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I throw a dice until sum of results is divided by 6. What's an expectation value of number of throws?

I thought that it will be a sum: $\sum_{k=1}^{\infty} k \cdot \left(\frac{6^k - 6^{k-1} - 6^{k-2}\cdot5 - ... -6\cdot5^{k-2}}{6^k}\right)$ but the sum in the brackets is bigger than 1 for large $k$ so I had to made a mistake in my way of thinking.

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HINT: The probability of the sum being $k$ modulo $6$ after any throw is $\frac 16$, so in particular you have $\frac 16$ probability that the sum will be divisible by $6$ after the $k$-th throw.

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  • $\begingroup$ so is it a sum: $\sum_{k=1}^\infty k* (\frac{5}{6})^{k-1} \frac{1}{6}$ ? $\endgroup$ – Filip Parker Dec 1 '17 at 22:13
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    $\begingroup$ @FilipParker Your writing is a bit messy, but I guess you meant: $$\frac 16 \sum_{k=1}^{\infty} k\left(\frac 56 \right)^{k-1}$$ If yes, then it's true. Now use the geometric progression formula to find the actual value. $\endgroup$ – Stefan4024 Dec 1 '17 at 22:18

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