Risk of Ruin for normally-distributed games Classic ruin theory assumes that the income is constant and that only the losses are random with an underlying distribution.
Suppose we want to determine the risk of ruin for a game (for example poker), where also the winnings are random.
Let $\psi(u)$ denote the risk of ruin, starting from initial surplus $u$. We assume that the winnings/losses in each game follow a normal distribution with mean $\mu>0$. The game is played until the player is broke or his surplus goes to infinity.
Classical ruin theory doesn't seem to apply because of the non-constant income. I am grateful for any advice on how to approach this problem.
 A: According to risk of ruin wikipedia page,
The formula for risk of ruin for such setting can be approximated by
$$\left( \frac{2}{1+\frac{\mu}{r}}-1 \right)^{u/r}=\left(\frac{1-\frac{\mu}{r}}{1+\frac{\mu}{r}} \right)^{u/r}$$
where $$r=\sqrt{\mu^2+\sigma^2}$$
It is described that the approximation formula is obtained by using binomial distribution and law of large numbers.
I have written the formula in the form of proposed by Perry Kaufman
$$\left( \frac{1-\text{edge}}{1+\text{edge}}\right)^{\text{capital units}}$$
A: I've  run a simulation in Julia programming language, using Normal(1,9) (mean = 1, SD = 3) starting from 10. 
The formula in previous answer gives 
 r = sqrt(mean ^2 + SD ^2)  = 3.162
 p = ( 2/(1+mean/r) - 1 )^(S/r) ~12.06%

I've run 10.000 simulations with 50.000 rounds each. 
Bankruptcy in ~ 7.4% of simulations with ~ 5000 average rounds  and standard deviation ~ 2900 rounds.
It's very different from theoric value in the Wikipedia entry and I cannot find any reference. 
This formula don't make sense for me. 
it's ok for SD=0 and mean>0 because 
 r = mean

so
 (2/(1+mean/r) - 1 )^(S/r)
 (2/2- 1)^(S/r) = 0

However if SD=0 and mean<0, the formula reduces to
r = abs(mean)

and 2 denominator turns
1+mean /abs(mean) ~ 0

However, it's clear that bankruptcy probability would be 100% in this case
Even if SD <>0  there may be  some problems. 
Let's calculate the condition for probability  = 100%.
In this situation, base = 1
2 / (1+ mean/r) - 1 = 1
1 =  (1+m/r)
r = r + m
mean = 0

So when mean = 0 the default probability is 100%. It's not true.
