Imagine if you will for a moment that for each specifically named card, you have a unique id number on it. I.e. you have an Ornithopter1, an Ornithopter2, an Ornithopter3, etc...
As there are 60 cards total in the deck and your handsize is 7 cards, the number of different starting hands is $\binom{60}{7}$.
Our ideal hand contains at least one of each of our five card categories, abbreviated hereout as $A,B,C,D,E$, which have 6,4,4,4, and 17 copies available respectively. There are then 25 other cards (not of the aforementioned types) in the deck.
We can approach the counting required directly or via inclusion-exclusion. Both methods at first glance will require similar amounts of work, so I will go directly.
$\begin{array}{c|c|c|c|c|l}
~A&B&C&D&E&\text{total hands}\\
\hline
1&1&1&1&1&6\cdot 4\cdot 4\cdot 4\cdot 17\cdot \binom{25}{2}\\
2&1&1&1&1&\binom{6}{2}\cdot 4\cdot 4\cdot 4\cdot 17\cdot 25\\
1&2&1&1&1&6\cdot \binom{4}{2}\cdot 4\cdot 4\cdot 17\cdot 25\\
1&1&2&1&1&6\cdot 4\cdot\binom{4}{2}\cdot 4\cdot 17\cdot 25\\
1&1&1&2&1&6\cdot 4\cdot 4\cdot \binom{4}{2}\cdot 17\cdot 25\\
1&1&1&1&2&6\cdot 4\cdot 4\cdot 4\cdot \binom{17}{2}\cdot 25\\
\color{red}{2}&\color{red}{2}&\color{red}{1}&\color{red}{1}&\color{red}{1}&\color{red}{\binom{6}{2}\cdot \binom{4}{2}\cdot 4\cdot 4\cdot 17}\\
2&1&2&1&1&\binom{6}{2}\cdot4\cdot\binom{4}{2}\cdot4\cdot17\\
\vdots\\
3&1&1&1&1&\binom{6}{3}\cdot4\cdot4\cdot4\cdot17\\
\vdots\\
\end{array}$
6*4^3*17*25*12 + 15*4^3*17*25 + 6*6*4^2*17*25*3 + 6*4^3*17*8*25 + 15*6*4^2*17*3 + 6^3*4*17*3 + 6^3*4*17*3 + 6^2*4^2*17*8*3 + 20*4^3*17+6*4^3*17*3 + 6*4^3*17*8*5 = 4479840 + 625600 = 5105440
The probability then is: $$\frac{5105440}{\binom{60}{7}}\approx 0.0132$$
or about one in seventy-five starting hands.
How to calculate one of the rows:
Taking the red row for example, we are in the situation where there are 2 of card A, two of card B, one of card C, one of card D, one of card E, and 0 cards of an unlisted type. By multiplication principle, pick which specific copies of card A you use: $\binom{6}{2}$ number of choices, pick which specific copies of card B you use: $\binom{4}{2}$ number of choices, pick which specific copy of card C you use: $\binom{4}{1}=4$ number of choices, etc... for a total number of $\binom{6}{2}\binom{4}{2}\cdot 4\cdot 4\cdot 17$ different hands in this specific case.
Note: I did not account for the cases where you have no lands and two Mox Opal yet, but those can be rather easily added in if you feel like stretching your muscles a bit. Just add the cases where $D\geq 2$ and $E=0$ and $A,B,C\geq 1$. This also does not take into account mulligan rules where you might be allowed to ditch a starting hand if it was particularly bad.
In particular with $A=B=C=1, D=2$ and $E=0$ it will add another $6\cdot4\cdot4\cdot\binom{4}{2}\cdot\binom{25}{2} = 172800$ additional "acceptable" hands, bumping the probability up a little higher. The other cases will add a rather insignificant amount in comparison.
