Patient Benefit When One Treatment Is Better

Why Allocation Matters When Response Rates Differ

If two treatments have the same response probability, changing how patients are divided between them does not change the expected total number of responses, provided the same number of patients is enrolled. When one treatment has a higher response probability, however, the allocation rule affects how many patients are expected to receive that treatment—and therefore the expected number of favorable outcomes. This is a patient-benefit comparison, not a claim that the realized outcomes in any one trial are certain.

The calculation below treats the response probabilities as known and stable. That makes it possible to isolate the effect of allocation itself. In practice, response probabilities are generally not known with certainty; the results here describe a benchmark in which the advantage is correctly known.

Definition: Let \(A\) and \(B\) denote two treatments, and let \(p_A\) and \(p_B\) be their probabilities of a favorable response for an enrolled patient. If \(N\) patients are enrolled, the expected number of favorable outcomes is the expected count of patients who respond favorably under a specified allocation rule. When \(p_B>p_A\), treatment \(B\) is the treatment with the known response advantage. The expected count is an average over the random outcomes and, for a random allocation rule, the random assignments.

The term favorable outcome refers here to a prespecified binary response, such as improvement by a stated assessment time. It does not automatically include other dimensions of benefit or harm. The calculation also treats patients as having the same response probability on a given treatment. These conditions let us compare allocation rules using a common set of response probabilities.

Expected Responses Under Fixed Allocation

Suppose \(n_A\) patients receive \(A\) and \(n_B\) receive \(B\), with \(n_A+n_B=N\). The expected number of favorable responses among patients assigned to \(A\) is \(n_Ap_A\); among patients assigned to \(B\), it is \(n_Bp_B\). Adding the two expectations gives the total:

$$ E[S]=n_Ap_A+n_Bp_B, $$

where \(S\) is the total number of favorable responses. This calculation uses linearity of expectation: the expected sum of the individual response indicators is the sum of their expectations. Independence among all responses is not needed for that identity, although the stable response-probability assumptions still matter for assigning each indicator its stated expectation.

For a fixed 1:1 allocation, each treatment receives \(N/2\) patients when \(N\) is even. Under equal-probability random assignment with probability one-half to each treatment, the expected allocation is also one-half to each treatment, even if the realized counts are not exactly equal. In either case, the expected total is:

E[S_{\text{1:1}}]=\frac{N}{2}p_A+\frac{N}{2}p_B =\frac{N(p_A+p_B)}{2}.

Worked Example: Equal Allocation as a Benchmark

Suppose a hypothetical trial enrolls 120 patients. The known response probability is \(p_A=0.35\) for treatment \(A\) and \(p_B=0.55\) for treatment \(B\). Under fixed 1:1 allocation, 60 patients receive each treatment. The assumptions for this calculation are that these response probabilities apply to all patients on their respective treatments and remain stable throughout enrollment.

The expected favorable responses are \(60(0.35)=21\) on \(A\) and \(60(0.55)=33\) on \(B\). Thus:

E[S_{\text{1:1}}] =60(0.35)+60(0.55) =21+33 =54.

The result is 54 expected favorable responses among 120 patients, or an expected response proportion of \(54/120=0.45\). It does not say that exactly 54 patients will respond in a particular trial.

Shifting Allocation Toward the Better Treatment

Let \(r\) be the proportion of patients assigned to \(B\), with the remaining proportion \(1-r\) assigned to \(A\). For a fixed allocation with those proportions, or for a random design whose expected allocation proportion to \(B\) is \(r\), the expected total is:

$$ E[S_r]=N(1-r)p_A+Nrp_B. $$

To compare this with equal allocation, subtract the 1:1 expectation and collect terms. The difference depends on both how far the allocation moves from one-half and the size of the response advantage.

\begin{aligned} E[S_r]-E[S_{\text{1:1}}] &=N\{(1-r)p_A+rp_B\} -N\left\{\frac{p_A+p_B}{2}\right\}\\ &=N\left(r-\frac12\right)(p_B-p_A). \end{aligned}

This identity gives a direct justification for the patient-benefit comparison. If \(B\) is better, so \(p_B-p_A>0\), and the allocation shifts toward \(B\), so \(r>1/2\), the expected difference is positive. If allocation shifts toward \(A\) instead, the difference is negative. If the response probabilities are equal, the difference is zero for every \(r\). These statements concern expected counts, not the direction of the realized difference in a particular trial.

