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Randomization Basics · Tutorial 27 of 501

Predictability From Fixed Block Sizes

Learn to calculate when a fixed-block allocation is merely more predictable and when the remaining assignments make the next treatment certain.

Beginner 11 min read

What You'll Learn

  • State the assumptions needed to calculate allocation probabilities within a fixed block.
  • Calculate the probability of the next assignment from the treatment slots still unused.
  • Identify when an observed sequence makes a future assignment certain.
  • Distinguish conditional predictability from the marginal probability of each treatment.
  • Explain how foreknowledge can compromise enrollment procedures even when the schedule was randomized.
  • Compare fixed-block predictability with complete randomization and concealed schedules.

Why a Randomized Schedule Can Become Predictable

Permuted blocks are used to keep the numbers assigned to treatment arms close as enrollment proceeds. But balance has a consequence: once some assignments in a block are known, the assignments still available are constrained by the block’s fixed composition. The next treatment may therefore be less than equally likely to be either arm, and at the end of a block it can be determined with certainty.

This is not evidence that the schedule was generated incorrectly. It is a consequence of sampling without replacement from a known set of treatment slots. The practical concern is whether someone involved in enrollment can use this information to anticipate an assignment before deciding which eligible participant to enroll.

Definition: In a fixed-size permuted block, a prespecified number of treatment assignments, with prespecified counts for each arm, are placed in a random order within each block. Predictability from fixed block sizes is the ability to infer, from the known block composition and assignments already observed in that block, the conditional probabilities of its remaining assignments. An assignment is deterministic given the observed history if its conditional probability for one arm is 1.

Conditional Probabilities Within a Block

Consider a block containing \(a\) assignments to arm A and \(b\) assignments to arm B, so its size is \(m=a+b\). Assume that all permutations of these \(m\) assignments are equally likely. Suppose the block boundary and composition are known, and the assignments already made in that block have been observed accurately. If \(r_A\) A assignments and \(r_B\) B assignments remain, then \(r_A+r_B\) slots remain. By symmetry among those remaining slots, the next assignment has probabilities

$$ P(\text{next assignment is A}\mid\text{observed assignments in the block}) =\frac{r_A}{r_A+r_B}, \qquad P(\text{next assignment is B}\mid\text{observed assignments in the block}) =\frac{r_B}{r_A+r_B}. $$

These are conditional probabilities: they apply after taking account of the known assignments in the current block. They are not necessarily \(1/2\), even in a block that contains equal numbers of A and B assignments. If only B slots remain, then \(r_A=0\), and the next assignment must be B.

The calculation depends on specific conditions. The observer must know which block is in progress, know the treatment counts assigned to that block, and know the assignments already used. The assignments must have been permuted uniformly, and there must be no unrecorded deviations from the schedule. If the block boundary is uncertain or the composition is concealed, this direct calculation is not available without additional information; uncertainty about the boundary does not, by itself, guarantee that prediction is impossible.

A Short Justification

Among the \(r_A+r_B\) remaining positions, each position is equally likely to be next under a uniform random permutation. Exactly \(r_A\) of those positions carry A. Equivalently, the remaining multiset can be ordered in \((r_A+r_B)!\) ways, while the orders with A first number \(r_A(r_A+r_B-1)!\). Therefore,

$$ P(\text{next is A}\mid\text{observed history}) =\frac{r_A(r_A+r_B-1)!}{(r_A+r_B)!} =\frac{r_A}{r_A+r_B}. $$

The same argument gives the probability for B. The formula is about the next position in the block, not about the probability that a particular patient benefits from treatment or about the probability that an allocation method is ethically acceptable.

Worked Example: A Four-Person, Equal-Allocation Block

Suppose a hypothetical trial uses blocks of four, each containing two A assignments and two B assignments. Assume all within-block orders are equally likely. The first three assignments observed in one block are A, A, B. What is the probability that the fourth assignment is A?

Two A assignments and one B assignment have been used. Thus, one A and one B remain among the two remaining positions. The conditional probability is

$$ P(\text{fourth is A}\mid A,A,B) =\frac{1}{1+1} =\frac{1}{2} =0.5000. $$

In this particular history the fourth assignment is not determined: it may be A or B. In contrast, if the first three assignments had been A, A, B in a block containing three A and one B, the remaining counts would be zero A and one B, and the fourth assignment would be forced to be B. Predictability must be calculated from the actual block composition as well as the observed sequence.

Worked Example: Updating Probabilities in a Six-Person Block

A hypothetical trial uses a block of six containing three A assignments and three B assignments. The first three observed assignments are A, A, B. Find the probability of A for the fourth position, and then update it if the fourth assignment is observed to be B.

After A, A, B, two A and one B have been used. One A and two B assignments remain among three positions, so

$$ P(\text{fourth is A}\mid A,A,B) =\frac{1}{3} \approx 0.3333, \qquad P(\text{fourth is B}\mid A,A,B) =\frac{2}{3} \approx 0.6667. $$

Suppose the fourth assignment is B. Then one A and one B remain among two positions. Consequently, the probability of A at the fifth position is \(1/2\), and the probability of B is also \(1/2\). Whichever arm is assigned fifth, the sixth and final assignment must be the other arm. For example, if the fifth is A, the sixth is B with conditional probability 1. The certainty arises from the remaining block counts, not from a special rule for the last participant.

