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Seamless Phase 2/3 Designs · Tutorial 500 of 799

Expected Sample Size Under Transition Rules

Calculate the probability-weighted enrollment and duration of a seamless trial, while accounting for pathway-specific sample sizes, recruitment constraints, follow-up, and transition delays.

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What You'll Learn

  • Define mutually exclusive transition pathways and their probabilities.
  • Calculate expected total enrollment when pathways have different stage 2 sample sizes.
  • Account for shared controls and unequal randomization without double counting participants.
  • Calculate expected calendar duration from pathway-specific recruitment and follow-up times.
  • Distinguish expected duration from duration calculated using expected enrollment.
  • Identify assumptions and operational details that can make a planning estimate misleading.

Planning Around More Than One Trial Path

At a seamless phase boundary, a trial may stop, continue with one treatment, or continue with another. Each decision leads to a different amount of future enrollment and may take a different amount of calendar time. Planning with only the largest possible sample size can overstate typical resource needs; planning with a single “average” pathway can conceal how much the actual trial may vary.

The expected enrollment and expected duration summarize the design across its possible pathways. They are useful for budgeting and operational planning, but they do not guarantee that a realized trial will match either expectation. The calculation begins by listing the possible transition outcomes, assigning a probability to each, and specifying the enrollment and timing that each outcome entails.

Definition: A transition pathway is one possible action at the phase boundary, such as stopping for lack of a treatment to advance or continuing with a selected treatment. Let \(\pi_j\) be the probability of pathway \(j\), where the pathways are mutually exclusive and exhaustive, so that \(\pi_j \geq 0\) and \(\sum_j \pi_j=1\). Let \(n_1\) be the fixed number of participants enrolled in stage 1, and let \(n_{2j}\) be the number of additional, unique participants enrolled in stage 2 on pathway \(j\). The expected total enrollment is the probability-weighted average of the pathway-specific total enrollments.

Here, enrollment means participants randomized into the trial. If a protocol also tracks screened patients, screen failures, or participants needed to replace those with unusable outcomes, those quantities require their own assumptions and calculations. The number enrolled in stage 1 is counted once on every pathway. Stage 2 enrollment is zero on a stop pathway.

Expected Enrollment From Transition Probabilities

Let \(N\) denote total enrollment. If pathway \(j\) occurs, the total is \(N_j=n_1+n_{2j}\). The expected total is therefore

$$ E[N]=\sum_j \pi_j(n_1+n_{2j}) =n_1+\sum_j \pi_j n_{2j}. $$

The second equality uses \(\sum_j\pi_j=1\): stage 1 is enrolled regardless of the later decision. This expression also works when some pathways stop, provided their \(n_{2j}\) is set to zero. It requires no assumption that the probabilities of the transition outcomes are independent of the stage 1 data; the probabilities describe the design's chance of taking each route.

Worked Example: Three Possible Transition Outcomes

Suppose a hypothetical seamless trial plans to enroll \(n_1=180\) participants in stage 1. At the phase boundary, the design stops with probability \(0.25\), continues with treatment A with probability \(0.45\), or continues with treatment B with probability \(0.30\). The stage 2 enrollments are 0, 360, and 420, respectively. Find expected total enrollment.

The probabilities are exhaustive because \(0.25+0.45+0.30=1.00\). The pathway-specific totals are 180, \(180+360=540\), and \(180+420=600\) participants. Thus,

$$ E[N]=0.25(180)+0.45(540)+0.30(600) =45+243+180=468. $$

Equivalently, the expected stage 2 enrollment is \(0.25(0)+0.45(360)+0.30(420)=288\), giving \(180+288=468\) participants in expectation. The actual trial enrolls 180, 540, or 600 participants under this simplified plan; 468 is a probability-weighted planning average, not a possible pathway total.

When treatment arms have different allocations, calculate \(n_{2j}\) as the number of unique participants randomized on pathway \(j\). A shared control group is not counted again for each active-treatment comparison. If the stage 2 total is specified through arm sizes, add the distinct arm allocations once, applying the intended randomization ratio and any planned control sharing.

Worked Example: Unequal Allocation and Shared Control

In a hypothetical design, stage 1 enrolls 150 participants. At the boundary, the trial stops with probability \(0.20\), advances treatment A with probability \(0.50\), or advances treatment B with probability \(0.30\). On the A pathway, 160 participants are assigned to A and 80 to a shared control; on the B pathway, 240 are assigned to B and 120 to that pathway's control. Find expected total enrollment.

