Randomization Versus Deterministic Target Assignment

Two Ways to Correct an Allocation Imbalance

A response-adaptive design can use current allocation and response information to set target shares, then use the current allocation imbalance to determine the next assignment. Even when the target is held fixed, there is a consequential choice: assign the next participant to the arm that appears most underrepresented with certainty, or randomize while making underrepresented arms more likely. Doubly adaptive biased coin design (DBCD) takes the latter approach.

The distinction is not between a rule that responds to imbalance and one that ignores it. Both approaches can respond to the same state. The distinction is whether that state determines the next arm uniquely or determines a probability distribution over arms. Randomization makes the sequence of assignments variable across trials, even when trials begin in the same state and use the same target.

Definition: At step $n$, let $N_{n,k}$ be the number assigned to arm $k$, and let $p_{n,k}=N_{n,k}/n$ be its current allocation proportion. Let $\rho_{n,k}$ be the target share for that step, with $\rho_{n,k}>0$ and $\sum_{k=1}^K \rho_{n,k}=1$. For a chosen $\gamma>0$, DBCD assigns the next participant to arm $k$ with probability $$ q_{n+1,k} = \frac{\left(\rho_{n,k}/p_{n,k}\right)^\gamma} {\sum_{j=1}^K\left(\rho_{n,j}/p_{n,j}\right)^\gamma}. $$ The probabilities are conditional on the information available before the next assignment. The rule requires positive current counts so that $p_{n,k}>0$; examples below assume such an initial allocation has already occurred.

The ratio $\rho_{n,k}/p_{n,k}$ compares an arm’s target share with its current share. A ratio above one indicates that the arm is below target; a ratio below one indicates that it is above target. Raising ratios to $\gamma$ controls how strongly the imbalance affects assignment probabilities. For finite $\gamma$, positive targets and positive current shares give positive assignment probabilities to every arm.

For a precise deterministic comparison, define a greedy target-relative rule to assign the next participant to an arm maximizing $\rho_{n,k}/p_{n,k}$. A tie requires a prespecified tie-breaking rule. This is a useful deterministic counterpart to DBCD because it selects the arm with the largest weight before normalization. It is not the only possible deterministic definition of “most underrepresented”: a rule based on the largest absolute difference $\rho_{n,k}-p_{n,k}$ can select a different arm when target shares are unequal.

What Randomization Changes

Write $I_{n+1,k}=1$ if the next participant is assigned to arm $k$, and $I_{n+1,k}=0$ otherwise. Conditional on the information available at step $n$, DBCD gives $I_{n+1,k}$ a Bernoulli distribution with success probability $q_{n+1,k}$. Thus $N_{n+1,k}=N_{n,k}+I_{n+1,k}$. The deterministic rule replaces $I_{n+1,k}$ with a fixed zero or one chosen from the current state.

This difference gives a simple conditional expectation result. The target and probabilities may be computed from the observed history, but once that history is fixed, $N_{n,k}$, $p_{n,k}$, and $q_{n+1,k}$ are fixed for the calculation.

Result: Under DBCD, the conditional expected next allocation proportion to arm $k$ is $$ E(p_{n+1,k}\mid\mathcal{F}_n) = \frac{n p_{n,k}+q_{n+1,k}}{n+1} = p_{n,k}+\frac{q_{n+1,k}-p_{n,k}}{n+1}, $$ where $\mathcal{F}_n$ denotes the information available before assignment $n+1$. The realized next proportion is random; the expression is its conditional mean, not a guarantee about the next participant.

The equation follows by substituting $N_{n,k}=np_{n,k}$ into $p_{n+1,k}=(N_{n,k}+I_{n+1,k})/(n+1)$, then taking the conditional expectation and using $E(I_{n+1,k}\mid\mathcal{F}_n)=q_{n+1,k}$. The factor $1/(n+1)$ also shows that one assignment changes the proportion by less when the trial is larger.

A correction in expectation is not the same as guaranteed step-by-step correction. DBCD can assign an arm already above target, because its probability is positive, and it can fail to assign an underrepresented arm on a particular step. A deterministic greedy rule avoids that uncertainty at the next assignment, but can also produce a rigid sequence and may alternate abruptly when small changes switch which arm ranks first.

