Probability of rolling a sum of at least 18 with 10 6-sided dice

Question

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.

Answer 1

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 $$p=1-q=1-\frac{\sum_{k=0}^{7}{9+k\choose k}-110}{6^{10}}$$

Answer 2

You want to calculate $1-P(\mbox{sum}\le 17)$. Now, $P(\mbox{sum}\le 17)=\sum_{k=10}^{17}P(\sum=k)$ (formally, it would be $P(\mbox{sum}\le 17)=\sum_{k=1}^{17}P(\sum=k)$, but the minimum sum is 10. Then for $k\le 10$, $P(\mbox{sum}=k)=0$).

Now, fix $k\in\{10,11,...,17\}$. You want to distribute $k$ identical balls into 10 distinguishable boxes such that each box has at least one ball and at most 6 balls. By generating functions method, it can be done with polynomial $p(x)=(x+x^2+...+x^6)^{10}=x^{10}(1+x+...+x^5)^{10}=x^{10}(1-x^6)^{10}(1-x)^{-10}$.

Now, the coefficient of $x^k$ in $p(x)$ is the same of $x^{k-10}$ on $(1-x^6)^{10}(1-x)^{-10}\\=\left(\binom{10}{0}-\binom{10}{1}x^6+\binom{10}{2}x^{12}-...+\binom{10}{10}x^{60}\right)\sum_{r=0}^\infty\binom{10+r-1}{r}x^r$.

But $k-10=0,1,2,...,7$. Thus, coefficients are

$x^0$) Is $\binom{10}{0}\binom{9}{0}$

$x^1$) Is $\binom{10}{0}\binom{10}{1}$

$x^2$) Is $\binom{10}{0}\binom{11}{2}$

$x^3$) Is $\binom{10}{0}\binom{12}{3}$

$x^4$) Is $\binom{10}{0}\binom{13}{4}$

$x^5$) Is $\binom{10}{0}\binom{14}{5}$

$x^6$) Is $\binom{10}{0}\binom{15}{6}-\binom{10}{1}\binom{9}{0}$

$x^7$) Is $\binom{10}{0}\binom{16}{7}-\binom{10}{1}\binom{10}{1}$

Thus, the total of sums of the dices for $\sum\le 17$ is $\binom{9}{0}+\binom{10}{1}+\binom{11}{2}+\binom{12}{3}+\binom{13}{4}+\binom{14}{5}+\binom{15}{6}+\binom{16}{7}-10\binom{10}{0}-10\binom{9}{1}=\binom{17}{7}-110$.

But the total of outputs that the 10 dices can be are $6*6*...*6=6^{10}$.

Thus, answer is $1-\frac{\binom{17}{7}-110}{6^{10}}=1-\frac{19338}{6^{10}}=0.99968$