Consider a simpler case: 4 boxes and balls numbered from 1 to 6. We still put 3 balls in each box. I have chosen this case, since it is possible to write out all combinations and check corresponding probabilities.
For each box, we now have $6\choose3$ possible combinations.
In the first box, we can select any ball, yielding a probability of 1.
For the second box, we have two options to select a valid ball (i.e. a ball not selected previously):
- A box without the ball that we already selected. The total number of possibilities for this case are ${1\choose0}{5\choose3}$. ${1\choose0}$ specifies the number of balls we select from the previously selected balls. ${5\choose3}$ specifies the number of combinations of choosing three balls from the remaining five that have not been selected previously. The probability of selecting a valid ball is 1, since the previously selected ball is not in the box.
- A box with the ball that we already have selected. Out of the $6\choose3$ possibilities, ${1\choose1}{5\choose2}$ have the already selected ball. Selecting a valid ball has a probability of $\frac{2}{3}$.
Thus, selecting a valid ball from the second box is given by $\frac{1\times {1\choose0}{5\choose3} + \frac{2}{3}\times{1\choose1}{5\choose2}}{6\choose3} = \frac{5}{6}$.
For the third box, we have the following cases:
- A box without the previously selected balls: ${2\choose0}{4\choose3}$ possibilities. The probability of selecting a valid ball is 1.
- A box with one previously selected ball: ${2\choose1} {4\choose2}$. The probability of selecting a valid ball is $\frac{2}{3}$.
- A box with two previously selected balls: ${2\choose2}{4\choose1}$ possibilities, with a probability of $\frac{1}{3}$ of not selecting a previously selected ball.
For the third box, the probability of selecting a valid ball is consequently:
$\frac{1 \times {2\choose0}{4\choose3} + \frac{2}{3} \times {2\choose1} {4\choose2} + \frac{1}{3}\times {2\choose2}{4\choose1}}{6\choose3} = \frac{4}{6}$.
For the fourth box, we have the following cases:
- No previously selected balls: ${3\choose0}{3\choose3}$ possibilities. We have a probability of 1 of selecting a valid ball.
- One previously selected ball: ${3\choose1}{3\choose2}$. Corresponding probability is $\frac{2}{3}$.
- Two previously selected balls: ${3\choose2}{3\choose1}$. Corresponding probability is $\frac{1}{3}$.
- Three previously selected balls: ${3\choose3}{3\choose0}$. Corresponding probability is 0.
For the fourth box, the probability of selecting a valid ball is thus given by:
$\frac{1 \times {3\choose0}{3\choose3} + \frac{2}{3}\times {3\choose1}{3\choose2} + \frac{1}{3} \times {3\choose2}{3\choose1} + 0 \times {3\choose3}{3\choose0}}{6\choose3} = \frac{3}{6}$.
We thus find the probability of valid combinations for this scenario as follows:
$\text{P(selected balls are distinct)} = \frac{6}{6}\times\frac{5}{6}\times\frac{4}{6}\times\frac{3}{6} = \frac{\frac{6!}{(6-4)!}}{6^4}$.
This pattern also arises in your scenario. In your scenario, we find a probability of:
$\text{P(Selected balls are all distinct)} = \frac{\frac{12!}{(12-10)!}}{12^{10}}\approx 0.003868$.