For a single die, let's find the expected value for the number of hits.
You have a $\dfrac{1}{2}$ chance of not getting any hits on one die.
You have a $\dfrac{2}{6} + \dfrac{1}{6}\cdot \dfrac{1}{2} = \dfrac{5}{12}$ chance of getting exactly 1 hit on one die.
You have a $\dfrac{1}{6}\cdot \dfrac{1}{3}+\dfrac{1}{6^2}\cdot \dfrac{1}{2} = \dfrac{5}{72}$ chance of getting exactly 2 hits.
In general, for $h>0$, the probability of exactly $h$ hits is $$\dfrac{5}{2\cdot 6^h}$$
I am not going to go into the proof of this, but I will demonstrate that the total probability is preserved.
$$\sum_{h\ge 1} \dfrac{5}{2\cdot 6^h} = \dfrac{5}{12}\sum_{h\ge 0} \left(\dfrac{1}{6}\right)^h = \dfrac{5}{12}\cdot \dfrac{6}{5} = \dfrac{1}{2}$$
Plus the probability of $h=0$, which is $\dfrac{1}{2}$.
So, this gives the expected value:
$$\dfrac{5}{12}\sum_{k\ge 0}(k+1)\left(\dfrac{1}{6}\right)^k = \dfrac{3}{5}$$
So, that is the expected value for the number of hits.
Continuing, let's create some notation.
Let $P(n,h)$ be the probability of rolling $n$ dice and getting exactly $h$ hits. Let $P(n,h^+)$ be the probability of rolling $n$ dice and getting at least $h$ hits.
$$P(n,h) = \sum_{\sum_{i=1}^n a_i = h}\prod_{i=1}^n\begin{cases}\tfrac{1}{2}, & a_i = 0 \\ \tfrac{5}{2\cdot 6^{a_i}}, & a_i>0\end{cases}$$
Basically, you are summing over all nonnegative integer solutions to the Diophantine equation:
$$a_1+\cdots + a_n = h$$
This tells you which dice rolled hits, and the total number of hits is $h$. Then, multiply the probabilities that the chosen die scores that many hits.
This is an impractical approach. Instead, we can try to break down the probabilities.
Example:
$$P(4,2^+) = 1-P(4,0)-P(4,1) = 1-\left(\dfrac{1}{2}\right)^4-\dbinom{4}{1}\left(\dfrac{5}{12}\right)\left(\dfrac{1}{2}\right)^3 \approx 72\%$$
What I am doing here is for $P(4,0)$, this is the probability that not a single die rolled a hit. For $P(4,1)$, this is the probability that exactly one die rolled exactly one hit.
$$P(4,5^+) = 1-P(4,0)-P(4,1)-P(4,2)-P(4,3)-P(4,4)$$
This is a bit trickier to calculate.
$$\begin{align*}P(4,4) & = \dbinom{4}{1}P(1,4)P(3,0)+\dbinom{4}{2}\dbinom{2}{1}P(1,3)P(1,1)P(2,0)+\dbinom{4}{2}P(1,2)^2P(2,0)+\dbinom{4}{3}\dbinom{3}{1}P(1,2)P(1,1)^2P(1,0)+P(1,1)^4 \\ & = 4\left(\dfrac{5}{2\cdot 6^4}\right)\left(\dfrac{1}{2}\right)^3+12\left(\dfrac{5}{2\cdot 6^3}\right)\left(\dfrac{5}{12}\right)\left(\dfrac{1}{2}\right)^2+6\left(\dfrac{5}{72}\right)^2\left(\dfrac{1}{2}\right)^2+12\left(\dfrac{5}{72}\right)\left(\dfrac{5}{12}\right)^2\left(\dfrac{1}{2}\right)+\left(\dfrac{5}{12}\right)^4 \\ & = \dfrac{865}{6912}\end{align*}$$
Essentially, what I am calculating is the probability of the following:
Only one die hits, but four times
2 dice have hits, one rolls 3 hits one rolls 1 hit
2 dice hit, both roll 2 hits
3 dice hit, one rolls 2 hits, two others roll 1 hit each
4 dice hit, one hit each
$$P(4,3) = \dbinom{4}{1}\left(\dfrac{5}{2\cdot 6^3}\right)\left(\dfrac{1}{2}\right)^3 + \dbinom{4}{2}\dbinom{2}{1}\left(\dfrac{5}{72}\right)\left(\dfrac{5}{12}\right)\left(\dfrac{1}{2}\right)^2+\dbinom{4}{3}\left(\dfrac{5}{12}\right)^3\left(\dfrac{1}{2}\right) = \dfrac{205}{864}$$
$$P(4,2) = \dbinom{4}{1}\left(\dfrac{5}{72}\right)\left(\dfrac{1}{2}\right)^3+\dbinom{4}{2}\left(\dfrac{5}{12}\right)^2\left(\dfrac{1}{2}\right)^2 = \dfrac{85}{288}$$
$$P(4,5^+) = \dfrac{55}{768}$$
Edit: How to build an Excel spreadsheet to perform these calculations:
On Sheet1 - Probability of getting exact number of hits:
Cell C1: Number of Dice
Cell C2: 0
Cell D2: 1
Cell E2: 2
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.
