Are sampling designs characterized by their first and second-order inclusion probabilities? If you have two sampling designs $p_1(\cdot)$ and $p_2(\cdot)$ (defined on the same set of samples $\mathcal{S}$ from the same population $\mathcal{U}$) such that they induced the same inclusion probabilities of first and second-order, then $p_1(s) = p_2(s)$ for all sample $s$ in $\mathcal{S}$. Is this sentence true?
 A: The assertion is not necessarily true, as the following counterexample demonstrates.
Let $\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\}$ be a population of four elements.  Consider two sampling plans.


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*Select $\omega_4$ no matter what.  Independently select $\omega_1,\omega_2,\omega_3$, each with a chance of $1/2$ of being included.  
There are eight possible samples, each with a chance of $1/8$.  A convenient way to write them is with the binary inclusion vector $s=(s_1,s_2,s_3,s_4)$ where $s_i=1$ when $\omega_i$ is in $s$ and $s_i=0$ otherwise.  The possible samples (all of which are equally likely) can be denoted $$\mathcal{S}_1=\{0001,0011,0101,1001,0111,1011,1101,1111\}.$$

*The second sampling plan draws one element uniformly from this set of possible samples: $$\mathcal{S}_2=\{0001,0111,1011,1101\}.$$
Regardless of the plan, write $\pi_i$ for the chance that $\omega_i$ is in a sample and $\pi_{ij}$ for the chance that $\omega_i$ and $\omega_j$ are both in a sample.  As you can readily check, these plans were constructed so that $\pi_i=1/2$ for $i=1,2,3$ and $\pi_4=1$.  Moreover, $\pi_{ij}=1/4$ for $i\lt j\lt 4$ and $\pi_{i4}=1/2$ in both plans.  Thus, both have identical first and second order inclusion probabilities, yet they are not the same: sampling plan $(1)$ can result in four samples that are impossible in sampling plan $(2)$.
