# 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.

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

$$\left( \frac{1-\text{edge}}{1+\text{edge}}\right)^{\text{capital units}}$$

• Thanks a lot, this seems to be exactly what I need. Do you have a reference or source, where the derivation of the formula is explained? – Babypopo May 30 '17 at 14:24
• Unfortunately, I don't. I just found the question interesting so I search around to see what has been done. I will be interested to look at the proof too. It is like a random walk with gaussian increment. – Siong Thye Goh May 30 '17 at 15:27
• In case you find something regarding the proof, please let me know! Thanks a lot anyways! – Babypopo May 31 '17 at 10:55

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.