7
$\begingroup$

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 $k$.

$\endgroup$
4
  • $\begingroup$ How is your $k$-sided die numbered? from $1$ to $k$, or from $0$ to $k-1$ or $2345$ to $2344+k$? $\endgroup$
    – user21436
    Commented Feb 9, 2012 at 3:51
  • $\begingroup$ @KannappanSampath, see edit $\endgroup$
    – soandos
    Commented Feb 9, 2012 at 3:53
  • $\begingroup$ So, you'd like to know that, $P(X=l)$ for a natural number $l$ and $X$ represents the sum of all outcomes, right? $\endgroup$
    – user21436
    Commented Feb 9, 2012 at 3:55
  • $\begingroup$ Quite honestly, I have no idea what that notation means. I want to know the number of ways that a number can be rolled. $\endgroup$
    – soandos
    Commented Feb 9, 2012 at 4:08

3 Answers 3

Reset to default
9
$\begingroup$

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

$\endgroup$
8
  • $\begingroup$ Perfect, and thanks for the references. $\endgroup$
    – soandos
    Commented Feb 10, 2012 at 2:56
  • $\begingroup$ @Byron Schmuland : I'm sorry to bother you, but there's something I'd like to ask. I'd like to find every $p$ such that $P(p,n,k)=P(p+1,n,k)$ where $P(p,n,k)$ is the expression you write in your answer. It seems easy to conjecture the answer, but I'd like to have a rigorous proof. If you have any ideas or references, could you please show me them? Thank you for taking your time to read this comment. $\endgroup$
    – mathlove
    Commented Aug 12, 2015 at 19:16
  • $\begingroup$ @mathlove I suppose you mean that $p\mapsto P(p,n,k)$ is symmetric about the midpoint $n(k+1)/2$ and strictly increasing for $p$ up to the midpoint if $n\geq 2$. The best way to prove this is with generating functions and induction on $n$. $\endgroup$
    – user940
    Commented Aug 12, 2015 at 22:36
  • $\begingroup$ @Byron Schmuland : Thank you very much for your reply. Yes, that's what I meant (So, my understanding was that I need to prove that (1) $P(p,n,k)=P(n+nk-p,n,k)$ for every $p$ such that $n\le p\le nk$ and that (2) $P(n,n,k)\lt P(n+1,n,k)\lt\cdots\lt P(\lfloor (n+nk)/2\rfloor,n,k)$ for $n\ge 2$.) From your hints, I thought a generating function $f(x)=\sum_{i=0}^{\infty}P(i,n,k)x^i$ where $P(i,n,k)=0$ for $p\lt n\ \text{or}\ p\gt nk$. But I've been facing difficulty in proving (1)(2) by induction on $n$. Could you please elaborate on that? $\endgroup$
    – mathlove
    Commented Aug 13, 2015 at 10:46
  • $\begingroup$ @mathlove The argument is a little long for a comment. You can email me directly and I will send you a (one page) proof. $\endgroup$
    – user940
    Commented Aug 13, 2015 at 22:31
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Just guessing, but perhaps $(\frac{k}{2})^{n}$? $\endgroup$
    – soandos
    Commented Feb 9, 2012 at 4:43
  • $\begingroup$ 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. $\endgroup$
    – Comic Book Guy
    Commented Feb 9, 2012 at 5:35
-1
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ see edit to question. $\endgroup$
    – soandos
    Commented Feb 9, 2012 at 5:33
  • $\begingroup$ 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. $\endgroup$
    – Comic Book Guy
    Commented Feb 9, 2012 at 5:36
  • $\begingroup$ In the "classic" 2 six sided dice, not all numbers can be rolled the same amount of ways... $\endgroup$
    – soandos
    Commented Feb 9, 2012 at 5:39
  • $\begingroup$ There are 36 possibilities... Also, I do not want to enumerate all the possibilities. I want a formula that will do that for me $\endgroup$
    – soandos
    Commented Feb 9, 2012 at 5:45
  • $\begingroup$ Honestly, I don't see how your approach is relevant to my question, nor is your answer (did you see the edit?). $\endgroup$
    – soandos
    Commented Feb 9, 2012 at 5:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .