Ways of getting a number with $n$ dice, each with $k$ sides Assume the dice are numbered from $1$ to $k$. My hunch is that this will form a normal distribution with a median at $n\cdot\frac{k}{2}$. However, I have no idea as to 


*

*turn this fact into an answer (I have a minimal knowledge of stats, but I know that I am missing the standard distribution)

*and this is probably the wrong approach


How can I approach and solve this problem? (Aside, this is not for a class, stats or other, so any and all approaches welcome).
*Edit: * I want to find the number of ways that the sum of the numbers that are rolled has a particular value, if $n$ dice are rolled, and each has $k$ sides, numbered $1$ to $k$.
 A: The exact probability is a bit complicated: the chance of getting a total of $p$ when you throw $n$ $k$-sided dice is:
$${1\over k^n}\sum_{j=0}^{\lfloor (p-n)/k\rfloor} (-1)^j {n\choose j}{p-kj-1\choose n-1}$$
You can see an explanation and examples here, see equation (10).
You are correct that for moderately large $n$ the distribution is well approximated by a 
normal curve with mean $n(k+1)/2$ and standard deviation $\sqrt{n(k^2-1)/12}$.  
A: As $n\to\infty$ with $k$ fixed, the sum of the outcomes divided by $\sqrt{n}$ approaches a normal distribution with mean $(k+1)/2$ if the numbers on the dice are $1,2,3,\ldots, k$.
As $k\to\infty$ with $n$ fixed, something quite different happens.
A: I am not certain what you mean to ask but based on your comment. The odds of ending up with a particular side out of your k sided die is 1/k, that's assuming your die is unbiased. Now you can throw it over and over the odds of a particular side showing up would be the same 1/k. 
your answer can be found at en.wikipedia.org/wiki/Dice#Probability
Cheers  
