Worked Example: Unequal Target Shares

Why Unequal Target Shares Need Careful Calculation

A two-arm response-adaptive randomization rule need not aim for a 50:50 allocation. A prespecified target might instead assign 70% of participants to Arm 1 and 30% to Arm 2. In that setting, seeing an assignment probability above one half does not by itself tell us whether the rule is correcting an imbalance. The relevant comparison is with the target share, not with one half.

This tutorial works through that comparison for a two-arm doubly adaptive biased coin design (DBCD). In each example, the target shares are treated as given for the next assignment. In an adaptive application, those shares may be obtained from the current response estimates through a target map; the same calculation applies once the current target has been specified.

The Two-Arm Rule

Definition: Let $n$ be the number of participants already assigned, and let $N_1$ and $N_2$ be the current allocation counts, with $N_1+N_2=n$. Write $x=N_1/n$ for the current proportion assigned to Arm 1. Let $\rho$ be the target share for Arm 1, so Arm 2's target share is $1-\rho$. For $0<x<1$, $0<\rho<1$, and tuning parameter $\gamma\geq 0$, the two-arm DBCD assigns the next participant to Arm 1 with probability $$ \pi_1 = \frac{\rho\left(\frac{\rho}{x}\right)^\gamma} {\rho\left(\frac{\rho}{x}\right)^\gamma+ (1-\rho)\left(\frac{1-\rho}{1-x}\right)^\gamma}. $$ The probability of assigning the next participant to Arm 2 is $\pi_2=1-\pi_1$.

The factors $\rho$ and $1-\rho$ encode the target shares. The powered ratios compare each target share with its current allocation proportion. If Arm 1 is below target, $\rho/x>1$, which increases its weight; if it is above target, $\rho/x<1$, which decreases its weight. The exponent $\gamma$ controls the strength of this adjustment. At $\gamma=0$, the rule reduces to $\pi_1=\rho$, regardless of the current proportions.

The formula requires both arms to have positive current counts, so $0<x<1$. It also requires target shares strictly between zero and one. These conditions ensure that the displayed ratios are defined and finite. Initialization, sparse counts, or targets on the boundary require additional handling; the calculations here do not replace those design choices.

Worked Calculations

Worked Example: Arm 1 Is Below a 70:30 Target

Suppose 10 participants have been assigned: 4 to Arm 1 and 6 to Arm 2. The current Arm 1 proportion is $x=4/10=0.40$. For the next assignment, the target is 70:30, so $\rho=0.70$; take $\gamma=1$. Both arms have positive counts, the target shares are positive and sum to one, and the tuning parameter is nonnegative, so the formula applies.

The Arm 1 weight is $0.70(0.70/0.40)=1.225$. The Arm 2 weight is $0.30(0.30/0.60)=0.1500$. Normalizing the weights gives

\pi_1 = \frac{0.70(0.70/0.40)} {0.70(0.70/0.40)+0.30(0.30/0.60)} = \frac{1.225}{1.375} =0.8909.

Thus the next participant has probability 0.8909 of being assigned to Arm 1 and probability $1-0.8909=0.1091$ of being assigned to Arm 2. Arm 1's current share, 0.40, is well below its target share, 0.70, so the rule assigns Arm 1 with probability above its target share. This is a probability for one assignment, not a promise about the eventual allocation proportion.

Worked Example: Arm 1 Is Above a 70:30 Target

Now suppose 20 participants have been assigned: 16 to Arm 1 and 4 to Arm 2. Then $x=16/20=0.80$, which is above the same Arm 1 target, $\rho=0.70$. Set $\gamma=2$. Both arm counts are positive, and the target and tuning parameter meet the formula's conditions.

The Arm 1 weight is $0.70(0.70/0.80)^2=0.5359$, and the Arm 2 weight is $0.30(0.30/0.20)^2=0.6750$. Therefore,

\pi_1 = \frac{0.5359}{0.5359+0.6750} = \frac{0.5359}{1.2109} \approx 0.4426.

The next participant is assigned to Arm 1 with probability about 0.4426, and to Arm 2 with probability about 0.5574. Although Arm 1 has the larger target share, its current proportion has risen to 0.80. Because that exceeds the target of 0.70, the rule gives Arm 1 a probability below 0.70. The relatively strong adjustment here reflects using $\gamma=2$, as well as the size of the current imbalance.

Worked Example: The Targeted Arm Is the Minority Arm

Suppose the target is 30:70, with Arm 1 targeted at $\rho=0.30$. After 10 assignments, let $N_1=2$ and $N_2=8$, so $x=0.20$. With $\gamma=1$, both counts and target shares are positive and the rule is defined. The weights are $0.30(0.30/0.20)=0.4500$ for Arm 1 and $0.70(0.70/0.80)=0.6125$ for Arm 2. Thus

\pi_1 = \frac{0.4500}{0.4500+0.6125} = \frac{0.4500}{1.0625} \approx 0.4235.

