Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 14 / b

For $0\le k\le n$, sum the marginal stopping probabilities over the selected tail:Equivalently it is $p(1-{\sum }_{j=0}^k{g}_j)+(1-p)(1-{\sum }_{j=0}^k{d}_j)$. At $k=n$, both empty tail sums are zero, as required.

The event includes stopping positions zero through $k$. Applying Bayes' theorem to their joint and marginal probabilities givesThis conditional probability is defined whenever its conditioning event has positive probability. At $k=n$, it reduces to $p$.

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 givesThus, for a fixed good-ball stopping law,Since $g_0\le 1$, the smallest possible production probability over admissible stopping laws isIf 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 $g_0\to 1$. 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 $1/(1+g_0)$.