Why Count Expected Successes?
A response-adaptive design changes future treatment assignments in response to outcomes already observed. To compare it with a fixed allocation plan on patient benefit, one simple quantity is the expected total number of favorable outcomes during the trial. This is an expectation over both the random assignments and the patient outcomes; it is not a guarantee about the result of a particular trial.
The calculation is especially direct when the response probabilities are known and stable. It separates two ingredients: how likely each treatment is to produce a success, and how many patients the design is expected to assign to each treatment. The allocation rule matters through the second ingredient.
Assume that each patient’s response probability depends on the assigned treatment, but not on the patient’s position in the trial or the outcomes of other patients. Assignment occurs before that patient’s outcome is observed, and the outcome probabilities are \(p_A\) and \(p_B\) regardless of the assignment history. For the calculations below, outcomes are fully observed and there is no interference between patients. The allocation rule may use earlier observed outcomes; it must not use a patient’s unobserved outcome to determine that patient’s assignment.
These are assumptions for evaluating a design in a hypothetical setting where the response probabilities are known. In an actual trial, those probabilities are usually unknown. A computed expected count therefore describes performance under the stated probabilities and assumptions; it does not establish that either treatment will have those response probabilities in practice.
The Allocation Formula
For a fixed plan assigning \(n_A\) patients to A and \(n_B=n-n_A\) to B, linearity of expectation gives the expected number of successes directly. For a response-adaptive rule, the numbers assigned can vary across possible trial paths, so their expected values enter the same calculation.
For fixed allocation, \(N_A=n_A\), so \(E(N_A)=n_A\). In particular, a fixed plan with exactly \(n/2\) patients per treatment, when \(n\) is even, has expected successes \(n(p_A+p_B)/2\). If the fixed plan randomizes each patient independently with probability \(1/2\) to A, then \(E(N_A)=n/2\) as well, and its expected success count is the same. The realized allocation counts may differ, but the expectations agree.
Why the Formula Holds
Let \(Z_i\) denote the treatment assigned to patient \(i\), and let \(I_i=1\) if \(Z_i=A\) and \(I_i=0\) otherwise. Under the response assumptions, conditional on the assignment and the information available before that patient’s outcome, the expected outcome is \(p_A\) when \(I_i=1\) and \(p_B\) when \(I_i=0\). Thus \(E(Y_i)=p_A P(Z_i=A)+p_B P(Z_i=B)\). Summing over patients yields the result.
The last equality follows because \(N_A=\sum_i I_i\), so \(E(N_A)=\sum_iP(Z_i=A)\), and \(N_B=n-N_A\). Substituting \(E(N_B)=n-E(N_A)\) gives the second form of the formula. This argument does not require independent outcomes: linearity of expectation is enough. It does require the stated treatment-specific response probabilities to hold at each assignment history.
Worked Comparisons
Worked Example: Two Fixed Allocation Plans
Suppose a hypothetical 12-patient trial has known response probabilities \(p_A=0.65\) and \(p_B=0.45\). Compare a fixed 6:6 allocation with a fixed 8:4 allocation, where the first number is the count assigned to A.
For the 6:6 plan, \(E(N_A)=6\) and \(E(N_B)=6\). Therefore,
The expected total is 6.60 successes out of 12 patients. For the 8:4 plan,
Under these specified probabilities, the 8:4 plan has 0.40 more expected successes. This calculation compares expected counts; it does not say that every trial using the 8:4 plan will produce more successes.
Worked Example: A Two-Patient Response-Adaptive Rule
In a hypothetical two-patient trial, let \(p_A=0.70\) and \(p_B=0.30\). Assign the first patient to A or B with probability \(1/2\) each. For the second patient, assign A with probability 0.90 if the first patient had a success and with probability 0.30 if the first patient had a failure. Otherwise assign B. The second assignment probability is chosen using only the first observed outcome.
The probability that the first patient succeeds is
Therefore the probability of assigning the second patient to A is
For patient 1, the probability of assignment to A is 0.50. Consequently, \(E(N_A)=0.50+0.60=1.10\), and \(E(N_B)=2-1.10=0.90\). The expected number of successes is
A fixed plan assigning one patient to each treatment has expected successes \(0.70(1)+0.30(1)=1.00\). The adaptive rule’s expected advantage is 0.04 successes. This also follows from the comparison formula: \((0.70-0.30)(1.10-1)=0.04\). The benefit here is specific to the stated response probabilities and rule.
