I'm reading Morris DeGroot's Probability and Statistics. In chapter 1.9 there's an example 1.9.3 says that
suppose that 12 dice are to be rolled. We shall determine the probability $p$ that each of the six different numbers will appear twice.
The answer in the book is that
$$p = \frac{12!}{2^6 \times 6^{12}} = 0.0034$$
But after I learnt Feller's star-and-bar, I thought there should be another way of thinking. That is to put 12 balls into 6 different boxes. Then the answer could be
$$p = \frac{1}{\binom{12+5}{5}} = 0.00016160$$
As the answer numbers are different, my thought should be wrong. But I could not figure it out why I'm wrong.
Your argument is this: There are $N:={12+5\choose5}$ different ways to put $12$ indistinguishable balls into $6$ boxes numbered from $1$ to $6$, and exactly one of these allocations has $2$ balls in each box. Therefore the probability of obtaining this outcome is given by $p={1\over N}$.
But the random mechanism producing this allocation does not choose each of the $N$ allocations with equal probability: The random mechanism has $6^{12}$ equiprobable elementary events. Only $1$ of these puts all balls in the first box, but ${12!\over2^6}=7\,484\,400$ elementary events realize the desired allocation of two balls in each box.