= Solution
For every estimator, its worst-case risk dominates its average risk under the <uniform prior>. Since the preceding <Bayes estimator> minimizes that average,
$$
\sup_{0<p<1}R(p,\delta)\ge\int_0^1R(p,\delta)\,dp\ge1/n.
$$
The estimator $X/n$ has risk exactly $1/n$ throughout the interior, so it attains this lower bound. Under either endpoint convention discussed above its supremum over $[0,1]$ is also $1/n$. Thus
$$
\boxed{\delta_{\rm minimax}=X/n,\qquad\inf_\delta\sup_pR(p,\delta)=1/n}.
$$
This constant-risk Bayes argument works also for $n=1$, when the only observed counts are endpoints and the posterior integrability argument still selects their corresponding decisions.
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