Worked Example: Equal Targets With One Arm Underrepresented

Suppose two arms have fixed targets $(0.5,0.5)$, and after $n=10$ assignments their counts are $N_{10,1}=7$ and $N_{10,2}=3$. Use $\gamma=1$. The current proportions are $p_{10,1}=0.7$ and $p_{10,2}=0.3$, so the unnormalized DBCD weights are $$ \frac{0.5}{0.7}=0.7143,\qquad \frac{0.5}{0.3}=1.6667. $$

After normalization, the probability of assigning to arm 2 is $$ q_{11,2} = \frac{1.6667}{0.7143+1.6667} = 0.7000, $$ and $q_{11,1}=0.3000$. The deterministic target-relative rule assigns to arm 2 with certainty, since its target-to-current ratio is larger. DBCD makes the same arm more likely, but does not make the assignment certain.

The conditional expected next proportion for arm 2 is $$ E(p_{11,2}\mid\mathcal{F}_{10}) = \frac{3+0.7000}{11} = 0.3364. $$ If arm 2 is assigned, its realized proportion is $4/11=0.3636$; if arm 1 is assigned, it is $3/11=0.2727$. The expected proportion moves upward from $0.3000$, but the realized proportion can move downward on this step.

Target-Relative and Absolute Deficits

The phrase “most underrepresented” needs care with unequal target shares. DBCD uses a target-to-current ratio. A different deterministic rule could maximize the absolute deficit $\rho_{n,k}-p_{n,k}$. The ratio scales an imbalance relative to the target share, whereas the absolute deficit measures the difference in percentage points. Neither phrase should be used as if the two criteria always select the same arm.

Worked Example: Unequal Targets Can Change the Greedy Choice

Consider three arms with target shares $(0.70,0.20,0.10)$. Suppose $n=50$ participants have been allocated with counts $(30,6,14)$, giving current shares $(0.60,0.12,0.28)$. The target-relative ratios are $$ \frac{0.70}{0.60}=1.1667,\qquad \frac{0.20}{0.12}=1.6667,\qquad \frac{0.10}{0.28}=0.3571. $$

The target-relative greedy rule selects arm 2, whose ratio is largest. In contrast, the absolute deficits are $0.10$, $0.08$, and $-0.18$, respectively, so the largest absolute deficit belongs to arm 1. These are distinct deterministic rules, not interchangeable descriptions.

For DBCD with $\gamma=1$, the sum of the ratios is $3.1905$. Therefore the assignment probabilities are approximately $$ (q_{51,1},q_{51,2},q_{51,3}) = \left(\frac{1.1667}{3.1905},\frac{1.6667}{3.1905},\frac{0.3571}{3.1905}\right) = (0.3657,0.5224,0.1119). $$ Arm 2 is most likely, though arm 1 remains a substantial possibility. The calculation makes explicit which measure of imbalance the allocation rule is using.

Stochastic Paths Versus a Greedy Path

A deterministic rule maps each observed allocation state to one next assignment. Given a fixed starting state and a fixed tie-breaking rule, its allocation path is determined, unless some other source of randomness enters the rule. DBCD instead defines a distribution over possible next assignments at each state. Repeating the design from the same starting state can therefore produce different allocation counts, even if the targets and $\gamma$ are unchanged.

The following example computes two successive DBCD assignments. It shows how the probabilities are recalculated after each possible first assignment; the second-step probabilities are not simply copied from the first step.

Worked Example: Two Randomized Assignments From the Same State

There are two arms with target shares $(0.5,0.5)$, $\gamma=2$, and current counts $(3,1)$ after $n=4$ assignments. The current shares are $(0.75,0.25)$. The DBCD weights are $$ (0.5/0.75)^2=0.4444,\qquad (0.5/0.25)^2=4, $$ so the probability of assigning to arm 2 first is $4/(4+0.4444)=0.9000$.

If arm 2 is assigned first, the counts become $(3,2)$ at $n=5$. The weights for the next assignment are $(0.5/0.6)^2=0.6944$ and $(0.5/0.4)^2=1.5625$. Hence the second assignment goes to arm 2 with probability $1.5625/(1.5625+0.6944)=9/13=0.6923$.

If arm 1 is assigned first, the counts become $(4,1)$, with shares $(0.8,0.2)$. The next weights are $(0.5/0.8)^2=0.3906$ and $(0.5/0.2)^2=6.2500$, so the probability of assigning to arm 2 second is $6.2500/(6.2500+0.3906)=16/17=0.9412$.

