= Solution
For the plug-in <Neyman allocation>, the estimated <Bernoulli distribution> variances are $6/25$ and $3/16$, so
$$
\widehat R^*=\sqrt{\frac{6/25}{3/16}}=\frac{4\sqrt2}{5},\qquad
\boxed{\mathbb P(\text{treatment }0)=\frac{\widehat R^*}{1+\widehat R^*}
=\frac{4\sqrt2}{5+4\sqrt2}.}
$$
This uses the target ratio directly as the allocation odds; a procedure that actively corrects previous allocation imbalances would need an additional rule.
For the <randomized play-the-winner rule> RPW$(1,1)$, return the drawn ball and add one ball of the same treatment after a success or of the opposite treatment after a failure. Treatment 0 has three successes and two failures, and treatment 1 has one success and three failures. The urn therefore has $1+3+3=7$ treatment-0 balls and $1+1+2=4$ treatment-1 balls. Hence
$$
\boxed{\mathbb P(\text{treatment }0)=\frac7{11}.}
$$
For <dynamic programming> with only the final patient left, there is no future value from learning. Independent uniform <prior distributions> and <Beta-binomial conjugacy> give posteriors $p_0\mid\mathcal D\sim\operatorname{Beta}(4,3)$ and $p_1\mid\mathcal D\sim\operatorname{Beta}(2,4)$. Their <posterior predictive probabilities> of success are $4/7$ and $2/6=1/3$. The optimal terminal action therefore gives
$$
\boxed{\mathbb P(\text{treatment }0)=1.}
$$
This is optimal for expected successes with no imposed lower bound on the <randomization> probabilities.
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