I want to work out the probability of at least x successes on n D10's where 7 an above is a success but 10 counts as two successes. Easy enough without the 10's counting as two successes with binomial distribution. Probability of success is 0.4 per dice and can work out the probability for any value of x and n using an online calculator, excel etc. But with a 10 on a d10 counts as 2 successes instead of 1, I'm lost. How do I work this out mathematically?
Edit
This is for a game. A friend is making an excel worksheet to work out the probabilities of success based on varying dice pools (n) and number of successes needed to pass (x). But he doesn't math. I kinda do... well once upon a time... I'm not after answers and explanations on how the math behind them work but the names of functions so I can work out how things work with the power of google would be appreciated. And it will all be plugged into excel eventually if possible so answers with that in mind would be appreciated.
Not a closed form, but here's a recursion . . .
For integers $x,n$ with $n \ge 0$, let $p(x,n)$ be the probability of at least $x$ successes in $n$ rolls.
Then $p$ satisfies the recursion $$p(x,n)=\frac{6}{10}p(x,n-1)+\frac{3}{10}p(x-1,n-1)+\frac{1}{10}p(x-2,n-1)$$ if $n > 0%$, together with the boundary conditions \begin{align*} p(x,0)&=0,\;\;\text{if $x > 0$}\\[4pt] p(x,n)&=1,\;\;\text{if $x \le 0$}\\ \end{align*}