= Solution
There are $99$ possible stopping points, so the equal defective-ball <probabilities> are $d_k=1/99$. With $g_0>0$, the desired <conditional probability> gives
$$
\frac{pg_0}{pg_0+(1-p)/99}=\frac{99}{100},
\qquad100pg_0=99pg_0+1-p.
$$
Thus, for a fixed good-ball stopping law,
$$
\boxed{p=\frac1{1+g_0}.}
$$
Since $g_0\le1$, the smallest possible production <probability> over admissible stopping laws is
$$
\boxed{p_{\min}=\frac12,\qquad\text{attained when }g_0=1.}
$$
If strict positivity of every $g_k$ is additionally imposed in this general case, then $g_0<1$ and one only has the infimum $1/2$, approached as $g_0\to1$. The paper states that strict-positivity assumption explicitly for part (a), not for the general case in part (b). Without choosing or bounding $g_0$, the exact fixed-law answer remains $1/(1+g_0)$.
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