Calculating the Probability of Sustained Loss
 

Could anyone tell me how to calculate the probability of a winning player sustaining a certain loss or greater over a certain period? Here's the example I wanna work out:

Player wins 1.5 BBs an hour. His standard deviation is 12.7 BBs an hour. What is the probability that he will be down 120 BBs or more after 106 hours? What if he wins only 1.0 BBs an hour?

Re: Calculating the Probability of Sustained Loss
 

By the central limit theorem, the sum of your hourly results over n hours ~ N(n*1.5, n*(12.7^2). So if n = 106, your results are ~ N(159, 131^2). Thus P(down 120 BB or more after 106 hours) = N((-120-159)/131) = N(-2.12) = 1.7%.
If you earn 1BB, it's N((-120-106)/131) = N(-1.72) = 4.2%. Someone smart correct me if I'm wrong.

Re: Calculating the Probability of Sustained Loss
 

Looks about right.

Re: Calculating the Probability of Sustained Loss
 

[ QUOTE ]
Could anyone tell me how to calculate the probability of a winning player sustaining a certain loss or greater over a certain period? Here's the example I wanna work out:

Player wins 1.5 BBs an hour. His standard deviation is 12.7 BBs an hour. What is the probability that he will be down 120 BBs or more after 106 hours? What if he wins only 1.0 BBs an hour?

[/ QUOTE ]
Now someone tell him how many hands are required to show an accurate hourly rate.