Imagine there are $N$ people throwing a party. For any two of them, the time before they meet each other and stick together thereafter is independent, and obeys an exponential distribution whose $\lambda = 1$. After they stick together, the newly formed group acts as an individual, and can merge with other individuals/groups following the same rule.
In other words, the time before any two groups (consisting of 1 or more people) meet each other and stick together thereafter is independent, and obeys an exponential distribution whose $\lambda = 1$. Obviously, these $N$ people will (almost surely) form a group of size $N$ in the end.
The question is, what's the average time an individual finds their first group? Here "find their first group" includes (1) form a group with another individual, and (2) join an existing group.
I have tried considering all people except this individuals as an whole, but the later is having internal merges, so it becomes a "compound non-homogeneous Poisson process" (not sure if that's the correct term) which I cannot solve.
Before the first pair joins there are $\frac 12N(N-1)$ possible pairs, so the time distribution is exponential with $\lambda=\frac {N(N-1)}2$. This takes on average $\frac 2{N(N-1)}$ and our person is part of the group with probability $\frac 2N$. If the person is not part of the first group, we have the same situation with $N-1$ groups, so it takes on average $\frac 2{(N-1)(N-2)}$ and our person joins with probability $\frac 2{N-1}$
We can write a recurrence. Let $T(n)$ be the expected time for a given individual to join a group assuming there are $n$ groups. We have $$T(n)= \frac 2{n(n-1)}+\frac {n-2}nT(n-1)\\T(2)=1$$ A quick calculation shows $T(n)=\frac 2n$. I am sure this can be proved by induction, but I haven't done so.