Why DBCD Does Not Hit Its Target at Every Step

Target Shares and Assignment Probabilities Are Different

A DBCD uses the current allocation and a target allocation to choose the probability of assigning the next participant to each arm. That probability is a randomized decision rule, not a command that makes the realized allocation share equal the target. One assignment changes an arm’s count by exactly one, while the target can be a non-integer fraction of the total sample size. Moreover, the randomization can select either arm, including one that is currently overrepresented.

To isolate this point, begin with two treatment arms, A and B, and a fixed target share $r$ for A, where $0<r<1$. After $n$ assignments, let $N_A(n)$ be the number assigned to A and let $x_n=N_A(n)/n$ be A’s observed allocation share. The target share for B is $1-r$. Suppose both arms have been assigned at least once, so both observed shares are positive.

Definition: Given the allocation history and data available before assignment $n+1$, a DBCD with tuning parameter $\gamma>0$ assigns the next participant to A with conditional probability $$ p_{n+1} = \frac{r\left(r/x_n\right)^\gamma} {r\left(r/x_n\right)^\gamma+(1-r)\left((1-r)/(1-x_n)\right)^\gamma}. $$ The conditional probability of assignment to B is $1-p_{n+1}$. The target $r$ describes the desired share; $p_{n+1}$ describes the chance of a particular next assignment at the current state.

This definition is conditional: the assignment probability is calculated using the current allocation and target before the next assignment is randomized. If the target is estimated from responses, its current value is also based on information available at that time. The fixed-target calculations below hold $r$ constant to make the effect of randomization clear. They assume the stated DBCD rule is used and that the current share is strictly between zero and one.

When $x_n<r$, A is underrepresented. For $\gamma>0$, the rule gives A a probability above $r$; when $x_n>r$, it gives A a probability below $r$. Thus the rule pushes its assignment probability in the direction of the imbalance. But a probability above $r$ is not certainty, and it need not be greater than one half. The realized assignment still has to be drawn.

What One Randomized Assignment Does to the Share

Let $I_{n+1}=1$ if participant $n+1$ is assigned to A, and $I_{n+1}=0$ if assigned to B. Given the history before that assignment, $I_{n+1}$ is a Bernoulli random variable with success probability $p_{n+1}$. The next observed share is therefore

x_{n+1} = \frac{N_A(n)+I_{n+1}}{n+1} = \frac{n x_n+I_{n+1}}{n+1}.

This expression makes the two possible outcomes explicit. If the next participant goes to A, the share becomes $(n x_n+1)/(n+1)$; if the participant goes to B, it becomes $n x_n/(n+1)$. Neither outcome is generally equal to $r$. In particular, an assignment to the arm that is already underrepresented can still leave its share below the target, while an assignment to the other arm can move it farther below.

Worked Example: An Assignment That Moves Away from the Target

Suppose $n=10$, $N_A(10)=5$, the fixed target for A is $r=0.60$, and $\gamma=1$. The current share is $x_{10}=5/10=0.50$, so A is below its target by $0.10$. Calculate the DBCD probability of assigning the next participant to A.

With $\gamma=1$, the unnormalized weights for A and B are $$ 0.60\left(\frac{0.60}{0.50}\right)=0.72, \qquad 0.40\left(\frac{0.40}{0.50}\right)=0.32. $$ Their sum is $0.72+0.32=1.04$, so $$ p_{11}=\frac{0.72}{1.04}=0.6923, \qquad P(\text{assignment to B}\mid\text{current history})=1-0.6923=0.3077. $$

If participant 11 is assigned to A, A’s share becomes $6/11=0.5455$, which is closer to $0.60$ than $0.50$ was. If participant 11 is assigned to B, A’s share becomes $5/11=0.4545$. Its distance from the target then increases from $|0.50-0.60|=0.10$ to $|0.4545-0.60|=0.1455$. This outcome has conditional probability $0.3077$. The assignment is consistent with the DBCD rule: A was favored, but B still had a positive chance of being selected.

Conditional Mean: A Directional Correction, Not a Guarantee

The mean of the next share averages over the two possible assignments, conditional on the history before assignment $n+1$. Since $E(I_{n+1}\mid\text{current history})=p_{n+1}$, the update equation gives

E(x_{n+1}\mid\text{current history}) = \frac{n x_n+p_{n+1}}{n+1}.

The conditional mean is a weighted average of the current share and the next-assignment probability. It is not, in general, the target $r$. Even if $p_{n+1}>r$ because A is underrepresented, the observed share can fall below $r$ on the next step. The conditional mean can also remain below $r$, depending on the current state and the probability.

Worked Example: Calculating the Conditional Mean and Variation

Continue with the state $n=10$, $x_{10}=0.50$, $r=0.60$, and $p_{11}=0.6923$ from the previous calculation. The next share is $6/11=0.5455$ with probability $0.6923$, or $5/11=0.4545$ with probability $0.3077$. Its conditional mean is $$ E(x_{11}\mid\text{current history}) = \frac{10(0.50)+0.6923}{11} = \frac{5.6923}{11} = 0.5175. $$

This mean is above the current share $0.50$ and therefore moves in the direction of the target $0.60$, but it does not equal the target. The conditional variance comes only from the randomized assignment: since $x_{11}=(10x_{10}+I_{11})/11$, $$ \operatorname{Var}(x_{11}\mid\text{current history}) = \frac{p_{11}(1-p_{11})}{11^2} = \frac{(0.6923)(0.3077)}{121} \approx 0.001761. $$

The conditional standard deviation is $\sqrt{0.001761}\approx0.04196$, or about 4.20 percentage points in A’s share. This summarizes the spread of the next share across repetitions at this same state; it does not predict the particular assignment in one trial. The actual next share is one of the two values, not their mean.

