TITLE: On average, how many times must I roll a dice until I get a $6$?
QUESTION [42 upvotes]: On average, how many times must I roll a dice until I get a $6$?
I got this question from a book called Fifty Challenging Problems in Probability. 
The answer is $6$, and I understand the solution the book has given me. However, I want to know why the following logic does not work: The chance that we do not get a $6$ is $5/6$. In order to find the number of dice rolls needed, I want the probability of there being a $6$ in $n$ rolls being $1/2$ in order to find the average. So I solve the equation $(5/6)^n=1/2$, which gives me $n=3.8$-ish. That number makes sense to me intuitively, where the number $6$ does not make sense intuitively. I feel like on average, I would need to roll about $3$-$4$ times to get a $6$. Sometimes, I will have to roll less than $3$-$4$ times, and sometimes I will have to roll more than $3$-$4$ times.
Please note that I am not asking how to solve this question, but what is wrong with my logic above.
Thank you!

REPLY [8 votes]: The probability of something happening in n rolls might be 1/2, and that number might be say '10' - what if there was then a situation where the probability of the same event happening between 1000 and 2000 times was 1/2 - so 1-10 is P=1/2 1000-2000 is 1/2
The above could all make sense, but you can see that the average is never going to be 10 only.
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After your first roll, you either get a 6 and finish in 1 (probability 1/6), or you get a non-six and are back in the same position you were in at the start, with an expectation of a further E rolls needed (plus the one you made) - probability (5/6)
E = 1/6 + 5/6(E + 1)
(1/6)E = 1
E = 6