A fair dice is rolled and the result and written down, Then the dice is rolled repeatedly (independently) untill the first occurence of a result bigger or equal to the result we've written down.
Find the expected value of the total number of dice rolls, including the first.
My attempt was to define a geometric RV $N$~$Geo(1-\frac {x-1}{6})$ , where $x$ is the result of the first roll.
So $N$ is the number of rolls we did untill we got a number bigger or equal to $x$, so we're looking for $E[N+1]$ in order to also count the first roll.
and with that: $$ E[N+1]=E[N]+E[1]=1+1\cdot (1-\frac {1-1}{6})+ \ldots +6\cdot (1-\frac {6-1}{6})=\frac {31}{3} $$
Which I'm not sure on how to confirm this answer, in general I feel like I'm missing something big due to the fact that $x$ takes on values $\in \\{1,\ldots,6\\}$ which directly affects the argument we're placing as $p$ in the geometric distribution.
This is a conditional expectation problem, where we condition the expected value based on the initial roll.
Let's call $X$ the total number of rolls and $R1$ the value rolled on the first roll.
$E[X | R1 = 1] = 1 + 1 = 2$
$E[X | R1 = 2] = 1 + \frac65 = \frac{11}5$
$E[X | R1 = 3] = 1 + \frac32 = \frac{5}2$
$E[X | R1 = 4] = 1 + 2 = 3$
$E[X | R1 = 5] = 1 + 3 = 4$
$E[X | R1 = 6] = 1 + 6 = 7$
All of these have a $\frac16$ chance of occuring thus: $E[X] = \frac16(2 + \frac{11}5 + \frac52 + 3 + 4 + 7) = 3.45$