Enrollment When Arms Stop at Different Stages

Why Different Stopping Stages Change Enrollment

In a multi-arm multi-stage design, treatment arms can stop at different interim stages. An arm that stops early contributes no participants to later stages, while another arm may continue. A common control group creates an additional accounting issue: control participants enrolled during a stage can serve the comparisons for all arms still active, but each person is enrolled only once. The total is therefore not obtained by adding a separate control group for every treatment arm.

The calculation in this tutorial concerns enrollment: the number of unique participants recruited under a specified stopping pattern or, when stopping is uncertain, the expected number recruited. It does not calculate the statistical information contributed by shared controls. To make the accounting precise, first state the stage sizes and the control-accrual rule used by the design.

Definition: Consider $K$ treatment arms followed for at most $J$ stages. Let $S_i$ be the last stage completed by treatment arm $i$, where $S_i\in\{1,\ldots,J\}$. A decision to stop an arm is made after that stage’s participants have been enrolled and assessed. Suppose each active treatment arm enrolls $e$ participants per stage. Suppose the design enrolls $c$ shared-control participants in each stage in which at least one treatment arm is active. The total enrollment for a realized stopping pattern is $$ N=e\sum_{i=1}^{K}S_i+c\max_{1\leq i\leq K}S_i. $$

This definition assumes equal stage increments and a fixed control accrual of $c$ participants per stage while any arm remains active. These are design assumptions, not universal MAMS rules: the allocation plan must specify how many controls are enrolled. If its rule changes control accrual according to the number of active arms, use that specified number at each stage instead. In either case, shared controls are counted once in total enrollment.

Deriving the Enrollment Formula

For stage $j$, treatment arm $i$ is active if $S_i\geq j$. Thus arm $i$ contributes $e$ participants at each of stages $1,\ldots,S_i$, for $eS_i$ treatment participants altogether. Summing across arms gives $e\sum_i S_i$. The shared control group accrues through the last stage reached by any arm. That stage is $\max_i S_i$, so the control contribution is $c\max_i S_i$. Adding these disjoint participant groups gives the formula.

A stagewise version makes the counting rule explicit. At stage $j$, the number of active treatment arms is $\sum_i I(S_i\geq j)$, where $I(\cdot)$ is 1 when its condition is true and 0 otherwise. The stage contributes $e$ participants for each active arm, plus $c$ shared controls if at least one arm is active:

N=\sum_{j=1}^{J}\left[e\sum_{i=1}^{K}I(S_i\geq j)+cI\!\left(\max_i S_i\geq j\right)\right].

The two expressions agree because $\sum_{j=1}^{J}I(S_i\geq j)=S_i$, and $\sum_{j=1}^{J}I(\max_i S_i\geq j)=\max_i S_i$. This also shows why controls cannot be counted as $c$ per active arm: at a stage with several active arms, the one shared-control cohort is still one cohort.

Worked Example: Three Arms Stop at Three Different Stages

Suppose a hypothetical design has three treatment arms, with 40 participants enrolled per active arm per stage and 50 shared-control participants enrolled per stage while at least one arm remains active. The observed stopping stages are $S_1=1$, $S_2=2$, and $S_3=3$. Find the total enrollment.

The treatment-arm enrollment is $40(1+2+3)=40(6)=240$ participants. The latest stopping stage is $\max(1,2,3)=3$, so the shared-control group enrolls $50(3)=150$ participants. Therefore, $$ N=240+150=390\text{ participants}. $$

Stage by stage, the design enrolls three active treatment groups and 50 controls at stage 1, two treatment groups and 50 controls at stage 2, and one treatment group and 50 controls at stage 3. This gives $3(40)+50+2(40)+50+1(40)+50=390$. Counting 50 controls separately for each active arm would incorrectly count the shared cohort multiple times.

Expected Enrollment When Stopping Is Uncertain

If the stopping stages are random, expected enrollment follows by taking the expectation of the stagewise formula. Define $q_{ij}=P(S_i\geq j)$, the probability that arm $i$ reaches stage $j$, and $r_j=P(\max_i S_i\geq j)$, the probability that at least one arm reaches stage $j$. By linearity of expectation,

E[N]=e\sum_{j=1}^{J}\sum_{i=1}^{K}q_{ij} +c\sum_{j=1}^{J}r_j.

This result does not require the arms’ stopping stages to be independent. Each arm’s continuation probabilities determine the expected treatment enrollment. The control term, however, depends on the probability that at least one arm continues to each stage. That probability generally requires joint stopping information. Marginal continuation probabilities alone do not determine it unless additional assumptions, such as independence, are justified.

Interpretation: $E[N]$ is the long-run average number of participants enrolled across repetitions of the same design under the assumed stopping probabilities and accrual rule. It is not the enrollment in every realized trial; an individual trial’s total is determined by its actual stopping pattern.