The A pathway needs \(160+80=240\) stage 2 participants, and the B pathway needs \(240+120=360\). Each control participant is counted once in the pathway on which the participant is randomized. Therefore,

$$ E[N]=150+0.20(0)+0.50(240)+0.30(360) =150+120+108=378. $$

The expected total is 378 randomized participants. Counting the A-pathway control group again as though it were a separate control group for each comparison would inflate enrollment. If a protocol instead specifies a control group common to both stage 2 comparisons on a single pathway, its participants should be included once in that pathway's total.

Expected Calendar Duration

Enrollment and duration are related, but duration is not always total enrollment divided by one accrual rate. Recruitment may be constrained by the slowest arm, the active treatment may change between pathways, or a phase transition and outcome follow-up may add calendar time. Define a pathway-specific duration from a clear start point to a clear endpoint, such as time from first enrollment to availability of the required primary outcome data.

Let \(T_j\) be that duration if pathway \(j\) occurs. If the stage 1 period is common to every pathway and stage 2 follows it, one useful decomposition is \(T_j=T_1+d_j+T_{2j}+f_j\), where \(T_1\) is stage 1 accrual time, \(d_j\) is any transition delay, \(T_{2j}\) is stage 2 accrual time, and \(f_j\) is additional follow-up required after the relevant accrual period. Include a component only if it is not already included in another term. Then

$$ E[T]=\sum_j\pi_jT_j. $$

For parallel arm recruitment with arm-specific enrollment targets \(n_{aj}\) and accrual capacities \(r_{aj}\) participants per month, a simple planning model for stage 2 is

$$ T_{2j}=\max_a\left(\frac{n_{aj}}{r_{aj}}\right). $$

This maximum applies when recruitment in all required arms proceeds concurrently, each rate is a stable capacity, and there are no other bottlenecks. If participants are recruited through one trial-wide pool at a stable rate \(r_j\), a different appropriate model may be \(T_{2j}=n_{2j}/r_j\). The operational assumptions determine which calculation is defensible.

Worked Example: Different Recruitment Times by Selected Treatment

A hypothetical trial enrolls 150 participants in stage 1 at a trial-wide rate of 30 participants per month, taking \(150/30=5\) months. After the boundary, the trial stops with probability \(0.20\), continues with A with probability \(0.50\), or continues with B with probability \(0.30\). On pathway A, 160 participants are recruited to A and 80 to control, with capacity for 20 participants per month in each arm. On pathway B, 240 are recruited to B and 120 to control, with the same arm-specific capacities. Assume concurrent arm recruitment, no transition delay, no extra follow-up, and that the five-month stage 1 period is common to all paths. Calculate expected duration.

On pathway A, the arm-specific durations are \(160/20=8\) months and \(80/20=4\) months. Concurrent recruitment means stage 2 takes the longer time, 8 months. On pathway B, the corresponding durations are \(240/20=12\) months and \(120/20=6\) months, so stage 2 takes 12 months. The total pathway durations are 5 months if the trial stops, \(5+8=13\) months on A, and \(5+12=17\) months on B. Thus,

$$ E[T]=0.20(5)+0.50(13)+0.30(17) =1+6.5+5.1=12.6\text{ months}. $$

This is an expected duration of 12.6 months under the stated timing model. For comparison, expected enrollment is 378 participants, but dividing that figure by 30 participants per month would give 12.6 months only by coincidence: stage 2 recruitment is arm-constrained here, and the stop pathway has a different duration.

Derivation and Checks

The expectation formulas follow from summing over the possible pathways. To see this directly for enrollment, define \(I_j\) to equal 1 if pathway \(j\) occurs and 0 otherwise. Exactly one pathway occurs, so \(N=n_1+\sum_j I_j n_{2j}\). Since \(E[I_j]=P(I_j=1)=\pi_j\), linearity of expectation gives

$$ E[N]=n_1+\sum_j E[I_j]n_{2j} =n_1+\sum_j\pi_j n_{2j}. $$

The same indicator argument gives \(T=\sum_j I_jT_j\), and therefore \(E[T]=\sum_j\pi_jT_j\). No independence between pathway indicators is required; indeed, because exactly one pathway occurs, they are mutually exclusive. These expectations do require a complete list of pathways and a duration or enrollment value specified for each one.

Worked Example: Adding Follow-Up and a Transition Delay

Consider a hypothetical trial with \(n_1=240\) participants and stage 1 accrual at 40 participants per month. The trial stops with probability \(0.40\), continues with A with probability \(0.35\), or continues with B with probability \(0.25\). Stage 2 requires 300 unique participants on A and 480 on B. Assume stage 2 recruitment takes 8 months on A and 12 months on B, based on the relevant arm capacities. Stage 1 outcomes require 2 months of follow-up after stage 1 accrual; a continuation requires a 1-month transition delay; and stage 2 outcomes require 3 months of follow-up after stage 2 accrual. No transition delay or stage 2 follow-up is needed if the trial stops. Calculate expected enrollment and time from first enrollment to the required outcomes.

