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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

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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

2 Answers 2

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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