These can be solved at the same time, the closed roll is just a special case of the open ended.
Let
$$d = \text{Number of sides on each die}$$
$$n = \text{Number of dice rolled}$$
$$s = \text{Score required for success}$$
$$m = \text{number of successes required}$$
Also, let
$$s = n^k+r$$
where
$$k\in 0,1,2,\dots$$
$$r\in 1,2,\dots,d$$
And, let
$$\begin{align}
p_o &= \text{chance of getting open ended}\\
&= \begin{cases}
\frac{1}{d}&,\text{if roll is open}\\
0&,\text{otherwise}\\
\end{cases}
\end{align}$$
So, for 1 die, we have a geometric distribution to get $k$ open ended results followed by a single roll to get $r$ (note that this is not exactly a geometric distribution since we need $k$ failures with $q=1-p_0$ followed by 1 success with a different probability). Also we will require $x^0=1$.
$$p_1=\left(p_0\right)^{k}\left(\frac{d-r+1}{d}\right)$$
To get $m$ success from $n$ dice, we have a binomial distribution as you noticed so
$$\begin{align}
p_{m,n}&=\sum_{i=m}^n\binom{n}{i}p_1^i\left(1-p_1\right)^{n-i}\\
\end{align}$$
Some examples:
- $d=6$, $n=1$, $s=4$, $m=1$ & open ended; so $k=0$, $r=4$
$$\begin{align}
p_0=\frac{1}{6}\\
\end{align}$$
$$\begin{align}
p_1&=\left(\frac{1}{6}\right)^{0}\left(\frac{6-4+1}{6}\right)\\
&=1\left(\frac{3}{6}\right)\\
&=1\times\frac{1}{2}\\
&=\frac{1}{2}\\
\end{align}$$
$$\begin{align}
p_{1,1}&=\sum_{i=1}^1\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{1}{1}\left(\frac{1}{2}\right)^{1}\left(1-\frac{1}{2}\right)^{1-1}\\
&=1\times\frac{1}{2}\times{1}\\
&=\frac{1}{2}\\
\end{align}$$
- $d=6$, $n=1$, $s=10$, $m=1$ & open ended; so $k=1$, $r=4$
$$\begin{align}
p_0=\frac{1}{6}\\
\end{align}$$
$$\begin{align}
p_1&=\left(\frac{1}{6}\right)^{1}\left(\frac{6-4+1}{6}\right)\\
&=\frac{1}{6}\times\frac{3}{6}\\
&=\frac{1}{12}\\
\end{align}$$
$$\begin{align}
p_{1,1}&=\sum_{i=1}^1\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{1}{1}\left(\frac{1}{12}\right)^{1}\left(1-\frac{1}{12}\right)^{1-1}\\
&=1\times\frac{1}{12}\times{1}\\
&=\frac{1}{12}\\
\end{align}$$
- $d=6$, $n=3$, $s=4$, $m=2$ & closed; so $k=0$, $r=4$
$$\begin{align}
p_0=0\\
\end{align}$$
$$\begin{align}
p_1&=\left(0\right)^{0}\left(\frac{6-4+1}{6}\right)\\
&=1\left(\frac{3}{6}\right)\\
&=\frac{1}{2}\\
\end{align}$$
$$\begin{align}
p_{3,2}&=\sum_{i=2}^3\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{3}{2}\left(\frac{1}{2}\right)^{2}\left(1-\frac{1}{2}\right)^{3-2}+\binom{3}{3}\left(\frac{1}{2}\right)^{3}\left(1-\frac{1}{2}\right)^{3-3}\\
&=3\times\frac{1}{4}\times\frac{1}{2}+1\times\frac{1}{8}\times1\\
&=\frac{4}{8}\\
&=\frac{1}{2}\\
\end{align}$$
- $d=6$, $n=3$, $s=10$, $m=2$ & open ended; so $k=1$, $r=4$
$$\begin{align}
p_0=\frac{1}{6}\\
\end{align}$$
$$\begin{align}
p_1&=\left(\frac{1}{6}\right)^{1}\left(\frac{6-4+1}{6}\right)\\
&=\left(\frac{1}{6}\right)\left(\frac{3}{6}\right)\\
&=\frac{1}{12}\\
\end{align}$$
$$\begin{align}
p_{3,2}&=\sum_{i=2}^3\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{3}{2}\left(\frac{1}{12}\right)^{2}\left(1-\frac{1}{12}\right)^{3-2}+\binom{3}{3}\left(\frac{1}{12}\right)^{3}\left(1-\frac{1}{12}\right)^{3-3}\\
&=3\times\frac{1}{144}\times\frac{11}{12}+1\times\frac{1}{1,728}\times1\\
&=\frac{34}{1,728}\\
&=\frac{17}{864}\\
&\approx0.02\\
\end{align}$$
When you roll 2 dice there are 36 equally likely options, starting at double one, $(1,1)$, and going all the way to double six, $(6,6)$. Of these, there are 11 equally likely ways to roll a $5$, $(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3),(5,4),(5,6)$, so the odds of rolling a 5 is $\frac{11}{36}$.
Similarly there are 18 equally likely ways to roll an odd number (I'll let you think about which combinations these are), so the odds of rolling an odd number are $\frac{18}{36}=\frac{1}{2}$.
At this point you may think the odds of rolling a $5$ or and odd number should be $\frac{11}{36}+\frac{18}{36}=\frac{29}{36}$, however this is not the case. Note that six of the rolls contain both a $5$ and are odd:
$(2,5), (4,5), (6,5), (5,2), (5,4),(5,6)$
If we just add the two probabilities, we'll count these situations twice, even though they're no more likely to occur than any other combination! In reality there are only $11+18-6=23$ dice rolls which contain a $5$ or are odd (see if you can list them), and so the odds of rolling either a $5$ or an odd combination is $\frac{11}{36}+\frac{18}{36}-\frac{6}{36}=\frac{23}{36}$.
This is a simple application of the "principle of inclusion and exclusion", which you may want to look up.
Best Answer
Let $p(t,d)$ be the probability that when you roll $d$ dice, the sum of their numbers is exactly $t.$
Then $p(t,d) = 0$ whenever $t < 0.$ Also $p(t,0) = 0$ when $t > 0,$ but $p(0,0) = 1.$
For $d \geq 0,$ $$p(t,d+1) = \frac16 p(t-1,d) + \frac13 p(t-2,d) + \frac13 p(t-3,d) + \frac16 p(t-4,d).$$
You can use that formula to find any value of $p(t,d)$ recursively. An efficient way to do this is to compute $p(t,1)$ for every relevant value of $t,$ then $p(t,2)$ for every relevant $t,$ and so forth until you have computed the necessary values of $p(t,d)$ in order to find the probability of rolling at least $x$; that probability is $$ \sum_{t\geq x} p(t,d).$$