Expected enrollment is

$$ E[N]=240+0.40(0)+0.35(300)+0.25(480) =240+105+120=465. $$

Stage 1 accrual takes \(240/40=6\) months. The stop pathway lasts \(6+2=8\) months. Pathway A lasts \(6+2+1+8+3=20\) months, and pathway B lasts \(6+2+1+12+3=24\) months. Therefore,

$$ E[T]=0.40(8)+0.35(20)+0.25(24) =3.2+7+6=16.2\text{ months}. $$

The 2-month stage 1 follow-up is included on every pathway because the boundary decision depends on stage 1 outcomes. The transition and stage 2 terms appear only on continuation pathways. Keeping these components explicit prevents a follow-up period from being omitted or counted twice.

Assumptions, Diagnostics, and Alternatives

A planning calculation is only as sound as its inputs. Transition probabilities may come from the assumed operating characteristics of the design under a specified scenario. They should represent the probabilities of the design taking each action, not probabilities that a treatment is truly effective. If probabilities are estimated from a set of scenarios, identify the scenarios and weighting used; different assumptions can yield different expected resource needs.

Before relying on the result, check that:

  • Every possible action at the boundary is represented, including stopping, and the probabilities sum to 1.
  • Stage 1 enrollment is counted once, and each \(n_{2j}\) counts unique randomized participants on that pathway.
  • Sample sizes reflect the planned randomization ratio and any shared control group.
  • Accrual rates, arm bottlenecks, transition delays, follow-up, and the duration endpoint match the operational plan.
  • The probabilities and sample sizes refer to the same design and decision rule.

For detailed budgeting, a single expectation may not be enough. Report the pathway-specific totals alongside the expected value, especially when a low-probability pathway has substantially higher enrollment or duration. A maximum or other upper planning value answers a different question from the expectation: it describes a more demanding scenario rather than the probability-weighted average. A fixed-sample-size plan is another useful comparator, but it does not represent the variable enrollment produced by transition rules.

Common pitfall: “The expected enrollment is 465, so divide 465 by the average monthly accrual rate to get expected duration.” This can be wrong when stop and continuation paths have different recruitment constraints, transition delays, or follow-up requirements. Calculate the calendar duration \(T_j\) for each pathway under its own operational assumptions, then use \(\sum_j\pi_jT_j\). In general, expected duration is not obtained by applying a single rate to expected enrollment.

Another diagnostic is to compare the calculated expectation with the possible pathway values. The expected enrollment must lie between the smallest and largest pathway enrollment, and expected duration must lie between the smallest and largest pathway duration when the pathway probabilities sum to one. A result outside those bounds usually signals a probability, arithmetic, or pathway-definition error. Large differences between pathways are not themselves errors; they are a reason to show the pathway breakdown rather than report only the average.

Check Your Understanding

Use the stated pathways and assumptions to answer each question. Solutions follow each problem.

  1. A trial enrolls 100 participants in stage 1. It stops with probability 0.30 or continues with probability 0.70, enrolling 200 additional participants if it continues. What is expected total enrollment?
    Solution: \(100+0.30(0)+0.70(200)=240\) participants.
  2. There are three exhaustive pathways with probabilities 0.2, 0.5, and 0.3. What check should be made before calculating an expectation?
    Solution: Confirm they are mutually exclusive and exhaustive and sum to 1: \(0.2+0.5+0.3=1.0\).
  3. On a continuation pathway, 120 participants are recruited to an active treatment at 15 per month and 60 to control at 10 per month. If the arms recruit concurrently, how long does stage 2 recruitment take?
    Solution: The arm durations are \(120/15=8\) months and \(60/10=6\) months. The stage 2 period is their maximum, 8 months.
  4. A stop pathway lasts 7 months with probability 0.4, and a continuation pathway lasts 15 months with probability 0.6. What is expected duration?
    Solution: \(E[T]=0.4(7)+0.6(15)=2.8+9=11.8\) months.
  5. Why should a shared control group not be counted once for every treatment comparison when calculating pathway enrollment?
    Solution: Enrollment counts unique randomized participants. A participant assigned to the shared control is one person, even if that control contributes to more than one comparison; counting the same participants repeatedly overstates the pathway total.
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