Under complete randomization, every assignment vector with exactly

n_{z}

patients receiving treatment

z

is equally likely. Thus, for

z = (z_{1}, \dots, z_{n})

and

N_{r} (z) = \sum_{i} 1_{{z_{i} = r}}

P (Z = z) = ⎩ ⎨ ⎧ \frac{n _{0} ! n _{1} ! n _{2} !}{n !}, 0, N_{r} (z) = n_{r} (r = 0, 1, 2), otherwise .

The reciprocal probability

n! / (n_{0}! n_{1}! n_{2}!)

is the multinomial coefficient counting assignments with those arm sizes.

b

Solution

Under independent assignment,

P (Z = z) = \prod_{i = 1}^{n} π_{z_{i}} = \prod_{r = 0}^{2} π_{r}^{N_{r} (z)} .

N_{r} = n_{r}

for

r = 0, 1, 2

, every compatible vector has the same factor

\prod_{r} π_{r}^{n_{r}}

, so

P (Z = z ∣ 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.

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