Worked Example: Adaptation Can Favor the Lower-Response Treatment
Now suppose a hypothetical five-patient trial has \(p_A=0.40\) and \(p_B=0.60\). The first patient is assigned to either treatment with probability \(1/2\). For each of the remaining four patients, the rule assigns A with probability 0.80 if the first patient succeeded and 0.40 if the first patient failed. The first outcome is used to set each later assignment probability.
The marginal probability of a first-patient success is \(0.5(0.40)+0.5(0.60)=0.50\). Thus, for each of patients 2 through 5,
It follows that \(E(N_A)=0.50+4(0.60)=2.90\) and \(E(N_B)=5-2.90=2.10\). Hence,
A fixed 2:3 allocation, favoring B, has expected successes \(0.40(2)+0.60(3)=2.60\). The response-adaptive rule has 0.18 fewer expected successes than that fixed plan. It allocates more patients to A in expectation, even though A has the lower response probability in this example. Being response-adaptive does not by itself mean that a rule directs more patients to the treatment with the higher response probability.
Interpreting the Comparison
The comparison formula provides a compact way to check a claimed benefit. If \(p_A>p_B\), then allocating more patients to A in expectation than a fixed comparator does raises the expected total success count under the assumptions. If \(p_A<p_B\), allocating more patients to A lowers it. If \(p_A=p_B\), the expected total is the same for every allocation rule with the same number of patients: \(E(S)=np_A\).
This is a comparison of the total number of favorable outcomes, not a measure of inferential performance. It does not describe how precisely the treatment response probabilities can be estimated, how often a statistical test will reject a null hypothesis, or how a particular patient fares. Those are different questions. Nor does the formula show that a treatment causes a favorable outcome; it evaluates a specified probability model and assignment rule.
Diagnostics, Alternatives, and Limitations
Before trusting a calculation, check that the expected allocation matches the actual rule. For a simple rule, calculate the marginal probability of each assignment by averaging over the relevant histories, as in the examples. For a more complicated rule, enumerate possible paths or use a simulation that records assignments and outcomes; verify that estimated allocation frequencies and success counts are consistent with the specified probabilities. A discrepancy can indicate a coding error, an incorrect history probability, or an outcome model that was applied inconsistently.
Fixed allocation is the direct benchmark when its treatment counts are specified. It makes \(N_A\) known, so only the treatment-specific response probabilities are needed for the expected count. Response-adaptive allocation instead requires averaging over possible allocation histories. If response probabilities are known, the identity above is often simpler and more transparent than enumerating every path. Enumeration or simulation can still be useful for checking the expected allocation generated by a complicated rule.
The calculation has important limits. If response probabilities vary by patient characteristics or change over time, a single \(p_A\) and \(p_B\) may not describe the trial; expected successes then require probabilities appropriate to each patient and assignment history. Missing outcomes, nonadherence, or interference also require a model beyond the one used here. Finally, when the response probabilities are not known, substituting estimates gives a plug-in calculation, not an exact known-probability expectation.
Common mistake: “The design is response-adaptive, so it must increase expected successes.” That conclusion does not follow. The correction is to calculate \(E(N_A)\) under the rule and compare it with the fixed plan. The direction depends on both \(p_A-p_B\) and the difference in expected allocation; adaptation can favor the lower-response treatment, as the third example shows.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- A fixed 4:6 allocation has \(p_A=0.55\) and \(p_B=0.35\). What is its expected number of successes?
- A response-adaptive rule has \(n=8\), \(p_A=0.60\), \(p_B=0.50\), and \(E(N_A)=5\). Find its expected success count and compare it with fixed 4:4 allocation.
- If \(p_A=p_B=0.25\), can changing the allocation rule change the expected total successes for a fixed trial size? Explain.
- A rule is response-adaptive. Is that alone enough to conclude that it raises expected successes relative to fixed 1:1 allocation? State what additional information is needed.
Solutions. (1) \(E(S)=0.55(4)+0.35(6)=2.20+2.10=4.30\) successes. (2) \(E(N_B)=8-5=3\), so \(E(S)=0.60(5)+0.50(3)=4.50\). Fixed 4:4 gives \(0.60(4)+0.50(4)=4.40\), so the adaptive rule has 0.10 more expected successes. (3) No. Every assignment has success probability 0.25, so \(E(S)=0.25n\) under any allocation rule satisfying the model. (4) No. One needs the treatment response probabilities and the rule’s expected allocation, as well as the fixed comparator’s allocation, to determine the direction and size of the difference.