Recent content by eViLtHeCaT

"That's a worst case scenario. If it were the best case scenario, there would be possibility of improvement. You are assuming you only catch single flawless Ditto. The result will actually be lower because if you catch a double flawless Ditto, you don't need to catch 5 other single flawless, you...

why do you multiply your 1/2^x * x with another x? i understand the 1/2^x bit, but not the * x ... edit: If i use your "2" in my equation i get an accuracy of 75%. Is that the mathematical definition of average? there's something missing here...

I did review your code but the leap from my basic C knowledge to JAVA was a bit too much in 2 lines of code. Also i'm not using the natures in the equation to keep it simple. ok, my train of thought is this: you have, lets say, a 50% chance of hitting heads or tails, right? how many times do...

you've lost me here already. if i want to hit a number other than the one I already hit then 5/6 = 0.83% of doing so. not hitting = 0.17%, chance of this happening at 1% (therefore 99% that the 0.83% will occur) is log0.01/log0.17 = 2.51 -> 3 rolls. 5/6 gives 1.2. I don't mean to be rude but...

I can't quite tell who your response is directed at. the 17.3% is a best case scenario because all it requires it that at least ONE of the IVs is maxed. anything above 1 is a bonus. the way you want to calculate it simply breaks down the 17.3% into its constituents. the logic applied above...

first post! Hi there. Now correct me if i'm wrong but your math is incorrect. I'm referring to the 170 dittos required for 6 dittos, each one with a differing maxed IV. the chance of finding a ditto with any given iv of 31 is, as you say, INITIALLY 17.3%. to have a 99% chance of finding that...