Worked Example: Expected Enrollment From Joint Stopping Patterns

Consider a hypothetical two-arm design with $e=30$ treatment participants per active arm per stage and $c=24$ shared controls per stage while any arm remains active. The joint stopping pattern has four possible outcomes:

Stopping stages $(S_1,S_2)$	Probability
(1, 1)	0.20
(2, 1)	0.20
(1, 3)	0.20
(3, 2)	0.40

The probabilities sum to $0.20+0.20+0.20+0.40=1.00$. For treatment arm 1, the stage-reach probabilities are $1$, $0.60$, and $0.40$: it reaches stage 2 in outcomes $(2,1)$ and $(3,2)$, and stage 3 only in $(3,2)$. For arm 2, they are $1$, $0.60$, and $0.20$. Thus expected treatment enrollment is $$ 30[(1+0.60+0.40)+(1+0.60+0.20)] =30(3.80)=114\text{ participants}. $$

At stage 1, at least one arm always remains active. At stage 2, at least one reaches that stage in all outcomes except $(1,1)$, so $r_2=0.80$. At stage 3, at least one reaches it in outcomes $(1,3)$ and $(3,2)$, so $r_3=0.20+0.40=0.60$. Expected control enrollment is therefore $24(1+0.80+0.60)=24(2.40)=57.6$ participants. The expected total is $$ E[N]=114+57.6=171.6\text{ participants}. $$

As a check, the total enrollments for the four outcomes are respectively 84, 138, 192, and 222 participants. Weighting them by their probabilities gives $0.20(84)+0.20(138)+0.20(192)+0.40(222)=171.6$ participants. This matches the stagewise calculation.

When the Control Accrual Rule Depends on Active Arms

A design may specify a control accrual amount that varies with the number of active arms. Let $c(m)$ be the number of shared controls enrolled at a stage with $m$ active arms, with no control enrollment when $m=0$. For a particular stopping pattern, define $m_j=\sum_i I(S_i\geq j)$. The total is then the sum, over stages, of treatment accrual $em_j$ and the single control accrual $c(m_j)$. This is an accounting extension of the same principle: apply the actual protocol rule at each stage and count every enrolled person once.

Worked Example: Control Accrual Changes as Arms Stop

Suppose a hypothetical three-arm design enrolls 20 participants per active treatment arm per stage. Its allocation plan enrolls 30 shared controls when three arms are active, 24 when two are active, and 18 when one is active. For the stopping pattern $(S_1,S_2,S_3)=(3,2,1)$, calculate enrollment.

At stage 1, all three arms are active: treatment accrual is $3(20)=60$, and control accrual is 30. At stage 2, two arms remain active: treatment accrual is $2(20)=40$, and control accrual is 24. At stage 3, one arm remains active: treatment accrual is 20, and control accrual is 18. The total is $$ N=(60+30)+(40+24)+(20+18)=192\text{ participants}. $$

The fixed-$c$ formula does not apply directly here because control accrual changes by stage. In particular, it would be wrong to use one fixed control number for all three stages or to multiply each stage’s control count by the number of active arms. The correct calculation follows the stated rule once per stage.

Checks, Alternatives, and Common Mistakes

A reliable calculation can be checked against the design’s stage-by-stage recruitment plan. Confirm that an arm contributes treatment participants through and including its stopping stage, that no arm contributes afterward, and that control accrual continues only while at least one arm remains active. For fixed $e$ and $c$, the total should also agree with both $e\sum_i S_i+c\max_i S_i$ and the stagewise sum.

There are two useful ways to calculate expected enrollment. Direct enumeration weights the total for each possible joint stopping pattern by its probability; it is transparent when the scenario list is short. The stagewise method sums expected stage contributions and is often more compact. Both should agree when they use the same stopping probabilities and accrual rule. The maximum-sample-size calculation answers a different question: it assumes every arm reaches the final stage and gives a design bound, not the expected enrollment under possible early stopping.

Mistake: counting shared controls once per active arm. This duplicates people who form one common control cohort. Count the cohort once at each stage, according to the allocation plan.
Mistake: stopping control accrual as soon as the first arm stops. If another arm remains active and the protocol continues shared-control recruitment, controls accrue through that later arm’s stages.
Mistake: using marginal continuation probabilities to infer control continuation. The chance that at least one arm reaches a stage depends on their joint stopping pattern. Do not assume independence without justification.
Mistake: treating expected enrollment as a guaranteed sample size. It is an average over possible trial paths; the realized enrollment can be above or below it.

Key takeaway: Treatment enrollment accumulates separately for each arm through that arm’s stopping stage. Shared-control enrollment accumulates once per stage for as long as the design has at least one active arm, using the control-accrual rule specified by the protocol.

Check Your Understanding

Use the assumptions stated in each question. The solutions follow the questions.

Two arms stop after stages 2 and 3. Each active arm enrolls 25 participants per stage, and 18 shared controls are enrolled per stage while any arm remains active. Find total enrollment.
In a two-arm, two-stage design, each active arm enrolls 10 participants per stage and 8 shared controls are enrolled per stage. The stopping outcomes are $(1,1)$ with probability 0.5 and $(2,2)$ with probability 0.5. Find expected enrollment.
Why are the marginal probabilities that each arm reaches stage 3 generally insufficient to calculate the probability of shared-control accrual at stage 3?
Three arms stop after stages 1, 1, and 2. Each active arm enrolls 12 participants per stage. The design enrolls 9 shared controls per stage while any arm remains active. Find total enrollment and identify the stage at which control accrual ends.

Solutions. (1) Treatment enrollment is $25(2+3)=125$; controls accrue through stage 3, giving $18(3)=54$. Total: $125+54=179$ participants. (2) The first outcome enrolls $2(10)+8=28$ participants; the second enrolls $4(10)+2(8)=56$. Expected enrollment is $0.5(28)+0.5(56)=42$ participants. (3) Shared controls accrue at stage 3 if at least one arm reaches it. The probability of this union depends on how the arms’ stopping events overlap, which marginal probabilities alone do not specify. (4) Treatment enrollment is $12(1+1+2)=48$. At least one arm reaches stage 2, so controls accrue for two stages: $9(2)=18$. Total enrollment is 66 participants, and control accrual ends after stage 2.