The probability of the path arm 2 then arm 2 is $0.9000(0.6923)=0.6231$. The probability of arm 2 then arm 1 is $0.9000(1-0.6923)=0.2769$. The probability of arm 1 then arm 2 is $0.1000(0.9412)=0.0941$, and the probability of arm 1 then arm 1 is $0.1000(1-0.9412)=0.0059$. These four probabilities sum to $1.0000$, up to rounding.

The target-relative deterministic rule assigns arm 2 at both steps in this example, ending with counts $(3,3)$. DBCD can reach that same endpoint, but it can also end with $(4,2)$ or $(5,1)$. Its probabilities favor correction without guaranteeing the deterministic path.

Diagnostics, Alternatives, and Limitations

The central diagnostic is to separate a probability rule from its realized sequence. A single assignment to an arm with a lower DBCD probability is not evidence that the rule failed; randomization can produce any arm with positive probability. Examine whether probabilities were computed from the intended target and current shares, whether they sum to one, and whether the observed assignment is compatible with the stated randomization mechanism. For a trajectory-level assessment, compare realized counts with their targets over the sequence, rather than demanding that every individual step reduce every discrepancy.

Common implementation or interpretation problems include the following:

Treating the largest weight as certainty. The weight must be normalized across arms. DBCD assigns using the resulting probabilities; only the defined deterministic comparator assigns to its maximizing arm with certainty.
Confusing absolute and relative imbalance. With unequal targets, the arm with the largest $\rho_{n,k}-p_{n,k}$ need not have the largest $\rho_{n,k}/p_{n,k}$. State which criterion a deterministic rule uses.
Reading a probability as a promise. A high probability is not a guaranteed next assignment, and a one-step deviation from target is not by itself a failure.
Ignoring the state used in the calculation. Probabilities must use the current shares and the target specified for that assignment. A stale state or a target from the wrong update can yield probabilities that do not implement the intended rule.

A deterministic greedy rule offers immediate, predictable correction under its chosen criterion, but removes assignment randomness and may be sensitive to ties or small state changes. DBCD retains positive chances for each arm under the assumptions above and softens the response to imbalance into probabilities. Fixed randomization is another contrast: it uses prespecified probabilities rather than recalculating them from current imbalance. DBCD’s probabilities are state-dependent, not independent of the allocation history.

These comparisons do not establish that one approach is universally best. A deterministic rule can be unsuitable when random assignment is required, while a stochastic rule produces allocation variability and need not hit a target at any particular step. The calculations here also assume usable positive target shares and current counts. They do not resolve how targets are estimated, how to handle zero counts, or how to impose additional assignment constraints; those choices change the design and require their own specification.

Key takeaway: DBCD uses allocation imbalance to change the probability of the next assignment, not to dictate it. Its conditional mean can move toward target while individual realized assignments remain random. A deterministic comparator must specify exactly how “most underrepresented” is measured.

Check Your Understanding

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

With two equal targets, current shares $(0.6,0.4)$, and $\gamma=1$, what is the DBCD probability of assigning to arm 2?
Explain why a DBCD probability of $0.8$ for an underrepresented arm does not mean that the next realized proportion must rise.
For target shares $(0.6,0.4)$ and current shares $(0.5,0.5)$, compare which arm is selected by the target-relative and absolute-deficit deterministic rules.
State one condition needed to apply the displayed DBCD formula as written, and explain what it ensures.

Solutions. (1) The weights are $0.6/0.6=1$ and $0.4/0.4=1$ only if the target shares match current shares; here they do not. The correct weights are $0.5/0.6=0.8333$ and $0.5/0.4=1.2500$, so the probability for arm 2 is $1.2500/(0.8333+1.2500)=0.6000$. (2) The assignment is random. The underrepresented arm can be missed, in which case its proportion may decline because the denominator increases while its count stays fixed. (3) The ratios are $0.6/0.5=1.2$ and $0.4/0.5=0.8$, while the absolute deficits are $0.1$ and $-0.1$; both rules select arm 1. (4) Current counts must be positive, so each current share is positive and the ratios in the formula are defined. Positive target shares additionally ensure positive weights and assignment probabilities.