An assembly of $n$ voters needs to decide on $t$ independent binary issues. Each voter has opinions about the issues, given by a $t$-bit vector. Anscombe's paradox shows that a policy following the majority opinion in each issue may not survive a vote by the very same set of $n$ voters, i.e., more voters may feel unrepresented by such a majority-driven policy than represented. A natural resolution is to come up with a policy that deviates a bit from the majority policy but no longer gets more opposition than support from the electorate. We show that a Hamming distance to the majority policy of at most $\lfloor (t - 1) / 2 \rfloor$ can always be guaranteed, by giving a new probabilistic argument relying on structure-preserving symmetries of the space of potential policies. Unless the electorate is evenly divided between the two options on all issues, we in fact show that a policy strictly winning the vote exists within this distance bound. Our approach also leads to a deterministic polynomial-time algorithm for finding policies with the stated guarantees, answering an open problem of previous work. For odd $t$, unless we are in the pathological case described above, we also give a simpler and more efficient algorithm running in expected polynomial time with the same guarantees. We further show that checking whether distance strictly less than $\lfloor (t - 1) /2 \rfloor$ can be achieved is NP-hard, and that checking for distance at most some input $k$ is FPT with respect to several natural parameters.