Deterministic Rounding as a DBCD Implementation Error

Why Rounding Is More Than a Computational Shortcut

A doubly adaptive biased coin design (DBCD) specifies a probability for assigning each eligible patient, conditional on the information available before that assignment. A computer may represent those probabilities as decimals, but the assignment is still meant to be generated by a random draw. Replacing the probabilities with a rounded number of patients per arm is not just a way to implement the same rule: it changes the randomization mechanism and, in a response-adaptive design, can also change when the allocation rule responds to new data.

The distinction matters even when rounded counts look close to their probability-based expected values. A fixed quota removes variation in the number assigned to each arm over the block. Randomly ordering patients within that quota can make the order uncertain, but it does not restore the original sequential assignment probabilities. The resulting design must be understood as a different assignment procedure.

Sequential DBCD Probabilities and Fixed Counts

Suppose there are \(K\) treatment arms and patient \(i\) is about to be assigned. Write \(N_{i-1,k}\) for the number assigned to arm \(k\) before patient \(i\), and let \(\hat{\rho}_{i,k}\) be the current estimated target share for that arm. When the target shares are positive and the current shares are defined, a common DBCD rule with tuning parameter \(\gamma \geq 0\) sets the assignment probability to

\pi_{i,k} = \frac{\hat{\rho}_{i,k} \left(\dfrac{\hat{\rho}_{i,k}}{N_{i-1,k}/(i-1)}\right)^\gamma} {\displaystyle\sum_{\ell=1}^{K}\hat{\rho}_{i,\ell} \left(\dfrac{\hat{\rho}_{i,\ell}}{N_{i-1,\ell}/(i-1)}\right)^\gamma}.

The probabilities sum to one. For the rule to be usable, its inputs must be defined and its output must give valid probabilities. Here the focus is not on how to handle initial zero counts or undefined estimates, but on what happens after a valid probability vector has been calculated.

Definition: In sequential DBCD, patient \(i\) is assigned to arm \(k\) with conditional probability \(\pi_{i,k}\), given the observed information available immediately before that assignment. A deterministic rounded-quota implementation instead converts a probability vector into integer arm counts for a group of patients and assigns those counts, rather than making the specified patient-by-patient random draws.

One way to express the sequential mechanism is to generate a fresh independent \(U_i\sim\operatorname{Uniform}(0,1)\) for each patient and use cumulative sums of \(\pi_{i,1},\ldots,\pi_{i,K}\) to select the arm. After assignment, the response and allocation information can be updated before the next patient’s probability is calculated. Rounding to counts intervenes between calculating probabilities and making assignments.

Worked Example: Rounding a DBCD Probability Vector

Suppose a hypothetical two-arm design has target shares \((0.60,0.40)\), current allocation shares \((0.50,0.50)\), and \(\gamma=1\). Calculate the next-patient probabilities, then convert them into counts for a fixed block of five using the largest-remainder rule: take the floors of the expected counts, then assign any remaining places to the largest fractional parts.

The unnormalized weights are

w_A=0.60\left(\frac{0.60}{0.50}\right)=0.72, \qquad w_B=0.40\left(\frac{0.40}{0.50}\right)=0.32.

Their sum is \(1.04\), so the DBCD probabilities are

\pi_A=\frac{0.72}{1.04}=0.6923, \qquad \pi_B=\frac{0.32}{1.04}=0.3077.

For five patients, the expected counts under a frozen probability vector are \(5(0.6923)=3.4615\) and \(5(0.3077)=1.5385\). Their floors are 3 and 1. One place remains; arm B has the larger fractional part, \(0.5385\) rather than \(0.4615\). The rounded quota is therefore 3 on A and 2 on B. That quota is an integer implementation choice, not a sequence of five draws with probabilities \(0.6923\) and \(0.3077\).

What the Quota Changes

For sequential DBCD, the arm assigned to one patient is random conditional on the pre-assignment information. The corresponding arm indicator has a nonzero conditional variance when its probability is strictly between zero and one. A fixed quota, by contrast, fixes the total number assigned to each arm within the block. Conditional on the quota, the block’s arm counts have no random variation at all.

This difference can be seen even in a simplified setting where the probability vector is held constant for the entire block. If each of \(m\) patients is independently assigned to arm A with probability \(q\), the number assigned to A has a binomial distribution with mean \(mq\) and variance \(mq(1-q)\). Rounding \(mq\) to one fixed integer preserves neither that distribution nor its variance. In actual DBCD, probabilities may also change as allocation counts and response estimates are updated, so a quota based on an earlier probability vector can diverge further from the intended path.

Worked Example: Variability Removed by a Fixed Quota

In a hypothetical four-patient block, suppose the probability of assignment to A is \(0.70\) for each patient and is deliberately held constant for this comparison. Under four independent draws, the number \(X\) assigned to A satisfies \(X\sim\operatorname{Binomial}(4,0.70)\). A largest-remainder quota rounds the expected counts \(2.8\) on A and \(1.2\) on B to 3 on A and 1 on B.

Under the random-draw rule, the mean and variance are

E(X)=4(0.70)=2.8, \qquad \operatorname{Var}(X)=4(0.70)(0.30)=0.84.

For example, the probability of exactly three assignments to A is

P(X=3)=\binom{4}{3}(0.70)^3(0.30) =4(0.343)(0.30)=0.4116.

The random rule also gives \(P(X=4)=(0.70)^4=0.2401\), so counts other than three occur with positive probability. The fixed quota makes \(X=3\) with probability 1 and has variance 0. Thus, the rounded quota matches a nearby expected count but removes the count variation generated by the stated randomization.

