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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).

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closed as unclear what you're asking by Hagen von Eitzen, Adam Hughes, studiosus, RecklessReckoner, user1729 Aug 27 '14 at 19:54

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

How many times, did you say??? – barak manos Aug 27 '14 at 11:57
I assume you mean rolling a die twice. If this is the case, you would divide the number of ways to roll a $10$ by the total number of outcomes when rolling a die twice. – danielson Aug 27 '14 at 12:03
I think it means: keep rolling the die and keep track of the sum, then ask: What is the probability that the sum of 10 is obtained in such a sequence of rolls. We don't know in advance how many rolls, in fact one could roll a lot of zeroes and have to keep going for example. But with probability 1 a given sequence eventually gets to 10 or more, after which one knows whether it counts toward the desired event or not. Is this the intent, Reka M? – coffeemath Aug 27 '14 at 12:10
But you can just ignore the zero throws, and pretend that the non-zero throws each have probability $\frac15$. And then the number of non-zero throws required is at most 10. – TonyK Aug 27 '14 at 13:09
The three revisions of your post are all completely different questions. Which one do you want answered? – AakashM Aug 27 '14 at 15:11

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.

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@Byron Schmuland: Thank you for your answer. I will try this approach as well. – Reka M Aug 27 '14 at 12:48

(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.

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I have replaced in the formula the specific probabilities for each value, but unfortunately, the results are not convincing. I've created a simulation of this problem, so I know approximately what my results should be. I just need the calculation nethod, and this doesn't seem to be it :( . – Reka M Aug 27 '14 at 14:39
What are the probabilities for each number then? – bobbym Aug 27 '14 at 15:43
@bobbym: the probabilities are: 0 - 92.77495997% ; 1 - 7.00000000% ; 2 - 0.22000000% ; 3 - 0.00500000% ; 4 - 0.00004000% ; 5 - 0.00000003% – Reka M Aug 27 '14 at 16:30
Hi Reka; I am getting 96.91% – bobbym Aug 31 '14 at 13:49

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