Bob has to find Alices hidden gold by questioning yes/no questions
Consider any strategy for Bob. For each item $v_i$, let $x_i$ be binary string composed of the answers that Alice gives to all questions (0 is No, 1 is Yes). You can check that $x_i$ is a prefix code. If $p_i$ is the probability of item $v_i$, then the optimal strategy is the one that minimizes
$$ \sum_i p_i |x_i| $$
over all prefix codes $x_i$ (every prefix code can be translated back to a strategy).

Huffman gave an algorithm which finds the best strategy. The cost $T(V)$ of this strategy satisfies
$$ H(V) \leq T(V) &lt; H(V) + 1. $$
If all $p_i$ are of the form $1/2^j$, then $T(V) = H(V)$, and otherwise $T(V) &gt; H(V)$.

Huffman's strategy is "bottom-up", whereas the strategy that you describe is "top-down". It is not optimal in general, but (if you implement halving properly) it is guaranteed to have cost at most $T(V) + 1$, even with an arbitrary ordering of the weights.

(Since you haven't described your halving strategy, I cannot describe an example in which Huffman coding beats your strategy.)

While prefix codes cannot in general reach the entropy exactly, they do reach the entropy in the limit when solving many instances at once. That is, suppose that Alice is actually hiding $N$ items, each chosen according to $V$. Huffman's algorithm produces a strategy whose cost per item is less than $H(V) + 1/N$. In the limit $N\to\infty$, you do get $H(V)$.