Under <complete randomization>, every assignment vector with exactly $n_z$ patients receiving treatment $z$ is equally likely. Thus, for $z=(z_1,\ldots,z_n)$ and $N_r(z)=\sum_i\mathbf1_{\{z_i=r\}}$,
$$
\mathbb P(Z=z)=
\begin{cases}
\dfrac{n_0!n_1!n_2!}{n!},&N_r(z)=n_r\quad(r=0,1,2),\\
0,&\text{otherwise}.
\end{cases}
$$
The reciprocal probability $n!/(n_0!n_1!n_2!)$ is the <multinomial coefficient> counting assignments with those arm sizes.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 221 / 1 / i / a / Solution

Under independent assignment,
$$
\mathbb P(Z=z)=\prod_{i=1}^n\pi_{z_i}
=\prod_{r=0}^2\pi_r^{N_r(z)}.
$$
The arm-count vector has a <multinomial distribution>. Conditional on $N_r=n_r$ for $r=0,1,2$, every compatible vector has the same factor $\prod_r\pi_r^{n_r}$, so
$$
\mathbb P(Z=z\mid N_0=n_0,N_1=n_1,N_2=n_2)
=\frac{n_0!n_1!n_2!}{n!}
$$
for compatible $z$, and zero otherwise. Conditioning independent assignment on its arm sizes therefore recovers <complete randomization>.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 221 / 1 / i / b / Solution

Complete randomization chooses uniformly among all treatment assignments having prespecified treatment-arm sizes.


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