The Effect of Early Outcomes on Allocation

Why Early Responses Can Matter

In a doubly adaptive biased coin design (DBCD), the probability of assigning the next participant to an arm depends on two evolving quantities: an estimated target allocation and the allocation already achieved. Early responses can affect both the estimate used to form the target and, as participants are assigned, the current arm counts. Consequently, two trials with the same number of participants on each arm can have different next-assignment probabilities if their early outcomes differ.

This is a feature of response-adaptive randomization, not evidence that a few observations have established which treatment is better. The calculations below use hypothetical data to isolate the arithmetic of the feedback. They do not establish a treatment effect or quantify uncertainty about one.

Model, Target, and DBCD Rule

For a simple two-arm illustration, suppose each response is binary, with value 1 for success and 0 for failure. Conditional on assignment to arm \(k\), responses are modeled as independent Bernoulli observations with success probability \(\theta_k\), for \(k=1,2\). Assume responses are observed before the next assignment is made, there are no missing outcomes, and the response probabilities are treated as stable over the sequence. These assumptions make the sample success proportions straightforward response estimates; a different response model or target map would produce different calculations.

Let \(N_{k,n}\) be the number assigned to arm \(k\) among the first \(n\) participants, and let \(\hat{\theta}_{k,n}\) be the observed success proportion on that arm. The current allocation share is \(\pi_{k,n}=N_{k,n}/n\). For this tutorial, use the prespecified target map that makes the target share for arm 1 proportional to its estimated success probability:

\hat{\rho}_{1,n}= \frac{\hat{\theta}_{1,n}}{\hat{\theta}_{1,n}+\hat{\theta}_{2,n}}, \qquad \hat{\rho}_{2,n}=1-\hat{\rho}_{1,n}.

This particular map is only an illustrative choice. It is defined when the two estimated success proportions have a positive sum. It is not a universal optimal-allocation rule: a different scientific objective can require a different prespecified target map.

Definition: At time \(n\), the plug-in target \(\hat{\rho}_{k,n}\) is the target share obtained by substituting current response estimates into a prespecified target map. The current share \(\pi_{k,n}=N_{k,n}/n\) records the allocation so far; it is not the same quantity as the target.

For a positive tuning parameter \(\gamma\), the two-arm DBCD assigns participant \(n+1\) to arm 1 with conditional probability

P(A_{n+1}=1\mid \mathcal{F}_n) = \frac{\left(\hat{\rho}_{1,n}/\pi_{1,n}\right)^\gamma} {\left(\hat{\rho}_{1,n}/\pi_{1,n}\right)^\gamma+ \left(\hat{\rho}_{2,n}/\pi_{2,n}\right)^\gamma},

where \(\mathcal{F}_n\) denotes the assignments and observed responses available before the next assignment. The probability for arm 2 is one minus this probability. The rule compares each target share with its current share: an arm below its target is pushed upward, while an arm above its target is pushed downward. The random assignment is made using these probabilities; it is not a deterministic instruction to send the next participant to whichever arm has the larger target.

How the Response Pattern Moves the Target

With binary responses, \(\hat{\theta}_{k,n}\) is the number of successes divided by the number assigned to arm \(k\). The target map then turns a difference in estimated response rates into a target-share difference. Because the estimate is updated as outcomes accrue, even one early success or failure can change the target, particularly when the arm has few observations.

Worked Example: Same Allocation Counts, Different Early Responses

Suppose six hypothetical participants have been assigned, three to each arm, so both current shares are \(3/6=0.5\). Compare two possible response histories. In history A, arm 1 has two successes and arm 2 has one. In history B, arm 1 has one success and arm 2 has two. Use the target map above and \(\gamma=2\).

History A: estimate the target. The observed response estimates are \(\hat{\theta}_{1,6}=2/3\) and \(\hat{\theta}_{2,6}=1/3\). Therefore,

\hat{\rho}_{1,6} =\frac{2/3}{2/3+1/3} =\frac{2}{3}, \qquad \hat{\rho}_{2,6}=\frac{1}{3}.

