Talk:Regiment Battle/@comment-177.141.176.148-20160118051032/@comment-26486269-20160118125101

Just for curiosity I googled the 4chan message to see the formula. They used 1 - (197/200) ^ n, where n is the number of drops. I found no explanation of the formula, but I think I understood:

Dropping rate of Higekiri is 1.5%. It means that you have 0.015 of chance to get Higekiri in one attempt, or 0.985 to get any other sword as a drop (not considering if victory A rank has higher probability than B rank, just suppose the best scenario).

Assuming X and Y events are independent, the probability of X and Y occurring (it can be at same time or in sequence) is X.Y. Furthermore, the probability of getting any other sword other than Higekiri in n tries is (0.985)^n (for some reason they used (197/200)^n instead lol). That means, if you receive 300 drops, the probability of all of them are any other sword but not Higekiri is very low, like 0.01073665873. The complementary case, that is, you receive 300 drops and at least one is Higekiri, stands as 1 - (197/200)^300 = 0.98926334126. Almost 99%.

I've seen this kind of formula applied to dices and such and it works, but in terms of a game we don't know exactly how the drop process was developed it's just assumptions.