[irchans]

I did some math with a lot of estimates and I got the following approximate formula

Prob of Winning = (1-R^n)/(1-R^u)

where

n = starting bank roll
u = winning bankroll
R = ( sigma^2 - e ) / ( sigma^2 + e )
sigma = 1 hour standard deviation
e = earn per hour

Plugging in the numbers

R = ( 13.11^2 - 1.05 ) / ( 13.11^2 + 1.05 ) = 0.987856
Pwin = ( 1 - R^50 ) / ( 1 - R^300 ) = 0.469158

Somebody posted a similar formula with u = Infinity a few weeks ago. (BruceZ maybe.)