probability or rolling 1 1 2 with six dice What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die?

Equivalently, to win a game, I need two 1s and a 2, (1,1,2). What is the probability of my rolling at least this with 6 dice. 

There are $6^6=46656$ possible rolls. I believe that there are 6896 'winning' rolls among these as I wrote some code to check every roll. This makes the probability around 0.1478

I'm not sure how to get this without the help of a computer though.
Approach via inclusion-exclusion and De Morgan's laws.

Let $A$ be the event that you roll at least two $1$'s.  Let $B$ be the event that you roll at least one $2$.

We are attempting to calculate then $Pr(A\cap B)$

$Pr(A\cap B)=1-Pr((A\cap B)^c)=1-Pr(A^c\cup B^c) = 1-Pr(A^c)-Pr(B^c)+Pr(A^c\cap B^c)$

We calculate each term on the right now that they are in simpler form.

$Pr(A^c)$ refers to the probability of strictly fewer than two $1$'s occurring, i.e. at most one $1$ occurring.  Either exactly one $1$ occurs or no $1$'s occur.  In the case of exactly one $1$, pick its location and then pick each remaining digit for a total of $6\cdot 5^5$ possibilities.  In the case of no $1$'s, pick each digit for a total of $5^6$ possibilities.  Taking the ratio of this compared to the $6^6$ equally likely dice rolls, we calculate $Pr(A^c)=\frac{6\cdot 5^5+5^6}{6^6}$.

$Pr(B^c)$ refers to the probability of strictly fewer than one $2$ occurring, i.e. no $2$'s.  This occurs with probability $Pr(B^c)=\frac{5^6}{6^6}$

$Pr(A^c\cap B^c)$ refers now to the probability of at most one $1$ and no $2$'s, which similarly to before we break into cases for either exactly one $1$ or no $1$'s.  For exactly one $1$, first pick the location, then pick each remaining digit for $6\cdot 4^5$ possibilities and for no $1$'s pick each digit for $4^6$ possibilities for a probability of $Pr(A^c\cap B^c)=\frac{6\cdot 4^5+4^6}{6^6}$

This gives a final probability of:

$$Pr(A\cap B) = 1-\frac{6\cdot 5^5+5^6}{6^6}-\frac{5^6}{6^6}+\frac{6\cdot 4^5+4^6}{6^6}$$

$$=1-\frac{39760}{46656} = \frac{431}{2916}\approx 0.147805$$
As I understood your question, you are asking about probability that between $6$ values will be at least two $1$'s and one $2$ (don't looking at other values and order).

Well, so the probability of $(1,1,2,x,x,x)$ (excluding order and remarking that there may be other $1$'s and $2$'s between $x$'s) is exactly the number of sets of six digits from $1$ to $6$ that start with $1,1,2$ divided by total number of sets of six digits from $1$ to $6$. There are as many (disordered) sets of six digits from $1$ to $6$ that start with $1,1,2$ as just (disordered) sets of three digits from $1$ to $6$. 

Thus the answer is $P = \frac{(3^3)/3!}{(6^6)/6!} = \frac{5}{72}$.