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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

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

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How is your $k$-sided die numbered? from $1$ to $k$, or from $0$ to $k-1$ or $2345$ to $2344+k$? –  user21436 Feb 9 '12 at 3:51
@KannappanSampath, see edit –  soandos Feb 9 '12 at 3:53
So, you'd like to know that, $P(X=l)$ for a natural number $l$ and $X$ represents the sum of all outcomes, right? –  user21436 Feb 9 '12 at 3:55
Quite honestly, I have no idea what that notation means. I want to know the number of ways that a number can be rolled. –  soandos Feb 9 '12 at 4:08

3 Answers 3

up vote 6 down vote accepted

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 $\sqrt{n(k^2-1)/12}$.

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Perfect, and thanks for the references. –  soandos Feb 10 '12 at 2:56

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.

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Just guessing, but perhaps $(\frac{k}{2})^{n}$? –  soandos Feb 9 '12 at 4:43
if k went to infinity would the probability of get anything out of k sides go to 0. I think we may need someone with rigrous probability theory to answer this one. –  Hardy Feb 9 '12 at 5:35

I am not certain what u 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 unbaised. 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


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see edit to question. –  soandos Feb 9 '12 at 5:33
that is easy but 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 is dependent on the sum. –  Hardy Feb 9 '12 at 5:36
In the "classic" 2 six sided dice, not all numbers can be rolled the same amount of ways... –  soandos Feb 9 '12 at 5:39
There are 36 possibilities... Also, I do not want to enumerate all the possibilities. I want a formula that will do that for me –  soandos Feb 9 '12 at 5:45
Honestly, I don't see how your approach is relevant to my question, nor is your answer (did you see the edit?). –  soandos Feb 9 '12 at 5:51

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