Least-Favorable Location for Binomial Top-$t$ Selection
Consider $k$ independent Bernoulli populations, each sampled $n$ times, and select the $t$ populations with the largest success counts, breaking ties uniformly. Classical monotonicity reduces the worst case over the preference zone with separation $\delta$ to the slippage family with levels $p$ and $p+\delta$, leaving only its absolute location $p\in[0,1-\delta]$ undetermined. A Gaussian approximation suggests the symmetric center $p_{\mathrm c}=(1-\delta)/2$, and the exact two-population problem is uniquely centered there for every $n\ge2$. For fixed $k,t$ and $\delta\in(0,1)$, we prove that exact eventual centering holds precisely when $k=2t$. When $k\ne2t$, the least-favorable location $p_{n,k,t}^*$ satisfies \[ p_{n,k,t}^*-p_{\mathrm c} =(k-2t)C_\delta n^{-1/2}e^{-n\Gamma_\delta}\{1+o(1)\}, \] where $C_\delta$ and $\Gamma_\delta$ are explicit and positive. In either case, the least-favorable location is eventually unique. The proof writes incorrect selection as a union of pairwise misrankings and applies inclusion--exclusion, yielding a bipartite graph expansion. A single misranking has its exact maximum at the symmetric center and determines the central curvature; two-edge intersections sharing one population determine the central slope through their multiplicity imbalance; all remaining graphs have higher large-deviation rates.