Say I'm sampling from a large population of size $N$ without replacement, and denote by $\pi_i$ the probability that unit $i$ is included in the sample, and $\pi_{ij}$ the probability that both $i$ and $j$ are included. The variance of the Horvitz-Thompson estimator $\hat{\tau}_\pi$ for the population total is given by $$Var(\hat{\tau}_\pi)=\sum\limits_{i=1}^N \left( \dfrac{1-\pi_i}{\pi_i}\right)y^2_i + \sum\limits_{i=1}^N \sum\limits_{j\neq i} \left( \dfrac{\pi_{ij}-\pi_i \pi_j}{\pi_i \pi_j}\right) y_i y_j$$
It seems that if I'm sampling a fixed number of units without replacement, there's no way to make the sampling probabilities independent, and so we are stuck with this messy second term. It's also quite inconvenient to sample this way, and difficult to compute the exact $\pi_i$'s.
But all of these problems seem to go away if we simply lift the requirement that a fixed number of units must be sampled. If instead we sample each unit independently with probability $\pi_i$, then the second term vanishes as $\pi_{ij}=\pi_i\pi_j$. Sampling this way is also much simpler to execute (I can just flip a weighted coin for each unit to decide whether to include it). And it even seems that the resulting variance may be smaller, since the second term is gone (although the $\pi_i$'s will be different so I'm not sure about this).
So my question is: is this a free lunch? Is there any downside to using independent sampling probabilities?
