The Two-Arm DBCD Allocation Rule

From an Allocation Target to a Randomization Probability

An allocation target and a next-patient assignment probability answer different questions. The target specifies the desired share of assignments to each arm; the assignment probability is the probability used for the next randomization, given the assignments and responses observed so far. A doubly adaptive biased coin design (DBCD) connects these quantities by adjusting the target according to how the current allocation compares with it.

This tutorial defines the standard two-arm rule. The estimated target may itself be updated as information accumulates, while the current allocation proportion changes after every assignment. The rule uses both: it gives weight to an arm in relation to its estimated target share and also accounts for whether that arm is currently under- or over-represented relative to that target.

Definition: At the point when $n$ patients have been assigned, let $N_{n,1}$ and $N_{n,2}$ be the numbers assigned to arms 1 and 2, with $N_{n,1}+N_{n,2}=n$. The current allocation proportions are $\pi_{n,k}=N_{n,k}/n$, for $k=1,2$. Let $\hat{\rho}_{n,1}$ be the estimated target share for arm 1, and set $\hat{\rho}_{n,2}=1-\hat{\rho}_{n,1}$. For tuning parameter $\gamma>0$, the two-arm DBCD assignment probability for arm 1 is $$ p_{n+1,1}= \frac{\hat{\rho}_{n,1}\left(\hat{\rho}_{n,1}/\pi_{n,1}\right)^\gamma} {\hat{\rho}_{n,1}\left(\hat{\rho}_{n,1}/\pi_{n,1}\right)^\gamma+ \hat{\rho}_{n,2}\left(\hat{\rho}_{n,2}/\pi_{n,2}\right)^\gamma}, $$ and $p_{n+1,2}=1-p_{n+1,1}$.

Here $\hat{\rho}_{n,k}$ is the target share currently supplied by the target-building procedure, not the observed allocation proportion. The distinction matters: the target can change as response estimates change, and the allocation proportions can differ from the target even when the target is fixed. The DBCD formula uses the current values of both.

Assumptions and Interpretation

The rule is applied after the first $n$ assignments and before assigning patient $n+1$. Write $\mathcal{F}_n$ for the trial history available at that point, including assignments and observed responses through patient $n$. The target estimate and current allocation proportions must be determined from this history, so the next randomization probability is not chosen using patient $n+1$'s unobserved outcome.

For the displayed formula to be defined in its standard form, assume $n>0$, $0<\hat{\rho}_{n,1}<1$, and $0<\pi_{n,1}<1$. Then both target shares and both current proportions are positive, and the two weights in the denominator are finite and positive. The initial assignments therefore need to provide observations on both arms before this formula is used. How to handle initialization is a design decision; the rule itself does not define a probability when a current arm proportion is zero.

Let $A_{n+1}$ denote the arm assigned to patient $n+1$. Conditional on the history, the rule means $P(A_{n+1}=1\mid\mathcal{F}_n)=p_{n+1,1}$. It does not mean that the next patient is certain to receive an arm with a low current allocation, nor that the eventual trial allocation will equal the target exactly. Randomization remains, and both the target and the allocation proportions may change at subsequent steps.

The exponent $\gamma$ controls how strongly the relative target-to-current-proportion terms enter the weights. In this tutorial it is treated as a prespecified positive design parameter. The formula can be implemented for any such value, but interpreting its full range of behavior is separate from calculating the probability for a specified value.

Why the Rule Has This Form

For each arm, first form an unnormalized weight: its estimated target share multiplied by its target-to-current-proportion ratio raised to $\gamma$. The target share supplies the target-based component; the ratio component modifies that weight according to the difference between the target and the current allocation. Normalizing the two weights converts them into probabilities.

w_{n,k}=\hat{\rho}_{n,k} \left(\frac{\hat{\rho}_{n,k}}{\pi_{n,k}}\right)^\gamma, \qquad p_{n+1,k}=\frac{w_{n,k}}{w_{n,1}+w_{n,2}}, \quad k=1,2.

This normalization is the complete justification that the weights define a two-arm randomization: under the stated positivity assumptions, $w_{n,1}>0$ and $w_{n,2}>0$, so each normalized value is between zero and one, and $p_{n+1,1}+p_{n+1,2}=(w_{n,1}+w_{n,2})/(w_{n,1}+w_{n,2})=1$. Thus the rule specifies a valid conditional probability distribution for the next assignment.

In practice, calculate both weights from unrounded inputs and normalize only after they have been computed. Rounding intermediate weights can alter the final probability, especially when the weights differ substantially. If an implementation reports probabilities to a fixed number of decimal places, it should check that the reported pair still sums to one to the stated precision.

Worked Calculations

Worked Example: The Current Allocation Exceeds the Estimated Target

Suppose, hypothetically, $n=20$ patients have been assigned, with $N_{n,1}=14$ and $N_{n,2}=6$. The target procedure supplies $\hat{\rho}_{n,1}=0.45$, so $\hat{\rho}_{n,2}=0.55$, and the design uses $\gamma=1$. Calculate the probability of assigning the next patient to arm 1.

The current proportions are $\pi_{n,1}=14/20=0.70$ and $\pi_{n,2}=6/20=0.30$. The two weights are $$ w_{n,1}=0.45\left(\frac{0.45}{0.70}\right) =0.289285714\ldots, \qquad w_{n,2}=0.55\left(\frac{0.55}{0.30}\right) =1.008333333\ldots. $$ Their unrounded sum is $1.297619048\ldots$, giving $$ p_{n+1,1} =\frac{0.289285714\ldots}{1.297619048\ldots} =0.22293578\ldots\approx 0.2229. $$ Thus the next patient has approximately a $22.29\%$ probability of assignment to arm 1, conditional on the stated history and target. The complementary probability for arm 2 is approximately $0.7771$, or $77.71\%$. The calculation uses unrounded weights before normalization.