A randomized order within fixed quotas does not generally repair the discrepancy. If five slots contain three A labels and two B labels and the order is uniformly shuffled, the first patient has probability \(3/5=0.60\) of receiving A. If that patient receives A, the next patient’s probability of A is \(2/4=0.50\); if the first patient receives B, it is \(3/4=0.75\). Those conditional probabilities arise from sampling without replacement from a fixed set of labels. They are not repeated draws from a fixed probability of \(0.60\), and they do not necessarily equal the probabilities that a response-updated DBCD would calculate.

Worked Example: A Random Order Does Not Restore the DBCD Rule

Consider a hypothetical five-patient block with a fixed quota of three A assignments and two B assignments, randomly ordered. Calculate the probability of A for the first two patients under two possible histories.

Before the first assignment, three of the five labels are A, so \(P(A\text{ first})=3/5=0.60\). If the first patient is assigned A, two A labels remain among four slots, giving \(P(A\text{ second}\mid A\text{ first})=2/4=0.50\). If the first patient is assigned B, all three A labels remain among four slots, giving \(P(A\text{ second}\mid B\text{ first})=3/4=0.75\).

A sequential rule that recalculates a DBCD probability after the first patient uses the updated information specified by that rule. The quota procedure instead uses the remaining fixed labels. The two mechanisms can have the same first-assignment probability and still differ on the second assignment.

Justification: Why Matching Expected Counts Is Not Enough

The key distinction is between matching an expectation and matching a probability law. For a fixed probability \(q\) across a block of size \(m\), let \(X\) be the number assigned to A by independent random draws. Then \(E(X)=mq\). A deterministic rounding function \(r(mq)\) may be chosen to be close to that expectation, but its count is constant conditional on the inputs. Therefore,

\operatorname{Var}\{r(mq)\}=0, \qquad \operatorname{Var}(X)=mq(1-q)>0 \quad\text{when }0<q<1.

Thus, even if \(r(mq)=mq\) when the expected count happens to be an integer, the random and deterministic procedures have different distributions unless the random count is itself degenerate. In DBCD, this fixed-\(q\) comparison is a simplified justification: the actual probabilities can depend on the assignments and responses observed along the way. A fixed quota also constrains the count path, while sequential draws allow the realized counts to affect subsequent probabilities.

Implementation Checks, Alternatives, and Limits

An implementation audit should check what the software does after it calculates the probability vector. If it turns probabilities into integer counts, inspect the rounding rule, tie-breaking rule, block size, and whether counts are fixed before outcomes are observed. Also check whether the probability vector is recomputed after each patient or only between blocks. These details determine the assignment mechanism; they are not cosmetic coding choices.

Look for deterministic assignment. Rounding each single-patient probability to 0 or 1, or choosing the arm with the largest probability, can remove randomization altogether.
Check whether quotas sum to the block size. Rounding each expected count separately may produce a total that is too small or too large. A rule such as largest remainder resolves the total, but it still defines a quota design.
Check for delayed adaptation. If probabilities are calculated once for a block, responses and assignments within that block cannot influence later assignments in the same way they would under sequential updating.
Check conditional probabilities, not just final totals. Similar end-of-study arm counts do not establish that two procedures used the same randomization rule.

The direct alternative is to draw an arm independently from the current DBCD probability vector at each assignment, then update the available information and recalculate for the next patient. A fixed-quota or permuted-block design may instead be appropriate when the design explicitly calls for fixed counts or constrained balance. It should then be specified and assessed as that design, not described as an equivalent implementation of sequential DBCD. Randomized rounding can sometimes be constructed to preserve expected integer counts, but it still does not, by itself, reproduce the patient-by-patient conditional probabilities and response updates of DBCD.

A practical limitation is that fixed quotas can be attractive operationally: they give predictable counts over a block and may simplify logistics. Those advantages do not make the procedure invalid. They do mean that conclusions about the intended DBCD cannot be assumed to apply unchanged. The allocation method actually used determines the randomization distribution and must be represented accurately in analysis and reporting.

Common Mistake

Common mistake: “The rounded quota is the nearest integer to the expected number of assignments, so it implements the same DBCD probability.” Correction: An expected count is not a realized count distribution. A fixed quota removes count variation and, if probabilities are not updated patient by patient, changes the adaptive assignment path. Similar expected or final counts do not imply the same conditional randomization.

Check Your Understanding

Answer each question, then compare with the solutions below.

A two-arm probability vector is \((0.62,0.38)\) for a block of ten. What expected counts does it imply, and what largest-remainder quota results?
For four independent assignments with a constant probability \(q=0.25\) of A, calculate the mean and variance of the number assigned to A. How do these compare with a deterministic quota of one A?
A fixed quota contains two A labels and two B labels, randomly ordered. What is the probability of A for the second patient given that the first received B?
Why is comparing only final arm totals inadequate for checking whether an implementation follows sequential DBCD?
Name one circumstance in which fixed quotas might be used appropriately, and state the necessary interpretation.

Solutions. (1) The expected counts are \(10(0.62)=6.2\) and \(10(0.38)=3.8\). Taking floors gives 6 and 3, with one place remaining; arm B has the larger fractional part, so the quota is 6 on A and 4 on B. (2) The mean is \(4(0.25)=1\), and the variance is \(4(0.25)(0.75)=0.75\). A deterministic quota of one has variance 0, despite matching the mean. (3) After B is first, two A labels remain among three places, so the probability is \(2/3=0.6667\). (4) Final totals do not reveal the probabilities conditional on each observed assignment history; different procedures can end with similar counts but use different randomization and updating. (5) A protocol may explicitly require fixed block counts for operational balance. In that case, describe and analyze the quota procedure as the specified design, not as an equivalent implementation of sequential DBCD.)