How would one calculate the expected duration of a game with asymmetric probabilities/payoffs (such as video poker), where total $p<q$ (probability of winning is less than probability of losing)? Assume the Player reinvests their winnings back into their bankroll ${b}$ each time.
We can find the probability of ruin by raising the below recursion to the power of $b$: $$\displaystyle{r}={\sum_{{{i}={0}}}^{n}}{{p_i}{r}^{u_i}}$$ where $p_i$ is the probability of prize $i$ times the recursion value to the power of earning $u$ units for prize $i$. However, I want to calculate the duration before ruin. So far I've come up with: $$\displaystyle\frac{b}{{{2}{q}-{1}}}$$ where ${b}$ represents the initial bankroll and ${q}$ represents the probability of losing any given turn.
But this seems far too generous for the asymmetrical probabilities and payoffs. With i.e. 12 different prize levels (0 to 100000) and 12 different probabilities, I don't believe the above accounts for the asymmetry. Example: let $b$ equal 100 units and $q$ equal $0.68$. Based on the formula above, the calculated duration would be approximately 278 turns. This is more than I expect. I understand this problem would likely need to be solved numerically, so please feel free to post examples.
Other Remarks:
The Player exits this game whenever $b < 1$ (must be able to bet 1 unit), or $b = 100000$ (wins top prize, which has very small probability of hitting). There are 12 levels of prizes/probabilities in all.
Thank You!
Compute the expected loss per round. The expected duration is the bankroll divided by the expected loss per round. This comes from the linearity of expectation.