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How many rolls of a fair six-sided die must one make, on average, until a 6 has been rolled precisely 6 times?

I worked out that on average number of rolls to roll a single 6 is 6, from the geometric formula of expected value.

However I am stumped on how to go about answering this question for 6 being rolled precisely 6 times.

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Let $X$ be the number of times it takes for a $6$ to be rolled 6 times. Then $X=W_1+\dotsb+W_{6}$ where the $W_i$ are geometric i.i.d random variables with $EW_1=6$. You can think of $W_{i}$ as the waiting time to see the next 6. Now use linearity of expectation.

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  • $\begingroup$ I got 46656, is this the correct answer $\endgroup$ – Maths student Nov 26 '19 at 23:42
  • $\begingroup$ It is $EX=\sum EW_i=6EW_1=6\times6=36$ $\endgroup$ – Sri-Amirthan Theivendran Nov 27 '19 at 1:16
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HINT

let $x_n$ denote the average number of times needed to roll a 6 exactly $n$ times. So you found that $x_1 = 6$. Can you write down a recurrence relationship between $x_{n+1}$ and $x_n$? Think about incorporating 2 outcomes after you rolled $n$ times -- either you rolled correctly or you didn't...

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