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Cell M2: 10 (go as high as you like)
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.
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Cell A3: Number of Hits
Cell B3: 0
Cell B4: 1
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Cell B13: 10 (go as high as you like)
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Cell C3: =POWER(0.5,C2)
Copy this cell and paste formula to all cells C3:M3
Set Cells C4:C13 to 0
Cell D4: =5/(2*POWER(6,B4))
Copy this cell and paste formula to all cells D4:D13
Cell E4: =E2*$D4*POWER(0.5,D2))
Copy this cell and paste formula to all cells E4:M4
Cell E5: =SUMPRODUCT($D$3:$D5,N(OFFSET(D5,ROW(D$3)-ROW(D$3:D5),0)))
Copy this cell and paste formula to all cells E5:M13
At this point, you should have a table filled out with all probabilities for getting the exact number of hits (0 to 10) on any number of dice (0 to 10).
Rename the spreadsheet: NumberHitsCalculator (or whatever else you want to name it, but I am using this name below).
Add a new worksheet. I called it ThresholdCalculator.
Copy everything from NumberHitsCalculator to get the same general layout. Change cell A3 to: Minimum Number of Hits
Change cell C3:M3 to be 1 across the board (you will always get at least zero hits).
Cell ThresholdCalculator!D4: =D3-NumberHitsCalculator!D3
Copy this cell and paste formula to all cells: ThresholdCalculator!D4:M13
This will give you all of the probabilities of hitting specific thresholds given the number of dice. Examples we have already gone over, like P(4,2+) gives 0.729167 as expected. P(4,5+) gives 0.071615 as expected, as well.
Representation via generating functions
This isn't satisfactory in the sense that we still cannot obtain a closed form, but the representation is concise and easily programmable. Suppose we have $(k_6, k_8, k_{10}, k_{12})$ dice of types d6, d8, d10, and d12 respectively. Let
\begin{align*}
f_6(x) &= \left(\frac{5}{6}+\frac{1}{6}x\right)^{k_6} \\
f_{8}(x) &= \left(\frac{5}{8}+\frac{2}{8}x+\frac{1}{8}x^2\right)^{k_8} \\
f_{10}(x) &= \left(\frac{5}{10}+\frac{2}{10}x+\frac{2}{10}x^2+\frac{1}{10}x^3\right)^{k_{10}}\\
f_{12}(x) &= \left(\frac{5}{12}+\frac{2}{12}x+\frac{2}{12}x^2+\frac{2}{12}x^3+\frac{1}{12}x^4\right)^{k_{12}} \\
f(x) &= f_6(x)f_8(x)f_{10}(x)f_{12}(x)
\end{align*}
Let $N$ be the random variable denoting the total number of successes (slightly different notation from your post, where you let $N$ represent the value of interest). Then, the probability of getting exactly $n$ successes is
\begin{align*}
P(N = n) =[x^n]f(x)
\end{align*}
where $[x^n]f(x)$ is the coefficient of $x^n$ of $f(x)$. The cumulative distribution function (i.e. the probability of getting $n$ successes or fewer) is
\begin{align*}
P(N \le n) = [x^n]\frac{f(x)}{1-x}
