= Solution
The <Rao-Blackwell theorem> says that if $S$ is a <sufficient statistic> and $\delta$ a finite-variance <estimator>, then $\delta^*=\mathbb E(\delta\mid S)$ is an <estimator> with the same expectation and no larger <variance>. Sufficiency ensures the conditional distribution, and hence this formula as a function of the data, does not involve the unknown parameter. The <tower property of conditional expectation> gives $\mathbb E\delta^*=\mathbb E\delta$, while the <law of total variance> gives
$$
\operatorname{Var}(\delta)=\operatorname{Var}(\delta^*)+\mathbb E\operatorname{Var}(\delta\mid S).
$$
This proves the theorem and shows that equality holds exactly when $\delta$ is already determined by $S$, almost surely. In particular, unbiasedness is preserved.
Assume $n\geq3$. Independence gives $\mathbb E\hat\theta=p_0p_1p_2=\theta$, so the proposed indicator is unbiased and has <variance> $\theta(1-\theta)$. The type counts $S=(n_0,n_1,n_2)$ are a <sufficient statistic>: conditional on them, every ordering of the labels is equally likely. Consequently,
$$
\boxed{\theta^*=\mathbb E(\hat\theta\mid S)=\frac{n_0n_1n_2}{n(n-1)(n-2)}.}
$$
For example, the conditional probability of the ordered first three labels is $(n_0/n)(n_1/(n-1))(n_2/(n-2))$. This is <Rao-Blackwellization of a multinomial probability product>.
If all three $p_i$ are positive, there is positive probability that all three counts are positive. On such an event the conditional indicator has probability strictly between zero and one, so its <conditional variance> is positive. Thus \b[$\operatorname{Var}(\theta^*)<\theta(1-\theta)$ for interior parameter values]. If any $p_i=0$, then $\theta=0$ and both <estimators> have zero <variance>; the printed strict inequality is impossible at those boundary parameters. The valid unrestricted statement is the non-strict inequality.
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