Sample Size and Expected Sample Size · Tutorial 218 of 499

Incorporating Shared-Control Data in Enrollment Calculations

Learn to calculate unique and expected enrollment when the same control participants contribute to multiple active-arm comparisons.

Advanced 10 min read

What You'll Learn

Distinguish pairwise comparison sample sizes from unique participant counts.
Calculate stage and cumulative enrollment with a shared control group.
Adjust enrollment calculations when treatment arms stop at different stages.
Derive expected enrollment using the probability that at least one arm continues.
Identify when shared-control observations cannot validly be counted as common data.

Why Shared-Control Enrollment Needs Care

In a multi-arm design, each active treatment may be compared with a common control group. The control participants contribute information to every relevant comparison, but they are still the same participants. Adding the sample sizes from the pairwise comparisons therefore counts some people more than once. This tutorial focuses on converting those comparison counts into unique participant enrollment, including when arms stop at different stages.

The calculations assume that participants are enrolled into one trial arm only, each control participant is eligible to contribute to each comparison designated by the protocol, and stage counts refer to new participants enrolled in that stage. For the shared-control formulas, the control cohort must genuinely be common to the comparisons; for example, its participants must meet the relevant eligibility and outcome-assessment requirements. If comparisons use distinct control cohorts, those cohorts cannot simply be treated as one shared group.

Definition: Let $n_{C,k}$ be the number of new control participants enrolled in stage $k$, and let $n_{j,k}$ be the number of new participants enrolled to active arm $j$ in that stage. If the same control participants are shared across the active-arm comparisons, the unique stage-$k$ enrollment is $$ N_k=n_{C,k}+\sum_{j\in A_k} n_{j,k}, $$ where $A_k$ is the set of active arms enrolling participants in stage $k$. The control count appears once, regardless of how many comparisons use it.

The distinction is between analysis counts and enrollment counts. A pairwise comparison of arm $j$ with control may use $n_{j,k}+n_{C,k}$ observations in a stage. Summing that quantity over arms gives the total number of observations across comparisons, not the number of distinct participants recruited. Earlier material on converting stage sizes to total enrollment uses the same basic principle: add new participants across stages, rather than repeatedly counting participants already enrolled.

From Pairwise Counts to Unique Participants

Suppose $m$ active arms each use the same control cohort in a stage. If $n_C$ controls and $n_j$ participants in active arm $j$ are enrolled, the sum of the $m$ pairwise sample sizes is $$ \sum_{j=1}^{m}(n_C+n_j)=m n_C+\sum_{j=1}^{m}n_j. $$ The unique participant count is instead $$ N=n_C+\sum_{j=1}^{m}n_j. $$ Thus the pairwise sum exceeds unique enrollment by $(m-1)n_C$: each control participant has been counted in $m$ comparisons, but should be counted once for enrollment.

This subtraction is valid only when the full control cohort is identical across all $m$ comparisons. More generally, count the distinct people in the union of the control cohorts, not the sum of their comparison-specific appearances. When cohorts overlap only partly, the overlap must be accounted for using that union. If the same control participant is observed at multiple stages, that person is likewise counted once in cumulative enrollment, even though the observation can contribute to more than one stage-specific analysis as specified by the design.

Worked Example: Two Comparisons With One Control Group

A hypothetical first stage enrolls 30 control participants, 30 participants to treatment A, and 30 to treatment B. Both treatments are compared with the same control group. Find the unique stage enrollment and compare it with the sum of the pairwise sample sizes.

The two pairwise sample sizes are $30+30=60$ for A versus control and $30+30=60$ for B versus control. Their sum is $60+60=120$, but this total counts the 30 control participants twice.

The unique enrollment is $$ N_1=30+30+30=90\text{ participants}. $$ Equivalently, subtract the duplicated control count: $120-(2-1)(30)=90$. The design has enrolled 90 distinct people, not 120.

An allocation ratio does not change the counting rule. It determines the arm-specific counts, after which the shared control cohort is included once. For example, if each active arm has $r$ participants for every control participant and there are $m$ equally allocated active arms, then $n_j=r n_C$ for every arm and $$ N=n_C+m r n_C=(1+mr)n_C. $$ The pairwise sample-size sum would be $m(1+r)n_C$, which is larger by $(m-1)n_C$. This relationship is an enrollment identity, not a method for determining the sample size required for a particular outcome or power target.

