What is the probability of rolling at least two 6's with 3 Dice and 2 Rolls? Question: What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls? (with your first roll you keep dice only if they are 6's and roll the remainder for your second roll).
(6*6*6 = 216 = outcomes when rolling 3 dice)
(6*6 = 36 = outcomes when rolling 2 dice)

1st roll: (3 dice)
A = 125/216 = 0 sixes
B = 75/216 = 1 six
C = 15/216 = 2 sixes
D = 1/216 = 3 sixes
Outcomes C and D fulfill requirement.

For outcome A: (first roll = no 6's, pick up all dice throw 3 dice again)
2nd roll:
a = 125/216 = 0 sixes
b = 75/216 = 1 six
c = 15/216 = 2 sixes
d = 1/216 = 3 sixes
Outcomes c and d fulfill requirements.

For outcome B: (first roll = one 6, pick up two non 6's and roll again)
2nd roll:
z = 25/36 = 0 sixes
y = 10/36 = 1 six
x = 1/36 = 2 sixes
Outcomes y and x fulfill requirements.

So would the formula below give me my answer? 
C + D + Ac + Ad + By + Bx = X

What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls?
If I substituted correctly and did the math correct the answer I got was 22.3%
Is this correct?
 A: An alternate approach would be to find the probability of the complementary event:
$\textbf{1)}$  The probability of getting no 6's is given by $\big(\frac{5}{6}\big)^3\cdot\big(\frac{5}{6}\big)^3=\big(\frac{5}{6}\big)^6$
$\textbf{2)}$ The probability of getting exactly one 6 is given by
$\hspace{.2 in}\big(\frac{5}{6}\big)^3\cdot3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2+3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2\cdot\big(\frac{5}{6}\big)^2=\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5+\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4$
Therefore the probability of getting at least two 6's is given by 
$\hspace{.2 in} 1-\big(\frac{5}{6}\big)^6-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4=\frac{5203}{23328}\approx.223$
A: Your approach and your calculations are correct.  Here is  a variation based upon    generating     functions.    We   encode  the roll  of  three  dice with
\begin{align*}
(5+t)^3
\end{align*}
marking an   occurrence of $6$ with $t$ and collecting all other five possibilities with $5$. The probability to get $j$ sixes $0\leq j \leq 3$ in the first roll can be written as
\begin{align*}
[t^j](5+t)^3\cdot\frac{1}{6^3}
\end{align*}

Encoding the second roll with $(5+u)^{3-j}$ we calculate
  \begin{align*}
\sum_{{0\leq j,k\leq 3}\atop{j+k\geq 2}}&[t^j](5+t)^3[u^k](5+u)^{3-j}\cdot\frac{1}{6^{6-j}}\\
&=[t^0](5+t)^3\left([u^2]+[u^3]\right)(5+u)^3\cdot\frac{1}{6^6}\\
&\qquad+[t^1](5+t)^3\left([u^1]+[u^2]\right)(5+u)^2\cdot\frac{1}{6^5}\\
&\qquad+[t^2](5+t)^3\left([u^0]+[u^1]\right)(5+u)^1\cdot\frac{1}{6^4}\\
&\qquad+[t^3](5+t)^3\\
&=\binom{3}{0}5^3\left(\binom{3}{2}5^1+\binom{3}{3}5^0\right)\cdot\frac{1}{6^6}\\
&\qquad+\binom{3}{1}5^2\left(\binom{2}{1}5^1+\binom{2}{2}5^0\right)\cdot\frac{1}{6^5}\\
&\qquad+\binom{3}{2}5^1\left(\binom{1}{0}5^1+\binom{1}{1}5^0\right)\cdot\frac{1}{6^4}\\
&\qquad+\binom{3}{3}5^0\cdot\frac{1}{6^3}\\
&=\frac{125\cdot16}{6^6}+\frac{75\cdot11}{6^5}+\frac{15\cdot6}{6^4}+\frac{1}{6^3}\\
&=\frac{5\,203}{23\,328}\\
&\doteq    0.223
\end{align*}

