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We roll a dice. If we get 6, then we roll again. What's the expected value of sum of rolls?

Let $X$ be the sum. In this game we can only obtain sum in form $6k + x$ where $k \in \mathbb{Z}$ and $x \in \{1,2,3,4,5\}$. If $n$ is in this form then probability of obtaining it is equal $(\frac{1}{6})^{[\frac{n}{6}]}\cdot\frac{1}{6}$. To get the expected value I would to sum only these $n$ which are in this form but I don't know how to properly do it. Or maybe my idea is wrong?

Nerwena
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    The expected number of $6$'s rolled in this game is $\frac{1}{5}$. The expected number of the final non-6 roll is $3$. This gives an overall expected value of $6\times \frac{1}{5}+3=4.2$ – JMoravitz Feb 03 '21 at 17:37
  • @JMoravitz I think that should be an answer rather than a comment. – saulspatz Feb 03 '21 at 17:40
  • @JMoravitz could you give more detailed answer? I don't understand where some of the values came from. I'd appreciate it. – Nerwena Feb 03 '21 at 17:42

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Here's a trick: let $E$ be the expected value. When you roll a $6$, you win $6+E$ points. Therefore $$E = \frac16\left(1+2+3+4+5+(6+E)\right)$$ Now solve for $E$.

Théophile
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