Sequential Assignment Probabilities From Trial Histories

Why the History at Assignment Time Matters

In a response-adaptive trial, the probability assigned to a participant can depend on information collected from earlier participants. To calculate that probability correctly, use the allocation rule and the information the rule could use immediately before the participant is randomized. A final trial database may contain more information than was available at that moment, particularly if some responses were delayed.

This calculation is a design record: it states the chance the participant would have been assigned to each treatment, given the specified rule and the history available at that point. It is not the probability that a treatment is effective, nor is it necessarily the probability of that treatment for a participant averaged over all possible histories.

Definition: Let \(A_i\) be the treatment assigned to participant \(i\), and let \(\mathcal H_{i-1}\) denote the recorded trial history available immediately before that assignment. The participant-level conditional assignment probability for treatment \(a\) is \(P(A_i=a\mid\mathcal H_{i-1})\), calculated by applying the specified randomization rule to that history. If the rule also uses participant \(i\)'s baseline information, include that known information in the conditioning set.

The history can include previous assignments, responses already observed, and any other recorded variables that the specified rule uses. It should not include a future response or a response that had not yet been reported when the assignment was made. The history is time-ordered: two participants with the same eventual database entries may have had different assignment probabilities if the information available before their assignments differed.

A Response-Adaptive Rule and How to Apply It

For a concrete rule, consider a hypothetical two-arm trial with treatments A and B and a binary response: success or failure. Suppose the specified rule calculates a smoothed success score for each arm, then allocates in proportion to those scores. For arm \(a\), let \(s_a\) be the number of successes observed and \(n_a\) the number of responses observed, both counted only among outcomes available before the next assignment. Define

\tilde q_a=\frac{s_a+1}{n_a+2}, \qquad P(A_i=a\mid\mathcal H_{i-1}) =\frac{\tilde q_a}{\tilde q_A+\tilde q_B}, \quad a\in\{A,B\}.

The added 1 and 2 smooth the observed success fraction, including when an arm has few or no observed responses. Numerically, \(\tilde q_a\) is the posterior mean under a Beta(1,1) prior and a Bernoulli response model. In this example, however, it is the specified allocation rule’s score; calculating it does not establish that the model is correct or that the score is the true success probability.

To apply the rule, first freeze the history at the assignment time, then count the eligible observed responses, calculate the scores, and normalize them. Normalizing means dividing each arm’s score by the sum of the scores. This makes the probabilities sum to one.

Fix the assignment time.
Identify which previous responses had actually been observed before the participant was randomized.

Calculate the rule’s inputs.
Count successes and observed responses for each arm, using the rule’s definitions and time cutoff.

Apply and check the rule.
Calculate each treatment probability and verify that the probabilities are between 0 and 1 and sum to 1.

Worked Example: Calculating a Probability From Observed Outcomes

All data here are hypothetical. Immediately before participant 6 is assigned, the recorded history contains three observed responses on A: two successes and one failure; and two observed responses on B: one success and one failure. Apply the smoothed-score rule above.

For A, \(s_A=2\) and \(n_A=3\), so

\tilde q_A=\frac{2+1}{3+2}=\frac{3}{5}=0.6000.

For B, \(s_B=1\) and \(n_B=2\), so

\tilde q_B=\frac{1+1}{2+2}=\frac{2}{4}=0.5000.

The sum of the scores is \(0.6000+0.5000=1.1000\). Therefore, the assignment probabilities for participant 6 are

P(A_6=A\mid\mathcal H_5)=\frac{0.6000}{1.1000} =\frac{6}{11}=0.5455, \qquad P(A_6=B\mid\mathcal H_5)=\frac{0.5000}{1.1000} =\frac{5}{11}=0.4545.

The values are nonnegative and sum to 1. Given this history and the stated rule, participant 6 had a 54.55% assignment probability for A and a 45.45% probability for B. The calculation does not say that A has a 54.55% chance of being more effective.

Why Normalization Gives the Assignment Probabilities

The probability calculation follows directly from the randomization step. Suppose the rule assigns a weight \(w_a\geq 0\) to each treatment \(a\), and then selects one treatment at random with probability proportional to its weight. The total weight is \(W=\sum_a w_a\). If \(W>0\), the probability of treatment \(a\) is \(w_a/W\). The probabilities are valid because

\sum_a \frac{w_a}{W} =\frac{\sum_a w_a}{W} =\frac{W}{W}=1.

For the two-arm smoothed-score rule, the weights are \(w_A=\tilde q_A\) and \(w_B=\tilde q_B\). If the implementation instead uses another specified mechanism—such as a fixed probability floor, a deterministic threshold, or a rule depending on covariates—the calculation must follow that mechanism. One should not replace the stated rule with a more familiar rule merely because the observed assignments look similar.

Worked Example: Applying a Minimum Allocation Probability

All data here are hypothetical. Suppose the design first calculates the raw probability for A using the smoothed-score rule, then enforces a minimum allocation probability of 0.10 for each arm. Before a new assignment, A has 0 successes among 19 observed responses, while B has 9 successes among 10 observed responses.

The raw scores are

\tilde q_A=\frac{0+1}{19+2}=\frac{1}{21}=0.04762, \qquad \tilde q_B=\frac{9+1}{10+2}=\frac{10}{12}=0.8333.