Worked Example: Applying an Allocation Ratio

A hypothetical stage has three active arms. Each arm is allocated 1.5 participants per control participant, and 40 control participants are planned. Calculate the number per active arm and the unique total.

For each active arm, $n_j=1.5(40)=60$ participants. Thus unique enrollment is $$ N=40+60+60+60=220\text{ participants}. $$ The pairwise sample size is $40+60=100$ for each comparison, so their sum is $3(100)=300$. The excess is $300-220=80=(3-1)(40)$, exactly the repeated counting of the shared controls.

Enrollment When Arms Stop at Different Stages

If an arm stops at an interim analysis, it contributes no new participants to later stages. A shared control group may nevertheless continue to accrue for arms that remain active. Therefore, stage enrollment must be calculated using the active arms in that stage, and each stage's control increment is counted once. Cumulative enrollment is the sum of these unique stage increments.

Identify the arms enrolling at each stage.
Let $A_k$ contain the arms still recruiting in stage $k$; omit arms that have stopped.

Count new participants by arm.
For stage $k$, add $n_{j,k}$ for every $j\in A_k$, together with the stage's new shared-control count $n_{C,k}$, if controls are recruited.

Add unique stage counts.
Calculate $N_k=n_{C,k}+\sum_{j\in A_k}n_{j,k}$, then sum $N_k$ over stages for cumulative enrollment.

Worked Example: Unequal Stages With One Arm Stopping

In a hypothetical three-arm design, stage 1 enrolls 24 controls and 24 participants to each of A, B, and C. After stage 1, A stops. Stage 2 then enrolls 16 new controls and 20 new participants to each of B and C. Assume the stage 2 controls are shared by the two continuing comparisons. Find the total unique enrollment and show the error from summing pairwise totals.

Stage 1 enrollment is $24+24+24+24=96$. Stage 2 enrollment is $16+20+20=56$, because A no longer recruits. Cumulative unique enrollment is therefore $96+56=152$ participants.

For comparison, the pairwise sample counts are 48 for A (24 in A and 24 stage 1 controls), 84 for B (44 in B and 40 controls), and 84 for C. Their sum is $48+84+84=216$. The 24 stage 1 controls appear in all three comparisons, while the 16 stage 2 controls appear in B and C. The pairwise counts therefore contain $2(24)+1(16)=64$ excess control appearances. Subtracting those appearances gives $216-64=152$, matching the direct enrollment calculation.

The stage-2 controls are not assigned to the stopped arm's comparison in this calculation: they were recruited for the continuing arms. This illustrates why arm-specific stopping and the design's rules for which control observations contribute to which comparison both matter.

Expected Enrollment With Stochastic Continuation

When continuation depends on interim results, enrollment is random. The key shared-control issue is that later control recruitment is needed if at least one active arm continues, not once for every arm that continues. Let $D_{j,k}$ equal 1 if arm $j$ continues to recruit in stage $k$, and 0 otherwise. Suppose the stage-$k$ control increment $n_{C,k}$ is recruited exactly when at least one arm continues, and the arm-specific increments are $n_{j,k}$. Then the stage-$k$ enrollment is $$ N_k=n_{C,k}\,I\!\left(\sum_j D_{j,k}>0\right)+\sum_j n_{j,k}D_{j,k}, $$ where $I(\cdot)$ is 1 when its condition is true and 0 otherwise. Taking expectations gives $$ E(N_k)=n_{C,k}P\!\left(\sum_j D_{j,k}>0\right)+\sum_j n_{j,k}P(D_{j,k}=1). $$

The probability multiplying the control increment is the probability of the union of continuation events. It is not generally the sum of the arm-specific continuation probabilities. For two arms, for example, $$ P(D_{1,k}=1\text{ or }D_{2,k}=1) =P(D_{1,k}=1)+P(D_{2,k}=1)-P(D_{1,k}=1,D_{2,k}=1). $$ Using the sum alone would count the event that both arms continue twice. Dependence between arm continuation decisions does not invalidate the formula, but it means that the joint probability, or the union probability directly, is needed. The continuation-probability material established earlier can supply these probabilities under the design's decision rules.

