The Randomization Process

Abstract

The randomized controlled clinical trial is the standard by which all trials are judged since other designs have certain undesirable features. In the simplest case, randomization is a process by which each participant has the same chance of being assigned to either intervention or control. An example would be the toss of a coin, in which heads indicates intervention group and tails indicates control group. Even in the more complex randomization strategies, the element of chance underlies the allocation process. Of course, neither trial participant nor investigator should know what the assignment will be before the participant’s decision to enter the study. Otherwise, the benefits of randomization can be lost. The role that randomization plays in clinical trials has been discussed in Chap. 5 as well as by numerous authors [1–12]. While not all accept that randomization is essential [11, 12], most agree it is the best method for achieving comparability between study groups and is the basis for statistical inference [2, 3].

Appendix

Adaptive Randomization Algorithm

Adaptive randomization can be used for more than two intervention groups, but for the sake of simplicity only two will be used here. To describe this procedure in more detail, a minimum amount of notation needs to be defined. First, let

$$\begin{array}{rl} {x}_{ik}=\hspace{-6pt}& \hbox{the number of participants already assigned intervention}\ k\\ &(k=1,2)\hbox{who have the same level of prognostic factor}\ i\\ &(i=1,2,\dots,f)\hbox{as the new participant}\end{array}$$

And define

$$ {x}_{ik}^{t}={x}_{ik}\rm{\hspace{1em}}\rm{if}t\ne k$$

$$={x}_{ik}+1\rm{\hspace{1em}}\rm{if}t\rm{\hspace{0.05em}}\rm{=}\rm{\hspace{0.05em}}k$$

The ${x}_{ik}^{t}$ represents the change in balance of allocation if the new participant is assigned intervention t. Finally, let

$$\begin{array}{l} B(t)=\hbox{function of the}{x}_{ik}^{t}\hbox{s, which measures the} \hbox{``lack of balance" over all}\\ \qquad\quad\, \hbox{prognosticfactors if the next participant is assigned intervention}\ t.\end{array}$$

Many possible definitions of B(t) can be identified. As an illustrative example, let

$$ B(t)={\displaystyle {\sum }_{i=1}^{f}{w}_{i}\ \text{Range}\ ({x}_{i1}^{t},{x}_{i2}^{t}})$$

where w _i = the relative importance of factor i to the other factors and the range is the absolute difference between the largest and smallest values of ${x}_{i1}^{t}\ and \ {x}_{i2}^{t}$.

The value of B(t) is determined for each intervention (t = 1 and t = 2). The intervention with the smaller B(t) is preferred, because allocation of the participant to that intervention will cause the least imbalance. The participant is assigned, with probability p > 1/2, to the intervention with the smaller score, B(1) or B(2). The participant is assigned, with probability (1−p), to the intervention with the larger score. These probabilities introduce the random component into the allocation scheme. Note that if p = 1 and, therefore, 1−p = 0, the allocation procedure is deterministic (no chance or random aspect) and has been referred to by the term “minimization” [51, 53].

As a simple example of the adaptive stratification method, suppose there are two groups and two prognostic factors to control. The first factor has two levels and the second factor has three levels. Assume that 50 participants have already been randomized and the following table summarizes the results (Table 6.2).

Table 6.2 Fifty randomized participants by group and level of factor (x _ik s)^a

In addition, the function B(t) as defined above will be used with the range of the x1 _ik s as the measure of imbalance, where w ₁ = 3 and w ₂ = 2; that is, the first factor is 1.5 times as important as the second as a prognostic factor. Finally, suppose p = 2/3 and 1−p = 1/3.

If the next participant to be randomized has the first level of the first factor and the third level of the second factor, then this corresponds to the first and fifth columns in the table. The task is to determine B(1) and B(2) for this participant as shown below.

• (a) Determine B(1)

  • Factor 1, Level 1

    	K	x 1k	${{x}^{1}}_{1k}$	Range (${x^{1}}_{11},{x^{1}}_{12}$)
    Group	1	16	17	|17–14| = 3
    	2	14	14

  • Factor 2, Level 3

    	K	x 2k	${x}_{2k}^{1}$ x 1 2k	Range (${x^{1}}_{21},{x^{1}}_{22}$)
    Group	1	4	5	|5–6| = 1
    	2	6	6

    Using the formula given, B(1) is computed as 3 × 3 + 2 × 1 = 11.

• (b) Determine B(2)

  • Factor 1, Level 1

    	K	x 1k	${x}_{1k}^{2}$	Range (${x^{2}}_{11}, {x^{2}}_{12}$)
    Group	1	16	16	|16–15| = 1
    	2	14	15

  • Factor 2, Level 3

    	K	x 2k	${x^{1}}_{1k}$	Range (${x^{2}}_{21},{x^{2}}_{22}$)
    Group	1	4	4	|4–7| = 3
    	2	6	7

    Then B(2) is computed as 3 × 1 + 2 × 3 = 9.

• (c) Now rank B(1) and B(2) from smaller to larger and assign with probability p the group with the smaller B(t).

  t	B(t)	Probability of assigning t
  2	B(2) = 9	p = 2/3
  1	B(1) = 11	1−p = 1/3

Thus, this participant is randomized to Group 2 with probability 2/3 and to Group 1 with probability 1/3. Note that if minimization were used (p = 1), the assignment would be Group 2.