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 There are 36 people split into 12 groups of 3.
If the same 36 people are split into 12 groups again, what is the probability that one or more of those groups will have the same 3 people as last time?
Edit: as recommended by a comment:
I don't have a lot of background beyond a first year stats course at university 10 years ago.
What I've tried so far:
There are $36 \choose 3$ = 7140 possible groups. 12 of those groups were used last time, so when picking the first group, there's a $\frac{12}{7140}$ chance that group appeared the first time.
I'm not really sure where to go from there though. There are $33 \choose 3$ = 5456 possibilities for second group, but I'm not sure how to go about figuring out the probability that second group appeared last time.
If this is really complicated and it's easier to show how to calculate it for 12 people in 4 groups of 3 that would at least help me understand the calculation.
Stuff I wrote out that may or may not be relevant:

*

*$\frac{36!}{(12! * (3!)^{12})}$ = possible arrangements of 12 groups of 3 people

*$36 \choose 3$ = 7140

*$33 \choose 3$ = 5456

*$30 \choose 3$ = 4060

*$27 \choose 3$ = 2925

*$24 \choose 3$ = 2024

*$21 \choose 3$ = 1330

*$18 \choose 3$ = 816

*$15 \choose 3$ = 455

*$12 \choose 3$ = 220

*$9 \choose 3$ = 84

*$6 \choose 3$ = 20

*$3 \choose 3$ = 1

I think I might need to find the probability that none of the groups are the same and then subtract that from 1 to find the probability that at least one is the same.
 A: 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 $p = 0.0199463$.
