Why Expected Enrollment Depends on the Interim Path
The maximum sample size of a multi-arm, multi-stage (MAMS) design describes the enrollment if every planned stage is reached and every planned treatment arm remains active. Actual enrollment can be smaller: a stage may not be reached, or an interim decision may stop one or more arms. Expected total enrollment accounts for these possibilities by weighting enrollment on each path by the probability of that path.
The key complication is that a later stage’s planned enrollment may depend on which arms continue. If three arms enter a stage, its enrollment may differ from enrollment when only one arm remains. The calculation must therefore average over both whether a stage is reached and which arms are active when it is reached. The continuation probabilities are inputs to this calculation; their derivation is a separate step.
Here, “enrollment at a stage” means the number of new participants enrolled there, not the cumulative number enrolled by that stage. If stage-specific sample sizes are already given as increments, they can be used directly. If they are cumulative totals, first convert them into increments; otherwise participants from earlier stages would be counted more than once.
The General Expectation Formula
For a design with \(K\) stages, the total enrollment on a particular path is the sum of the increments for stages reached on that path. This can be written using the indicator \(R_k\):
Taking expectations and using linearity of expectation gives:
The sum over \(S\) includes the possible active-arm sets at stage \(k\). If the set of active arms is fixed whenever a stage is reached, this reduces to the familiar stage-weighted formula:
More generally, separate the probability of reaching a stage from the distribution of active sets among paths that reach it:
This factorization applies when \(P(R_k=1)>0\). At stage 1, the design ordinarily begins with its planned initial enrollment, so \(P(R_1=1)=1\). An empty active set is not treated as a reached treatment stage: if no treatment arm continues and the trial stops, no later-stage enrollment is added.
Why the Formula Works
For any realized path, a stage contributes its enrollment increment if and only if it is reached. The product \(R_kN_k(A_k)\) is therefore \(N_k(A_k)\) when \(R_k=1\), and 0 otherwise. To find its expectation, sum that contribution across the possible active sets, multiplying each stage enrollment by the joint probability of reaching the stage with that set active. Summing those expected stage contributions gives expected total enrollment.
No independence assumption between stages or between arm-level continuation decisions is needed for this result. The joint probabilities already describe the relevant paths. In particular, probabilities for different arms generally cannot be treated as independent merely because their marginal continuation probabilities are known.
Worked Example: Fixed Enrollment at Each Reached Stage
Suppose a hypothetical design has three stages. The first stage enrolls 120 participants. If reached, stage 2 enrolls 90 participants; if reached, stage 3 enrolls 60 participants. Suppose the supplied probabilities are \(P(R_1=1)=1\), \(P(R_2=1)=0.65\), and \(P(R_3=1)=0.30\). The stage sizes are fixed and do not depend on which arms continue.
The model assumptions are that the stated stage sizes are new-participant increments and that the stated reach probabilities apply to the scenario of interest. No further assumptions about how those probabilities were obtained are needed for the expectation calculation.
Apply the fixed-stage formula:
Expected total enrollment is 196.5 participants under the specified scenario. This is a long-run average across repetitions of the design, not a possible realized enrollment requirement: an actual trial enrolls a whole number of participants. The maximum enrollment is \(120+90+60=270\) participants, which would occur if both later stages were reached.
Allowing the Active-Arm Set to Change
When different arm sets imply different stage sizes, the enrollment for stage \(k\) is not a single fixed number. Calculate the average stage enrollment among paths that reach the stage, then weight that average by the probability of reaching it. Equivalently, sum over the joint probabilities \(P(R_k=1,A_k=S)\). The two expressions are the same; the latter makes clear that both stage reach and the active set matter.
Worked Example: Stage Enrollment Depends on the Continuing Arms
Consider a hypothetical three-arm design with a first-stage enrollment of 150 participants. The probability of reaching stage 2 is 0.70. Among paths that reach stage 2, the active sets and probabilities are: all three arms, probability 0.40; two arms, probability 0.40; and one arm, probability 0.20. Stage 2 enrolls 120, 90, or 60 participants for those respective active sets.
Suppose the probability of reaching stage 3 is 0.28. Among paths that reach stage 3, the active set contains two arms with probability 0.25 and one arm with probability 0.75. Stage 3 enrolls 80 participants with two active arms and 50 with one. These inputs can describe nested arm sets: the stage-3 two-arm paths can arise from stage-2 all-arm paths, while stage-3 one-arm paths can arise from stage-2 two-arm or one-arm paths. The probabilities supplied for each stage are conditional on reaching that stage, so each set of conditional probabilities sums to 1.
