On average, how many times must I roll a dice until I get a $6$? 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!

The probability of the time of first success is given by the Geometric distribution.
The distribution formula is:
$$P(X=k) = pq^{n-1}$$
where $q=1-p$.
It's very simple to explain this formula. Let's assume that we consider as a success getting a 6 rolling a dice. Then the probability of getting a success at the first try is
$$P(X=1) = p = pq^0= \frac{1}{6}$$
To get a success at the second try, we have to fail once and then get our 6:
$$P(X=2)=qp=pq^1=\frac{1}{6}\frac{5}{6}$$
and so on.
The expected value of this distribution answers this question: how many tries do I have to wait before getting my first success, as an average? The expected value for the Geometric distribution is:
$$E(X)=\displaystyle\sum^\infty_{n=1}npq^{n-1}=\frac{1}{p}$$
or, in our example, $6$.
Edit: We are assuming multiple independent tries with the same probability, obviously.
