Tutorials › Response-Adaptive Randomization › Stratified Permuted-Block Randomization

Randomization Basics · Tutorial 31 of 501

Stratified Permuted-Block Randomization

Construct balanced permuted blocks separately within each stratum, then combine their assignment probabilities correctly using independence.

Beginner 10 min read

What You'll Learn

  • Define a permuted-block schedule separately within each stratum.
  • Check whether a block size is compatible with the intended allocation ratio.
  • Calculate the probability of an exact full-block assignment sequence.
  • Find probabilities for specified partial-block patterns and allocation counts.
  • Combine probabilities across independent strata without confusing counts and ordered sequences.
  • Identify assumptions, practical diagnostics, and limitations of stratified blocking.

Why Block Within Strata?

Stratification aims to keep allocation balanced within important groups, rather than only across the trial as a whole. Complete randomization within each stratum can still produce appreciable short-term imbalance, especially when enrollment is small. Permuted blocks address that issue by fixing the treatment counts in each completed block. The construction is carried out separately in every stratum, and the schedules are randomized independently.

The distinction between a completed block and an unfinished block matters. A completed block has its prespecified treatment counts; a partial block at an interim enrollment time need not. Probabilities for an interim allocation must therefore account for the random order of assignments within the current block.

Definition: In stratified permuted-block randomization, participants are assigned to prespecified strata, and a separate randomization schedule is generated within each stratum. Within a stratum, each block has a prespecified size and treatment counts compatible with the allocation ratio. The assignments in a block are a uniformly selected permutation of those treatment labels. When the design specifies independent randomization, block permutations in different blocks and strata are generated independently.

For a two-treatment design, write \(T\) for the experimental treatment and \(C\) for control. In a block of size \(m\) with \(r\) assignments to \(T\) and \(m-r\) to \(C\), there are \(\binom{m}{r}\) distinct assignment sequences. Under uniform permutation, each complete sequence has probability \(1/\binom{m}{r}\). A 1:1 allocation uses an even block size; a 1:2 allocation uses a block size divisible by three, with one-third of the assignments to the first treatment and two-thirds to the second.

Constructing the Schedules

The design must state the strata, allocation ratio, block size or permitted block sizes, and how independent schedules are generated. For each stratum, list the required treatment labels in each block and randomly permute them. Continue to the next block when enrollment in that stratum reaches the block boundary. The same participant order within a stratum must be used when reading the schedule as when assigning enrolled participants.

Consider the following hypothetical design: stratum North uses blocks of size 4 with two \(T\) and two \(C\) assignments; stratum South uses blocks of size 6 with two \(T\) and four \(C\) assignments. One possible realization is:

StratumBlock contents before permutationOne possible randomized sequenceCompleted-block counts
NorthT, T, C, CC, T, C, T2 T; 2 C
SouthT, T, C, C, C, CC, C, T, C, T, C2 T; 4 C

Both rows meet their block requirements. The chance of these two particular complete sequences, under the independent uniform-permutation design, is \(1/6 \times 1/15=1/90\), since North has \(\binom{4}{2}=6\) possible sequences and South has \(\binom{6}{2}=15\). The table is an illustrative hypothetical randomization, not a claim about an actual study.

Worked Example: Checking a Two-Stratum Schedule

Suppose a hypothetical study uses blocks of size 4 with two assignments to each treatment in stratum A, and blocks of size 6 with two \(T\) and four \(C\) assignments in stratum B. Check the candidate blocks A: \(T,C,T,C\) and B: \(C,T,C,C,T,C\), and find the probability of this exact pair of completed-block sequences.

In A, the sequence contains two \(T\) and two \(C\), so it is a valid block. There are \(\binom{4}{2}=6\) equally likely sequences, giving probability \(1/6\).

In B, the sequence contains two \(T\) and four \(C\), so it is also valid. There are \(\binom{6}{2}=15\) equally likely sequences, giving probability \(1/15\).

The block randomizations are independent by design. Therefore the probability of this exact pair is

$$ P(\text{specified A and B sequences}) =\frac{1}{6}\times\frac{1}{15} =\frac{1}{90} \approx 0.01111. $$

This probability refers to the exact ordered sequences shown, not merely to their treatment counts. Every valid complete block in A has the same counts, but only one of its six possible orders is the specified sequence.

Probabilities for Partial Blocks

An interim allocation often concerns only the first \(k\) positions of a block. If a block has \(r\) treatment assignments among \(m\) total assignments, a particular ordered prefix containing \(x\) \(T\)'s and \(k-x\) \(C\)'s has probability

$$ \frac{(r)_x(m-r)_{k-x}}{(m)_k}, $$

where \((a)_j=a(a-1)\cdots(a-j+1)\), with \((a)_0=1\). The formula follows by multiplying the conditional probabilities for the ordered draws without replacement. It applies when the prefix is feasible; otherwise its probability is zero.