First find the mean stage-2 enrollment conditional on reaching that stage:
The stage-2 contribution to expected total enrollment is \(0.70(96)=67.2\) participants. For stage 3:
The stage-3 contribution is \(0.28(57.5)=16.1\) participants. Including the initial stage:
Expected enrollment is 233.3 participants under the stated path probabilities. The maximum under the listed stage sizes is \(150+120+80=350\) participants. The weighted average at stage 3 is based on the active sets among paths that reach stage 3, not on all paths that began the trial.
Shared Controls and Arm-Level Contributions
A shared control group requires particular care. If several treatment arms continue, the same control participants may contribute to comparisons with more than one arm. They are enrolled once, not once per comparison. For this reason, an arm-by-arm sum is valid only if it keeps shared-control participants from being counted repeatedly.
Worked Example: Counting a Shared Control Group Once
In a hypothetical three-arm design, stage 1 enrolls 40 participants per treatment arm and 60 shared-control participants, for 180 participants total. At stage 2, each active treatment arm requires 30 new participants. A shared control group of 45 participants is enrolled if at least one arm continues. At stage 3, each active treatment arm requires 20 participants, and a shared control group of 30 is enrolled if at least one arm continues.
Suppose the probabilities that arms A, B, and C reach stage 2 are 0.60, 0.50, and 0.30. The probability that at least one arm reaches stage 2 is 0.70. At stage 3, the respective arm-level reach probabilities are 0.25, 0.20, and 0.10, and the probability that at least one arm reaches stage 3 is 0.35. These probabilities are assumed to be supplied for the same scenario. The shared-control probabilities are supplied directly because the marginal probabilities for individual arms alone do not determine a union probability.
By linearity of expectation, the expected number of stage-2 treatment participants is \(30(0.60+0.50+0.30)=42\). The shared control contributes \(45(0.70)=31.5\), counted once. Thus the expected stage-2 contribution is \(42+31.5=73.5\).
At stage 3, the expected number of treatment participants is \(20(0.25+0.20+0.10)=11\). The shared control contributes \(30(0.35)=10.5\). The expected stage-3 contribution is \(11+10.5=21.5\). Therefore:
Expected total enrollment is 275 participants. The calculation does not require independence among arm-level continuation events: sums of expected treatment contributions use their marginal probabilities, while shared-control enrollment uses the probability that at least one arm continues. The assumed probabilities must still be mutually coherent with the design’s interim paths.
Practical Checks and Common Failure Modes
Use new participants at each stage. If the inputs are cumulative sample sizes, subtract earlier enrollment before applying the formula.
For every possible set of continuing arms, state the stage enrollment, including any shared-control allocation exactly once.
Use \(P(R_k=1)\) for stage reach and conditional active-set probabilities among paths reaching stage \(k\), or use the corresponding joint probabilities.
Weight each stage’s enrollment by the probability of the event that triggers it, and add the stage contributions.
Expected enrollment must be between the minimum and maximum possible total enrollment under the design. It is an average under the stated scenario, not a guaranteed enrollment or a substitute for the maximum required sample size.
A common mistake is to multiply a stage size by the probability that any one treatment arm continues, even when enrollment at that stage depends on how many arms continue. The correction is to average stage enrollment over the active-set distribution, then multiply by the probability of reaching the stage. Another mistake is to add each arm’s implied control allocation. With a shared control, count its enrollment according to the design’s actual shared-control rule, not once for every active comparison.
The maximum-sample-size calculation answers how many participants could be enrolled on the largest planned path. Expected enrollment answers the average across paths under specified probabilities. A full path enumeration can be used instead of stage-by-stage averaging, and simulation can estimate the same expectation when there are many possible paths. These approaches should agree when they use the same path probabilities and enrollment rules. Expected enrollment alone does not establish that a design has adequate power, controls its error rate, or meets a recruitment cap.
Check Your Understanding
Use the stated probabilities as inputs; do not try to derive them.
- A two-stage design enrolls 100 participants in stage 1 and 80 more if stage 2 is reached with probability 0.40. What is expected total enrollment?
- Stage 2 is reached with probability 0.50. Conditional on reaching it, one active arm has probability 0.30 and two active arms have probability 0.70. Their stage-2 enrollments are 40 and 70. What is the expected stage-2 contribution?
- Why can marginal continuation probabilities for two arms be insufficient to determine expected shared-control enrollment?
- Stage sizes are given as cumulative enrollments of 120 at stage 1 and 200 at stage 2. Stage 2 is reached with probability 0.25. What stage-2 increment should be weighted, and what is expected total enrollment?
Solutions. (1) \(100+0.40(80)=132\) participants. (2) The conditional mean stage-2 enrollment is \(0.30(40)+0.70(70)=61\), so the expected contribution is \(0.50(61)=30.5\) participants. (3) The probability that at least one arm continues depends on the joint behavior of the continuation events, not just their separate marginal probabilities; the shared-control rule is triggered by that union event. (4) The stage-2 increment is \(200-120=80\), not 200. Expected total enrollment is \(120+0.25(80)=140\) participants.