Consider the following game: Suppose the initial value of the pot is $ S $. Our player Josephine then rolls a fair $n$-sided die. If the roll is not $1$, then the pot is multiplied by that roll, and she rolls again; if it's $1$, then she receives whatever was in the pot, and the game is over. The game continues until Josephine sees a $1$. Am I right that the "expected value" of this game should be infinite?
My reasoning is about this: Payment only occurs when the sequence of rolls hits $1$ for the first time. So we can divide things up into the families $W_{k}$ sequences $w_{1} \ldots w_{k - 1} 1$, where $w_{j} \in \{2, 3, , \ldots, n \}$. Since the die is fair, these are all equiprobable. So \begin{align*} E & = \sum_{k = 1}^{\infty} \frac{\sum_{w \in W_{k}} w_{1} \cdots w_{k - 1}}{n^{k}} \\ & = \sum_{k = 1}^{\infty} \frac{\sum_{w_{1}, \ldots, w_{k - 1} = 2}^{n} w_{1} \cdots w_{k - 1}}{n^{k}} \\ & = \sum_{k = 1}^{\infty} \frac{(2 + \cdots + n)^{k}}{n^{k}} \\ & = \frac{1}{2} \sum_{k = 1}^{\infty} \frac{(n(n + 1) - 1)^{k}}{n^{k}} , \end{align*} which should diverge.
But at the same time, I have a sneaking suspicion I'm doing something wrong. Can anyone make sure?
Moreover, if I'm right, and the payout is infinite, then could an unfair die get it under control?
It's even easier than that. When $n = 2$ the expected payout of the game is $2^{-1}\sum_{k=0}^{\infty} 2^k / 2^k = \infty$, and this is a lower bound on the expected payout for general $n$.