average number of rolls of a die between appearance of a side I saw this might have been duplicated in places here -- I think this might be a variation on the coupon collector problem -- but I wanted to be sure and understand how to do the calculation. 
I have an n-sided die. I want to know what the average number of rolls between the appearance of a number on the die, k is. 
I thought that the binomial distribution would be appropriate here. The way I originally approached it was to say that we have a 1/n chance of getting a number. The chance of getting any other number is (n-1)/n. I know that if I wanted to know the odds of getting the same number several times in a row is $\left(\frac{1}{n}\right)^m$ with m being the number of rolls.  But beyond that I was a bit stumped. I know that there's a binomial distribution or a Harmonic number involved somehow, and I read the coupon collector's problem but honestly that explanation seemed to make things less clear rather than more. 
Anyhow, if someone could point me to either a duplicate question or a better explanation that would be much appreciated. 

Let n denote any face number other than k.  At the outset or after a $k$ has turned up, we roll the die until a $k$ reappears. The possibilities are:
$$k, nk, nnk, nnnk, ...$$
If $p$ is the probability of $k$ and $q = (1-p)$ is the probability of $n$, then the expected waiting time is
$$E(N) =  \sum_{j=1}^{\infty} jq^{j-1}p = p \frac{d}{dq} \sum_{j=0}^{\infty} q^j = \frac{p}{(1-q)^2} = \frac{1}{p}$$
