Worked Example: Three-Arm Allocation Weights

From Three-Arm Weights to Assignment Probabilities

In a three-arm doubly adaptive biased coin design (DBCD), the next assignment depends on both the desired allocation and the allocation already observed. The calculation has two parts: first compute a nonnegative weight for each arm, then divide each weight by the total weight. The resulting probabilities—not the unnormalized weights—are the chances used to randomize the next patient.

This tutorial focuses on the arithmetic at one allocation step. Assume the target shares for that step have already been specified. They may come from a prespecified target rule using current response estimates, but those estimates are not recalculated here. The examples use equal target shares to isolate the three-arm weight calculation.

Definition: At step $n+1$, let $\rho_j$ be the target share for arm $j$, and let $N_j(n)$ be the number already assigned to that arm, for $j=1,2,3$. The current allocation share is $q_j(n)=N_j(n)/n$, where $n=N_1(n)+N_2(n)+N_3(n)$. For positive target shares, positive current counts, and tuning parameter $\gamma\geq 0$, define the DBCD weight and next-patient probability by $$ w_j(n+1)=\rho_j\left(\frac{\rho_j}{q_j(n)}\right)^\gamma, \qquad p_{j,n+1}=\frac{w_j(n+1)}{\sum_{k=1}^3 w_k(n+1)}. $$

The factor $\rho_j$ preserves the target allocation as the baseline: when $\gamma=0$, $w_j=\rho_j$, so the next patient is assigned according to the target shares. When $\gamma>0$, the ratio $\rho_j/q_j(n)$ adjusts the weight according to how the current share compares with its target. A share below target gives a ratio above one and increases its weight relative to the baseline; a share above target gives a ratio below one. The strength of this adjustment depends on $\gamma$.

If all target shares are equal, the common factor $\rho_j=1/3$ multiplies every weight by the same amount. It can then be omitted when calculating normalized probabilities: the weights proportional to $(\rho_j/q_j(n))^\gamma$ give exactly the same probabilities. This simplification is specific to equal target shares.

A Calculation Sequence

Before using the formula, check that the target shares are positive and sum to one, that the counts are positive, and that the current counts sum to $n$. Positive counts matter because the displayed ratio uses $q_j(n)$ in its denominator. Initialization and zero-count handling require separate design choices; the formula here is for a state where every arm has already received patients.

Find current shares.
Divide each arm count by the total $n$, producing $q_1(n),q_2(n),q_3(n)$.

Calculate each weight.
For each arm, form the target-to-current ratio, raise it to $\gamma$, and multiply by the target share.

Normalize once.
Add the three weights and divide each individual weight by that same sum.

Check and interpret.
Confirm that the probabilities sum to one and describe them as chances of assignment for the next patient.

Worked Example: Weights With Counts of 12, 9, and 6

Suppose the target shares are equal, $\rho_1=\rho_2=\rho_3=1/3$, the current counts are $(12,9,6)$, and $\gamma=1$. All arms have positive counts, the total is $n=12+9+6=27$, and the target shares sum to one, so the stated calculation applies.

The current shares are $q_1=12/27=4/9$, $q_2=9/27=1/3$, and $q_3=6/27=2/9$. The weights are $$ w_1=\frac13\left(\frac{1/3}{4/9}\right)=\frac14,\qquad w_2=\frac13\left(\frac{1/3}{1/3}\right)=\frac13,\qquad w_3=\frac13\left(\frac{1/3}{2/9}\right)=\frac12. $$ Their sum is $1/4+1/3+1/2=13/12$. Therefore, $$ p_{1,28}=\frac{1/4}{13/12}=\frac3{13}\approx0.2308,\quad p_{2,28}=\frac{1/3}{13/12}=\frac4{13}\approx0.3077,\quad p_{3,28}=\frac{1/2}{13/12}=\frac6{13}\approx0.4615. $$ The probabilities sum to $13/13=1$. The next patient has approximately a 23.08% chance of assignment to arm 1, a 30.77% chance to arm 2, and a 46.15% chance to arm 3. Arm 3 is currently furthest below its one-third target share, and has the largest probability in this calculation.

What Changing Gamma Does at a Fixed State

The count state and target shares can remain fixed while a different $\gamma$ changes the weights. The following comparison uses the same valid state as the preceding example: equal targets and current shares $4/9,1/3,2/9$. It changes only the tuning parameter.

Worked Example: Recalculating the Same State at Gamma 2

With $\gamma=2$, the weights are $$ w_1=\frac13\left(\frac{3}{4}\right)^2=\frac3{16},\qquad w_2=\frac13(1)^2=\frac13,\qquad w_3=\frac13\left(\frac32\right)^2=\frac34. $$ Their sum is $3/16+1/3+3/4=61/48$. Dividing through by this sum gives $$ p_{1,28}=\frac{9}{61}\approx0.1475,\qquad p_{2,28}=\frac{16}{61}\approx0.2623,\qquad p_{3,28}=\frac{36}{61}\approx0.5902. $$

These probabilities sum to $61/61=1$. Relative to $\gamma=1$, the arm with a current share above target receives less probability, while the under-target arm receives more. This is a comparison at one fixed state, not a claim that larger $\gamma$ always improves a trial’s performance; the calculation only shows how the weight rule responds to that state.

Exact Forms Help Prevent Rounding Errors

A useful practice is to retain fractions or exact radicals through normalization, and round only the final probabilities. In a three-arm calculation, premature rounding can alter the displayed sum or make a small but consequential error in one arm’s probability.