Key result: When \(p_B>p_A\), allocating an expected proportion \(r>1/2\) of the \(N\) patients to \(B\) changes expected favorable outcomes relative to 1:1 allocation by \(N(r-1/2)(p_B-p_A)\). The gain grows with the number of patients, the increase in allocation to \(B\), and the size of its response advantage.

Worked Example: Quantifying the Expected Gain

Use the hypothetical response probabilities from the previous calculation, \(p_A=0.35\) and \(p_B=0.55\), for \(N=120\) patients. Suppose a design assigns an expected 65% of patients to \(B\), so \(r=0.65\). This might be a planned unequal allocation or the average allocation proportion produced by a random rule. It is not necessary for every realized trial to assign exactly 65% to \(B\).

The expected counts assigned to \(A\) and \(B\) are \(120(0.35)=42\) and \(120(0.65)=78\). Therefore:

E[S_{0.65}] =42(0.35)+78(0.55) =14.7+42.9 =57.6.

Relative to the 1:1 benchmark of 54 expected responses, the difference is \(57.6-54=3.6\). The identity gives the same result:

$$ 120(0.65-0.50)(0.55-0.35) =120(0.15)(0.20) =3.6. $$

Under the stated assumptions, assigning more patients to \(B\) increases the expected count by 3.6 favorable responses. The decimal is an expectation, not a possible observed count that must occur.

When Allocation Depends on Earlier Responses

A response-adaptive rule may choose each new assignment using information from earlier patients. Its allocation proportion is then random: it can differ across possible trial histories. The same expected-benefit calculation still applies if each new patient’s response probability on a treatment is stable and is not changed by the assignment history, beyond the treatment received.

To see this, let \(T_i\) be the treatment assigned to patient \(i\), and let \(Y_i\) equal 1 for a favorable response and 0 otherwise. Let \(H_{i-1}\) denote the assignment and response history before patient \(i\) is assigned. Assume that, conditional on the history and current treatment, the response probability is \(p_A\) for \(A\) and \(p_B\) for \(B\). Also assume that assignment occurs before that patient’s response is observed. Then the expected response for patient \(i\), conditional on the history, is the response probability averaged over that patient’s assignment probabilities. Taking expectations across histories and summing over patients yields:

\begin{aligned} E[S] &=\sum_{i=1}^{N}E[Y_i]\\ &=\sum_{i=1}^{N}\left\{p_A P(T_i=A)+p_B P(T_i=B)\right\}\\ &=p_A E[N_A]+p_B E[N_B], \end{aligned}

where \(N_A\) and \(N_B\) are the final numbers assigned to \(A\) and \(B\). Since \(E[N_A]+E[N_B]=N\), the expected-benefit comparison depends on the expected number assigned to each treatment, not on the exact path by which the adaptive rule reaches those counts. Adaptation can affect the distribution of allocations across trials; for this expected response count, its expected allocation is what enters the expression.

Worked Example: A Simple Adaptive Allocation Schedule

Consider a hypothetical 10-patient trial with \(p_A=0.40\) and \(p_B=0.60\). For the first four patients, the assignment probability is one-half to each treatment. After that, the rule assigns each of the remaining six patients to \(B\) with probability 0.75. To keep the arithmetic focused on expected benefit, suppose these probabilities are specified in advance for the two stages; this example illustrates the expectation calculation rather than a rule that updates from observed responses.

The expected number assigned to \(B\) is \(4(0.50)+6(0.75)=2+4.5=6.5\), so the expected number assigned to \(A\) is \(10-6.5=3.5\). The expected favorable-response count is:

$$ E[S] =3.5(0.40)+6.5(0.60) =1.4+3.9 =5.3. $$

Under equal randomization, the expected number assigned to each treatment would be five, giving \(5(0.40)+5(0.60)=5.0\) expected favorable responses. The schedule therefore increases the expected count by \(0.3\), also calculated as \(10(0.65-0.50)(0.60-0.40)=0.3\). The expected allocation share to \(B\) is 0.65 because its expected count is 6.5 out of 10.

