assume_R said:
Is there an equation in which you can calculate the probability of being ahead after X hands?
I assume that N0 is actually X for a certain probability (something high like 95% I'd guess).
It is a simple statistic problem. If you know the standard deviation and win rate per hand...
mean = X * win rate
total standard deviation = SQRT(X) * standard deviation
Now, the probability of being ahead would be the probability of being at 0 or greater. To find this, you need to know how many standard deviations from mean this would be (z-score). This would be...
z-score = (0-Mean)/total standard deviation
Then, using a statistical calculator or the internet, find the cumulative probability of being at that z-score or higher.
Rules of thumb:
z-score of -1 would have a probability of ~84%
z-score of -2 would have a probability of ~97.5%
z-score of -3 would have a probability of ~99.85%
Hopefully I didnt screw something up.
PS: Since N0 is defined as when the mean = SD, this is equal to a z-score of -1 aka 84% chance of being ahead after N0 rounds.