Probability of rolling a sum of at least 18 with 10 6-sided dice I'm trying to work out how to do this, and I'm very stuck. My grasp of probability is shaky to begin with, and generally with probability questions, I list out cases. Due to the number of possibilities, this one is beyond my grasp. I can't even think of a quick way to figure out which combinations would lead to a sum of 18, without listing all possible cases, so this post isn't so helpful right no. 
I searched here first, but unfortunately, this post is beyond me, and the link in this post isn't working. 
 A: Lets mark the result of each dice roll with $x_i$
$$x_1+x_2+\dots+x_{10}\geq 18$$
And
$$1\leq x_i\leq6$$
Which is the same as
$$x_1+x_2+\dots+x_{10}\geq 8$$
And
$$0\leq x_i\leq 5$$
The complement
Lets have a look at the complement:
$$x_1+x_2+\dots+x_{10}< 8$$
Where $0\leq x_i\leq 5$
There are 2 case:
Case 1: $x_1+x_2+\dots+x_{10}\in \{0,1,2,3,4,5\}$
No need to account for the bounds on $x_i$
The number of solution to the equation is
$$c_1=\sum_{k=0}^{5}{9+k\choose k}$$
Case 2: $x_1+x_2+\dots+x_{10}\in \{6,7\}$
Same as the above, but we need to subtract the illegal solutions of $x_i\geq 6$
There is one illegal solutions for $k=6$, which is $x_i=6$
And two illegal solutions for $k=7$, which are $x_i=6$ or $x_i=7$
$$c_2={9+6\choose 6}-10{8+0\choose 0} + {9+7\choose 7}-10{8+1\choose 1}-10{8+0\choose 0}$$
Summary
The complementary probability is
$$q=\frac{c_1+c_2}{6^{10}}$$
Thus the probability of rolling a sum of at least 18 with 10 6-sided dice is
$$p=1-q=1-\frac{\sum_{k=0}^{7}{9+k\choose k}-110}{6^{10}}$$
