Consider this probability excercise:
You roll a six-sided die. If you roll a six, you keep rolling until you roll a first non-six. Then you win $6^n$ dollars, where n is the amount of times you rolled a six before you failed.
It is clearly a geometric distribution. We have $n\sim Geo(\frac{5}{6})$. We have that $E(n)=\frac{6}{5}$. So $E(6^{n-1})=6^{\frac{1}{5}}$. Thus answer is $\sqrt[5]{6}$
However I saw this (clearly incorrect solution):
The probability that you lose in $n^{th}$ trial is $\frac{5}{6}\times \left(\frac{1}{6}\right)^{n-1}$ and if you lose on the $n^{th}$ trial you get $6^{n-1}$ dollars. Thus the expected winning should be $\sum_{n=1}^{\infty} \left(\frac{5}{6}\times \left(\frac{1}{6}\right)^{n-1}\times 6^{n-1}\right) = \sum_{n=1}^{\infty} \frac{5}{6}$, which diverges.
Now this obviously doesn't give the right answer, but I don't know where particularly this reasoning fails. I am not a statistician and what I suspect is wrong is that this formula (sum overvalues times the probability they occur) does not apply here, I do however wonder why.
Edit: I updated details fory solution. Turns out the other one was wrong, so the follow up question: what is wrong with the first solution?
Appreciate any help.
An expected value that does not converge, is the correct answer. This does happen.
$$\begin{align}n&\sim\mathcal{Geo}_0(5/6)\\ \Pr(n\,{=}\,k) &= (5/6)(1/6)^k~\mathbf 1_{k\in[[0..\infty)]}\\\mathsf E(6^n) &= \sum_{k=0}^\infty 6^k(5/6^{k+1})\\&=\sum_{k=0}^\infty\frac 56\end{align}$$