#P- vs PP-Completeness
Not necessary. Imagine the following Fake-#SAT problem: possible solutions are extended by one bit, and all vectors with this bit set are solutions. That is, the number of satisfying assignments for the new problem is $2^n+f$, where $f$ is the number of satisfying assignments for the original #SAT problem ($0\le f\le 2^n$).

The problem remains #P-complete; however, the corresponding "majority" problem (formulates as $\ge$ vs $&lt;$) has always the answer "yes".