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If I roll a 10-sided dice and I get 6 or more as a result, I get a "success". But if I roll the dice and I get 1, it cancels one success, so that If I roll two times the dice and I get (7,1), I have 0 success, if I roll it and I get (7,2), I have one success, if I roll it and I get (1,1) I have -2 success. I am trying to compute the probability of to get at least one success. And yeah, I can have negative successes.

What I thought so far is to call A="get 6 or more" and B="to not get 1" and compute $P(A \cap B)$. But I am pretty sure is wrong since it doesn't depend of the number of times I roll the dice.

I also tried to compute it rolling the dice one time but it changes dramatically as I increase the number of times I roll the dice, and I don't understand how to describe this change.

Best regards.

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Adding information

You can choose the number of times we roll the dice. Obviously I am interested in the general case ($n$ times) but I suspect is not possible, so I would like to know the case of rolling it 5 times.

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  • $\begingroup$ How many times do you roll the die? $\endgroup$ – callculus May 26 '18 at 4:38
  • $\begingroup$ Hi. I added aditional info. Thanks for your reply. $\endgroup$ – Topologicalife May 26 '18 at 4:41
  • $\begingroup$ Is the ten sided die labeled 0 through 9 (like many are) or 1 through 10? $\endgroup$ – user1390 May 26 '18 at 4:53
  • $\begingroup$ What is the shape of your dekahedron? Are all faces equally likely to be landed on? Or are you using an icosahedron and giving adjoining faces the same number, in pairs? $\endgroup$ – Lubin May 26 '18 at 6:10
  • $\begingroup$ Is labeled 1 through 10 (I don't see the exact diferences between 0-9 and 1-10). The dice is "ideal". $\endgroup$ – Topologicalife May 27 '18 at 5:01
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In general we can use the multinomial distribution. Let $s$ and $l$ the indices for a success and a loss respectively. Then one condition is thaAt $s>l$. And let $n$ be the number of tosses. Then the number of tosses where we neither win nor loose is $n-s-l$. This has to be greater or equal to $0$

$n-s-l\geq 0 \Rightarrow s\leq n-l$

And additionally the number of losses has to be smaller than the half of number of tosses. Thus the probability that we have more successes than losses is

$$\sum_{0\leq l<n/2} \ \ \sum_{s=l+1}^{n-l} \frac{n!}{s!\cdot l!\cdot (n-l-s)!}\cdot \left( \frac{1}{10} \right)^l\cdot \left( \frac{5}{10} \right)^s\cdot \left( \frac{4}{10} \right)^{n-l-s}$$

Let $n=3$

The probability to get more successes than losses is the sum of

$l=0$

$ s=1:\frac{3!}{1!\cdot 0!\cdot 2!}\cdot \left( \frac{1}{10} \right)^0\cdot \left( \frac{5}{10} \right)^1\cdot \left( \frac{4}{10} \right)^{2}$

$ s=2:\frac{3!}{2!\cdot 0!\cdot 1!}\cdot \left( \frac{1}{10} \right)^0\cdot \left( \frac{5}{10} \right)^2\cdot \left( \frac{4}{10} \right)^{1}$

$ s=3:\frac{3!}{3!\cdot 0!\cdot 0!}\cdot \left( \frac{1}{10} \right)^0\cdot \left( \frac{5}{10} \right)^3\cdot \left( \frac{4}{10} \right)^{0}$

$l=1$

$ s=2:\frac{3!}{2!\cdot 1!\cdot 0!}\cdot \left( \frac{1}{10} \right)^1\cdot \left( \frac{5}{10} \right)^2\cdot \left( \frac{4}{10} \right)^{0}$

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Isn't it just a random sum? Where you have $S=\sum_{i=1}^nX_i $ for $X_i$ distributed like you described with $P(X_i=-1)= \frac 1{10}$, $P(X_i=0)= \frac 4{10}$ and $P(X_i=1)= \frac 5{10}$ with $n$ the number of experiments. I must mention $X_i$ are assumed independent since clearly each roll does not depend on the previous one (or any other for that matter).

All the remains is to find the distribution of the sum. Plenty of ways to go about this. Best method would be to calculate the Probability Generating Function (PGF), then find the Cumulative Distribution from there using standard methods.

i.e. you find the PGF of S and the $P(S=i) \forall i \in (-n,n)$ since at th extremes we can roll all 1s or all greater than 6 in n goes, correct?

EDIT 1

Let me just give you an example of what I mean:

Suppose $n=2$; that is we roll 2 times. $P(S=-2)$ is stating the probability that we roll $X_1=-1$ and $X_2=-1$. By law of probability this is just $P(X_1=-1) \times P(X_2=-2)=\frac 1{10} \frac 1{10}=\frac 1{100}$. Likewise for each of possible occurences of $X_i$ we can find the resulting sum $S$. And the best part is we can find a general closed form solution for any number of rolls n :)

Please let me know if I can clarify!

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