Worked Sample-Size Calculation for a Three-Arm Design

From a Stopping Rule to an Enrollment Calculation

A MAMS sample-size plan is not described by its maximum enrollment alone. If an interim rule can stop the trial or drop one active arm, the number of participants actually enrolled depends on the outcomes that trigger those decisions. A useful planning calculation therefore reports both the maximum enrollment and the expected enrollment under clearly specified treatment-effect scenarios.

This tutorial works through that calculation for a two-stage design with two active arms and one shared control. The stage sizes and continuation rule are treated as fixed design inputs; the task is to translate them into enrollment under the global null, a global alternative and a partial-benefit scenario. The calculation is illustrative, not a recommendation for a particular trial.

Definition: Maximum enrollment is the number of participants enrolled if every planned stage and arm is used. Expected enrollment under a scenario is the average enrollment over repeated trials when the treatment effects and probability model in that scenario hold. It is not the enrollment of a typical individual trial, nor is it guaranteed to be the realized enrollment.

Design and Assumptions for the Calculation

Suppose stage 1 enrolls \(n_1\) participants in each of control, active arm A and active arm B. At the interim analysis, each active arm is assessed against the same control group. An arm continues if its stage-1 standardized treatment-control statistic exceeds a prespecified threshold \(c\). If neither active arm continues, the trial stops. If one or both continue, stage 2 enrolls \(n_2\) new participants in each continuing active arm and \(n_2\) new control participants. Thus, stage-2 control enrollment is shared: it is one control group of size \(n_2\), not a separate control group for each continuing arm.

For the numerical example, assume independent normally distributed participant outcomes with common known variance \(\sigma^2\), equal allocation within each stage, no missing outcomes, and independent observations across participants and stages. Let \(\delta_A\) and \(\delta_B\) denote the active-minus-control mean differences. These assumptions let us describe the two interim statistics jointly. In an actual design, the outcome model, variance treatment, allocation, stopping rules and any additional recruitment constraints must match the protocol.

Let \(Z_A\) and \(Z_B\) be the stage-1 treatment-control statistics, each standardized by its standard error. Because both contrasts use the same stage-1 control mean, their errors are correlated. With equal stage-1 group size \(n_1\),

\operatorname{Var}(\bar{X}_A-\bar{X}_C) = \operatorname{Var}(\bar{X}_B-\bar{X}_C) = \frac{2\sigma^2}{n_1}, \qquad \operatorname{Cov}(\bar{X}_A-\bar{X}_C,\bar{X}_B-\bar{X}_C) = \frac{\sigma^2}{n_1}, \qquad \rho(Z_A,Z_B)=\frac{1}{2}.

The covariance is positive because a control mean that happens to be low makes both active-minus-control differences look larger. The standardized means are \(\mu_A=\delta_A\sqrt{n_1/2}/\sigma\) and \(\mu_B=\delta_B\sqrt{n_1/2}/\sigma\), so the joint model is bivariate normal with unit marginal variances and correlation \(1/2\). For this worked calculation, take \(c=0\): an arm continues when its stage-1 statistic is positive. This simple screening boundary makes the enrollment arithmetic transparent; it is not proposed as a generally suitable efficacy boundary.

Enrollment Formulas

Let \(K\) be the number of active arms that continue, so \(K\) can be 0, 1 or 2. Stage 1 always enrolls \(3n_1\). Stage 2 enrolls nothing when \(K=0\), and enrolls \(n_2(1+K)\) when \(K\geq1\): one shared control group plus one group for each continuing active arm. Equivalently, stage-2 enrollment is \(n_2\{I(K\geq1)+K\}\), where \(I(\cdot)\) is 1 if its condition holds and 0 otherwise. Taking expectations gives the key calculation.

N_{\max}=3n_1+3n_2, \qquad E(N)=3n_1+n_2\{P(K\geq1)+E(K)\}.

The maximum uses all three groups at both stages. The expectation needs two probabilities: the chance at least one arm continues, which determines whether stage-2 control is recruited, and the expected number of continuing active arms, which determines stage-2 active-arm enrollment. If \(q_A=P(Z_A\leq c)\), \(q_B=P(Z_B\leq c)\), and \(p_{00}=P(Z_A\leq c,Z_B\leq c)\), then

P(K\geq1)=1-p_{00}, \qquad E(K)=P(Z_A>c)+P(Z_B>c) =(1-q_A)+(1-q_B).

