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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.

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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

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

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  • $\begingroup$ 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? $\endgroup$ – Babypopo May 30 '17 at 14:24
  • $\begingroup$ 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. $\endgroup$ – Siong Thye Goh May 30 '17 at 15:27
  • $\begingroup$ In case you find something regarding the proof, please let me know! Thanks a lot anyways! $\endgroup$ – Babypopo May 31 '17 at 10:55
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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.

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