Given $n$ fair coins, we define round $i$ as flipping all of them and obtain one observation $\vec{o_i}$.
We know the domain $O$ has $2^n$ variations.
If the game is designed as terminate as whenever all $2^n$ variations have being observed. What's the expected number of rounds $I$ for the game to terminate?
Even though all the coins are fair, but not necessarily independent. Assume we design the procedure of producing $\vec{o}$ and we know the correlation $\Sigma^{n \times n}$.
The expected number of rounds corresponds to a coupon collector's problem. But as the question's requirement that coins are not necessarily independent from each other. So it's not going to be a standard coupon collector's problem but a non-uniform coupon collector's problem.
The expected number of rounds is going to depend on the probability of rarest observation $P(\vec{o_r})$ and the most common observation $P(\vec{o_c})$.
Based on this paper.
The expected number is going to be bounded between $O(\frac{P(\vec{o_c})}{P(\vec{o_r})} * 2^n * log(2^n))$ and $O(\frac{P(\vec{o_r})}{P(\vec{o_c})} * 2^n * log(2^n))$