\end{align*}
And so
\begin{align*}
P(N \ge n) = 1 - [x^{n-1}]\frac{f(x)}{1-x}
\end{align*}
Finite-Sample Upper Bound
Let
\begin{align*}
K = k_6 + k_{8} + k_{10} + k_{12}
\end{align*}
and so the proportion of the $K$ dice which are d6, d8, d10, and d12 are respectively
\begin{align*}
(p_6, p_8, p_{10}, p_{12}) = (k_6, k_8, k_{10}, k_{12})/K
\end{align*}
Let $N_k \in \{0, \cdots, 4\}$ ($k = 1, \cdots, K$) be the random variable denoting the success number for each die, and
\begin{align*}
X_m = \sum_{k=1}^{K}\mathbb{I}(N_k = m)
\end{align*}
denote the number of successes produced from the $K$ dice. Then the proportion of the $K$ dice falling in each $m$ ($m = 0, \cdots, 4$), is
\begin{align*}
q_0 &= \frac{5}{6}p_6 + \frac{5}{8}p_8 + \frac{5}{10}p_{10} + \frac{5}{12}p_{12} \\
q_1 &= \frac{1}{6}p_6 + \frac{2}{8}p_8 + \frac{2}{10}p_{10} + \frac{2}{12}p_{12} \\
q_2 &= \frac{0}{6}p_6 + \frac{1}{8}p_8 + \frac{2}{10}p_{10} + \frac{2}{12}p_{12} \\
q_3 &= \frac{0}{6}p_6 + \frac{0}{8}p_8 + \frac{1}{10}p_{10} + \frac{2}{12}p_{12} \\
q_4 &= \frac{0}{6}p_6 + \frac{0}{8}p_8 + \frac{0}{10}p_{10} + \frac{1}{12}p_{12}
\end{align*}
So, $(X_0, \cdots, X_4) \sim \text{Multinomial}(K, (q_0, \cdots, q_4))$.
Finally,
\begin{align*}
P(N \ge n) &= P\left(\sum_{m=0}^{4} mX_m \ge n\right) \\
&= P\left(\exp\left(t\sum_{m=0}^{4} mX_m\right) \ge \exp(tn)\right) & z \mapsto e^{tz} \text{ is increasing for } t>0\\
&\le \frac{E\left[\exp\left(t\sum_{m=0}^{4} mX_m\right)\right]}{e^{tn}} & \text{Markov's inequality} \\
&= e^{-nt}\left(\sum_{m=0}^{4}q_m e^{mt}\right)^K \\
&= \left(\sum_{m=0}^{4}q_m e^{t(m - K^{-1}n)}\right)^K
\end{align*}
and so we can form the Chernoff bounds
\begin{align*}
P(N \ge n) \le \left(\inf_{t>0}\sum_{m=0}^{4}q_m e^{t(m - K^{-1}n)}\right)^K
\end{align*}
Example
Let's suppose we have $(k_6, k_8, k_{10}, k_{12}) = (5, 7, 11, 13)$ and want to find $P(N \ge 30)$. Then
\begin{align*}
P(N \ge 30) = 1 - [x^{29}]\frac{f(x)}{1-x} = 1- \frac{56649270689104302470179125877}{148888471031133469396697088000} \approx 0.6195
\end{align*}
Using the Chernoff bound with
\begin{align*}
K = 36, \mathbf{q} = (0.5405, 0.1931, 0.1456, 0.0907, 0.0301)
\end{align*}
We find that the infimum is attained at $t^* = 0.0894$ giving us $P(N \ge 30) \le 0.8453$.
Best Answer
You can use generating functions.
I presume D2 means dice with numbers 1 and 2.
In which case the probability generating function is
$$(x/2 + x^2/2)(x/6 + x^2/6 + x^3/6 + x^4/6 + x^5/6 + x^6/6) = \frac{x^2(x^2-1)(x^6 - 1)}{12(x-1)^2}$$
You need to find the coefficient of $x^k$ in this to get the probability that the sum is $k$.
You can use binomial theorem to expand out $\frac{1}{(x-1)^2}$ in the form $\sum_{n=0}^{\infty} a_n x^n$
You can generalized it to any number of dice with varying sides.
I will leave the formula to you.