Calculating probability of total score S in n dice rolls without brute force

Question

I came upon a problem where you're asked the probability of scoring 10 or higher in three dice(fair, six-sided) rolls.

The solution I found was 2/9, which was correct according to the source of the problem, but getting that answer required me to work through all possible cases. I hoped the solution provided by my resource would show how to get the answer in a simpler way but it did the same thing(and so did several other resources I've found online). But that brute force method of solving this problem is unsatisfying to me, and inelegant insofar as it can't be extended to intractably large numbers of rolls.

I've been enjoying taking my probability problems and writing python scripts that can calculate them for more general cases. In this case, I'd be calculating the probability of a total score S after n rolls. I want to learn how one would solve this problem without using brute force.

Thanks in advance, been learning a ton from perusing this website!

Edit: Every answer provided has been tremendously useful to me and I am very grateful for everyone's effort. I realize now that I have reworded the original problem incorrectly.

The exact wording of the original problem is: "Your game piece is on the starting space, and the finish is 10 more spaces away. During each of your turns, you roll a fair six-sided die, and move that many spaces forward. However, if the result would put you past the finish, you do not move. What is the probability that you will reach the finish in 3 or fewer turns?"

In addition to still puzzling over the problem that I intended to ask about, I am now very interested in studying the problem I've inadvertently asked about, and I'm also now trying to work out exactly why my rephrasing was incorrect.

Again, thanks to everyone who responded, I cannot adequately express how much I appreciate the collective knowledge and generosity of this community.

Answer 1

Let us consider the expansion of $(x^6 + x^5 + x^4 + x^3 + x^2 + x)^3$. When I calculate this I get

$$x^{18}+3x^{17}+6x^{16}+10x^{15}+15x^{14}+21x^{13}+25x^{12}+27x^{11}+27x^{10}+25x^{9}+21x^{8}+15x^{7}+10x^{6}+6x^{5}+3x^{4}+x^{3}$$

This says to me there are 10 ways to choose three items in $(1,2,3,4,5,6)$ which sum to 15, since the coefficient $10$ is the number of ways to form the product $x^{15}$ from the product of $x^a x^b x^c = x^{a+b+c}$ where $a, b, c \in [1,6]$.

Is this brute force?

This expression tells us that there is a $\frac{1 + 3 + 6 + 10 + 15 + 21 + 25 + 27}{6^3} = \frac{5}{8}$ chance of getting a sum of 10 or more.

Answer 2

I don't know if you'd consider this an improvement, but you could take a recursive approach:

Let $P(n,k)$ be the probability of getting a sum of $n$ with $k$ rolls.

Then $$P(n,k)=\frac{1}{6}\cdot[P(n-1,k-1)+P(n-2,k-1)+P(n-3,k-1)+p(n-4,k-1)+P(n-5,k-1)+P(n-6,k-1)]$$

Answer 3

Perhaps this won't be useful to you as it is a brute force approach, but brute force can be used to find patterns. There are reoccurring patterns when considering sums of $n$ dice rolls. The columns indicate the number of dice rolls and the numbers underneath denote the number of ways to get a given total.

\begin{array}{|c|c|c|c|c|} \hline Total&1& 2 & 3 & 4 &5\\ \hline 1& 1& & &\\ \hline 2& 1& 1&&\\ \hline 3& 1& 2&1&\\ \hline 4& 1& 3&3&1\\ \hline 5& 1& 4&6&4&1\\ \hline 6& 1& 5&10&10&5\\ \hline 7& & 6&15&20&15\\ \hline 8& & 5&21&35&35\\ \hline 9& & 4&25&56&70\\ \hline 10& & 3&27&80&126\\ \hline 11& & 2&27&104&205\\ \hline 12& & 1&25&125&305\\ \hline 13& & &21&140&420\\ \hline 14& & &15&146&540\\ \hline 15& & &10&140&651\\ \hline 16& & &6&125&735\\ \hline 17& & &3&104&780\\ \hline 18& & &1&80&780\\ \hline 19& & &&56&735\\ \hline 20& & &&35&651\\ \hline 21& & &&20&540\\ \hline 22& & &&10&420\\ \hline 23& & &&4&305\\ \hline 24& & &&1&205\\ \hline 25& & &&&126\\ \hline 26& & &&&70\\ \hline 27& & &&&35\\ \hline 28& & &&&15\\ \hline 29& & &&&5\\ \hline 30& & &&&1\\ \hline \end{array}