Partition $k$ people into $t$ groups of sizes $k_1,\ldots,k_t$ where $k = \sum_{i=1}^t k_i$ randomly. What is the probability that two people belong to two different groups?
I find that the answer is roughly $$1- \sum_{i=1}^t\ \frac{k_i^2}{k^2}~. $$
However, can we have a more precise formula?
The number of ways to assign two people to the same group is $$(k-2)!\sum_{i=1}^t\binom{k_i}{2}$$ since there are $\binom{k_i}{2}$ ways to pick their positions in group $i$. The total number of ways to assign the people is $\frac{k!}{2}$. The probability works out to be $$1-\sum_{i=1}^t\frac{k_i(k_i-1)}{k(k-1)},$$ very close to your estimate.