Avg Value of Dependent Events If I have 26 bins and on a given "turn" each bin can take on one of many values, or no value at all (null) with probability that varies by bin. Let's call the average of the values that can occur A-Z, and the probability that a non-null value occurs for a given bin a-z. You start with 5 turns, and in the event none of the bins get a value on a given turn, you lose a turn. If you do get one or more values on a given turn, the values stay there until you are out of turns, the number of turns is reset to 5.
Is there a simplish way to calculate the average value for this series of events using some clever combinatorics? Even though each bin outcome is independent I called the events dependent, because the number of turns depends on getting or not getting a value in the other bins. Please let me know if I can clarify anything, I know it sounds strange. 
 A: For $i\in[1,n]$ ($n=26$) you have geometrically distributed variables $G_i\sim Geom(p_i)$ that describe the time of the first fill of that bucket. Let $G_{(i)}$ be their order statistics with $G_{(0)}=0$ and $G_{(n+1)}=\infty$. Then the process ends at time $T=\min\{s:\exists i\in[0,n]:G_{(i)}+k\le s<G_{(i+1)}\}$ and you only need $E(T)$.
I don't know how to solve this version, but you can Poissonize and instead imagine a non-homogeneous Poisson process of "new fill in bucket $i$ "with $\lambda_i(t)=\lambda_ie^{-\lambda_i t}$ for $\lambda_i=-\log(1-p_i)$ for a total rate of "new fills" of $g(t)=\sum_i^n\lambda_i(t)$. Then for $t>k$, $$-\bar F'_{T}(t)=\bar F_T(t-k)g(t-k)\exp(-\int_{t-k}^{t}g(x)\,dx)$$ and $\bar F_T(t)=1$ for $t\le k$ which you could integrate in the piece-wise fashion. Adding $\frac 1 2$ (or $\log 2$) to $T$ would address discretization issues but I have no idea how good of approximation this would be (not that the delayed equation is that tractable analytically either).
