Calculating Stagewise Continuation Probabilities

From Interim Decisions to Stage-Reach Probabilities

In a multi-arm multi-stage (MAMS) design, later-stage enrollment occurs only if interim decisions allow the trial to continue. The probability of reaching a stage therefore depends on the decisions made at all preceding interims. These probabilities are useful for calculating expected enrollment under a specified treatment-effect scenario: scenarios that lead to different continuation patterns can have different expected enrollment even when the maximum planned sample size is unchanged.

The key input here is the joint probability of the relevant interim decisions under a specified scenario. Those probabilities might be supplied as design operating characteristics. This tutorial uses them to find stage-reach probabilities and expected enrollment; it does not derive them from an interim data model.

Definition: Let $C_j$ denote the event that the design continues from stage $j$ to stage $j+1$. Let $R_j$ denote the event that stage $j$ is reached. Stage 1 is reached with probability 1, so $P(R_1)=1$. For $j\geq 2$, reaching stage $j$ requires continuing at every preceding interim: $$ R_j=\bigcap_{k=1}^{j-1} C_k, \qquad P(R_j)=P\left(\bigcap_{k=1}^{j-1} C_k\right). $$

For three stages, this gives $P(R_2)=P(C_1)$ and $P(R_3)=P(C_1\cap C_2)$. The second-stage continuation decision matters for stage 3 only on paths that first reach stage 2. Consequently, $P(C_2)$ on its own does not generally give the probability of reaching stage 3. If the supplied quantity is a conditional probability, the chain rule gives $P(R_3)=P(C_1)P(C_2\mid C_1)$. Replacing $P(C_2\mid C_1)$ with $P(C_2)$ requires independence, which should not be assumed without justification.

Expected Enrollment When Stage Sizes Are Fixed

Suppose stage $j$ enrolls $n_j$ participants if that stage is reached, and $n_j$ is a fixed incremental enrollment—not cumulative enrollment through stage $j$. Let $I(R_j)$ equal 1 if stage $j$ is reached and 0 otherwise. Total enrollment is then the sum of the enrollments contributed by the stages that occur:

N=\sum_{j=1}^{J} n_j I(R_j).

Taking expectations and using $E[I(R_j)]=P(R_j)$ gives the expected-enrollment formula:

E[N]=\sum_{j=1}^{J} n_jP(R_j).

This derivation uses linearity of expectation; it does not require the stage-reach events to be independent. It does require that the stated $n_j$ is the enrollment added at stage $j$ whenever that stage is reached. The result is an average over repeated hypothetical trials under the specified scenario, not a prediction that any one trial will enroll a fractional number of participants.

Worked Example: Three Stages With Joint Continuation Probabilities

Consider a hypothetical three-stage trial with incremental planned enrollments of 120, 90, and 60 participants. Under one specified treatment-effect scenario, suppose the supplied joint decision probabilities are $P(C_1)=0.70$ and $P(C_1\cap C_2)=0.45$. Find the stage-reach probabilities and expected enrollment.

Stage 1 is always reached, so $P(R_1)=1$. Stage 2 is reached when the first interim permits continuation, giving $P(R_2)=P(C_1)=0.70$. Stage 3 requires continuation at both interims, so the supplied joint probability gives $P(R_3)=P(C_1\cap C_2)=0.45$.

The expected enrollment is

E[N]=120(1)+90(0.70)+60(0.45) =120+63+27 =210\text{ participants}.

The maximum planned enrollment is $120+90+60=270$ participants. The expected enrollment is lower because later stages are not reached on every trial. The calculation uses the joint probability for reaching stage 3; it does not need separate marginal probabilities for both interims.

When Continuation Is Arm-Specific

In a MAMS design, an interim can stop some treatment arms while allowing others to continue. In that setting, “the trial continues” can mean either that at least one arm continues or that a particular arm continues. These are different events. The probability of reaching a later stage at all is based on the union of arm-specific continuation events. Expected enrollment may instead require the probability that each individual arm remains active, because each active arm contributes its own planned enrollment.

For two arms, if $C_{1A}$ and $C_{1B}$ mean that arms A and B, respectively, continue after the first interim, then the probability that stage 2 is reached is

P(R_2)=P(C_{1A}\cup C_{1B}) =P(C_{1A})+P(C_{1B})-P(C_{1A}\cap C_{1B}).

The subtraction prevents trials in which both arms continue from being counted twice. If stage enrollment is charged arm by arm, its expected contribution can instead be calculated by summing each arm’s enrollment multiplied by that arm’s probability of reaching the stage. This also works when arms’ continuation events are dependent, by linearity of expectation.

Worked Example: Two Arms With Different Continuation Probabilities

In a hypothetical three-stage trial, the planned enrollment for each active treatment arm is 50 participants in stage 2 and 60 in stage 3. Suppose the supplied probabilities under a specified scenario are $P(C_{1A})=0.80$, $P(C_{1B})=0.65$, $P(C_{1A}\cap C_{1B})=0.55$, and

P(C_{1A}\cap C_{2A})=0.32,\quad P(C_{1B}\cap C_{2B})=0.27,\quad P(C_{1A}\cap C_{2A}\cap C_{1B}\cap C_{2B})=0.18.

Assume stage 1 enrolls 100 participants total, and that each arm’s stated stage enrollment is additional to the other arm’s enrollment. Find the probability of reaching each later stage and expected total enrollment.