The calculation also explains why a statement such as “DBCD moves the allocation toward its target at every step” is too strong. The rule can move the assignment probability toward correcting the imbalance, while the observed share moves either way. Even the conditional mean being closer to the target would not imply that every realized outcome is closer.

A Short Derivation of the Conditional Update

The result follows directly from the count update. Let $I_{n+1}$ indicate assignment to A. Then $N_A(n+1)=N_A(n)+I_{n+1}$, and the new total is $n+1$. Dividing the count by that total and taking the conditional expectation yields

\begin{aligned} x_{n+1} &=\frac{N_A(n)+I_{n+1}}{n+1},\\ E(x_{n+1}\mid\text{current history}) &=\frac{N_A(n)+E(I_{n+1}\mid\text{current history})}{n+1}\\ &=\frac{n x_n+p_{n+1}}{n+1}. \end{aligned}

The second equality uses the defining conditional probability of the DBCD: $E(I_{n+1}\mid\text{current history})=p_{n+1}$. The realized share replaces $p_{n+1}$ with the actual indicator $I_{n+1}$, which can be either zero or one. That difference between a conditional probability and a realized indicator is the source of one-step deviations.

Worked Example: Two Consecutive Assignments to the Underrepresented Arm’s Opponent

Use the same fixed target $r=0.60$, initial state $n=10$, $N_A(10)=5$, and $\gamma=1$. The first assignment goes to B, an event with conditional probability $0.3077$. The updated state is $n=11$, $N_A(11)=5$, so $x_{11}=5/11=0.4545$.

At this updated state, the unnormalized weights are $$ 0.60\left(\frac{0.60}{5/11}\right)=0.7920, \qquad 0.40\left(\frac{0.40}{6/11}\right)=0.2933. $$ Thus the next probability of assignment to A is $$ p_{12}=\frac{0.7920}{0.7920+0.2933}=0.7297, $$ and the conditional probability of another assignment to B is $1-0.7297=0.2703$. If that occurs, A’s share becomes $5/12=0.4167$, farther from $0.60$ than $5/11=0.4545$ was. The conditional probability of this two-assignment path, B followed by B, is approximately $$ (0.3077)(0.2703)=0.0832. $$

After the first B assignment, the rule increases A’s chance from $0.6923$ to $0.7297$, strengthening the correction in its next probability. Yet a second B assignment remains possible. A path that temporarily falls farther behind is therefore compatible with a rule that responds to the imbalance at each step.

When the Target Changes

In response-adaptive designs, the target may be recalculated as response information accumulates. In that case, denote the target available before assignment $n+1$ by $r_n$, and use that value to calculate the conditional assignment probability. The next observed share still changes according to the count update above. After the new participant’s response is observed, the target for a later assignment may differ from $r_n$.

Consequently, a comparison such as “the share moved closer to the target” must specify which target is being used. A useful one-step check holds the current target fixed and compares $x_n$ with $r_n$, then $x_{n+1}$ with that same $r_n$. Comparing $x_{n+1}$ with a newly updated target answers a different question: both the allocation share and the reference point may have changed. This distinction does not require the target to be wrong or the assignment to be incorrectly implemented.

Diagnostics, Alternatives, and Limitations

A non-monotone allocation path is not, by itself, evidence of a faulty DBCD. At each step, the assignment should be compared with the probability calculated from the state immediately before that assignment. If an apparent discrepancy needs investigation, check the arm counts, the target used at that time, and the resulting normalized probabilities. In a two-arm design, the probabilities should sum to one, and an assignment to B should be evaluated using $1-p_{n+1}$, not by treating $p_{n+1}$ as a required fraction of assignments.

A different approach is to choose assignments deterministically to keep observed counts as close as possible to a target. That can avoid some one-step random deviations, but it is not the DBCD rule described here: DBCD randomizes according to conditional probabilities. The comparison is useful because it separates two goals. A target describes a desired allocation proportion; a randomization rule describes how to generate assignments while responding to the current allocation.

The examples also have limits. The numerical calculations hold the target and $\gamma$ fixed, and assume both arms already have positive counts. If the target changes, the relevant reference share changes too. If a current allocation share is zero, the displayed ratio formula is not defined without an initialization or boundary convention. Those issues must be handled by the design specification rather than by interpreting a one-step deviation as a target failure.

Key takeaway: A DBCD target is not a per-step quota. The rule assigns probabilities conditional on the current state; the next observed share depends on the randomized assignment actually drawn and may temporarily move farther from the target.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

At $n=10$, A has 5 assignments and target share $0.60$. If the next assignment is to B, what is A’s new share and how does its distance from the target change?
In the same state, suppose the probability of assigning to A is $0.6923$. Write the formula for the conditional mean of the next share and calculate it.
Why does an assignment probability for A that is above the target share not ensure that the next observed share increases?
What should be held fixed when checking whether a one-step allocation change moved closer to a target that may be recalculated over time?

Solutions. (1) A’s new share is $5/11=0.4545$. Its absolute distance from $0.60$ increases from $0.10$ to $0.1455$. (2) The formula is $E(x_{n+1}\mid\text{current history})=(n x_n+p_{n+1})/(n+1)$, giving $(10(0.50)+0.6923)/11=0.5175$. (3) The probability describes a randomized choice, not a guaranteed assignment; the next participant can still be assigned to B. (4) Use the target available before the assignment for both the current and next observed shares, so the comparison does not also change its reference point.