LovinItAll said:
N0 = ~1 SD, right? Does that mean 2 SD's = N0*2?
nope.
You need to calculate as following:
N=N0 is the number of hands N, for which EV(N) = SD(N) after those N hands.
Since EV(N) is proportional to N, and SD(N) is proportional to sqrt(N),
if you ask for the number of hands (lets call it N2) where you are ahead at least by 2*SD, one will immediatly find:
EV(N2) = 2*SD(N2)
N2/N0 EV(N0) = 2* sqrt(N2/N0) SD(N0)
and hence
N2 = 4 * N0.
The interpretation of N0 is rather simple: it is the "scale" at which your winnings should show up.
After 1*N0 hands you are down with probability of 15.8%.
After 4*N0 hands you are down with probability of 2.2%.
After 9*N0 hands you are down with probability of 0.1%.
Depending on the level of certainty, if you want to be 99.9% sure that you are playing a winning game, you should play at least 10*N0 hands before evaluating your results.