Suppose I have a 3-sided die with each side contains different number of 1, 2, and 3. In this problem, each time I roll the die I will add up the number that shows up with previous rolls. The target number I need is either 11 or 12. Once it reaches either of those number I won't roll the die again. What is the probability to get a value of 11 or 12 in 10 rolls or less?
I tried to calculate myself but the result is more than 100%.
Sum of 11 or 12 in 4 rolls: 0,061728395
Sum of 11 or 12 in 5 rolls: 0,308641975
Sum of 11 or 12 in 6 rolls: 0,366255144
Sum of 11 or 12 in 7 rolls: 0,195244627
Sum of 11 or 12 in 8 rolls: 0,057613169
Sum of 11 or 12 in 9 rolls: 0,010211858
Sum of 11 or 12 in 10 rolls: 0,001100781
Total: 1,000795949
Let $p(X)$ denote the probability of reaching the sum $X$. Then $p(13) = p(10)/3$. We note the following: $$p(1) = 1/3, p(2) = 4/9, p(3) = 16/27$$(assuming a fair dice) and $$p(n) = (p(n-1)+p(n-2)+p(n-3))/3$$for $3<n\le10$. Using this, we find: $$p(10) = 29404/59049 = 29404/3^{10}$$ Hence, $p(13) = 29404/177147 = 29404/3^{11}$. Now, the probability that you get $11$ in 11 rolls is $(1/3)^{11}$, that you get $12$ in $11$ rolls is $(1/3)^{11}$. So, the probability you're looking for is $$1 - 29404/3^{11} - 2\cdot(1/3)^{11} = 49247/3^{10} \approx \boxed{0.834}$$