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I read On average, how many times must I roll a dice until I get a $6$? but the answers all used geometric series, which I know how to do. I was wondering if there was a more "elegant" solution that uses linearity of expectation?

Will Kanga
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    I find the question confusing. One solution: the expectation $x$ is the solution of the equation $x=1+\frac56x$. The other solution uses linearity of expectation and leads to a geometric series. If $X$ is the number of rolls then $X=\sum_{n=1}^\infty I_n$ where $I_n=1$ if $X\ge n$ and $I_n=0$ otherwise, and $$E(X)=\sum_{n=1}^\infty E(I_n)=\sum_{n=1}^\infty\left(\frac56\right)^{n-1}.$$ Which one do you consider the "elegant" solution? – bof Jul 16 '21 at 23:48
  • @DavidC.Ullrich I haven't formally studied probability, so it is unclear to me whether the following argument is valid. If you roll the die $(6000)$ times, you can expect to get $(1000)$ of those rolls to be a $(6)$. Therefore, the average number of rolls per the occurrence of a $(6)$ is $\frac{6000}{1000} = 6.$ – user2661923 Jul 16 '21 at 23:52
  • @user2661923 - If you flip a fair coin $1001$ times you expect a quarter of the $1000$ consecutive pairs to be HH. But if you flip a coin until you get HH as a consecutive pair, the expected number of flips is $6$. So your argument would need a little more justification – Henry Jul 17 '21 at 00:00
  • Re my previous comment, upon reflection, I think that the argument is incomplete. The question of the average number of rolls per the occurrence of a $(6)$ is different from the average number of rolls per the first occurrence of a $(6)$. The issue is therefore whether the argument can (somehow) be validly taken to completion. – user2661923 Jul 17 '21 at 00:00
  • @Henry Yes, I agree - re see my last comment; it just occurred to me. Unclear if my approach can be salvaged. – user2661923 Jul 17 '21 at 00:02
  • @user2661923 possibly using the memorylessness property of the geometric distribution – Henry Jul 17 '21 at 00:05

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