The joint probability \(p_{00}\) is essential: the two decisions are not independent when they share a control group. Under the bivariate normal model above, it is the bivariate normal probability at the two failure thresholds, using the stated means, unit variances and correlation \(1/2\).

Worked Example: Maximum Enrollment and the Enrollment Formula

Suppose \(n_1=80\) participants per group at stage 1 and \(n_2=60\) per group at stage 2. Confirm the maximum, then express expected enrollment in terms of the continuation probabilities.

The maximum occurs when both active arms continue. There are three groups at each stage, so

N_{\max}=3(80)+3(60)=240+180=420\text{ participants}.

If both arms continue, stage 2 adds \(60\) in control and \(60\) in each active arm, or \(180\) participants. If exactly one continues, it adds \(60\) in control and \(60\) in that active arm, or \(120\). If neither continues, it adds zero.

Writing \(p_{00}\) for the probability that both fail the continuation rule, and \(q_A,q_B\) for the marginal failure probabilities, expected enrollment is

E(N)=240+60\{(1-p_{00})+(1-q_A)+(1-q_B)\}.

This expression counts shared-control enrollment once whenever at least one arm continues. Counting a new control group for each continuing arm would instead describe a different allocation rule and would overstate enrollment for this design.

Three Treatment-Effect Scenarios

Use the same design inputs in all three scenarios: \(n_1=80\), \(n_2=60\), \(c=0\), and correlation \(1/2\). To specify the alternatives, take standardized stage-1 means \((\mu_A,\mu_B)\) as shown below. With \(n_1=80\), a mean of 1 corresponds to a treatment-control difference \(\delta/\sigma=1/\sqrt{40}\approx0.1581\); the numerical effects are hypothetical. Denote the standard normal cumulative distribution function by \(\Phi\), and the bivariate standard normal cumulative distribution function with correlation \(1/2\) by \(\Phi_2(\cdot,\cdot;1/2)\). In each case, \(q_j=\Phi(-\mu_j)\) and \(p_{00}=\Phi_2(-\mu_A,-\mu_B;1/2)\).

Worked Example: Global Null

Under the global null, neither active treatment differs from control, so \(\mu_A=\mu_B=0\). Each marginal statistic is standard normal, giving \(q_A=q_B=\Phi(0)=0.5\). For two standard normal variables with correlation \(1/2\), the probability both are below zero is \(p_{00}=1/4+\arcsin(1/2)/(2\pi)=1/3\approx0.3333\).

Therefore, the probability at least one arm continues is \(1-p_{00}=2/3\approx0.6667\), and the expected number of continuing active arms is \(2(1-0.5)=1\). Substitution into the enrollment formula gives

E(N)=240+60\{0.6667+1\} \approx240+100.0 =340.0\text{ participants}.

The design has maximum enrollment 420 and expected enrollment 340 under this global-null model. Notice that the probability both arms continue is

1-2(0.5)+\frac{1}{3}=\frac{1}{3}\approx0.3333.

This probability is not the product of two independent continuation probabilities: shared-control statistics are positively correlated.

Worked Example: Global Alternative

Now assume the same beneficial effect in both active arms, represented by \(\mu_A=\mu_B=1\). Each marginal failure probability is \(q_A=q_B=\Phi(-1)\approx0.1587\). Numerical evaluation of the bivariate normal probability gives \(p_{00}=\Phi_2(-1,-1;1/2)\approx0.0625\).

Thus, \(P(K\geq1)=1-0.0625=0.9375\), and \(E(K)=2(1-0.1587)=1.6827\). Using the more precise probabilities in the arithmetic,

E(N) =240+60\{(1-0.06254)+2(1-0.158655)\} \approx240+157.2 =397.2\text{ participants}.

The probability both arms continue is \(1-2(0.158655)+0.06254\approx0.7452\). The expected enrollment is close to the maximum because, in this scenario, each arm has a relatively high probability of passing the continuation rule. It remains below the maximum because stopping or dropping an arm can still occur.

