I do admit I'm wrong sometimes, if I didn't I wouldn't post HH if I thought I did everything perfectly. On this thread though I'm right and that is why I maintain my position.
OP will be a losing player originally since he appears to be a new player. If you drew a graph of OPs poker with amount of time spent on poker on the X-axis and BR on Y-axis. it would form a basic X^2 graph (obv shifted across to where x = 0, y = 20. obviouslt at latter ends on the right hand side curve would become a straight line as OP maximises productivity. **** it, I'll draw a basic one of what I mean in paint so I can illustrate things better. It's poorly drawn but I only need it to illustrate my point. This is rpesuming variance falls within nomral values and nothing too freaky happens, obv doesn't encapsulate all circumstances but gets my point across sufficiently.
X-Axis = poker time
Y-axis = BR
I drew X^2 graph on with dashed lines to show my point that I made above
So if Player does not run -(A) above expectation he will bust.
However if player recieves BR boost in a tourney that takes him above20 -(A) then assuming natural variance without random downsing then he becomes winning player before hitting the bust line.
We can express OPs graph as the function f(h)
IMO
(f(h)<0 ) < (0.15*(f(h)-b)
I can express it as a mathematical function but not with numbers as I don't have any true ones. f(h)<0 = amount of times (expressed as frequency) f(h) dips below x-axis.
Yes the standard deviations probably do overlap of the two so it's hard to say anything significant is occuring as you'll never have a big enough sample size to crunch and find true numbers. but to say it's plain dumb because you believe one sign sohuld be the opposite way round is ridiculous and taking the piss to be quite frank.