TITLE: expected value of a sum of a 10 sided die
QUESTION [12 upvotes]: 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?

REPLY [12 votes]: 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.