Originally Posted by

**salohcin** Yes that is very helpful, I am still stuck on where to go next though,

Then, if the payoff were constant, the total win would be $\displaystyle Ny$ because eventually the player wins $\displaystyle N$ times before the game ends.

Taking the geometric decrease of the payoff into account, the total win is $\displaystyle S=(c^{T_1}+\cdots +c^{T_N})y$ where $\displaystyle T_1,\ldots,T_N$ are the winning times. Note that $\displaystyle T_{k+1}-T_k$ is a geometric random variable of parameter $\displaystyle 1-\frac{k}{N}$ (after k wins, the next win has probability $\displaystyle 1-\frac{k}{N}$ to happen), and these random variables are independent.

If $\displaystyle T$ is a geometric r.v. with parameter $\displaystyle p$, then one computes $\displaystyle E[c^T]=\frac{cp}{1-c(1-p)}$. Using this, we get (up to possible mistakes)

$\displaystyle E[S]=y \Big(\frac{c(1-\frac{1}{N})}{1-\frac{c}{N}}+\frac{c(1-\frac{1}{N})}{1-\frac{c}{N}}\frac{c(1-\frac{2}{N})}{1-\frac{2c}{N}}+\cdots$ $\displaystyle +\frac{c(1-\frac{1}{N})}{1-\frac{c}{N}}\cdots \frac{c(1-\frac{N-1}{N})}{1-\frac{(N-1)c}{N}}\Big)$.

This is not very nice, but I'm not sure this can be turned to something nicer...