Trials until First Success

Let $p$ be the probability of a 6 on a given trial. Then the probabilities of success for the first time on each trial are (let $q=1-p$ ):

The sum of probabilities is $p+pq+pq^2+...=1$ .

The mean numbers of trials, $m$ , is by definition,

$m=p+2pq+3pq^2+4pq^3+...$ .

Multiplying by $q$ gives

$qm=pq+2pq^2+3pq^3+...$ .

Subtracting the second expression from the first gives

$m-qm=p+pq+pq^2+pq^3+...$ ,

$m(1-q)=1$

Consequently,

$mp=1$ , and $m=1/p$ .

Here, $p=\frac{1}{6}$ , and so $m=6$ .

There are $6$ possibilities when rolling the die. $\frac{1}{6}$ is the chance to roll a $6$ . On average, if you roll the die $6$ time, $\frac{1}{6}$ will be a six. So, that concludes the answer for the question to be, $6$ .