The unnormalized DBCD weights are \(((2/3)/(1/2))^2=16/9\) for arm 1 and \(((1/3)/(1/2))^2=4/9\) for arm 2. Thus,

P(A_7=1\mid\mathcal{F}_6) =\frac{16/9}{16/9+4/9} =\frac{16}{20} =0.8000.

History B: the estimates and target shares are reversed, so the arm 1 weight is \(4/9\) and the arm 2 weight is \(16/9\). Hence \(P(A_7=1\mid\mathcal{F}_6)=4/20=0.2000\). The current allocation counts are identical in the two histories; the difference in the next-assignment probability comes from the response-dependent target.

These are conditional probabilities for the next randomization, not guarantees about the next assignment. In history A, arm 1 is more likely to receive the next participant, but assignment to arm 2 still has probability \(0.2000\).

The Current Counts Also Matter

A moving target is only part of the calculation. The DBCD rule also accounts for whether current allocation shares already exceed or fall short of that target. The odds form makes the two influences especially clear. Dividing the arm 1 weight by the arm 2 weight gives

\frac{P(A_{n+1}=1\mid\mathcal{F}_n)} {P(A_{n+1}=2\mid\mathcal{F}_n)} = \left( \frac{\hat{\rho}_{1,n}}{\hat{\rho}_{2,n}} \frac{\pi_{2,n}}{\pi_{1,n}} \right)^\gamma = \left( \frac{\hat{\rho}_{1,n}N_{2,n}} {\hat{\rho}_{2,n}N_{1,n}} \right)^\gamma.

The first equality follows by dividing the two normalized weights; their common normalizing denominator cancels. The second follows from \(\pi_{k,n}=N_{k,n}/n\), so the common factor \(1/n\) cancels in the share ratio. Thus, an arm can have the larger target and still have a next-assignment probability below one half if it is sufficiently overrepresented relative to that target.

Worked Example: A Higher Target but a Lower Assignment Probability

Suppose four hypothetical participants have been assigned to arm 1 and two to arm 2. There are three successes among the four arm 1 responses and one success among the two arm 2 responses. Use \(\gamma=1\).

The estimated success proportions are \(\hat{\theta}_{1,6}=3/4=0.75\) and \(\hat{\theta}_{2,6}=1/2=0.50\). The target is

\hat{\rho}_{1,6} =\frac{0.75}{0.75+0.50} =0.6000, \qquad \hat{\rho}_{2,6}=0.4000.

Arm 1 has the higher target share, \(0.6000\), but its current share is \(4/6=0.6667\), already above that target. The weights are \((0.6000/0.6667)^1=0.9000\) for arm 1 and \((0.4000/0.3333)^1=1.200\) for arm 2. Therefore,

P(A_7=1\mid\mathcal{F}_6) =\frac{0.9000}{0.9000+1.200} =0.4286.

Despite the higher estimated success rate on arm 1, the next participant is less likely to be assigned there under this rule. The probability reflects the target-to-current-share comparison, not a direct ranking of the two response estimates.

One Outcome Can Alter the Next Probability

The following example holds the allocation path fixed through the next assignment and compares the two possible outcomes after that assignment. It shows the immediate effect of an early response, without suggesting that either outcome is a reliable estimate of the long-run response probability.

Worked Example: Updating After One Early Response

At \(n=4\), suppose two hypothetical participants are on each arm and each arm has one success. Both estimated response proportions are \(1/2\), giving target shares \((0.5,0.5)\) under the stated map. With \(\gamma=2\), both current shares are also \(0.5\), so the probability of assigning participant 5 to arm 1 is \(0.5000\). Suppose participant 5 is assigned to arm 1.

If the response is a success: after this response, arm 1 has two successes among three participants and arm 2 has one among two. The target for arm 1 is \(\hat{\rho}_{1,5}=(2/3)/(2/3+1/2)=4/7\); the arm 2 target is \(3/7\). The current shares are \(3/5\) and \(2/5\). The DBCD weights are \[ \left(\frac{4/7}{3/5}\right)^2=\frac{400}{441}, \qquad \left(\frac{3/7}{2/5}\right)^2=\frac{225}{196}. \] Expressing both weights with denominator \(1764\) gives \(1600/1764\) and \(2025/1764\), respectively. Thus,

P(A_6=1\mid\mathcal{F}_5,\text{success}) =\frac{1600}{1600+2025} =0.4414.

