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Is there a simple form for the probability of rolling at least $n$ on $k$ 6-sided dice? Of course you can do it by recursion (see here). But is there a way to do it with just a few binomial coefficients, without the recursion?

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    $\begingroup$ If you're willing to sacrifice accuracy, you can use the normal approximation, if $k$ is large. There isn't much alternative apart form doing the calculation. $\endgroup$
    – Calvin Lin
    Commented Jan 27, 2013 at 19:18

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What you want to know is here.

The formula for a homogeneous and fair dice from 1 to D values is this

$$P(S,n,D)=\frac{1}{D^n}\sum_{k=0}^{\lfloor(S-n)/D\rfloor}(-1)^k\binom{n}{k}\binom{S-Dk-1}{n-1}$$

where S is the sum that you want, n is the number of dice you roll and D is the number of sides of the dice (D6, D8, etc.)

To know the sum of at least S just need to sum what you want over S.


A better way: in the book of generatingfunctionology there is an exercise in the chapter one asking just this but the answer is nicer.

We set $P(x):=\sum_{k=1}^{m}\frac1m x^k=x(1-x^m)/m(1-x)$, then $P$ is the generating function of the probabilities of a fair dice of $m$ sides and $[x^\ell]P(x)^n/(1-x)$ is the probability to get at lest a sum of $\ell$ throwing $n$ times the dice. The closed form is $$ \begin{align}[x^\ell]P(x)^n/(1-x)&=m^{-n}[x^\ell]x^n(1-x^m)^n(1-x)^{-n-1}\\ &=m^{-n}[x^\ell]\sum_{j,k\ge 0}\binom{n}{k}\binom{-n-1}{j}(-1)^{k+j}x^{km+j+n}\\ &=m^{-n}[x^\ell]\sum_{j,k\ge 0}\binom{n}{k}\binom{n+j}{j}(-1)^kx^{km+j+n},\quad km+j+n=\ell\\ &=m^{-n}\sum_{k\ge 0}\binom{n}{k}\binom{\ell-km}{n}(-1)^k\end{align} $$

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