= Solution
\b[Index correction:] the printed definition only permits $i\leq n$, although the required average includes $i=n+1$. Extend the same deletion formula to $1\leq i\leq n+1$. Without that extension the final summand is undefined; there is no new sample or different estimator to supply.
For a fixed outcome of the sample and a fixed symbol $x$, put $A_j=\mathbf1_{\{X_j=x\}}$. The leave-one-out <empirical distribution> then satisfies
$$
\begin{aligned}
\sum_{i=1}^{n+1}\widehat q_{n,i}(x)
&=\frac1n\sum_{i=1}^{n+1}\left(\sum_{j=1}^{n+1}A_j-A_i\right)\\
&=\frac1n\left((n+1)\sum_{j=1}^{n+1}A_j-\sum_{i=1}^{n+1}A_i\right)
=\sum_{j=1}^{n+1}A_j.
\end{aligned}
$$
Dividing by $n+1$ gives \b[$\widehat p_{n+1}=(n+1)^{-1}\sum_{i=1}^{n+1}\widehat q_{n,i}$.] Each of the $n+1$ <indicator functions> was counted exactly $n$ times before division by $n$. This <leave-one-out identity for empirical distributions> is deterministic; <independence> plays no part in this part.
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