<s>[INST] Expected number of rolls required to get sum greater than n for n faced die? Suppose a guy has a die with $n$ faces. He can go on rolling it as many times as possible and add the sum of each outcome. What is the expected number of rolls after which the sum is at least $n$?
 [/INST] I'm assuming the OP means what is the expected number of rolls until the sum is at least $n$ (as Alecos assumes in his answer).
Let $E(m)$ be the expected number of rolls until the sum is at least $n$, starting with a sum of $m$.
Then we have, for a start:
\begin{align*}
E(n) &=0 \\
E(n-1) &=1 \\
E(n-2) &=1+\frac{1}{n}E(n-1) = 1+\frac{1}{n} \\
E(n-3) &=1+\frac{1}{n}E(n-2)+\frac{1}{n}E(n-1) \\
    &=1+\frac{1}{n}(1+\frac{1}{n}E(n-1))+\frac{1}{n} \\
    &=1+\frac{2}{n} + \frac{1}{n^2} \\
E(n-4) &=1+\frac{1}{n}E(n-3) + \frac{1}{n}E(n-2)+\frac{1}{n}E(n-1) \\
    &=1+\frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3}\\
\end{align*}
We observe a pattern here, and you can prove with induction that
$$
E(n-k) = \sum_{i=0}^{k-1} \frac{\binom{k-1}{i}}{n^i}$$
for $1 \le k \le n$.
The value we seek is $E(0)$:
$$
E(0)=\sum_{i=0}^{n-1} \frac{ \binom{n-1}{i}}{n^i} = \left( 1+\frac{1}{n} \right)^{n-1}.
$$
Note that as $n \rightarrow \infty$, $E(0) \rightarrow e$.
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