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As stated the first person that can help me answer this question I will give credit to them if and when I get the game I'm working on published!

I am using a pool of $10$-sided dice and counting a success when a die rolls a 9 or a 10. Each die acts independently of the others to achieve a success so that if you had a pool of $5$ dice, you could achieve $5$ successes. I need to know the probability of rolling $1$, $2$, or $3$ successes in a pool of dice of sizes ranging from $1$ to $12$. Obviously you have a $20\%$ chance to roll $1$ success with a dice pool of $1$. Sadly, that is about as far as my limitation of probability mathematics can get me.

Any help would be greatly appreciated!

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  • $\begingroup$ Die is singular; dice is plural. $\endgroup$ – N. F. Taussig May 15 '18 at 8:43
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Guide:

You are right that the probability of success is $p=\frac15$.

If you have $n$ dices, then it should follows the binomial distribution $Bin(n,p)$. You can then use the formula of binomial distribution to compute it.

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With a $10$-sided die, the probability of rolling a $9$ or $10$ is $20$% as stated because there are only $10$ possibilites, $2$ of which are successes.

With $2$ $10$-sided dice, there are $100$ possibilities of outcomes. If only die $1$ is a success, die $1$ has values {$9,10$} and die $2$ has values {$1,2,3,4,5,6,7,8$}, so there are $16=2(8^1)$ possible roles for success for just die $1$, likewise for die $2$

With $n$-dice, there are $10^n$ outcomes, in this case, $1\leq n\leq12$.

For $1$ success, only $1$ of the $n$ dice has values {$9,10$} and the $n-1$ dice have values {$1,2,3,4,5,6,7,8$}, so there are $2(8^{n-1})$ ways to succeed, likewise for the other $n-1$ dice.

$$P_{n,1}=\frac{2n(8^{n-1})}{10^n}$$

This gives back $20$% for $P_{1,1}$ gives back the $32$% for $P_{2,1}$, and decays to $0$ as $n$ increases because it becomes increasingly more difficult to roll a {$9,10$} with only $1$ die among many

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  • $\begingroup$ Die is singular; dice is plural. Your answer for one success is correct, but you did not address how to find the probability of two or three successes. $\endgroup$ – N. F. Taussig May 15 '18 at 8:45
  • $\begingroup$ @N.F.Taussig Good catch, thanks. I didn't post 2 or 3 because I couldn't decide if repeats were being counted or not in the formulation $\endgroup$ – AEngineer May 15 '18 at 15:55
  • $\begingroup$ The probability of just one dice out of a pool of dice coming up with a 9 or 10 on a 10-sided dice using pools of dice from 1 to 12. It looks like it's a 20% chance in a pool of one dice and a 32% chance in a pool of 2 dice? $\endgroup$ – D. Ada May 21 '18 at 23:56
  • $\begingroup$ Right. I worded the last sentence poorly. The probability increases until $n=4$ and then $P_{4,1}=P_{5,1}$. It decays to $0$ as $n$ increases for $n>5$ $\endgroup$ – AEngineer May 22 '18 at 1:14

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