Worked Example: A Target Above the Current Proportion

In a hypothetical trial, $n=40$ patients have been assigned evenly: $N_{n,1}=N_{n,2}=20$. Suppose the estimated target shares are $0.60$ for arm 1 and $0.40$ for arm 2, and $\gamma=2$. Find both assignment probabilities.

The current proportions are $\pi_{n,1}=\pi_{n,2}=0.50$. Substitution gives $$ w_{n,1}=0.60\left(\frac{0.60}{0.50}\right)^2 =0.60(1.20)^2=0.864, \qquad w_{n,2}=0.40\left(\frac{0.40}{0.50}\right)^2 =0.40(0.80)^2=0.256. $$ The total weight is $0.864+0.256=1.120$. Therefore, $$ p_{n+1,1}=\frac{0.864}{1.120}=0.7714286\ldots\approx0.7714, \qquad p_{n+1,2}=\frac{0.256}{1.120}=0.2285714\ldots\approx0.2286. $$ These probabilities sum to one. Under these inputs, assignment to arm 1 is more likely than assignment to arm 2; this is a probability statement for the next patient, not a guarantee about the final allocation.

Worked Example: Fractional Tuning Parameter

Suppose, hypothetically, that $n=50$, with $N_{n,1}=20$ and $N_{n,2}=30$. The estimated target is $0.70$ for arm 1 and $0.30$ for arm 2, and the prespecified tuning parameter is $\gamma=0.5$. Calculate the next assignment probability.

The current proportions are $0.40$ and $0.60$. The weights are $$ w_{n,1}=0.70\left(\frac{0.70}{0.40}\right)^{0.5} =0.70\sqrt{1.75} =0.926012959\ldots, $$ $$ w_{n,2}=0.30\left(\frac{0.30}{0.60}\right)^{0.5} =0.30\sqrt{0.50} =0.212132034\ldots. $$ The total is $1.138144993\ldots$, so $$ p_{n+1,1}=\frac{0.926012959\ldots}{1.138144993\ldots} =0.813616\ldots\approx0.8136, \qquad p_{n+1,2}\approx0.1864. $$ The next patient is assigned to arm 1 with probability about $81.36\%$, conditional on this history and these design inputs. The fractional power is applied to each positive ratio before multiplying by its target share.

Implementation Checks, Alternatives, and Limitations

A useful implementation check is to compute the current proportions directly from the arm counts, confirm that they sum to one, and confirm the same for the target shares. Then calculate both positive weights, normalize them, and verify that the resulting probabilities lie in $[0,1]$ and sum to one. Independent calculation or a small test script can catch index reversals, use of $N_{n,k}$ in place of $N_{n,k}/n$, or accidental use of a target from after the next assignment.

The most direct alternative is target-only randomization, which assigns to arm 1 with probability $\hat{\rho}_{n,1}$, without using the observed allocation proportions. That probability follows the current estimated target but does not correct for a current allocation that has drifted away from it. The DBCD instead uses both the target and the current allocation. Neither procedure makes the target itself correct: if the target estimate is unsuitable for the scientific or ethical objective, accurately implementing the DBCD formula does not repair that problem.

The standard rule also has practical limits. It requires positive current proportions and interior target shares, so a zero allocation or boundary target makes its ratios undefined or degenerate. Further, without additional restrictions, the formula does not guarantee a minimum assignment probability for either arm. Very small probabilities can arise from the inputs, and the method does not by itself establish that a particular statistical analysis is valid under adaptive assignment. The design should retain the assignment probabilities and the information used to calculate them, and the analysis plan should reflect the actual randomization procedure.

A common mistake is to say, “The target for arm 1 is $0.45$, so the next patient must have a $45\%$ chance of receiving arm 1.” That is target-only randomization, not the DBCD rule. In the first worked example, the target was $0.45$, but the current proportion was $0.70$, and the DBCD probability was approximately $0.2229$. The correction is to distinguish the desired target share from the probability generated for the next assignment, and to calculate the latter using both arm weights and their normalization.

Key takeaway: At each eligible assignment step, compute each arm’s weight as its estimated target share times its target-to-current-proportion ratio raised to $\gamma$, then divide each weight by the sum of both weights. The result is a conditional probability for the next assignment, not the target share or a promise about the final allocation.

Check Your Understanding

For each question, identify the relevant quantities and calculate or explain the result. Solutions follow the questions.

At $n=30$, suppose $N_{n,1}=18$, the target for arm 1 is $0.50$, and $\gamma=1$. What is $p_{n+1,1}$?
Why must the target shares and current allocation proportions be distinguished?
In the standard formula, what happens to its definition if $N_{n,1}=0$?
Suppose the two computed weights are $0.30$ and $0.70$. What are the normalized assignment probabilities?

Solutions. (1) The current proportions are $0.60$ and $0.40$, and both target shares are $0.50$. The weights are $0.50(0.50/0.60)=0.4166667$ and $0.50(0.50/0.40)=0.625$. Thus $p_{n+1,1}=0.4166667/(0.4166667+0.625)=0.4000$. (2) The target describes a desired allocation share, while the current proportion describes assignments already made; the DBCD uses both to determine a next-patient probability. (3) Then $\pi_{n,1}=0$, so the target-to-current-proportion ratio is undefined. The standard formula cannot be applied until both current proportions are positive. (4) The weights sum to $1.00$, so the normalized probabilities are $0.30$ for arm 1 and $0.70$ for arm 2.