Why Enumerate Continuation Patterns?
In a multi-arm multi-stage (MAMS) design, the trial may stop early or continue with only some treatment arms. As a result, the final number enrolled is not fixed: it depends on which arms satisfy the continuation rules at each interim analysis. An analytical expected-enrollment calculation summarizes this uncertainty, but a small design can also be checked directly by listing every possible continuation pattern.
Enumeration is especially useful as a validation check. It makes explicit which outcomes of the interim decisions have been included, whether their probabilities add to one, and how many participants each outcome entails. It is not a replacement for a general calculation when the number of arms or stages is large; it is a transparent audit that can expose mistakes in a calculation or its assumptions.
The scenario matters: continuation probabilities depend on the assumed treatment effects, outcome variability, decision boundaries, and other design assumptions. Enumeration checks the arithmetic conditional on those probabilities; it does not establish that the scenario assumptions or decision rules are scientifically appropriate.
Two Equivalent Ways to Calculate the Expectation
Suppose a design has a finite set of possible continuation patterns, indexed by \(r\). Let \(N_r\) be the total enrollment under pattern \(r\), and let \(\pi_r\) be its probability under the scenario being evaluated. The patterns must be exhaustive and mutually exclusive: exactly one occurs. Therefore, \(\sum_r \pi_r=1\), and the expected enrollment is
A second calculation works stage by stage. Write total enrollment as the sum of enrollment increments at each stage. If stage \(j\) adds \(n_j\) participants whenever it is reached, and \(R_j\) is the event that it is reached, then
where \(I(R_j)\) equals 1 if stage \(j\) is reached and 0 otherwise. If the number enrolled at a stage depends on which arms remain active, the increment \(n_j\) must instead be represented by the relevant arm-specific and shared-control increments. In either form, the expected total follows by adding the expected increments; independence between stage decisions is not required.
A Short Justification of the Stage-Reach Check
For each continuation pattern \(r\), let \(I_{jr}\) equal 1 when stage \(j\) is reached under that pattern and 0 otherwise. The enrollment on pattern \(r\) is \(N_r=\sum_j n_j I_{jr}\), when increments \(n_j\) are fixed whenever a stage is reached. Substitution into the enumeration formula gives
The last equality holds because the pattern probabilities for which stage \(j\) is reached sum to \(P(R_j)\). This establishes why the enumeration and stage-reach calculations must agree when they describe the same design and scenario. If increments vary by pattern, use each pattern’s actual increment in the first calculation, or sum the corresponding arm-specific expected increments in the second.
Worked Example: One Treatment Arm and Two Stages
Consider a hypothetical trial with one treatment arm and a control arm. Stage 1 enrolls 10 participants per arm, for 20 total. If the trial continues after stage 1, stage 2 enrolls 6 more per arm, for an additional 12. Under a specified scenario, the probability of continuing is 0.65. Assume these are the only possible outcomes: stop after stage 1 or continue to stage 2.
The two patterns are exhaustive. Their probabilities are 0.35 for stopping and 0.65 for continuing. Their enrollment totals are respectively 20 and \(20+12=32\). The enumeration gives
The stage-reach calculation gives the same answer. Stage 1 is always reached, and stage 2 is reached with probability 0.65:
Thus the hypothetical expected enrollment is 27.8 participants. It need not be an attainable realized total: each actual trial enrolls either 20 or 32 participants under this simplified design.
Enumerating Arm-Specific Continuation With Shared Control
For multiple treatment arms, a pattern must record which arms continue, not merely whether the trial continues. Shared control requires particular care. If a single group of control participants serves all active treatment comparisons, those participants are counted once in total enrollment, even though their data contribute to more than one comparison.
Worked Example: Two Arms With One Shared-Control Cohort
Consider a hypothetical two-stage design. Stage 1 enrolls 18 participants in treatment A, 18 in treatment B, and 18 in a shared control group, for 54 participants. At the interim, either, both, or neither treatment arm may continue. At stage 2, each continuing treatment arm adds 8 participants, and if at least one arm continues, the design adds one shared-control cohort of 12 participants. Under a specified scenario, the probabilities are: both arms continue, 0.35; A only continues, 0.20; B only continues, 0.15; neither continues, 0.30.
First check the pattern probabilities: \(0.35+0.20+0.15+0.30=1.00\). The patterns are mutually exclusive because they specify distinct continuation statuses for A and B. Stage 2 adds 28 participants if both arms continue (\(8+8+12\)), 20 if just one continues (\(8+12\)), and none if neither continues. The corresponding total enrollments are 82, 74, 74, and 54.
Check this result by stage increments. A continues with probability \(0.35+0.20=0.55\), and B continues with probability \(0.35+0.15=0.50\). The shared-control cohort is added whenever at least one arm continues, an event with probability \(0.35+0.20+0.15=0.70\). Therefore,
The two methods agree. Counting a separate 12-person control cohort for each continuing arm would instead add too many participants when both arms continue; the design specifies one shared cohort.
Joint Patterns Matter: Do Not Assume Independence
An enumeration uses the joint probabilities of complete patterns. Marginal continuation probabilities alone do not determine the probability that at least one of several arms continues unless their dependence is also accounted for. Shared-control enrollment is often tied to precisely that “at least one arm” event.
