How to find the expected value of infinite discrete random variable? Question: A fair six-sided die is rolled repeatedly. Let $X$ be the number of times it is rolled until two different numbers are seen (so, for example, if the first two rolls were each $4$ and the third roll was $5$ then $X=3$).
How to I find the expected value? I realize that this is a geometric distrition and this is the expression I have for $\mathsf{E}[X]$:
$\mathsf{E}[X]=2\left(\frac 56\right) + 3\left(\frac 56\right)\left(\frac 16\right) + 4\left(\frac 56\right)\left(\frac 16\right)^2 + 5\left(\frac 56\right)\left(\frac 16\right)^3 + ...$
I wrote down an expression in terms of summation and this seems to be a airthmetico-geometric series but there has got be an easier way to compute the expected value. What am I missing here?
 A: There are faster ways, but one relatively simple approach is 
$\mathsf{E}[X]=2\left(\frac 56\right) + 3\left(\frac 56\right)\left(\frac 16\right) + 4\left(\frac 56\right)\left(\frac 16\right)^2 + 5\left(\frac 56\right)\left(\frac 16\right)^3 + \cdots$
so $\frac16 \mathsf{E}[X]=2\left(\frac 56\right)\left(\frac 16\right) + 3\left(\frac 56\right)\left(\frac 16\right)^2 + 4\left(\frac 56\right)\left(\frac 16\right)^3 + 5\left(\frac 56\right)\left(\frac 16\right)^4 + \cdots$
and subtracting $\frac56 \mathsf{E}[X]=2\left(\frac 56\right) + \left(\frac 56\right)\left(\frac 16\right) + \left(\frac 56\right)\left(\frac 16\right)^2 + \left(\frac 56\right)\left(\frac 16\right)^3 + \cdots$
i.e. $\mathsf{E}[X]=2 + \left(\frac 16\right) + \left(\frac 16\right)^2 + \left(\frac 16\right)^3 + \cdots$
This is essentially $1$ more than a geometric series but you can in fact do the same thing again
so $\frac16 \mathsf{E}[X]=2\left(\frac 16\right) + \left(\frac 16\right)^2 + \left(\frac 16\right)^3 + \left(\frac 16\right)^4 + \cdots$
and subtracting $\frac56 \mathsf{E}[X]=2 -\frac 16 = \frac{11}{6}$
and thus  $\mathsf{E}[X] = \frac{11}{5}=2.2$
