<s>[INST] 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.
 [/INST] You can consider the possibility to find probability directly without inclusion-exclusion principle. This is a longer way, but there are those who like it. Consider all possibilities to roll at least two ones and at least one two with six 6-sided die. We can have exactly two $1$ and one $2$. Denote this event by $A_{21}$. Or we can have exactly two $1$ and two $2$: event $A_{22}$. There are $10$ disjoint events total: $A_{21}$, $A_{22}$, $A_{23}$, $A_{24}$, $A_{31}$, $A_{32}$, $A_{33}$, $A_{41}$, $A_{42}$, $A_{51}$. 
Find and then add their probabilities. The probability of $A_{21}$ is
$$
\mathbb P(A_{21})=\frac{\binom{6}{2}\binom{6-2}{1}4^{6-2-1}}{6^6}=\frac{15\cdot 4\cdot 4^3}{6^6}.
$$
By first multiplier $\binom{6}{2}=15$ we choose the rolls which result in $1$, next multiplier $\binom{6-2}{1}=4$ we choose the roll among the rest which results in $2$, and the last multiplier $4^3$ is number of possible outcomes in the rest $6-2-1=3$ rolls.
Similar, 
$$\mathbb P(A_{22})=\frac{\binom{6}{2}\binom{4}{2}4^{2}}{6^6}=\frac{15\cdot 6\cdot 16}{6^6},$$
$$\mathbb P(A_{23})=\frac{15\cdot 4\cdot 4}{6^6},\quad \mathbb P(A_{24})=\frac{15}{6^6},\quad \mathbb P(A_{31})=\frac{20\cdot 3\cdot 16}{6^6},$$
$$\mathbb P(A_{32})=\frac{20\cdot 3\cdot 4}{6^6}, \quad \mathbb P(A_{33})=\frac{20}{6^6}, \quad \mathbb P(A_{41})=\frac{15\cdot 2\cdot 4}{6^6},$$
$$\mathbb P(A_{42})=\frac{15}{6^6}, \quad \mathbb P(A_{51})=\frac{6}{6^6}.$$
The total probability of rolling at least two ones and at least one two with six 6-sided die is equal to the sum of probabilities above:
$$
\frac{6896}{6^6}=0,147805213.
$$
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