The raw probability for A is

p_A^{\mathrm{raw}} =\frac{1/21}{1/21+10/12} =\frac{2}{37} =0.05405.

The specified floor replaces any raw probability below 0.10 with 0.10; the other arm receives the remaining probability. Thus \(p_A=0.1000\) and \(p_B=0.9000\). The floor changes the allocation rule’s raw result, so it must be included in the calculation rather than added after reporting the design probability.

Delayed Responses and the Assignment-Time Cutoff

Response-adaptive rules often face delays: an earlier participant may have been assigned, but their outcome may not be available yet. The correct treatment of that participant depends on the specified rule. If the rule updates from outcomes observed by the assignment time, a pending response contributes to neither the success count nor the observed-response count. It is not silently coded as a failure, and it is not included as though its eventual value were already known.

Worked Example: Excluding a Pending Outcome

All data here are hypothetical. Immediately before a new assignment, three outcomes on A are available, with two successes. One outcome on B is available and is a success; a second participant assigned to B has a response that is still pending. The rule uses only observed responses. Calculate the assignment probabilities.

For A, \(\tilde q_A=(2+1)/(3+2)=3/5=0.6000\). For B, only the one reported success counts, so \(\tilde q_B=(1+1)/(1+2)=2/3=0.6667\). The pending outcome is not included in \(n_B\) or \(s_B\). Therefore,

P(A_i=A\mid\mathcal H_{i-1}) =\frac{0.6000}{0.6000+0.6667} =\frac{9}{19}=0.4737, \qquad P(A_i=B\mid\mathcal H_{i-1}) =\frac{10}{19}=0.5263.

These probabilities describe the rule using the information available at that assignment time. Replacing the pending response with its eventual outcome would answer a different, hindsight-based question and could produce a different probability.

Checks, Alternatives, and Common Errors

A useful calculation check is to recompute the scores and normalization from the recorded history, then verify that each final probability is in \([0,1]\) and the probabilities over all possible treatments sum to 1. Also verify that the response coding, assignment order, and outcome-availability cutoff match the rule. If the recorded assignment had probability zero under the stated rule and history, something is inconsistent: the rule may have been transcribed incorrectly, the history may be incomplete, or the recorded assignment may not have followed that rule.

For a rule with a floor or ceiling, check the probability after applying the constraint. A floor on each arm must be feasible: for \(K\) treatments, a common lower bound \(\epsilon\) requires \(K\epsilon\leq 1\). If probabilities are transformed, renormalized, or rounded in the actual design, reproduce that specified operation; seemingly small implementation differences can change the design record.

A fixed-allocation design uses the same stated probabilities at each assignment, regardless of response history. A response-adaptive design instead uses probabilities conditional on the current history. A marginal probability such as \(P(A_i=A)\), without conditioning on a particular history, averages over possible histories and is not generally the participant-level probability calculated here. These quantities answer different questions.

Wrong: “The observed assignment was A, so the probability of A was 1.” Correction: The observed treatment is the result of the random draw; the assignment probability is the chance of that result before the draw, given the rule and history.
Wrong: “A pending response can be entered as a failure to keep the counts balanced.” Correction: Include it only if the specified rule explicitly uses pending outcomes in that way. Otherwise, use only information available under the rule.
Wrong: “The allocation probability for A is the probability that A is the better treatment.” Correction: The allocation probability is a randomization probability. It need not equal a posterior probability of effectiveness or any other probability about treatment effects.

The calculation is only as reliable as the recorded rule and history. A probability computed from an incomplete or misordered record may be arithmetically correct but not the probability that governed the assignment. Nor does calculating probabilities by itself establish that a particular response model is scientifically appropriate or that the adaptive design has desirable operating characteristics.

Key takeaway: For each participant, calculate the treatment probabilities by applying the specified allocation rule to the history available immediately before that assignment. Keep conditional probabilities distinct from observed assignments, probabilities averaged over histories, and probabilities about treatment effectiveness.

Check Your Understanding

Use the smoothed-score rule unless a question states otherwise. All data are hypothetical.

Before an assignment, A has 1 success among 2 observed responses and B has 2 successes among 3 observed responses. Calculate both assignment probabilities.
For the history in question 1, what should be done with an additional B response that is still pending if the rule uses only observed outcomes?
A rule calculates raw probabilities of 0.04 for A and 0.96 for B, then imposes a minimum probability of 0.10 for each arm. What are the final probabilities?
Explain why the fact that a participant was assigned to A does not mean that their probability of assignment to A was 1.

Solutions. (1) The scores are \(\tilde q_A=(1+1)/(2+2)=0.5000\) and \(\tilde q_B=(2+1)/(3+2)=0.6000\). Thus \(P(A)=0.5/1.1=5/11=0.4545\) and \(P(B)=0.6/1.1=6/11=0.5455\). (2) Exclude the pending response from both \(s_B\) and \(n_B\); it was not observed at assignment time. (3) Apply the floor to A and assign the remainder to B: \(P(A)=0.10\), \(P(B)=0.90\). (4) A is the realized outcome of the random assignment draw. The probability is the chance of that draw before it occurred, conditional on the specified rule and the available history; it can be less than 1 even when A is ultimately selected.