Why Predictability Matters During Enrollment

The statistical calculation becomes a conduct concern when someone can use a likely or known assignment to influence enrollment. For example, a person who can see the allocation history might delay enrolling an eligible patient, accelerate enrollment, or affect when eligibility is confirmed in the hope of obtaining a preferred arm for that patient. Even if the randomization schedule was generated correctly, using assignment foreknowledge to shape who enters at a particular point can undermine the intended random allocation.

The relevant protection is not merely that the original list was random. The allocation needs to remain concealed from people who could alter enrollment or eligibility decisions until the participant has been irrevocably entered and the assignment is issued. In practice, this supports procedures such as restricted access to the schedule and a secure allocation service that reveals an assignment only when an eligible participant is enrolled. The appropriate safeguards depend on who can see the schedule, who controls enrollment, and how the allocation system operates.

Predictability and concealment are related but distinct. Predictability describes what can be inferred from available information. Concealment describes whether the people making enrollment decisions can access that information before enrollment. A schedule can have mathematically predictable late-block assignments while strong access controls prevent recruiters from knowing them. Conversely, an assignment process can be vulnerable even when no one can calculate the next assignment with certainty, if people can exploit a substantial shift in its probabilities.

Worked Example: Fixed Blocks Compared With Complete Randomization

Consider two hypothetical allocation schedules for eight participants, each with four A and four B assignments overall. In one schedule, the first block has four positions containing two A and two B; in the other, all eight assignments are uniformly permuted together. In both cases the first three observed assignments are A, A, B. Compare the probability of A for the fourth assignment.

Fixed blocks. In the first four-person block, two A and one B have already been used. One A and one B remain, so

$$ P(\text{fourth is A}\mid A,A,B,\text{ first block of size 4}) =\frac{1}{2} =0.5000. $$

Complete randomization of all eight positions. Across the full schedule, two of the four A assignments and one of the four B assignments have been used. Two A and three B assignments remain among five positions. Therefore,

$$ P(\text{fourth is A}\mid A,A,B,\text{ total counts }4\text{ A and }4\text{ B}) =\frac{2}{5} =0.4000. $$

The conditional probabilities differ because the information about the first block imposes an additional local constraint. Neither schedule makes A certain in this example. In other observed histories, however, the fixed block can make a late assignment certain, whereas the complete-randomization schedule still has unused assignments of both arms. This comparison concerns predictability, not a claim that one design is always preferable.

Diagnostics, Alternatives, and Limits

A useful diagnostic is to reconstruct the current block’s remaining counts after each assignment and calculate the conditional probabilities. If the sequence is being monitored operationally, the review should check whether the block size and boundary are visible, who can access the assignment history, and whether any assignment was skipped, delayed, or otherwise deviated from the schedule. A calculation based on incorrect counts or an assumed block boundary can give a false impression of certainty.

Fixed permuted blocks trade unrestricted allocation for local balance. Complete randomization avoids fixed local block boundaries but can allow greater chance imbalance during enrollment. Randomizing the block size can make it harder for an observer to know which block position is current, although it does not make every allocation unpredictable in every circumstance. The central comparison is therefore not “predictable versus random”: each approach is randomized under its design, but the information revealed by past assignments differs.

Fixed blocks also do not guarantee equal allocation at every point in the trial. They guarantee the specified counts at the end of each completed block, assuming the schedule is followed. During a block, the running totals can differ. Nor does arithmetic alone quantify the practical risk of enrollment manipulation: that assessment also depends on access, timing, staff roles, and compliance procedures.

Common Mistake

A common error is to reason, “The block was randomized, so the next assignment still has probability \(1/2\) for each arm.” Randomization gives probabilities before the relevant assignments are observed; conditioning on the observed history can change those probabilities. For a six-position block with three A and three B assignments, after observing A, A, B, one A and two B assignments remain, so the next probability of A is \(1/3\), not \(1/2\). The correction is to count the unused assignments in the current block and divide by the number of remaining positions.

A second error is to say that every fixed-block sequence is predictable from its first assignment. It is not. Some histories leave a balanced mix of arms and therefore do not identify the next assignment; certainty occurs only when all remaining slots belong to one arm. The correct statement is conditional: predictability changes as assignments are observed and depends on the remaining block composition.

Key takeaway: With a known fixed block and a uniformly permuted assignment order, the conditional probability for an arm equals its number of unused assignments divided by the number of unused positions. Local balance creates late-block constraints, so allocation concealment must protect enrollment decisions from access to the schedule and its revealing history.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

  1. A block contains three A and three B assignments. After observing A, B, A, how many assignments of each arm remain, and what is the probability that the next assignment is B?
  2. A block contains two A and two B assignments. The first two assignments are A, B. Is the third assignment determined? Calculate its probability of being A.
  3. In a block of four containing three A and one B, the first three observed assignments are A, A, A. What is the probability of B for the final position?
  4. Why can fixed-block predictability matter even if the schedule was generated uniformly at random?
  5. State one difference in conditional predictability between fixed-block randomization and complete randomization with fixed overall totals.

Solutions. (1) Two A and one B have been used, so one A and two B remain. The probability of B is \(2/3\approx0.6667\). (2) One A and one B remain, so the third assignment is not determined; its probability of A is \(1/2=0.5000\). (3) No B has been used, so the only remaining slot is B; its probability is \(1/1=1\). (4) A person with access to the schedule or assignment history may use a likely or certain next assignment to influence enrollment or eligibility decisions. (5) A known fixed block adds local remaining-count constraints; complete randomization conditions on the overall remaining totals instead, without those fixed within-block counts.

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