The worst-case scenario(s) for MAX3E SAT
In fact, we are able to perform "clause counting".  The total number of satisfied clauses for all remaining truth assignments is proportional to the expected number of satisfied clauses for a random truth assignment, which is the sum of the probabilities $P(C_i | \{x, y, ...\})$, where $C_i$ is the $i$-th clause, and $\{x,y,...\}$ are the literals that have been assigned to be true.  In particular, if there are three distinct literals per clause, and we are promised a satisfying truth assignment (so a variable and its negation never occur in the same clause), and a single literal $x$ has been assigned to be true, then the probability that a particular clause is satisfied by a random assignment of the remaining variables is $3/4$ if the clause contains $\neg{x}$, or $7/8$ if the clause contains neither $x$ nor $\neg{x}$, or $1$ if the clause contains ${x}$.  That is,
$$P(C_i | \{x\}) = \frac{7}{8} + \frac{1}{8}n_i(x),$$
where $n_i(x) \in\{-1, 0,1\}$.  So the literal picked out by "clause counting" will be the one that maximizes $\sum_{i}n_i(x)=N(x)$, where the sum is over all clauses, and $N(x)$ is the (net) number of occurrences of the literal $x$ in the entire formula.

To use this to construct a counterexample, we need a literal that is false (say) in all satisfying assignments, but that maximizes the above count.  We can force a literal to be false with four occurrences of its negation, as follows:
$$
\begin{eqnarray}
\neg{a} &amp;\equiv&amp; \neg{a} \wedge (b \vee \neg{b}) \wedge (c \vee \neg{c}) \\
&amp;\equiv&amp; (\neg{a} \vee b \vee c) \wedge (\neg{a} \vee b \vee \neg{c}) \wedge (\neg{a} \vee \neg{b} \vee c) \wedge (\neg{a} \vee \neg{b} \vee \neg{c}) \\
&amp;\equiv&amp; \Phi(\neg{a};b,c),
\end{eqnarray}
$$
for any additional literals $b$ and $c$; this subformula has $N(a)=-4$ and $N(b)=N(c)=0$.  We then need to include enough occurrences of $a$ in the rest of the formula to make $N(a)$ the largest count.  This can be done using
$$
\Omega(a;d,f) \equiv (a \vee d \vee f) \wedge (a \vee d \vee \neg{f}) \wedge (a \vee \neg{d} \vee f),
$$
which is equivalent to $a \vee (d \wedge f)$, and has $N(a)=3$ and $N(d)=N(f)=1$.  So a counterexample with four variables and ten (unique) clauses is
$$
\Phi(\neg{a}; x_1, x_2) \wedge \Omega(a; x_1, x_3) \wedge \Omega(a; x_2, x_3),
$$
which has $N(a)=2$ and $N(x_1)=N(x_2)=N(x_3)=1$, and a unique satisfying assignment: $\{\neg{a}, x_1, x_2, x_3\}$.