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@ryogokusama I will try to explain this to you one more time, because it’s not a case of looking for a single item. It’s looking for a class of items. In that case the frequency at which you draw a set item isn’t important since you would assume that you have equal chance of drawing any item on any spin. So over a number of spins you get a certain number of set items. The number of those set items, when sorted into classes, should match the frequency at which each class is present in the total number of set items. So, for *any* of the items for the 3 new sets, there are six in total. There are 5 old sets, so there are ten old set items. So the probability of drawing *any* item from the old set is 10 out of 16. The probability of drawing *any* of the new set items is 6 out of 16. If you draw 50 set items, and *none* of them are from any of the new sets, the likelhood of that happening is (10/16) raised to the 50th power. This is a very small number. It would be roughly equivalent to flipping heads (or tails) 50 times in a row. That is just not going to happen unless you are using a loaded coin.
So the situation is what I described, I’ve drawn 50 set items, none of which are for *any* of the new sets. *If* I had drawn say 10 new set items and 40 old set items, then what you say would be true since there is a relatively large probability of that happening (I think it’s like 0.3%). Even if there had been only five new set items I could accept that as happening by chance (0.001% chance). But 0 out of 50 is *so* improbable that it almost certainly represents a broken game.
If you don’t understand that, or how to calculate binomial probabilities, and what they represent, then you shouldn’t be commenting on this subject. Either way, I’m finished discussing this with you since you don’t have anything useful to contribute.