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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?

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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 $n$.

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  • $\begingroup$ Thanks. What does it look like if the die is made to prefer lower numbers? $\endgroup$
    – AJY
    Jul 4, 2015 at 16:54
  • $\begingroup$ In what way exactly? As long as the probability of not rolling a 1 is at least 1/2 the above argument would still work, but it's possible you could force the game to have finite expectation. $\endgroup$
    – dsaxton
    Jul 4, 2015 at 17:17
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    $\begingroup$ 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 $\endgroup$
    – Tryss
    Jul 4, 2015 at 17:26
  • $\begingroup$ I like 1 sided die. And 2,3 sided die for that matter. ;D $\endgroup$ Dec 23, 2015 at 0:54

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