What a Completed Block Guarantees
Permuted blocks constrain the order in which assignments occur while prescribing how many assignments of each treatment appear in a block. The guarantee is exact but local: if the block is completed and its assignments are implemented as specified, its treatment counts equal the prescribed counts. That fact does not say that every prefix of the block is balanced, or that patients assigned to different treatments have similar characteristics.
To make the claim precise, suppose there are \(K\) treatments. A block has \(B\) assignment slots, with \(m_j\) slots prescribed for treatment \(j\), where \(m_j\) is a nonnegative integer and \(\sum_{j=1}^{K}m_j=B\). A permuted block is a random ordering of those prescribed assignments. The order is randomized; the composition is fixed.
The relevant assumptions are straightforward. The block composition must be specified correctly for the intended allocation ratio; the block must reach all \(B\) assignment slots; and the assignments must be implemented without substitutions, omissions, or duplications. Randomly ordering a prescribed composition changes the order, not the counts. If the block is not completed, the completed-block guarantee cannot be applied to its partial sequence.
Why the Guarantee Holds
Let \(I_{ij}\) equal 1 if slot \(i\) in the block assigns treatment \(j\), and 0 otherwise. The construction places exactly \(m_j\) copies of treatment \(j\) among the \(B\) slots. Therefore, for each treatment \(j\), summing its slot indicators gives its prescribed count.
This argument does not depend on which permitted ordering was selected. If the ordering is selected uniformly from all distinct arrangements, there are \(B!/\prod_{j=1}^{K}m_j!\) such arrangements, and each has the same prescribed counts. Uniform randomization affects the probability of different orders; it cannot change the count in a completed block.
The result is an accounting identity rather than a probabilistic approximation. Once all slots have been used, the count is exactly \(m_j\), not merely expected to be \(m_j\). Before completion, however, a block can have any prefix compatible with the allowed arrangements. That prefix need not have the target ratio.
Worked Example: A 1:1 Block Can Be Imbalanced Midway
Suppose a hypothetical two-treatment trial uses a block of size 4, prescribing 2 assignments to treatment A and 2 to treatment B. One permitted randomized order is A, B, B, A. The assignment counts after each slot are shown below.
| Slots assigned | Sequence so far | A count | B count | A minus B |
|---|---|---|---|---|
| 1 | A | 1 | 0 | 1 |
| 2 | A, B | 1 | 1 | 0 |
| 3 | A, B, B | 1 | 2 | -1 |
| 4 | A, B, B, A | 2 | 2 | 0 |
The prescribed counts are met at completion: \(2\) assignments to A and \(2\) to B. The cumulative counts are equal after slots 2 and 4, but not after slots 1 and 3. Thus “balanced within a completed block” does not mean “equal after every assignment.” The maximum observed absolute difference in this particular order is 1 participant.
Tracking Balance Relative to an Allocation Ratio
For unequal allocation, raw differences between treatment counts can be misleading. For example, with a target ratio of 2:1, equal counts are not the target. Let \(q_j=m_j/B\) denote treatment \(j\)'s target share within a block. After \(s\) assignments of that block, let \(C_j(s)\) be the number assigned to treatment \(j\). A useful signed deviation from the target is \(D_j(s)=C_j(s)-q_js\). A positive deviation means that treatment \(j\) has received more than its target share of the assignments made so far; a negative deviation means it has received less.
At the end of a completed block, \(C_j(B)=m_j\), so the deviation is zero. At an intermediate point, it need not be. Because the target share can be fractional, the deviation is often fractional even though the treatment count is an integer. It is a measure of departure from the target proportion, not a number of fractional patients.
Worked Example: A 2:1 Block and Its Cumulative Deviations
Consider a hypothetical block of size 6 prescribing 4 assignments to A and 2 to B. Its target shares are \(q_A=4/6=2/3\) and \(q_B=2/6=1/3\). Suppose the randomized order is A, B, A, A, B, A. For each prefix, calculate \(D_A(s)=C_A(s)-(2/3)s\) and \(D_B(s)=C_B(s)-(1/3)s\).
| Slots \(s\) | A count | B count | \(D_A(s)\) | \(D_B(s)\) |
|---|---|---|---|---|
| 1 | 1 | 0 | \(1-2/3=1/3\) | \(0-1/3=-1/3\) |
| 2 | 1 | 1 | \(1-4/3=-1/3\) | \(1-2/3=1/3\) |
| 3 | 2 | 1 | \(2-2=0\) | \(1-1=0\) |
| 4 | 3 | 1 | \(3-8/3=1/3\) | \(1-4/3=-1/3\) |
| 5 | 3 | 2 | \(3-10/3=-1/3\) | \(2-5/3=1/3\) |
| 6 | 4 | 2 | \(4-4=0\) | \(2-2=0\) |
At slot 6, the block contains exactly the required 4 A assignments and 2 B assignments, and both deviations are zero. At slot 5, the cumulative counts happen to be 3 and 2, while the target shares of 5 assignments are \(10/3\) and \(5/3\). The deviations, \(-1/3\) for A and \(1/3\) for B, show that the prefix has one fewer A and one more B than those target shares. The block guarantee applies at completion, not to that prefix.
From One Block to a Trial Sequence
If successive blocks all have the same prescribed composition, every completed-block boundary returns the overall allocation to the target ratio. For example, after \(L\) completed blocks each prescribing \(m_j\) assignments to treatment \(j\), the total for that treatment is \(Lm_j\), and the total number assigned is \(LB\). Its overall share is therefore \(Lm_j/(LB)=m_j/B=q_j\).