If the event concerns only the number of \(T\)'s in the first \(k\) positions, the order is not specified. There are \(\binom{k}{x}\) possible orders with \(x\) \(T\)'s. Equivalently, counting possible block sequences gives

$$ P(\text{\(x\) assignments to \(T\) in the first \(k\) positions}) =\frac{\binom{r}{x}\binom{m-r}{k-x}}{\binom{m}{k}}. $$

The numerator counts ways to select the \(x\) treatment labels and \(k-x\) control labels that appear in the prefix; the denominator counts all ways to select the \(k\) labels. This is the without-replacement structure of a uniformly permuted block.

Worked Example: A Mixed Prefix in One Stratum

In a hypothetical 1:1 block of size 4, there are two \(T\) and two \(C\) assignments. What is the probability that the first two assignments contain one of each treatment, in either order?

For the ordered prefix \(T,C\), the first assignment is \(T\) with probability \(2/4\). Given that first assignment, two \(C\)'s remain among the three positions, so the next assignment is \(C\) with probability \(2/3\). Thus

$$ P(T,C)=\frac{2}{4}\times\frac{2}{3}=\frac{1}{3}. $$

For the other order, \(C,T\), the same calculation gives \(P(C,T)=(2/4)(2/3)=1/3\). These two ordered prefixes are mutually exclusive, so

$$ P(\text{one of each in the first two}) =P(T,C)+P(C,T) =\frac{1}{3}+\frac{1}{3} =\frac{2}{3} \approx 0.6667. $$

The probability is not \(1/3\): that value applies to either one specified order, whereas the question allows both.

Combining Probabilities Across Strata

Independent randomization means that events determined by separate strata have probabilities that multiply. For example, if \(E_A\) is a specified event in stratum A and \(E_B\) is a specified event in stratum B, then \(P(E_A\text{ and }E_B)=P(E_A)P(E_B)\). The same logic extends across independent blocks.

It is important to define the event precisely before multiplying. An exact assignment vector specifies the treatment of each participant in order. An allocation-count event specifies only how many participants receive each treatment. A count event can include many assignment vectors, so its probability is generally the sum of their probabilities. If an event links outcomes across blocks—for example, it specifies a total count across two blocks—one must sum over the compatible combinations of block outcomes; one cannot multiply marginal count probabilities as if the event were a separate condition in each block.

Worked Example: A Combined Partial Allocation

Suppose independent hypothetical schedules use one size-4 block with two \(T\) and two \(C\) assignments in East, and one size-6 block with two \(T\) and four \(C\) assignments in West. Find the probability that East's first two assignments contain one of each treatment, while West's first three assignments are exactly \(C,T,C\).

The East event allows two ordered prefixes, \(T,C\) and \(C,T\). As shown by the without-replacement calculation, each has probability \(1/3\), so the East event has probability \(2/3\).

For West's specified ordered prefix \(C,T,C\), the successive probabilities are \(4/6\), then \(2/5\), then \(3/4\). After drawing \(C\), four positions remain, including two \(T\)'s; after then drawing \(T\), three positions remain, including three \(C\)'s. Hence

$$ P(C,T,C)=\frac{4}{6}\times\frac{2}{5}\times\frac{3}{4} =\frac{24}{120} =\frac{1}{5}. $$

Because the strata are randomized independently, the combined event probability is

$$ P(\text{East event and West prefix}) =\frac{2}{3}\times\frac{1}{5} =\frac{2}{15} \approx 0.1333. $$

The West requirement is one exact ordered prefix, but the East requirement includes two orders. Treating East as one specified order would incorrectly halve the combined probability.

Justification for the Multiplication Rule

A full schedule consists of one permissible sequence for each block in each stratum. Under the design, each block's sequence is selected independently and uniformly from its own set of valid permutations. Thus every joint collection of block sequences has probability equal to the product of the reciprocal numbers of valid permutations. For events that constrain individual blocks separately, adding the probabilities of all compatible joint schedules yields the corresponding product of event probabilities.

For a more explicit argument, suppose event \(E_A\) is satisfied by \(a\) of the \(M_A\) possible schedules in stratum A, and \(E_B\) by \(b\) of the \(M_B\) schedules in stratum B. There are \(a b\) joint schedules satisfying both, out of \(M_A M_B\) equally likely joint schedules. Therefore

$$ P(E_A\text{ and }E_B) =\frac{ab}{M_A M_B} =\frac{a}{M_A}\frac{b}{M_B} =P(E_A)P(E_B). $$

This argument depends on independent schedule generation. If a design deliberately coordinates the randomizations across strata, the product rule need not hold.

