sum of independent variables Suppose you have a sum of IID random variables (uniformly distributed in [0, 1])
$$S = \sum_{i=1}^N X_i$$
if I want to have a rough idea of the average value of $N$ such that the sum is equal to some number $S_0$ is it safe to say that $N$ is going to be around $S_0/{\rm E}[X_i]$ (assuming I'm not under the conditions of using Martingale).  
UPDATE:
I reformulate the problem. Suppose $N$ is a random variable defined as the smallest integer such that $$ \sum_{i=1}^N X_i \geq S_0  $$  where $X_i$ are IID and uniformly distributed in $[0, x_0]$, with $x_0 < S_0$. What is $E[N]$?
 A: $S_N \ge S_0$ (horrid notation) so I would expect $E[N]$ to be greater than $\frac{S_0}{E[X_i]}$.  
With $X_i$ iid uniform on $[0,1]$, then $E[X_i]=\frac{1}{2}$ and $\frac{S_0}{E[X_i]}=2S_0$.  
For $S_0=1$ I seem to recall from this Ponder This question $E[N]=\exp(1) \approx 2.718$ (it must be more than $2$). 
A slightly weaker recollection is that as $S_0$ increases: $E[S_N-S_0] \rightarrow \frac{1}{3}$ and $E[N]-2S_0 \rightarrow \frac{2}{3}$.
Added: my recollection is now stronger, and I gave some detail at a sci.math thread.  For $N\ge 3$ the $\frac{2}{3}$ convergence is very close, and this is related to $$\lim_{n->+\infty}\left(\sum_{i=1}^{n}\frac{(-i)^{n-i}e^i}{(n-i)!}\right)-2n=\frac{2}{3­} $$
There I gave the argument: Consider the total distance $N+E$ when the total first hits or exceeds 
$N$, with a final step $F$. So both the excess distance $E$ and the final 
step $F$ are in $[0,1]$ with $F\gt E$.   $F$ will tend to be big because it is 
conditioned on passing $N$, while $E$ will tend to be small since it is 
conditioned on $N$ having been passed. 
For large $N$, the probability density of $F$ is close to $2t$ while the 
density of E is close to $2-2t$.  This makes the expected value of $E$ be $1/3$ and so the expected total 
distance is about $N+1/3$.  Since for large $N$ the average distance per 
step is close to $1/2$ and as this is a stopping time, the expected number of steps is about $2N+2/3$. 
I justified the previous paragraph with: the joint distribution of the final step $F$ and the excess $E$ tends 
toward being flat for large $N$ subject to the constraint 
$0 \le E \lt F \le 1$. 
We have $\int_{x=0 }^1 \int_{y=0 }^x dy dx = 1/2$ so the joint 
probability density must be close to $2$ for large $N$. 
So $\Pr(F\le t) = \int_{x=0 }^t \int_{y=0}^x 2 dy dx = t^2$ 
and the marginal probability density for $F$ is (close to) $2t$ while $\Pr(E \le t) = \int_{y=0}^t \int_{x=y}^1 2 dx dy = 2t - t^2$ 
and the marginal probability density for $E$ is (close to) $2 - 2t$. 
