Tutorials › Response-Adaptive Randomization › Complete Randomization as Uniform Sampling

Randomization Basics · Tutorial 19 of 501

Complete Randomization as Uniform Sampling

Learn how fixed-count complete randomization makes every valid assignment equally likely, and how to calculate the probability of one specified assignment.

Beginner 9 min read

What You'll Learn

  • Define the assignment set for complete randomization with fixed treatment counts.
  • Count valid assignments using a binomial coefficient.
  • Calculate the probability of a specified valid assignment.
  • Distinguish uniform complete randomization from assigning each participant independently.
  • Check whether a randomization procedure really produces a uniform distribution over its intended assignment set.

What Complete Randomization Means

When a trial plans fixed numbers of participants in each treatment group, randomization is a way to choose which participants receive each treatment. Complete randomization means choosing uniformly from all assignments that satisfy those fixed counts. “Uniformly” is the defining feature: every permitted assignment has the same probability of selection.

Suppose there are \(N\) participants and two treatments, A and B. As in the assignment-vector notation, represent an assignment by a vector whose \(i\)th entry records the treatment assigned to participant \(i\). Write 1 for A and 0 for B. If the design specifies \(n_A\) participants for A and \(n_B=N-n_A\) for B, a valid vector contains exactly \(n_A\) ones and \(n_B\) zeros. The participants and the required counts are taken as fixed before the assignment is drawn.

Definition: Under complete randomization with fixed treatment counts, the random assignment is selected uniformly from the set of all assignment vectors containing the specified number of entries for each treatment. Thus each valid vector has probability equal to one divided by the number of valid vectors, and every vector not in that set has probability zero.

The definition concerns the probability of a whole assignment vector. It does not mean that the final group sizes vary: the counts are fixed by design. Nor does it mean that a treatment is assigned independently to each participant. Those distinctions matter when calculating assignment probabilities.

Counting the Valid Assignments

To construct a valid assignment, choose which \(n_A\) of the \(N\) participant positions receive A. All remaining positions receive B, so choosing the A positions determines the entire vector. The number of choices is the binomial coefficient \(\binom{N}{n_A}\). Therefore, if \(z\) is any specified valid assignment vector, its probability is

$$ P(Z=z)=\frac{1}{\binom{N}{n_A}}. $$

If a proposed vector has the wrong treatment counts, it is not in the assignment set, and its probability under this design is zero. The counting argument also explains why a particular participant order, such as A, B, A, is not a different probability model from specifying the positions that receive A: each valid vector is one distinct member of the set.

A practical way to implement the design is to select \(n_A\) of the \(N\) labeled participant positions uniformly without replacement and assign A to those positions; assign B to the rest. The procedure must make every subset of \(n_A\) positions equally likely. Merely using a random-number generator is not enough if the selection rule gives some subsets greater chance than others.

Worked Example: Six Participants and Equal Group Sizes

A hypothetical trial has six labeled participants, with three assigned to A and three to B. What is the probability of the specified vector \(z=(1,0,1,0,0,1)\)?

Check the counts. The vector contains three ones and three zeros, so it is valid for the design.

Count the assignment set. Choose the three positions assigned to A from the six participant positions:

$$ \binom{6}{3}=\frac{6!}{3!3!} =\frac{6\cdot5\cdot4}{3\cdot2\cdot1}=20. $$

Calculate the probability. Each valid vector has probability \(1/20\):

$$ P(Z=(1,0,1,0,0,1))=\frac{1}{\binom{6}{3}} =\frac{1}{20}=0.05000. $$

Thus this one specified assignment has a 5% chance of being selected. This is a probability for the complete vector, not a statement that the design has a 5% treatment rate.

A Second Assignment and an Invalid Vector

The order in which participants appear in a vector identifies who receives each treatment. For example, changing which positions contain ones generally gives a different assignment, even if the treatment totals remain the same. Each such vector is counted once.

Worked Example: Two of Seven Assigned to A

In a hypothetical study, seven participants are labeled 1 through 7, and exactly two will receive A. Consider the specified assignment in which participants 2 and 6 receive A and all others receive B.

There are

$$ \binom{7}{2}=\frac{7\cdot6}{2\cdot1}=21 $$

valid assignments. The specified vector has exactly two entries for A, so its probability is

$$ P(Z=z)=\frac{1}{21}=0.04762. $$

Now consider a proposed vector that assigns A to participants 2, 6, and 7. It has three entries for A, not two. It is outside the design’s assignment set, so its probability under this fixed-count procedure is \(0\), not \(1/21\).

The distinction is useful when checking a written schedule: the probability formula applies only after confirming that the specified assignment has the required counts.

Why the Uniform Probability Has This Form

The result follows directly from two facts: the number of valid assignment vectors is \(\binom{N}{n_A}\), and the design assigns equal probability to every one of them. Let that common probability be \(q\). The probabilities over all valid vectors must sum to 1, so

$$ \binom{N}{n_A}q=1. $$

Solving gives \(q=1/\binom{N}{n_A}\). This short argument is also a check on the formula: assigning the same probability to each member of a finite assignment set must make the total probability of the set equal to 1.

