Probability of selecting a ball from a box. Suppose we have $12$ identical balls. Think of the balls as having distinct ID numbers, $1-12$. We have $3$ identical boxes, each of which can contain exactly $4$ balls. One after the other the balls are thrown at random into one of the three boxes.
Suppose Mr. A will buy one ball from each box, so he will buy a total of three balls from the three boxes. Habit of Mr. A is to buy that ball from the box which ID is smallest of all other balls in that box.
For example, the randomly thrown balls in a box have ID $4$, $8$, $9$, and $12$. So Mr. A will buy the ball with ID $4$ from that box.
Then Mr. A goes to another box and here the balls have ID $1$, $3$, $6$, and $11$. He will buy the ball with ID $1$ from that box.
At last, Mr. A goes to the last box and here the balls have ID $2$, $5$, $7$, and $10$. He will buy the ball with ID $2$ from that box.
My question is:

What is the probability that Mr. A will buy the ball with ID $i$, where $i=1,2,\ldots,12$?

What I think is that we need to consider the following two things to calculate the probability:

$(1)$ the position of a ball in the box. Mr. A will buy the ball with ID number $2$ if and only if the box does not contain the ball with ID number $1$.
$(2)$ If Mr. A buy a ball with smallest ID from the first box in front of him, then are there $11$ balls remaining to buy another two balls or are there $8$ balls (excluding all $4$ balls from the first box)?

 A: I got the following answer:

The probability that Mr. A will buy the ball with id $i$ is,
  $$p(i) := \left( \begin{array}{c} 8 \\ i-1 \end{array} \right) \frac{(12-i)!(i-1)!}{11!}\,. $$

If you know this answer to be incorrect, please feel free to stop reading :)
Total number of ways the balls can be distributed among three boxes is $$n := \frac{12!}{(4!)^3}\,.$$
Let us assume that there are $12$ numbered slots, one slot is to be occupied by exactly one ball, slots $1$ to $4$ constitute box $1$, slots $5$ to $8$ constitute box $2$ and the rest constitute box $3$. At the beginning we will consider all the slots to be distinguishable but at the end we will consider all the slots in the same box to be indistinguishable from each other.
We will count the number ways the $i$'th ball can be bought from box $1$. The $i$'th ball can be placed in either of the $4$ slots in box $1$. The balls with id $1$ to $(i-1)$ can be placed in box $2$ and $3$ in $\left( \begin{array}{c} 8 \\ i-1 \end{array} \right)(i-1)!$ ways. Balls with id $(i+1)$ to $12$ can be arranged in the rest of slots in $(12-i)!$ ways. Multiplying all these numbers we get $4 \left(\begin{array}{c} 8 \\ i-1 \end{array}\right) (i-1)! (12-i)!$, in order to impose indistinguishability of the slots in the same box we must divide this number by $(4!)^3$. So, the number ways the $i$'th ball can be bought from box $1$ is:
$$n_1(i) := \frac{4}{(4!)^3} \left(\begin{array}{c} 8 \\ i-1 \end{array}\right) (i-1)! (12-i)!\,.$$
Therefore, the probability of the $i$'th ball being bought from box one is:
$$p_1(i) := \frac{n_1(i)}{n}\,.$$
The $i$'th ball can be bought from either of the three boxes, so the net probability of the $i$'th ball being bought is:
$$p(i) := 3 p_1(i) = \left( \begin{array}{c} 8 \\ i-1 \end{array} \right) \frac{(12-i)!(i-1)!}{11!}\,.$$
A: Ball $i$ is thrown into a box with $3$ other balls. There are $\binom{11}3$ possible combinations for the three balls in the same box with ball $i,$ and all of them are equally likely. The man will buy ball $i$ if the three other balls have numbers greater than $i$. There are $12-i$ balls with numbers greater than $i,$ so there are $\binom{12-i}3$ favorable cases, and the probability that the man buys ball $i$ is
$$\frac{\binom{12-i}3}{\binom{11}3}=\frac{(12-i)(11-i)(10-i)}{990}.$$
This is equivalent to the answer posted earlier by Nafiz Ishtiaque Ornob.
A: For $1\lt j\lt k$, there are $3!\binom{12-k}{3}\binom{8-j}{3}$ ways to buy $1,j,k$.
$3!$ due to the arrangements of $1,j,k$. $\binom{12-k}{3}$ due to the number of ways to choose $3$ balls greater than $k$. $\binom{8-j}{3}$ due to the number of ways to choose $3$ balls greater than $j$ from those left over from choosing $3$ greater than $k$.
The number of ways for $n$ to be the greatest is
$$
\begin{align}
3!\binom{12-n}{3}\sum_{j=2}^{n-1}\binom{8-j}{3}
&=3!\binom{12-n}{3}\left[\binom{7}{4}-\binom{9-n}{4}\right]\\
&=3!\binom{7}{4}\binom{12-n}{3}-3!\binom{7}{4}\binom{12-n}{7}
\end{align}
$$
The number of ways for $n$ to be the middle is
$$
\begin{align}
3!\binom{8-n}{3}\sum_{k=n+1}^9\binom{12-k}{3}
&=3!\binom{8-n}{3}\binom{12-n}{4}\\
&=3!\binom{7}{4}\binom{12-n}{7}
\end{align}
$$
Since the total number of arrangements is $\frac{12!}{(4!)^3}$, and $\frac{3!\binom{7}{4}}{\frac{12!}{(4!)^3}}=\frac1{165}$, the probability of buying the ball labeled $n$ would be
$$
p(n)=\frac1{165}\binom{12-n}{3}
$$
