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On average, how many times must a 6-sided die be rolled until a 6 turns up twice in a row?

This a question from "a collection of dice problem", the original explanation I couldn't understant. Can someone explain it from the expectation (sum of an infinite series) point of view?

Stephen Ge
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  • The most straightforward way, I think, is to condition on the result of the first toss and (if it a $6$) conditioning on the result of the second toss. Is this the type of approach you are referring to, that you don't understand? – André Nicolas Aug 08 '16 at 20:02
  • The solution depends on what you were first taught and are most comfortable with. You can describe this as a three-state absorbing markov chain with states (Second6inarow),(First6inarow),(Not6) and corresponding transition matrix $\begin{bmatrix}1& \frac{1}{6}&0 \0 &0&\frac{1}{6} \0& \frac{5}{6}& \frac{5}{6}\end{bmatrix}$. The fundamental matrix will tell you information about the expected time it takes to roll the die to reach the absorbing state. This is the way I personally think. – JMoravitz Aug 08 '16 at 20:04
  • Have you read about any of the probability distributions? Expected values? – RonaldB Aug 08 '16 at 20:09
  • Hi RonaldB, Could you explain it from Expected value point of view (sum of infinity series) – Stephen Ge Aug 10 '16 at 03:02

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