Worked Example: Checking an Independence Shortcut
Use the hypothetical two-arm design from the previous example, with continuation-pattern probabilities 0.35 for both arms, 0.20 for A only, 0.15 for B only, and 0.30 for neither. The marginal continuation probabilities are 0.55 for A and 0.50 for B. The actual probability that at least one arm continues is \(0.35+0.20+0.15=0.70\).
If someone incorrectly assumes independence, the probability that neither continues would be \((1-0.55)(1-0.50)=0.225\), giving an “at least one” probability of \(1-0.225=0.775\). That contradicts the enumerated probability 0.70, because the listed pattern probabilities do not represent independent continuation events. Using this incorrect shortcut would give
rather than the correct 70.8 participants. The error is 0.9 participants in expected enrollment. The marginal probabilities correctly calculate the expected treatment-arm increments, but they do not justify replacing the joint probability of any continuation with an independence formula.
How to Audit an Enumeration
A useful validation is systematic: define the pattern set, verify its probabilities, calculate each total from the actual enrollment rules, and independently reconstruct the expectation from stage or arm increments. A discrepancy should be investigated, not averaged away. Common sources include missing or duplicated patterns, confusing conditional probabilities with unconditional probabilities, and miscounting shared control.
State the assumed scenario, interim continuation criteria, stage increments, and how shared-control participants are allocated and counted.
Record every possible set of continuing arms at each interim. Make the patterns mutually exclusive and collectively exhaustive.
Check that pattern probabilities are nonnegative and sum to one, then compute enrollment for each pattern from the stated rules.
Find the probability-weighted total, then compare it with the sum of expected stage or arm-specific increments. Investigate any disagreement.
Diagnostics should include more than checking the final sum. Confirm that the initial-stage enrollment appears in every pattern; a later stage contributes only when reached; and each participant group is counted exactly once. When probabilities are conditional on earlier continuation, multiply along the relevant branch to obtain a complete-pattern probability. For example, if stage 2 is reached with probability 0.75 and, conditional on reaching it, stage 3 is reached with probability 0.60, then \(P(\text{stage 3 reached})=(0.75)(0.60)=0.45\).
Enumeration is most practical when the number of patterns is small enough to inspect. For many arms or stages, the number of branches can become unwieldy; a stage-reach calculation or a carefully checked computational calculation is more efficient. Simulation can estimate expected enrollment when direct probabilities are difficult to obtain, but it introduces Monte Carlo error. Enumeration, when feasible and supplied with correct pattern probabilities, calculates the scenario-specific expectation exactly up to the precision of those probabilities and arithmetic.
Common Mistakes and Limitations
- Adding probabilities that describe overlapping events. “A continues” and “B continues” overlap when both continue. To calculate the probability that at least one continues, sum the mutually exclusive patterns that satisfy that condition, or use a valid joint-probability calculation.
- Using conditional probabilities as if they were pattern probabilities. A conditional probability must be multiplied by the probability of reaching the condition. The probability of two successive continuations is not generally the second conditional probability alone.
- Double-counting shared control. Count the actual control cohort specified by the design, not a separate copy for every comparison it supports.
- Checking arithmetic but not assumptions. Agreement between two calculations cannot validate incorrect continuation probabilities, an incorrectly specified enrollment rule, or an unsuitable scenario.
Enumeration is a validation tool, not a general solution to uncertainty about the design. It does not replace evaluating expected enrollment across relevant scenarios, assessing whether the assumed model is credible, or considering the uncertainty in simulation-based estimates. Its value is that, in a small example, every contribution to the expected total can be traced and checked.
Check Your Understanding
Work through the questions, then compare your reasoning with the solutions below.
- A hypothetical trial enrolls 30 participants at stage 1 and 15 more if it continues. The continuation probability is 0.40. What is expected enrollment by enumeration?
- Two arms have continuation-pattern probabilities 0.25 for both, 0.30 for A only, 0.10 for B only, and 0.35 for neither. What is the probability that at least one continues?
- In question 2, each continuing arm adds 5 participants, and a single shared-control cohort of 9 is added if any arm continues. What is expected stage-2 enrollment?
- If stage 2 is reached with probability 0.80 and stage 3 is reached with probability 0.50 conditional on reaching stage 2, what is the probability that stage 3 is reached?
Solutions. (1) Stop with probability \(0.60\) at total enrollment 30; continue with probability \(0.40\) at total 45. Thus \(E(N)=(0.60)(30)+(0.40)(45)=18+18=36\) participants. (2) The patterns with at least one continuing arm are disjoint, so the probability is \(0.25+0.30+0.10=0.65\). (3) A continues with probability \(0.25+0.30=0.55\); B continues with probability \(0.25+0.10=0.35\); the shared-control cohort is added with probability 0.65. The expected increment is \((5)(0.55)+(5)(0.35)+(9)(0.65)=2.75+1.75+5.85=10.35\) participants. (4) Multiply the probability of reaching stage 2 by the conditional probability of reaching stage 3: \((0.80)(0.50)=0.40\).