Probability of dice with a cumulative successes Needing some Math help!
I'm working out part of a system for a game I'm working on, and I need to know if I need to tweak some things, so I'm checking to see if it all adds up (no pun intended).
defender has a base of 1, 3, 5, 7, or 11 for the target successes that the attacker has to roll; if the attacker achieves the # of successes required the attacker wins the roll, otherwise the defender wins.
the attacker rolls 1, 2, 3, 4, or 5 10-sided dice based on his/her increasing skill, however only 8-10 count as a success, and each 9-10 granting an additional die to add to the pool. That means, there is a 3/10 chance of at least one success, and 1/5 chance to add another die, to increase the chance for the attacker to hit the defender's number.
I feel like this should be really easy, but I can't figure out how to get started...
What is the chance (expressed as a ratio please...) for the attacker to win the roll against a defender with a target of 1? a target of 3, 5, 7, 11?
 A: For a single die this is pretty straightforward.
To get $N$ successes you have only two options:
1. $N-1$ consecutive 9-10's followed by an 8.
2. $N$ consecutive 9-10's followed by a 1-7, of which I'll think as $N-1$ consecutive 9-10's followed by a sequence of 9-10 and 1-7.
So this leads to:
$P(n=N) = (\frac{2}{10})^{N-1}(0.1+0.2\times 0.7)$, and $P(n=0)=0.7$ of course.
For a larger set of dice you should only sum such single die distributions, each of the original dice can be analyzed independently of the others.
If you are only interested in this for practical reasons then I wrote a script which gives me pretty good estimates.
A: Let the skill (i.e. the numbers to be rolled) be $n$.
We have the event $X_n$ that's the number of successes with skill $n$.
Now $P(X_n\geq 1)$ is pretty easy to calculate, it's as simple as $$P(X_n\geq 1) = 1-P(X_n=0)=1-(\frac{5}{8})^n$$ As you see it's quite difficult to express it as a ratio unless an $n$ is given.
$P(X_n\geq a)$ for $a> 1$ is much more complicated. Let's begin with $P(X_1\geq a)$. For that you need to roll $n-1$ times $9$ or $10$, and then $8$, $9$ or $10$. So $$P(X_1\geq a)=(\frac{2}{10})^{a-1}\times \frac{3}{10}$$ 
Now the fun/most hard part is left... $$P(X_n\geq a)$$
I'll think about it a bit more, but post it here so people can use it for further discussion and work upon it.