If the response is a failure: the arm 1 estimate becomes \(1/3\), while the arm 2 estimate remains \(1/2\). The target shares are \(2/5\) and \(3/5\). With current shares still \(3/5\) and \(2/5\), the weights are \((2/3)^2=4/9\) and \((3/2)^2=9/4\). Therefore,

P(A_6=1\mid\mathcal{F}_5,\text{failure}) =\frac{4/9}{4/9+9/4} =\frac{16}{97} =0.1649.

The success history produces a next-assignment probability of \(0.4414\), while the failure history produces \(0.1649\). In both cases the probability is below one half because arm 1 currently has three of the five assignments. The changed response estimate moves the target, but the current imbalance continues to influence the result.

Interpreting the Feedback and Its Limits

The examples separate three components that should be checked whenever an early outcome appears to produce a large allocation change:

The response estimate: how many observations support the estimated response rate on each arm?
The target map: how does that estimate translate into a target share? The same response data can imply different targets under different prespecified maps.
The current allocation and \(\gamma\): how far are current shares from target shares, and how strongly does the chosen tuning parameter weight that discrepancy?

For a fixed state with target shares strictly between zero and one and positive current counts on both arms, increasing \(\gamma\) magnifies the ratio of the two DBCD weights. It does not change the target estimate itself. Thus, a probability change can arise because outcomes moved the estimated target, because allocation counts changed, or because the design uses a different \(\gamma\); these causes should not be conflated.

A useful diagnostic is to record the response estimates, target shares, current shares, and next-assignment probabilities together at each update. If a reported probability cannot be reconstructed from those quantities, check that the response was incorporated at the intended time and that the target shares and current shares have not been interchanged. Also inspect whether an estimate is based on very few responses or lies at a boundary. In such cases, the plug-in target can be highly sensitive or may fail to be defined; the calculation should not be treated as stable merely because it is arithmetically valid.

Compared with fixed allocation, DBCD lets observed outcomes influence subsequent randomization, while a fixed rule does not update assignment probabilities in response to those outcomes. Compared with assigning everyone directly according to the current target, DBCD uses the ratio of target shares to current shares, retaining a correction for allocation already achieved and assigning probabilistically. These approaches answer different design needs; a moving estimated target alone does not determine the next probability.

The principal limitation of this small-sample illustration is also its teaching point: early response estimates are based on little information. A few responses can cause a conspicuous shift in a plug-in target even when the underlying response probabilities have not changed. DBCD translates the state supplied to it into a randomization probability; it does not make that state estimate precise or guarantee that the estimated target is the scientifically appropriate one.

Key takeaway: An early response changes a DBCD assignment probability only through the design state it helps create: the response estimate and resulting target, together with current allocation shares and \(\gamma\). Always calculate and interpret all of these components in context.

Check Your Understanding

Answer each question, then compare with the solutions below.

With equal current shares, target shares \(0.6\) and \(0.4\), and \(\gamma=1\), what is the next-assignment probability for arm 1?
Why can an arm with the larger estimated success rate have a next-assignment probability below one half?
If both target shares and current shares are equal, what is the next-assignment probability for arm 1 for any positive \(\gamma\)?
For a fixed state with positive target and current shares, what does increasing \(\gamma\) change, and what does it leave unchanged?

Solutions. (1) The weights are \(0.6/0.5=1.2\) and \(0.4/0.5=0.8\), so the probability is \(1.2/(1.2+0.8)=0.6000\). (2) The arm may already have a current share larger than its target; the DBCD rule compares target with current share, rather than simply ranking estimated response rates. (3) The two weights are identical, so the probability is \(0.5\). (4) Increasing \(\gamma\) strengthens the weight contrast between the target-to-current-share ratios; it does not change the response estimates, target shares, or current shares for that fixed state.