expected value of a sum of a 10 sided die Suppose you have a fair die with 10 sides with numbers from 1 to 10. You roll the die and take the sum until the sum is greater than 100. What is the expected value of this sum?
 A: Using generating functions, there is one way to make each of the numbers $1$ through $10$ on each roll: $$G(x)=1x^1+1x^2+\cdots 1x^{10}=x\frac{x^{10}-1}{x-1}$$ To find the possible ways to get to any number $a$ after $n$ rolls, we need to find the coefficient of $x^a$ in the expansion of $G(x)^n$. 
So to answer your question, we must find the number of ways to get $101-110$ total, and the number of ways to get each of the outcomes specifically. Thus, we examine the coefficients of $x^{101}$ through $x^{110}$ in the sum of all powers of $G$ (since we don't care about how many rolls it takes). $$G+G^2+\cdots=\frac{G}{1-G}=\frac{x(x^{10}-1)}{(x-1)-x(x^{10}-1)}$$
From here you would have to use a CAS to find the exact result. I got the numbers on Wolfram Alpha, but I have no way to copy them (and they're quite large) so I can't provide the exact answer, but as I mentioned in the comments it will be around $104$.
Alternatively, you could try looking at special (or general) cases to find a pattern or formula which would possibly apply to this case. Depending on where you got this question, there is a good chance that it has a much more elegant solution.
A: An argument like the one  in this answer, shows that the distribution 
of the final position is very closely approximated 
by $[10/55,9/55,8/55,\dots ,1/55]$ on the 
states $[101,102,103,\dots, 110]$. Therefore the 
average position at the end of the game is very close 
to $$\sum_{i=1}^{10} (100+i)(11-i)/55=104. $$

Added: Here is some further information on the approximate hitting distribution.
Let's express the hitting distribution of $100,101,102,\dots$ as 
$\sum_{i=0}^9 \pi_i \,\delta_{100+i}$, and the hitting distribution
 of $101,102, 103,\dots$ as 
$\sum_{i=0}^9 \pi^\prime_i\, \delta_{101+i}$, where 
$\pi=(\pi_0,\dots,\pi_9)$ and  $\pi^\prime=(\pi^\prime_0,\dots,\pi^\prime_9)$
are distributions on $\{0,1,2,\dots,9\}$.
By the strong Markov property, we have 
$$\pi^\prime=\pi_0 U+\sum_{i=1}^9 \pi_i\delta_{i-1} $$
where $U$ is the uniform distribution on $\{0,1,2,\dots,9\}$.
On the other hand, since the process has been running for 
a long time, we have $\pi\approx \pi^\prime$. If you take 
this as equality, write $\pi=\pi_0 U+\sum_{i=1}^9 \pi_i\delta_{i-1}$,
 and solve for $\pi$ you get the required pattern. 