Stage 2 is reached if at least one arm continues after the first interim. By inclusion-exclusion,

$$ P(R_2)=0.80+0.65-0.55=0.90. $$

For stage 3, the probability that at least one arm reaches it is the probability of the union of the two events that each arm survives both interims. Thus,

$$ P(R_3)=0.32+0.27-0.18=0.41. $$

To find expected enrollment, count the stage 2 enrollment separately for each arm. Arm A contributes $50(0.80)=40$ expected participants and arm B contributes $50(0.65)=32.5$. For stage 3, the corresponding probabilities are 0.32 and 0.27, so the expected contributions are $60(0.32)=19.2$ and $60(0.27)=16.2$. Therefore,

E[N]=100+40+32.5+19.2+16.2 =207.9\text{ participants}.

The value $0.41$ is appropriate for answering whether stage 3 is reached at all. It is not, by itself, sufficient to calculate stage 3 enrollment when the stage cost depends on how many arms remain active. In that calculation, the arm-specific reach probabilities are needed.

Comparing Specified Treatment-Effect Scenarios

Continuation probabilities are scenario-specific: changing the assumed treatment effects can change the probability of passing each interim decision. To compare expected enrollment, use the probabilities belonging to each scenario with the same planned stage sizes. This comparison describes expected enrollment conditional on each specified scenario; it does not average across scenarios or assign probabilities to which scenario is true.

Worked Example: Expected Enrollment Under Two Scenarios

A hypothetical three-stage design plans incremental enrollments of 100, 80, and 60 participants. Suppose joint continuation probabilities have been provided for two treatment-effect scenarios. Under scenario A, $P(C_1)=0.80$ and $P(C_1\cap C_2)=0.60$. Under scenario B, $P(C_1)=0.50$ and $P(C_1\cap C_2)=0.20$. Calculate and compare expected enrollment.

Under scenario A, the stage-reach probabilities are $1$, $0.80$, and $0.60$. Thus,

E[N\mid\text{scenario A}] =100+80(0.80)+60(0.60) =100+64+36 =200\text{ participants}.

Under scenario B, the stage-reach probabilities are $1$, $0.50$, and $0.20$. Thus,

E[N\mid\text{scenario B}] =100+80(0.50)+60(0.20) =100+40+12 =152\text{ participants}.

For these supplied probabilities and stage sizes, expected enrollment is 48 participants lower under scenario B. This is a comparison of two conditional expectations under specified scenarios. It does not establish which scenario is more likely, nor does it say what the enrollment will be in any particular trial.

Checks, Alternatives, and Limitations

Before using the calculation, verify that the probability inputs refer to the same treatment-effect scenario and the same decision rules, that joint probabilities describe the required path of decisions, and that stage sizes are incremental enrollments. For arm-specific decisions, check whether the question asks for reaching a stage at all or for expected enrollment within each surviving arm. A probability table should also be coherent: event probabilities must lie between 0 and 1, and a joint event cannot have greater probability than any event it intersects.

A common mistake is to multiply marginal continuation probabilities as if interim decisions were independent. For example, if $P(C_1)=0.70$ and $P(C_2)=0.60$, then $0.42$ is $P(C_1)P(C_2)$, but it equals $P(C_1\cap C_2)$ only if $C_1$ and $C_2$ are independent. When the joint probability is supplied, use it directly. When a conditional probability is supplied, use $P(C_1\cap C_2)=P(C_1)P(C_2\mid C_1)$.

Counting cumulative enrollment as a stage increment. The formula weights the number newly enrolled at a stage, not the cumulative total through that stage.
Using an “any arm” probability for an arm-specific cost. If stage enrollment varies with the number of arms retained, calculate expected contributions for the possible active arms or arm sets.
Treating an expected value as a guaranteed enrollment. Expected enrollment is an average across repetitions under the scenario. The realized enrollment in one trial is an integer and may differ substantially.
Mixing scenarios. Do not combine reach probabilities from different treatment-effect scenarios in one calculation unless a separate, explicitly specified scenario-averaging step is intended.

With fixed stage increments, the stagewise probability-weighted sum is direct and does not require independence. With arm-dependent costs, arm-specific probabilities—or probabilities for each active-arm set—are more informative than a single probability of reaching each stage. Either way, the calculation is limited by its inputs: incorrect or mismatched decision probabilities produce an incorrect expected enrollment, even when the arithmetic is correct.

Check Your Understanding

Try each question, then compare your result with the solutions below.

A three-stage design has incremental stage enrollments of 90, 70, and 50. Under a specified scenario, $P(C_1)=0.60$ and $P(C_1\cap C_2)=0.30$. Find expected enrollment.
If $P(C_1)=0.75$ and $P(C_2\mid C_1)=0.40$, what is the probability of reaching stage 3?
Two arms have first-interim continuation probabilities 0.70 and 0.60, with joint continuation probability 0.45. What is the probability that at least one arm continues?
Explain why multiplying two marginal continuation probabilities may give the wrong probability of reaching stage 3.

Solutions. (1) Stage reach probabilities are $1$, $0.60$, and $0.30$. Expected enrollment is $90+70(0.60)+50(0.30)=90+42+15=147$ participants. (2) By the conditional probability rule, $P(R_3)=P(C_1)P(C_2\mid C_1)=0.75(0.40)=0.30$. (3) Inclusion-exclusion gives $0.70+0.60-0.45=0.85$. (4) The multiplication of marginal probabilities equals the joint probability only under independence. If interim decisions are dependent, use their joint probability or multiply the first-interim probability by the appropriate conditional probability.