Worked Example: Partial Benefit

Suppose arm A has \(\mu_A=1\) while arm B has no benefit, \(\mu_B=0\). Then \(q_A=\Phi(-1)\approx0.1587\) and \(q_B=\Phi(0)=0.5\). The joint failure probability is \(p_{00}=\Phi_2(-1,0;1/2)\approx0.1274\), evaluated under the same shared-control correlation.

The chance at least one arm continues is \(1-0.1274=0.8726\). The expected number of continuing active arms is \((1-0.158655)+(1-0.5)=1.341345\). Hence

E(N) =240+60\{(1-0.1274)+(1-0.158655)+(1-0.5)\} \approx240+132.8 =372.8\text{ participants}.

Although the beneficial arm often continues, the null arm continues about half the time under this zero-threshold rule. This explains why the expected enrollment lies between the global-null and global-alternative calculations. It does not mean that the trial will enroll exactly 372.8 participants; expected enrollment is an average over repeated trials under the specified model.

Comparison, Diagnostics and Limitations

The maximum and expected enrollment answer different planning questions. Maximum enrollment is the cap implied by the stage sizes and allocation rule. Expected enrollment describes resource use under an assumed scenario and stopping rule. A scenario grid is useful because expected enrollment changes with the assumed effects; there is no single expected value that applies regardless of the treatment effects.

A direct probability calculation, as above, is efficient when the joint distribution of interim statistics is known and the stopping rule is simple. Simulation is a practical alternative when continuation boundaries, stage sizes, allocation or other design features make analytic probabilities cumbersome. In either approach, the simulated or calculated enrollment must implement the actual shared-control rule and the same stopping paths. Earlier material on expected-enrollment simulation and enumeration provides tools for checking more complex calculations.

Several checks help reveal errors before a result is used for planning:

Expected enrollment must lie between the minimum enrollment \(3n_1\) and maximum enrollment \(3n_1+3n_2\) for this design.
Check the stage-2 count on every path: zero continuing arms means zero extra participants; one means \(2n_2\); two means \(3n_2\).
Check that the probability of at least one arm continuing is \(1-p_{00}\), while the expected number continuing is the sum of the two marginal continuation probabilities.
Do not multiply marginal probabilities as if the arms were independent. Shared control creates correlation, which affects joint stopping probabilities.
Compare results across scenarios: with this positive-direction rule, larger beneficial effects should generally increase continuation and expected enrollment.

These calculations are conditional on the assumed model and design. If the variance is estimated, outcomes are non-normal, allocation is unequal, control recruitment differs by path, or stopping rules use more information, the joint probabilities and possibly the enrollment formula must be revised. Nor does expected enrollment describe statistical power, type I error or treatment-effect estimation: those are separate design properties. The introductory material on reporting MAMS sample-size assumptions and results emphasizes that the assumptions accompanying an enrollment figure are part of its interpretation.

Common pitfall: Treating the two interim decisions as independent because they concern different active arms. That reasoning ignores their common control mean. Under the global null here, independence would give \(P(K=0)=0.5^2=0.25\); the correct correlated value is \(1/3\). Use the joint distribution induced by the shared control, or a simulation that reproduces it, before calculating expected enrollment.

Check Your Understanding

Use the design in this tutorial unless a question says otherwise. Solutions follow each question.

If exactly one active arm continues, how many stage-2 participants are enrolled? Solution: One shared control group and one active group are enrolled, so \(60+60=120\) participants.
Under the global null, what is the probability neither arm continues, and why is it not \(0.25\)? Solution: It is \(p_{00}=1/3\), because the two statistics have correlation \(1/2\) from the shared control; multiplying the marginal failure probabilities would incorrectly assume independence.
Under the global null, verify the expected number of continuing active arms. Solution: Each arm continues with probability \(1-0.5=0.5\), so \(E(K)=0.5+0.5=1\).
Why can expected enrollment be non-integer? Solution: It is an average across repeated hypothetical trials, not a count that a single trial must realize. Each realized enrollment is an integer.
What changes in the calculation if stage-2 control enrollment is separately sized for each continuing active arm? Solution: The control contribution becomes \(n_2K\), rather than \(n_2I(K\geq1)\). The expected stage-2 enrollment would then be \(2n_2E(K)\), so the formula and maximum must reflect that different allocation rule.