Assumptions, Diagnostics, and Limits

The arithmetic is simple, but its interpretation depends on the assumptions. The response advantage must be real and correctly ordered: if \(p_B\) is not greater than \(p_A\), a rule that favors \(B\) need not increase expected favorable outcomes. The response probabilities must also be applicable to the patients enrolled under the allocation rule. For example, if the trial enrolls patients with different prognoses over time, a single pair of response probabilities may not describe every assignment.

A useful diagnostic is to check both parts of the gain calculation separately: is the expected allocation truly shifted toward \(B\), and is \(p_B-p_A\) positive under the model being used? Then check that \(N\), \(r\), and the response probabilities refer to the same patient population, outcome definition, and follow-up period. If the rule’s assignments vary with prior responses, use the expected allocation generated by that rule—not an allocation proportion from one unusually favorable or unfavorable trial history.

This comparison measures only the expected number of favorable responses. It does not account for adverse events, response severity, duration of benefit, treatment cost, or patient preferences. It also does not establish that an adaptive design is preferable overall. The earlier material on patient benefit and inferential efficiency distinguishes patient response benefit from the statistical objectives of a trial; the two objectives can point in different directions.

The main alternatives have different allocation goals. Equal fixed allocation provides a stable 1:1 benchmark. Unequal fixed allocation can deliberately place more patients on a treatment believed to be better. A response-adaptive rule changes assignment probabilities using accumulating trial information, so its realized allocation can vary with the observed history. Under the assumptions above, compare these approaches for expected favorable-response count by comparing their expected allocation proportions. This does not compare their inferential precision or guarantee the same realized benefit in each trial.

Common pitfall: “Treatment \(B\) is better, so any trial that allocates more patients to \(B\) will observe more responses than an equal-allocation trial.” This is too strong: it turns a statement about expectations into a guarantee about realized outcomes. The correct conclusion is that, under the stated stable-response assumptions, shifting expected allocation toward \(B\) increases the expected response count by \(N(r-1/2)(p_B-p_A)\). Random outcomes can still make a particular trial’s observed count lower.

Check Your Understanding

Answer each question using expected counts, and keep the distinction between an expectation and a realized trial result clear.

For 80 patients, let \(p_A=0.30\) and \(p_B=0.50\). What is the expected favorable-response count under 1:1 allocation?
With the same values, what is the expected count if 60% of patients are assigned to \(B\), and how much does it differ from 1:1 allocation?
If \(p_A=0.45\) and \(p_B=0.45\), what is the expected-count difference between allocation shares \(r=0.50\) and \(r=0.70\)? Explain why.
A 20-patient adaptive design has expected counts \(E[N_A]=8\) and \(E[N_B]=12\). If \(p_A=0.25\) and \(p_B=0.40\), calculate the expected favorable-response count.
True or false: if an allocation rule favors a treatment with a higher response probability, the observed response count must exceed the count from 1:1 allocation in every trial. Correct the statement if needed.

Solutions. (1) The expected count is \(80(0.30)+80(0.50)\) divided by two for each arm: \(40(0.30)+40(0.50)=12+20=32\). (2) The expected allocation counts are 32 to \(A\) and 48 to \(B\), so the expected count is \(32(0.30)+48(0.50)=9.6+24=33.6\). The gain is \(33.6-32=1.6\), matching \(80(0.60-0.50)(0.50-0.30)=1.6\). (3) The difference is \(N(0.70-0.50)(0.45-0.45)=0\); equal response probabilities make allocation irrelevant to the expected total. (4) The expected count is \(8(0.25)+12(0.40)=2+4.8=6.8\). (5) False. Under the assumptions, favoring the better treatment increases the expected response count, but random outcomes mean the observed count need not be higher in every trial.