I throw a 6-sided dice (with values: 0,1,2,3,4,5) multiple times and add each value to a sum, which is 0 in the beginning. What is the probability of my sum reaching exactly 10, 11, 12, 13, 14? After reaching a requested sum, the sum will return to it's original 0 value.
E.g: 5 + 5 = 10, and afterwards the sum returns to 0. Also, the probability for each number on the dice is different (it's not a fair dice).
On
(Note: This is the solution to the problem in its original form.)
Denote by $q(n)$ the probability that we hit $10$, given that the momentary sum is $n$ and we have not hit $10$ before. Then $$q(10)=1;\qquad q(n)=0\quad(11\leq n\leq14)\ .$$ Furthermore we have the following backwards recursion: $$q(n)=\sum_{k=1}^5 {1\over5} q(n+k)\qquad(n=9,8,7,\ldots)\ .$$ This formula reflects the fact that the next move forward is one of $\{1,2,3,4,5\}$ with equal probability.
Performing the recursion gives $$q(0)={3327696\over9765625}\ ,$$ as determined by Byron Schmuland with another argument.
It pays to generalize. Let's calculate the probability $p(n)$ that we ever reach $n$ for *any integer * $n$.
Since we start at zero, we have $p(0)=1$, while $p(n)=0$ for $n<0$.
For larger $n$, by conditioning on the previously taken value we get $$p(n)=\sum_{j=0}^5 p(n-j)/6,$$ and if you solve this recursive equation for $n=10$ you get $$p(10)={3327696\over 9765625}=.34076.$$
For large values of $n$ the probability $p(n)$ will be very close to $1/3$, since each die throw adds three (on average) to the total.