# Does this game have infinite expected payout?

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?

• Can I play your game? Dec 23, 2015 at 0:52

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 $n$.
• I think you can show that the expectation is finite if the probability to stop $\times$ the average multiplication of the pot is strictly lower than 1 Jul 4, 2015 at 17:26