more shiny math
In a surprising result, if the chance of getting a shiny is 1/x, then the average number of attempts required to see one shiny is precisely x. This is not an approximation, but an exact calculation.
more shiny math
@wallhackio kind of yes, kind of no.
yes the average (expected value) number of shinies after x attempts is, but that's not the same as your number
your chances of actually seeing at least one after x attempts is either approximately 1/e or 1 - 1/e (i can't remember whether @aescling and i's calculations from april were for chance of not seeing a shiny or chance of seeing at least one shiny)
(1/e is the limit as x goes to infinity of.... whichever probability we were looking at, with the value for a given x monotonically decreasing as x increases)
re: more shiny math
@alyssa @wallhackio this is why he is saying the expected value calculation was surpurrising; we had already gone over the 1-1/e (~63%) calculation (i think twin posted about it a couple days ago?)