Worked Example: Probability of an Exact Multi-Block Schedule

A hypothetical design has two completed size-4 blocks in stratum A, each with two \(T\) and two \(C\), and one completed size-4 block in stratum B with the same counts. Find the probability of the exact sequence \(T,C,T,C\) in A's first block, \(T,T,C,C\) in A's second block, and \(C,C,T,T\) in B.

Each specified sequence is a valid permutation of two \(T\)'s and two \(C\)'s. Each block has \(\binom{4}{2}=6\) possible sequences, so each specified block sequence has probability \(1/6\). The two A blocks and the B block are independent under the design. Therefore

$$ P(\text{exact three-block schedule}) =\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6} =\frac{1}{216} \approx 0.004630. $$

If the question instead asked only whether each completed block had two assignments to each treatment, that event would have probability 1: the block construction guarantees those counts. Randomness concerns the order within each block.

Assumptions, Diagnostics, and Alternatives

These calculations assume that strata are defined before treatment assignment, participants are assigned to the correct stratum, block sizes and treatment counts are compatible with the allocation ratio, and each valid permutation is equally likely. The multiplication rule additionally assumes independent randomization across blocks and strata. The calculations also presume that the enrollment order within each stratum is clear and that assignments are used in that order.

Basic schedule checks can reveal design or implementation errors. Within every completed block, verify the prescribed size and treatment counts. Verify that each assignment belongs to the stratum's schedule, that no participant is assigned twice or skipped, and that the intended block boundaries are respected. For probability calculations, check that event categories are mutually exclusive before adding their probabilities, and check whether a stated condition describes an exact order or a set of orders.

Stratified complete randomization is the main alternative introduced in the earlier material on stratified complete randomization. It fixes treatment totals within each stratum across the full allocation, but does not guarantee small imbalances at intermediate enrollment points. Permuted blocks provide that short-term balance at completed-block boundaries; they do not guarantee balance inside an unfinished block. Compared with complete randomization, fixed block sizes can also make upcoming assignments more predictable, particularly near a block's end if its size and treatment ratio are known. Randomly varying block sizes can address some predictability concerns, but is a different design choice.

Blocking has practical limits. Many strata with low enrollment can leave blocks incomplete for long periods, and the workload of managing separate schedules increases with the number of strata. A block's fixed treatment counts may also be less useful if enrollment is extremely slow or if participants are not enrolled in the planned order. These are reasons to choose strata and block sizes carefully, not reasons to ignore the specified randomization procedure.

Common pitfall: “The first two assignments are mixed, so the probability is the probability of \(T,C\).” This omits the equally possible order \(C,T\). For two treatments, distinguish an exact ordered prefix from a count condition: add the probabilities of all disjoint orders allowed by the count condition. Then multiply across strata only when the events are defined on independently randomized schedules.

Check Your Understanding

Answer each question by identifying whether it concerns an exact sequence or an allocation count. Full solutions follow the questions.

  1. A size-6 block has two \(T\)'s and four \(C\)'s. Is the sequence \(T,C,T,C,C,C\) valid, and what is the probability of that exact sequence?
  2. In a size-4 block with two \(T\)'s and two \(C\)'s, calculate the probability that the first two assignments are both \(T\).
  3. Two independent strata each use a size-4 block with two assignments to each treatment. What is the probability that each specified block has the exact sequence \(C,T,C,T\)?
  4. In a size-4 block with two assignments to each treatment, find the probability that the first two assignments are one \(T\) and one \(C\), in either order.

Solutions. (1) The sequence is valid because it contains two \(T\)'s and four \(C\)'s. There are \(\binom{6}{2}=15\) valid block sequences, so its exact-sequence probability is \(1/15\approx0.06667\). (2) The first \(T\) has probability \(2/4\), and the second \(T\), given the first, has probability \(1/3\). Thus \(P(T,T)=(2/4)(1/3)=1/6\approx0.1667\). (3) Each exact sequence has probability \(1/\binom{4}{2}=1/6\). Independence gives \((1/6)(1/6)=1/36\approx0.02778\). (4) The two allowed prefixes, \(T,C\) and \(C,T\), each have probability \((2/4)(2/3)=1/3\). They are disjoint, so the count-event probability is \(1/3+1/3=2/3\approx0.6667\).

Clinical Biostats