Talk:Uberfest (Gacha Event)/@comment-34112061-20180131203050

Assuming each trial has a 9% chance and is independent of the trial previous, you have 64.56% chance of getting at least one uber in an11 draw

Let P(A) be the probability of getting 1 uber in one draw

P(A) = 0.09 P(!A) = 1 - P(A) = 0.91, P(x =11 draws, at least one uber) = 1 - P(no ubers) =

1 - (0.91)^11 = 0.6456....

Expected number of ubers per 11 draw = np = 11 * 0.09 = 0.99 ~1

So in the long run (say if you have a 100,000 cat food), you should expect 1 uber per 11 draw

Theoretically if your luck is somehow really, really bad, to have a 95% chance of getting at least one uber would require 32 draws (but you should expect to get 2.88 ubers) and to have a 99% chance of getting at least one uber would rquire 49 draws (but you should expect 4.41)