Shortest way to answer probability / game question? 
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?

What is the shortest way to get this answer? I know the answer is 2/9, but curious what is the most efficient method of answering this? 
EDIT: The question (and the given answer / method to obtain in a long way can be found at this link: Check out this problem I found on Brilliant!
https://brilliant.org/practice/conditional-probability-casework-calculations/?problem=discrete-mathematics-problem-109432&chapter=conditional-probability-2
I would have simply copied and pasted the answer but couldn't get the graphics to display properly and it will be better displayed in the associated link.
 A: To win in exactly 3 turns without "overshooting" can be done in $[x^{10}]((x+x^2+x^3+x^4+x^5+x^6)^3)$ number of ways where $[x^n]$ represents taking the coefficient of the $x^n$ term in the expansion.  Here, that would correspond to $27$ different ways.  This occurs with probability $\frac{27}{6^3}$
To win in exactly 2 turns can be done in $[x^{10}](x+x^2+\dots+x^6)^2$ number of ways, here $3$ ways.  This occurs with probability $\frac{3}{6^2}$ and is mutually exclusive with the earlier case.
To win in exactly 3 turns while overshooting once can only occur as the sequence of rolls 5-6-5, 6-5-4, and 6-6-4, occurring with probability $\frac{3}{6^3}$.
This gives us a final total probability of $\frac{27+6\cdot 3+3}{6^3}=\frac{48}{216}=\frac{2}{9}$
A: Win in 2 turns
Roll exactly 10 on 2 rolls.
Win in 3 turns.
Not roll exactly 2,3 or 10 on the first two rolls.
If you roll a 2 or a 3, you are still not in striking distance, if you roll a 10 you won in two rolls.
Roll exactly what you need to cross the goal with the 3rd roll.