Arm 1 is assigned next with probability about 0.4235, even though its target share is only 0.30. That is the intended direction: its current share of 0.20 is below target. The result also shows why “Arm 1 should get less than one half because it has the smaller target” is faulty reasoning. The current allocation matters alongside the target.

Why the Adjustment Changes Sides at the Target

A useful check is to compare the DBCD assignment odds with the target odds. Dividing the Arm 1 weight by the Arm 2 weight in the rule gives

\frac{\pi_1}{1-\pi_1} = \left(\frac{\rho}{1-\rho}\right)^{\gamma+1} \left(\frac{1-x}{x}\right)^\gamma.

The target odds are $\rho/(1-\rho)$. Dividing the assignment odds by those target odds yields

\frac{\pi_1/(1-\pi_1)}{\rho/(1-\rho)} = \left(\frac{\rho(1-x)}{x(1-\rho)}\right)^\gamma.

For $\gamma>0$, the expression in parentheses exceeds one exactly when $\rho>x$: multiplying by the positive denominator $x(1-\rho)$ reduces the comparison to $\rho(1-x)>x(1-\rho)$, or $\rho>x$. Since the odds function $u/(1-u)$ increases for $0<u<1$, assignment odds above target odds imply $\pi_1>\rho$. Reversing the inequality gives the other case.

Result: For $0<x,\rho<1$ and $\gamma>0$, if $x<\rho$, then $\pi_1>\rho$; if $x>\rho$, then $\pi_1<\rho$; and if $x=\rho$, then $\pi_1=\rho$. When $\gamma=0$, $\pi_1=\rho$ for every $x$.

This result checks the direction of the adjustment, not its exact magnitude. In particular, it does not say that one assignment will bring the observed allocation proportion to the target. The coin can assign either arm, and a single assignment changes the counts by only one.

Checks, Alternatives, and Limitations

Before interpreting a calculation, check that the target shares sum to one, that the current proportions are calculated from the allocation counts rather than from outcome counts, and that the two weights are both included in the denominator. A final probability must be between zero and one, and the two assignment probabilities must sum to one. If Arm 1 is below its target and $\gamma>0$, a computed $\pi_1$ below $\rho$ signals an arithmetic error, a reversed arm label, or an incorrect formula.

The main contrast is with assigning at the target probability every time. That rule uses $\pi_1=\rho$ even when the current allocation is far from target; it corresponds to the $\gamma=0$ case here. A DBCD with positive $\gamma$ adjusts the next coin probability according to the current imbalance. At the other extreme, assigning deterministically to whichever arm is under target is not the same rule: DBCD retains randomization for interior states and finite $\gamma$. The purpose of this comparison is to interpret the calculation, not to claim that one procedure is best for every design.

These calculations also have limits. A probability can be correct given the inputs but still be inappropriate if the target share is misspecified or, in a response-adaptive design, the estimated target is unstable. The formula alone does not determine the target, guarantee an exact realized allocation, or resolve how to initialize the design when one arm has no participants. Those are separate design questions. For the worked examples, the targets are treated as fixed for the next assignment, and the observed counts are assumed to be correct.

Common Mistake

Pitfall: “The target is 70:30, so Arm 1's next-assignment probability should always be 0.70.” That is true only when $\gamma=0$, or when the current share already equals the target. With positive $\gamma$, compare the current share $x$ with $\rho$: if $x<\rho$, the DBCD gives Arm 1 a probability above $\rho$; if $x>\rho$, it gives Arm 1 a probability below $\rho$. The target is the allocation objective, not necessarily the next-coin probability.

Check Your Understanding

Answer each question using the two-arm rule and the comparison with target odds. Solutions follow.

With target $\rho=0.60$, current share $x=0.75$, and $\gamma=1$, should $\pi_1$ be above or below 0.60?
For $\rho=0.25$, $x=0.25$, and $\gamma=3$, calculate $\pi_1$.
With $\rho=0.40$, $x=0.20$, and $\gamma=1$, calculate $\pi_1$ and interpret it relative to the target.
Why is it incorrect to call a computed $\pi_1=0.42$ “an allocation of 42% to Arm 1” without qualification?

Solutions. (1) Since $x>\rho$ and $\gamma>0$, the result gives $\pi_1<\rho=0.60$. (2) At $x=\rho$, both target-to-current ratios equal one, so $\pi_1=0.25$, for any $\gamma$. (3) The weights are $0.40(0.40/0.20)=0.8000$ and $0.60(0.60/0.80)=0.4500$. Hence $\pi_1=0.8000/(0.8000+0.4500)=0.6400$, above the target share of 0.40 because Arm 1 is currently below target. (4) The value 0.42 is the probability for the next participant’s assignment conditional on the current state. It is not a guarantee or prediction that 42% of all participants will ultimately be assigned to Arm 1.