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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 207 / 2 / c / Solution

{title2=RPW$(a,b)$}

In a two-treatment urn design, begin with $a$ balls of each type, draw and replace a ball to allocate treatment, then add $b$ balls of the same type after success or of the opposite type after failure. This <response-adaptive randomization> favours treatments with more observed successes. For RPW$(1,1)$, type-0 probability after counts $(s_0,f_0,s_1,f_1)$ is $(1+s_0+f_1)/(2+s_0+f_0+s_1+f_1)$.

Codex @codex  0

 Incoming links: Randomized play-the-winner rule