# Expected Duration of Game with Asymmetric Probabilities and Payoffs

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!

• Your formula does not account for the variable payoffs. If the payoffs are large enough the player is winning and the expected duration is infinite (though this doesn't happen in video poker). Is the bet size fixed so the player loses any time the bankroll is less than the bet size? Commented Jun 4, 2017 at 18:04
• Yes, the bet size is always 1 unit; thus if the bankroll is less than the bet size then the player is ruined. The example I am looking at has a top payoff of 100000 units with small probability.
– LMY
Commented Jun 4, 2017 at 18:18

• That is why I picked $1000$. I made a sheet with $n$ vertical and $1005$ lines. It takes some scrolling, but isn't so bad. Then I just made a column with $D(n)$ computed with values computed from the previous column. Copy right $20$ times and you have $20$ iterations. You have to make up some starting values, but they shouldn't matter. I used $\frac {45}{43}n$ and didn't update $0$ or above $1000$. Copy the last column for $n=1$ to $n=999$ and paste special values into the first and you get the next $20$ iterations. Commented Jun 4, 2017 at 18:47
• The expected duration for $n=1$ was up to $26$ after a few hundred iterations. If I used a programming language I would make large $n$ much larger. Commented Jun 4, 2017 at 18:48
• The first column just has numbers $0$ through $1005$, the bankrolls. The second column has a guess at the duration for each bankroll, starting with $45/2$ times the bankroll. The rows for $0$ and $1000+$ are just copied right. In D4 is written 1+31*C3/45+C6/5+C9/9 and all the middle has that equation copied right and down with relative addressing. Make as many columns as you want, you get one iteration per column. Bankroll of $6$ then starts at 135. After one iteration I have 142, then 177.1454 Commented Jun 4, 2017 at 22:49