Probability that the sum of N dice is at least X

Question

If N fair 6 sided dice are thrown, what is the probability that the sum of the thrown dice is at least X?

Answer 1

The chance of hitting exactly sum $p$ with $n$ $s-$sided dice is:

$$P(p,n,s) = \frac{1}{s^n}\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{p-sk-1}{n-1}$$

Summing from $x$ to $6n$ and substituting $s = 6$ gives:

$$\sum_{p=x}^{6n} \frac{1}{6^n}\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{p-6k-1}{n-1}$$

$$\sum_{p=x}^{6n} \binom{p-6k-1}{n-1} = \frac{1}{n}\left((n+6k-x)\binom{x-6k-1}{n-1} + (5n-6k+1)\binom{6n-6k}{n-1}\right)$$

Giving the somewhat ugly closed form, but very efficient to evaluate:

$$ \frac{1}{n6^n}\sum_{k=0}^n(-1)^k\binom{n}{k}\left((n+6k-x)\binom{x-6k-1}{n-1} + (5n-6k+1)\binom{6n-6k}{n-1}\right)$$

Answer 2

Assuming fair cubical dice with numbers 1 through 6, the PDF and CDF for the sum $X$ on two dice is given in the table below.

     k:      2     3     4     5     6      7     8     9    10    11    12
P(X=k):   1/36  2/36  3/36  4/36  5/36  6/36   5/36  4/36  3/36  2/36  1/36
P(X<=k):  1/36  3/36  6/36 10/36 15/36  21/36 26/36 30/36 33/36 36/36   1

The CDF $P(X \le x)$ gives the probabilities that $X$ is at most $k,$ for $k = 2, 3, \dots, 12.$

I will leave it to you to add a line to the table that gives the probabilities $P(X \ge k)$ for the relevant values of $k$. Then maybe you can deal with $N > 2$ dice.

You don't provide much context about the source of this problem. So it is difficult to say whether you are supposed to use normal approximations, symbolic software such as Mathematica, simulations in statistical software, or a convolution formula.