For a multi-stage design, expected total enrollment is the sum of expected unique stage enrollments. This follows from linearity of expectation and does not require continuation indicators at different stages to be independent. If some controls are recruited under a different rule, replace the “at least one arm continues” indicator by the indicator for that actual control-recruitment rule.

Worked Example: Expected Enrollment for Two Continuing Arms

A hypothetical design enrolls 20 controls and 20 participants to each of two treatments in stage 1, for 60 participants. In stage 2, it recruits 12 controls if either treatment continues, and 15 participants to each treatment that continues. Suppose the continuation probabilities are 0.40 for A and 0.50 for B, and the probability that both continue is 0.20. Find expected total enrollment.

The probability at least one arm continues is $$ P(A\text{ or }B\text{ continues})=0.40+0.50-0.20=0.70. $$ Expected stage 2 control enrollment is $12(0.70)=8.4$ participants. Expected active-arm enrollment is $15(0.40)+15(0.50)=6+7.5=13.5$ participants. Thus $$ E(N)=60+8.4+13.5=81.9\text{ participants}. $$ This expected value need not be an integer: it is an average over possible trial paths, not a planned fractional enrollment on any one path. Counting the controls separately for each arm would instead give $12(0.40+0.50)=10.8$ expected controls and an incorrect total of 84.3.

Assumptions, Diagnostics, and Alternatives

These calculations rely on a clear accounting definition of a participant and the recruitment increment. Check that stage counts represent new enrollment rather than cumulative sample sizes, that a control participant is not enrolled into multiple arms, and that the same control cohort is truly available to each comparison to which it is credited. When arms stop at different stages, verify that later controls are counted only for the comparisons they can validly support.

A useful diagnostic is to make a stage-by-stage table with one row for each stage and separate columns for shared controls and each active arm. The row total should equal the sum of distinct new recruits in that stage. The cumulative total should equal the sum of those row totals. Separately list the comparison-specific sample sizes; their sum is not expected to match unique enrollment when control data are shared. In an expected-enrollment calculation, check that the probability for control recruitment is between the largest arm-specific continuation probability and 1, and use the probability of the union rather than a sum that could exceed 1.

The main alternative is to recruit a separate control cohort for each active treatment. In that design, the control participants are distinct, so pairwise sample sizes can be added to obtain total enrollment. This avoids shared-cohort bookkeeping, but generally requires more control participants. With a shared group, control observations contribute to multiple comparisons and the resulting comparison statistics are dependent. Correctly counting enrollment does not by itself address that dependence, multiplicity, or the statistical validity of a design; those require the corresponding design and analysis methods.

Common pitfall: “There are three comparisons, so I should add three control sample sizes.” That reasoning counts the same shared participants three times. For enrollment, count the control cohort once at each stage; for expected enrollment, multiply its stage increment by the probability that at least one eligible arm continues. Add a control increment more than once only if the design recruits genuinely distinct control participants.

Check Your Understanding

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

A stage enrolls 25 controls and 25 participants to each of two active arms, with one shared control group. What is unique enrollment, and what is the sum of the pairwise sample sizes?
Three arms use a shared cohort of 30 controls. Each has 45 active-arm participants. How many unique participants are enrolled, and how much does the sum of the pairwise counts exceed that number?
In a later stage, two arms have continuation probabilities 0.30 and 0.60, with joint continuation probability 0.15. What is the probability that shared controls are required, if controls are recruited whenever at least one arm continues?
Give one reason that two comparison-specific control cohorts should not be counted as a single shared cohort.

Solutions. (1) Unique enrollment is $25+25+25=75$. The pairwise sample sizes are 50 each, summing to 100; the 25 excess appearances are the duplicated controls. (2) Unique enrollment is $30+3(45)=165$. Pairwise counts sum to $3(30+45)=225$, exceeding unique enrollment by $225-165=60=(3-1)(30)$. (3) The union probability is $0.30+0.60-0.15=0.75$, so the probability shared controls are required is 0.75. (4) For example, the cohorts may contain different participants, or one cohort may not meet the eligibility or outcome-assessment requirements for the other comparison; in either case their distinct enrollment must be counted.

Back to Tutorials