What is the probability of my sum reaching exactly 10? 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).
 A: 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. 
A: (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.
