Calculate probability of random numbers adding up to or being greater than another number What's a simple formula I can use for calculating the probability of a set of random numbers adding up to or being greater than another number? Where $W$ is the amount of random numbers picked, $X$ is the lower limit, $Y$ is the upper limit and $Z$ is the number I need to reach. 
Example 1: 2 numbers are picked randomly from 500 to 800, what is the probability they will total 1300 or greater. 
Example 2: 3 numbers are picked randomly from 400 to 600, what is the probability they will total 1500 or greater.
The random numbers can include the upper and lower limit and will only be whole numbers.
Edit 1: I should also add the exact same number can be picked multiple times (so they are replaced) and that any number is equally as likely as another.
Edit 2: Could the formula be any of these?  
1) $P=\left(\frac{Y-(Z/W)}{Y-X}\right)^{W}$
2) $P=\frac{Y-(Z/W)}{Y-X}$
3) $P=\frac{\frac{Y-(Z/W)}{Y-X}}{W}$
Do any of those work? If not, why not?
 A: @whuber and @Erosennin have already provided you with some helpful ideas here for the general case.  A simple (depending on your definition of simple) is available for the case when the number of draws, $n=2$.  In that case, let the value of the first draw be $X_1$ and the second, $X_2$, and their sum, $X_1+X_2=Z.$  Then, the probability that $Z=z$ is given by:
\begin{eqnarray*}
P\left(Z=z\right)=f_{Z}(z) & = & \begin{cases}
\frac{4}{(b-a+1)(b-a+2)} & ,\,\mathbf{if}\,2a+2\le z\le2b-2\\
\frac{2}{(b-a+1)(b-a+2)} & ,\,\mathbf{if}\,2b-2<z\le2b\\
\frac{2}{(b-a+1)(b-a+2)} & ,\,\mathbf{if}\,2a\le z<2a+2\\
0, & \mathbf{otherwise}
\end{cases}
\end{eqnarray*}
where $a$ is the number of the lower bound and $b$ is the number of the upper bound.  For more general cases, @whuber's suggestion to use sums from continuous uniform distributions is likely to be as accurate as you'd need in most practical applications.
None of the formulae that you provide work.  They can be easily shown to be incorrect through a counterexample.  Take the relatively simple instance where you draw 3 numbers from 1 to 3 and want to determine the probability that their sum is 9.  In this case each of the formulae that you provide involve the numerator term $Y-(Z/W)$ (using your notation).  In this case, that $Y-(Z/W)=3-(9/3)=0$, which implies the probability is zero, but this can't be the case since we know that if you draw, {3, 3, 3}, the numbers sum to 9, which implies $P(Z=9)>0$.
A: Rename lower limit to $L$ and upper limit to $U$, and denote sum of draws as $S$. 
Example 1: We denote first draw by $X$ and second draw by $Y$, and $Z$ the number you need to reach:
$P(S\geq Z) = P(X+Y\geq Z) = \sum_{x=L}^U P(X=x) P(Y\geq Z -x)$.
But, 
if $Z-U > L$, then there is a lower limit $L_1$ in $(L,U)$ that the first draw needs to be greater than, so we must have: 
$P(S\geq Z) = P(X+Y\geq Z) = \sum_{x=L_1}^U P(X=x) P(Y\geq Z-x)$
I do not know if you are able to find an explicit formula, though. The probabilities totally depend on your limits in each case. The reasoning is to me analogous for Example 2.
