Odds of specific scenario after $4$ dice rolls I have started studying for a statistics competition but seem to be a bit rusty and unsure about the answer (which I can't check). 
"A fair die is rolled $4$ times. What's the probability of each of the numbers $1, 2$ and $3$ appearing at least once each?"
I tried thinking that I can get the numbers $1, 2$ and $3$ then a fourth roll providing $6$ different numbers. That would leave me $6$ possibilities of getting $1, 2$ and $3$ on the $3$ first rolls, in that order.
Since I don't need the $1,2$ and $3$ necessarily on the $3$ first rolls, I thought they can actually be placed in $C_{4,3} =4$ "positions" of rolls.
And also, they can come in $3!=6$ possible orders. 
Adding all that up equals to ${\rm{6}}{\rm{.4}}{\rm{.6 = 144}}$ possibilities of getting $1, 2$ and $3$ at least once, from a total of ${6^4} = 1296$ different outcomes.
$$P = {{144} \over {1296}} \approx 0.11$$
It is a multiple choice and the answer is one from: $${5 \over {36}},{1 \over {12}},{5 \over {12}},{1 \over {24}}$$
And none of the above is equivalent to 0.11, unfortunately. Can anyone help figuring out which step I got wrong?
 A: Brute-forcing:
$$\begin{array}{c|c|c|c}
123(6)&12(5)3&1(4)23&(3)123\\
\hline
132(6)&13(5)2&1(4)32&(3)132\\
\hline
213(6)&21(5)3&2(4)13&(3)213\\
\hline
231(6)&23(5)1&2(4)31&(3)231\\
\hline
312(6)&31(5)2&3(4)12&(3)312\\
\hline
321(6)&32(5)1&3(4)21&(3)321\\
\hline
3!\cdot 6&3!\cdot 5&3!\cdot 4&3!\cdot 3&108\end{array}$$
Hence, the required probability is:
$$\frac{108}{6^4}=\frac1{12}.$$
A: You're double-counting cases like $1,2,3,3$. You need to find how many of those there are and subtract that, since you've counted each of them twice. That yields one of the solutions offered.
A: One approach is via the Principle of Inclusion / Exclusion (PIE).
Say the outcome of four rolls has "Property $i$" if there is no roll of $i$, for $i=1,2,3$.  Let $S_j$ be the sum of the probabilities of the outcomes with $j$ of the properties, for $j=1,2,3$.
Then
$$\begin{align}
S_1 &= \binom{3}{1} \left( \frac{5}{6} \right)^4 \\
S_2 &= \binom{3}{2} \left( \frac{4}{6} \right)^4 \\
S_3 &= \binom{3}{3} \left( \frac{3}{6} \right)^4 
\end{align}$$
By PIE, the probability that the four rolls have none of the properties, i.e. each number from 1 to 3 appears at least once, is
$$1-S_1 + S_2 - S_3 = \frac{1}{12}$$
A: Let's just count the number of ways it could happen.
If they each appear once each, so the last number is a 4,5,6:
* pick if the last number is 4,5,6 (3)

* order everything (4!)

If one of them appears twice:
* pick which appears twice (3)

* order everything, but unorder the numbers that appear twice (4!/2!)

Add those products and then divide by 6^4.