More generally, the same conclusion holds when completed blocks have different sizes, provided each block has the same target share \(q_j\). If block \(b\) has size \(B_b\) and prescribes \(m_{bj}=q_jB_b\) assignments to treatment \(j\), then across \(L\) completed blocks the count is \(\sum_{b=1}^{L}m_{bj}=q_j\sum_{b=1}^{L}B_b\). This is exact balance at completed-block boundaries relative to that share. A partial block contributes a deviation of its own, so it can move the overall cumulative allocation away from the target until it is completed.
Worked Example: Two Completed Blocks and an Unfinished Block
Suppose a hypothetical 1:1 trial uses blocks of size 4, each with 2 assignments to A and 2 to B. The first completed block is A, A, B, B; the second is B, A, B, A. A third block has begun, and its first assignment is A. Track the overall counts.
After the first block, the cumulative counts are \(A=2\) and \(B=2\), for a total of 4. The signed difference is \(2-2=0\). After the second block, the counts are \(A=4\) and \(B=4\), for a total of 8, and the difference is \(4-4=0\). This is the completed-block result applied twice.
After the first assignment in the third block, the totals are \(A=5\) and \(B=4\), for a total of 9. The overall difference is \(5-4=1\), so the cumulative counts are not equal at this point. If that block is later completed with its prescribed two assignments per treatment, its remaining three slots must include one A and two B assignments. The final totals will then be \(A=6\) and \(B=6\), restoring equality at the completed-block boundary.
This calculation also shows what happens if enrollment or observation stops before a block is complete: the final observed totals need not satisfy the block's prescribed counts. The guarantee is conditional on completing the block, not a claim about every possible stopping point.
Justification Across Blocks
The cumulative result follows directly by adding the counts from each completed block. For a fixed target share \(q_j\), block \(b\) contributes \(m_{bj}=q_jB_b\) assignments to treatment \(j\). Therefore, after \(L\) completed blocks, the total deviation is
If the next block is only \(s\) slots complete, its contribution to the deviation is \(C_j(s)-q_js\). The earlier completed blocks contribute zero, so the overall deviation at that point is exactly the current block's deviation. This proves why exact target-share balance returns at the boundary but need not hold between boundaries.
What the Guarantee Does Not Say
Permuted blocks are one way to constrain allocation counts. Under complete randomization, assignments are selected for the full allocation set rather than organized into blocks; when the total treatment counts are fixed, the full set has those counts, but there is no corresponding guarantee at each short segment of the sequence. Permuted blocks add a local count constraint at completed-block boundaries. Neither count-based statement alone guarantees balance in age, disease severity, site, or any other patient characteristic. Those are separate features of the realized sample, not components of the treatment-count identity.
The earlier material on choosing block sizes for an allocation ratio establishes the need for prescribed counts compatible with the intended ratio. This tutorial's guarantee depends on that choice being valid: if a block's counts do not represent the intended target share, completion guarantees the specified counts, not the intended ratio. The guarantee also does not establish that the realized sequence has no implementation errors, that a partial block is balanced, or that treatment outcomes will be comparable in every respect.
A useful diagnostic is to audit the schedule at each completed-block boundary. Compare the observed treatment counts within that block with the prescribed \(m_j\) values, then check the cumulative counts against the target shares. A mismatch at a completed boundary suggests that the wrong block composition was used, a slot was skipped or duplicated, or the recorded assignment differs from the schedule. A mismatch during a partial block is not, by itself, evidence of an error.
- Wrong reasoning: “The target ratio is 1:1, so the counts must be equal after every participant.”
Correction: A permuted block guarantees the prescribed ratio when the block is complete; its prefixes can be unequal. - Wrong reasoning: “The completed block has the right counts, so it must also have balanced patient characteristics.”
Correction: The count guarantee concerns treatment labels only. It does not force covariate distributions to match. - Wrong reasoning: “The study stopped during a block, so the completed-block guarantee was violated.”
Correction: An unfinished block is not a completed block. Its observed counts need not equal the full prescribed counts.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- A completed block has 3 assignments to A and 3 to B. What counts are guaranteed, and what can be said about the counts after the first four assignments?
- A block of size 6 prescribes 4 assignments to A and 2 to B. After five slots, the counts are 3 and 2. Calculate both target-share deviations.
- Two completed blocks each prescribe 2 assignments to A and 2 to B. What are the cumulative counts and A-minus-B difference at the second block boundary?
- A trial stops after the first assignment in a 1:1 block, and that assignment is to A. Has the completed-block guarantee failed? Explain.
- Does exact within-block treatment balance imply balance in a baseline covariate? Why or why not?
Solutions. (1) Completion guarantees \(A=3\) and \(B=3\). The first four assignments need not be equal by treatment; their counts depend on the randomized order, subject to what remains available in the block. (2) The target shares are \(q_A=2/3\) and \(q_B=1/3\). Thus \(D_A(5)=3-(2/3)(5)=3-10/3=-1/3\), and \(D_B(5)=2-(1/3)(5)=2-5/3=1/3\). (3) Each completed block contributes 2 to each treatment, so the cumulative counts are \(A=4\), \(B=4\), and \(A-B=0\). (4) No. The block is unfinished, so the completed-block guarantee does not apply; the observed counts are \(A=1\), \(B=0\). (5) No. The guarantee fixes treatment counts, not the distribution of patient characteristics across treatments.