Worked Example: Radical Weights and a Sum Check

Suppose the target shares are equal, the current counts are $(3,2,1)$, and $\gamma=1/2$. The total is $6$, so the current shares are $1/2,1/3,1/6$. The target-to-current ratios are $2/3,1,2$. Since the common target factor $1/3$ can be omitted for equal targets, proportional weights are $$ \sqrt{\frac23},\qquad 1,\qquad \sqrt{2}. $$ The total proportional weight is $\sqrt{2/3}+1+\sqrt2\approx3.2307101$. Thus $$ p_{1,7}\approx\frac{0.8164966}{3.2307101}=0.2527,\quad p_{2,7}\approx\frac{1}{3.2307101}=0.3095,\quad p_{3,7}\approx\frac{1.4142136}{3.2307101}=0.4377. $$ Using more digits, the probabilities are approximately $0.2527298$, $0.3095295$, and $0.4377408$, which sum to one up to rounding. In particular, the third probability rounds to $0.4377$, or $43.77\%$, not $0.4378$. The first two are $25.27\%$ and $30.95\%$, respectively.

Worked Example: Rescaling Weights Before Normalizing

Consider equal targets, $\gamma=1$, and counts $(35,5,3)$, totaling $n=43$. The current shares are $35/43,5/43,3/43$. Because targets are equal, the normalized probabilities are unchanged if the common target factor is dropped. Weights proportional to the inverse current shares are therefore proportional to $1/35,1/5,1/3$. Multiplying all three by the same positive constant $35/63$ gives the equivalent weights $1/21,1/3,5/9$. Rescaling every weight by one common positive constant does not alter the normalized probabilities.

The sum of these rescaled weights is $$ \frac1{21}+\frac13+\frac59=\frac3{63}+\frac{21}{63}+\frac{35}{63}=\frac{59}{63}. $$ Consequently, $$ p_{1,44}=\frac{1/21}{59/63}=\frac3{59}\approx0.05085,\quad p_{2,44}=\frac{1/3}{59/63}=\frac{21}{59}\approx0.3559,\quad p_{3,44}=\frac{5/9}{59/63}=\frac{35}{59}\approx0.5932. $$ As percentages, these are approximately $5.085\%$, $35.59\%$, and $59.32\%$. The largest weight is for arm 3, which has the smallest current count; the final probabilities sum to $59/59=1$.

Why Normalization Works, and What to Check

The normalization step is more than a convenient scaling operation. If all three weights are positive and finite, their sum $W=\sum_{k=1}^3 w_k$ is positive and finite. Hence every $p_{j,n+1}=w_j/W$ is positive, and $$ \sum_{j=1}^3p_{j,n+1} =\sum_{j=1}^3\frac{w_j}{W} =\frac{\sum_{j=1}^3w_j}{W} =\frac{W}{W}=1. $$ So the normalized values form a valid probability distribution over the three mutually exclusive arm assignments. Also, for any two arms $j$ and $k$, $p_{j,n+1}/p_{k,n+1}=w_j/w_k$: normalization changes the scale but preserves the relative weights.

At the end of a hand calculation, check that the counts sum to $n$, that the target shares sum to one, and that each current share is calculated using the same $n$. Then verify that the three weights are nonnegative and that the probabilities sum to one, allowing for small rounding differences. A probability greater than one, a negative probability, or a probability sum far from one signals an arithmetic or formula error.

A frequent mistake is to treat the weights themselves as assignment probabilities. For example, weights $1/4,1/3,1/2$ sum to $13/12$, so they cannot directly be probabilities. The correction is to divide each by $13/12$, giving $3/13,4/13,6/13$. Another error is to normalize each weight using a different or incorrectly computed total. All three probabilities must use the same denominator, namely the sum of the three weights.

At a high level, this calculation differs from assigning every patient independently according to fixed target shares: a DBCD uses the current allocation shares as well as its target. It also differs from deterministic assignment to whichever arm is most underrepresented: the normalized weights retain random assignment chances for each arm in these positive-count examples. The arithmetic alone does not establish any broader performance guarantee. It assumes the target shares and $\gamma$ are already chosen, and it does not account for uncertainty in estimated responses, special constraints, or a rule for zero counts.

Key takeaway: Compute each arm’s target-to-current adjustment, multiply by its target share, and normalize all three weights by their common sum. Preserve exact arithmetic when practical, and interpret the results as next-patient assignment probabilities.

Check Your Understanding

Try each problem first, then compare your work with the solutions below.

For equal targets, counts $(4,2,2)$, and $\gamma=1$, find the weights and next-patient probabilities.
With equal targets and positive current counts, what probabilities result when $\gamma=0$? Explain using the weight formula.
For equal targets, counts $(35,5,3)$, and $\gamma=1$, explain why weights $1/21,1/3,5/9$ can be used and give the normalized probabilities.
A calculation produces three positive weights that sum to 2.4. Is that, by itself, an error? What must be done next?

Solutions. (1) Here $n=8$ and the current shares are $1/2,1/4,1/4$. The weights are $(1/3)(2/3)=2/9$, $(1/3)(4/3)=4/9$, and $4/9$. Their sum is $10/9$, giving probabilities $1/5,2/5,2/5$. (2) With $\gamma=0$, each adjustment ratio to the zero power is 1, so $w_j=\rho_j=1/3$; normalization gives $1/3$ for each arm. (3) Those weights are a common rescaling of weights proportional to $1/35,1/5,1/3$. Their sum is $59/63$, so the probabilities are $3/59,21/59,35/59$, approximately $0.05085,0.3559,0.5932$. (4) No. Positive weights need not sum to one. Divide each by the common total 2.4; the resulting probabilities will sum to one.