from their respective list. The event selection has equal probabilities for each element in its authorized lists, disregarding their list. Example: assuming Z1 is the very first event executed, it can with equal odds select an element in [A, B, B, B, E]. Assuming the very first Z1 randomly selects E, and we execute Z2 this time, Z2 will with equal odds select an element in [A, B, B, B, C, C, D, D].

Precision: What I'm precisely looking for is to understand the correct terminology here, and to maximize the selection of A and C elements (= successes). An ideal output would be: "For X=4 and Y=2, this combination maximize successes: [Z1, Z1, Z1, Z1, Z2, Z2] with in average P successes". And an understanding to generalize this for different X,Y, lists and events.

Trad
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  • Clarifications requested: [1]: "Z1 is a pick from L1 and L3": is this sampling done with or without replacement. For example, if the marble from L3 is selected, after event Z1, is the marble from L3 put back into the L3 bag? [2]: "With a limited amount X of event Z1 and limited amount Y of event Z2, in what order to use Z1 & Z2 to maximise the amount of A and C picked": Is an illustration of your intent that $X=4$ and $Y=2$, and you are asking which of the $(6)$ positions that event Z2 should be placed? ...see next comment – user2661923 Aug 06 '22 at 23:02
  • In my previous comment/question, if the answer to question [1] is sampling with replacement, and if I have my question [2] interpretation is what you intend, then the answer is that it is irrelevant what order the Z2 events occur, with respect to the Z1 events. The reason is that by having the sampling be with replacement, a Z2 event that (for example) occurs after a Z1 event is independent of the Z1 event. Alternatively, if, for question [1], you intend sampling without replacement, what happens if a bag is emptied? ...see next comment – user2661923 Aug 06 '22 at 23:06
  • Assuming that you intend sampling without replacement, suppose for example that you have two consecutive Z1 events, with the first event selecting E from L3. Does this automatically imply that the second Z1 event will be forced to select from L1, because L3 is now empty? Also, I just thought of a separate unresolved question: Suppose that the very first event is a Z1 event. Is the probability that L3:E is selected $~\displaystyle \frac{1}{2}~$ or is it $~\displaystyle \frac{1}{5}~?~$ ...see next comment – user2661923 Aug 06 '22 at 23:11
  • That is, do you select a bag at random, and then select one of the marbles from that bag at random, or do you instead (somehow) select one of the $(5)$ marbles at random? – user2661923 Aug 06 '22 at 23:12
  • You're right I should have precised! I will edit the post accordingly, but also to answer you: the question is without replacement. The selection is done amongst all available elements in authorized lists: if for example Z1 is the very first event to be used, it has 1/5 chance to pick E - "you instead (somehow) select one of the (5) marbles at random" is the wanted behaviour. Continuing the example, the first Z1 had [A, B, B, B, E] available and selected E. If we then execute another Z1, the possible choices are [A, B, B, B] – Trad Aug 07 '22 at 00:31
  • Regarding your first question, yes, my objective is to maximize A+C selection with the event execution order. Ideally, the result would be "For X=4 and Y=2, this combination maximize successes: [Z1, Z1, Z1, Z1, Z2, Z2] with probability P" – Trad Aug 07 '22 at 00:38

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