Worked Example: A Larger Fixed-Count Assignment

A hypothetical trial plans to assign four of ten participants to A and six to B. A particular vector assigns A to participants 1, 3, 6, and 8. It is valid because it contains exactly four entries for A.

The number of valid vectors is

$$ \binom{10}{4} =\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2\cdot1} =210. $$

Therefore the probability of that particular assignment is

$$ P(Z=z)=\frac{1}{210}=0.004762. $$

This is the probability of that exact allocation of all ten labeled participants. The answer does not depend on which four labels appear in the vector, provided the vector is valid, because the design treats all four-participant subsets equally.

Assumptions and Implementation Checks

The calculation is justified when the trial’s assignment mechanism has a fixed, known set of participants and treatment counts, and selects uniformly from all vectors meeting those counts. In practice, the design should specify the participant labels or enrollment positions to which assignment applies, the required count for each treatment, and a procedure that selects among the permitted vectors without favoring some vectors.

For example, a procedure that selects \(n_A\) distinct positions uniformly from the \(N\) positions meets the uniformity requirement. A procedure that first favors certain participants, or that selects positions with unequal chances and then adjusts the counts, need not produce a uniform distribution over assignment vectors. The intended counts alone do not establish that complete randomization was implemented.

Common checks include confirming that every generated vector has the required counts, that no vector is repeated as though it were a new participant assignment, and that the selection procedure does not use participant characteristics to favor particular allocations. If a schedule is generated by software, the design description and the actual selection algorithm should agree. A fixed random seed makes a computation reproducible; by itself, it does not establish that the algorithm samples uniformly.

How It Differs From Other Assignment Approaches

The key comparison is whether the total counts are fixed and what assignment set is sampled. These are differences in randomization mechanisms, not differences in how the treatments are defined.

ApproachWhat is selectedAre final counts fixed?
Complete randomizationUniformly from all vectors with the prespecified countsYes
Independent assignmentA treatment is drawn separately for each participant according to a stated probabilityNot generally
Restricted assignmentFrom a constrained subset of possible vectorsMay be, depending on the restrictions

Under independent assignment, the number assigned to A can vary from one realized trial to another. That approach is not complete randomization with fixed counts. Restricted procedures can also maintain fixed counts, but if they exclude some fixed-count vectors, they use a different assignment set; probabilities then depend on the rule for choosing within that set. The defining uniformity here applies to all vectors with the prescribed counts.

Common Mistakes and Limitations

  • Confusing equal assignment probabilities with independent assignments. In complete randomization, the whole vector is drawn from a fixed-count set. The fixed counts link the entries of the vector; they are not selected separately without regard to the total.
  • Using the formula for an invalid vector. First count the treatment labels in the proposed vector. If its counts do not match the design, its probability is zero under that design.
  • Counting only treatment totals. The quantity \(\binom{N}{n_A}\) counts assignments to labeled participants, not just possible group sizes. Different choices of participant positions give different vectors.
  • Assuming fixed counts imply uniformity. A procedure might always produce the right totals but favor some allocations. Uniform selection is a separate requirement.

Complete randomization controls the treatment totals, but it does not guarantee that participant characteristics will be balanced between groups in a particular realized assignment. It also does not, on its own, address how treatment assignment should respond to accumulating outcome data. Its simplicity makes the assignment mechanism easy to define, while the lack of further restrictions leaves chance variation among valid assignments.

Key takeaway: With \(N\) labeled participants and a fixed count \(n_A\) for treatment A, complete randomization selects uniformly from \(\binom{N}{n_A}\) valid assignment vectors. Any one valid vector therefore has probability \(1/\binom{N}{n_A}\).

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. A trial has nine participants and will assign three to A. How many valid assignment vectors are there, and what is the probability of one specified valid vector?
  2. For a design assigning two of six participants to A, what is the probability of a specified vector that assigns A to three participants?
  3. Why does \(\binom{N}{n_A}\) count assignment vectors rather than merely treatment totals?
  4. A procedure always assigns exactly four of eight participants to A, but some four-person subsets are deliberately given higher selection probabilities. Is it complete randomization as defined here? Explain.
  5. In one or two sentences, distinguish complete randomization with fixed counts from independent assignment.

Solutions. (1) The number is \(\binom{9}{3}=9\cdot8\cdot7/(3\cdot2\cdot1)=84\), so one specified valid vector has probability \(1/84=0.01190\). (2) The vector is invalid because it assigns three rather than two participants to A; its probability is \(0\). (3) The participants are labeled, so each choice of \(n_A\) positions for A determines a distinct vector, even though every vector has the same treatment totals. (4) No. Fixed counts are satisfied, but the assignments are not selected uniformly from all vectors with those counts. (5) Complete randomization selects one vector uniformly from the fixed-count assignment set, whereas independent assignment draws a treatment separately for each participant and does not generally fix the final counts.

Clinical Biostats