A die is thrown repeatedly until a six comes up. If I write the Sample Space for the above event, this would be a very long series. 
What I wanted to know is that, whether is would ever get complete or end. 
Or simply, is it finite or infinite?
The sample space is like- {6, (1,6),(2,6)....}
 A: The sample space is infinite. You could have any number of "non-six" (say "$\bar6$") results before you get your first six, so $\# \Omega=\infty$. You could actually define the following sample space:
$$\{(6),(\bar6,6),(\bar6,\bar6,6),(\bar6,\bar6,\bar6,6),\ldots\}.$$
Anyway, the probability of $X=$number of tries being $x$ is given by the geometric distribution, that is:
$$P(X=x)=p(1-p)^{x-1},$$
where $p$ is the success probability each time (here $p=\frac16$), so
$$P(X=x)=\frac16 \left(\frac56\right)^{x-1}.$$
This is so,because the variable $X$ is such that
$$X\big((6)\big)=1,$$
$$X\big((\bar6,6)\big)=2,$$
$$X\big((\bar6,\bar6,6)\big)=3,$$
and so on. So the probability of $\{X=1\}$ is just the probability of a $6$ in the first try, that is $\frac16$. The probability of $\{X=2\}$ is the probability of not getting a $6$ in the first try and a $6$ in the second try; by independence, that is
$$P(X=2)=\frac56 \cdot \frac16.$$
Similarly, you have
$$P(X=3)=\frac56\cdot\frac56 \cdot \frac16=\left(\frac56\right)^2 \cdot \frac16$$
and
$$P(X=4)=\frac56\cdot\frac56 \cdot \frac56 \cdot \frac16=\left(\frac56\right)^3 \cdot \frac16,$$
which generalizes to the mentioned formula.
A: Sample space will be infinite, but you can calculate the probability of six coming up.
$$P(Rolling\ a\ 6\ on\ \mathbf{I}\ roll )=\frac{1}{6}\\
P(Rolling\ a\ 6\ on\ \mathbf{II}\ roll )=\frac{5}{6}\cdot\frac{1}{6}\\
P(Rolling\ a\ 6\ on\ \mathbf{n^{th}} roll) =\bigg(\frac{5}{6}\bigg)^{n-1}\cdot\frac{1}{6}\\
P(Rolling\ a\ 6) = (Rolling\ a\ 6\ on\ \mathbf{I}\ roll )+(Rolling\ a\ 6\ on\ \mathbf{II}\ roll )+...P(Rolling\ a\ 6\ on\ \mathbf{n^{th}} roll)\\
=\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6}+\bigg(\frac{5}{6}\bigg)^2\cdot\frac{1}{6}...$$
This is an infinite geometric series,which is of the form
$$a+a\cdot r+a\cdot r^2+a\cdot r^3$$
Sum of this series is given by the formula,
$$Sum=\frac{a}{1-r}\qquad (\ |r| < 1\ ) $$
So,
$$ P(Rolling\ a\ 6 )=\frac{\frac{1}{6}}{1-\frac{5}{6}}\\ \qquad \hspace{1cm} =1$$
This means at some point a six will show up.
Infinite geometric series is often used to calculate the probability of winning of a person, when two or more people takes turns and throw. Dice roll game probability.
