Thx
This is pretty much how I figured the "$splt = 1" case
(although ∑a^k is a / (1 - a) not 1 / (1 - a) as k starts at 1 since $chance=0 causes the return of "zero" value !)
but how can it be implemented to solve for "$splt = 2" ?
Yeah I know "sum over value times odds" but it seems like that approach goes requires some very complicated combinatorial problem solving.
if(you see something I'm missing){please share ^^}

bry: while checking for "$splt=2", I discovered the value of "1" is returned when "$chance" is about 0.35793
for which I can't find any special properties
(maybe that exp(0.35793^0.35793) ~ 2, but not counting on it being the actual exact value ^^)

In reply to Re^2: Average random depth formula required by palkia
in thread Average random depth formula required by palkia

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