What is the probability of at least one group having the same members when 36 people are randomly split into 12 groups of 3, twice

In order to simplify discussion, let's assume the groups are distinguishable, and the first time the people are split into groups, the groups are $\{1,2,3\}, \{4,5,6\}, \dots , \{34,35,36\}$. The second time the people are split into groups, this can be done in $N = 36! / (3!)^{12}$ ways, all of which we assume are equally likely. We want to count the number of arrangements in which at least one of the original groups appears in the new groupings, although maybe not in the same spot as initially. We will use the Principle of Inclusion and Exclusion.

Let's say a new arrangement has "Property $i$" if the group numbered $i$ in the initial arrangement is found somewhere in the new arrangement, for $1 \le i \le 12$, and let $S_j$ be the number of arrangements with $j$ of the properties, for $1 \le j \le 12$. If $j$ of the original groups appear in the new arrangement then the initial groups can be chosen in $\binom{12}{j}$ ways, and their new locations can be chosen in $\binom{12}{j} j!$ ways. Then the remaining people can be grouped in $(36-3j)!/(3!)^{12-j}$ ways. So $$S_j = \binom{12}{j}^2 j! \frac{(36-3j)!}{(3!)^{12-j}}$$ By inclusion/exclusion, the number of arrangements with at least one of the properties, which is the number of arrangements with at least one group having the same members in the first and second splits, is $$N_1 = S_1 - S_2 + S_3 - \dots -S_{12}$$ and the probability of this event is $$p = \frac{N_1}{N}$$ Calculation results in $p = 0.0199463$.