From the Current Trial State to One Assignment Probability
A response-adaptive allocation rule is applied at a particular moment in a trial: the patients already assigned and the response information available so far determine the probabilities for the next assignment. A hand calculation makes each ingredient visible. It also helps catch errors that can be hard to spot when the rule is implemented in software, especially errors in the arm weights or their normalization.
For this update, the current state consists of the arm counts, the response estimates, and the target shares obtained from those estimates under a prespecified target map. The counts describe how many patients have already been assigned; the target shares describe the desired allocation at this state. The doubly adaptive biased coin design (DBCD) combines both pieces of information. This tutorial works with two arms, labelled A and B, and computes just the next-patient assignment probabilities.
The response estimates enter through the target shares: a prespecified target map takes the estimates available at the current state and returns \((\rho_A,\rho_B)\). Once those shares are supplied, the weight calculation uses the counts, targets, and \(\gamma\). In particular, response estimates are not themselves the assignment probabilities. Nor does the formula treat a target share as the current share.
For the direct calculation below, assume that both arms have positive counts, so \(\pi_A,\pi_B>0\), and that both target shares are positive. These conditions make the ratios in the formula well-defined and give positive weights. The estimates and target shares are treated as the values available before the next assignment; the next patient’s response is not yet observed and cannot enter this update. If a target is on a boundary or an arm has count zero, the direct formula may not be defined as written; an initialization or other specified safeguard is needed before applying it.
Why the Weights Must Be Normalized
The two weights measure relative allocation pressure, not probabilities. The factor \(\rho_i\) reflects the target share for arm \(i\); the factor \((\rho_i/\pi_i)^\gamma\) adjusts for the gap between that target and the arm’s current share. If an arm’s current share is below its target, its ratio \(\rho_i/\pi_i\) exceeds one. If it is above its target, the ratio is below one. Normalizing the weights turns these relative quantities into probabilities that sum to one.
A short algebraic check shows how the computation relates the two arms. Dividing their probabilities cancels the common normalizing denominator, so the odds of the next assignment are the ratio of the weights:
This identity is also a useful arithmetic check: the normalized probabilities must have the same ratio as the two unnormalized weights. To obtain the probabilities themselves, divide each weight by their sum; do not stop at the weights or treat them as probabilities without checking their sum.
Worked Example: An Underrepresented Arm With Equal Targets
Suppose a hypothetical trial has assigned 40 patients to A and 60 to B. The available response estimates are \(\hat p_A=0.70\) and \(\hat p_B=0.70\). For this example, the prespecified target map assigns shares in proportion to these estimated response rates, so \(\rho_A=\hat p_A/(\hat p_A+\hat p_B)=0.50\) and \(\rho_B=0.50\). Use \(\gamma=1\) to compute the next-patient probabilities.
There are \(N=40+60=100\) patients, giving current shares \(\pi_A=40/100=0.40\) and \(\pi_B=60/100=0.60\). Both counts are positive, and the targets are positive and sum to one, so the direct rule applies.
The weights are \(w_A=0.50(0.50/0.40)^1=0.50(1.25)=0.625\) and \(w_B=0.50(0.50/0.60)^1=0.50(5/6)=0.4167\). Their sum is \(0.625+0.4167=1.0417\), with the unrounded sum equal to \(25/24\).
Therefore, \(P(A_{101}=A\mid\mathcal H_{100})=0.625/1.0417=0.6000\) and \(P(A_{101}=B\mid\mathcal H_{100})=0.4167/1.0417=0.4000\). The next patient has a 60% conditional chance of assignment to A and a 40% chance of assignment to B. These probabilities are not the current shares, which are 40% and 60%, and they do not guarantee that the next assignment will be to A.
Worked Example: Unequal Targets and Unequal Current Shares
In a second hypothetical trial, 30 patients have been assigned to A and 70 to B. The available estimates are \(\hat p_A=0.75\) and \(\hat p_B=0.50\). Suppose the prespecified target map again sets each target proportional to its estimated response rate. With \(\gamma=1\), find the next-patient probabilities.
The total is \(N=100\), so \(\pi_A=0.30\) and \(\pi_B=0.70\). The target shares from the stated map are \(\rho_A=0.75/(0.75+0.50)=0.60\) and \(\rho_B=0.50/(0.75+0.50)=0.40\). They are positive and sum to one; the current counts are also positive.
Substitution gives \(w_A=0.60(0.60/0.30)^1=0.60(2)=1.20\). For B, \(w_B=0.40(0.40/0.70)^1=0.40(4/7)=0.2286\). The total weight is \(1.20+0.2286=1.4286\) (exactly \(10/7\)).
Normalizing, \(P(A_{101}=A\mid\mathcal H_{100})=1.20/1.4286=0.8400\) and \(P(A_{101}=B\mid\mathcal H_{100})=0.2286/1.4286=0.1600\). Thus this update assigns an 84% conditional probability to A and a 16% probability to B. A’s current share is below its target, while B’s current share is above its target; the calculated probabilities reflect that state and the specified rule. They do not mean that 84% of all patients will necessarily have been assigned to A after this one assignment.
Worked Example: A Non-Equal Target With a Different Gamma
Consider a third hypothetical trial with 50 patients assigned to each arm. The response estimates are \(\hat p_A=0.66\) and \(\hat p_B=0.54\), and the stated target map assigns shares in proportion to these estimates. Let \(\gamma=2\). Calculate the next-patient probabilities.
The current shares are \(\pi_A=\pi_B=50/100=0.50\). The targets are \(\rho_A=0.66/(0.66+0.54)=0.55\) and \(\rho_B=0.54/(0.66+0.54)=0.45\). Both arms have positive counts and both targets are interior, as required for direct use of the formula.
For A, the target-to-current ratio is \(0.55/0.50=1.10\), so \(w_A=0.55(1.10)^2=0.55(1.21)=0.6655\). For B, the ratio is \(0.45/0.50=0.90\), so \(w_B=0.45(0.90)^2=0.45(0.81)=0.3645\). The total is \(0.6655+0.3645=1.0300\).
The normalized probabilities are \(P(A_{101}=A\mid\mathcal H_{100})=0.6655/1.0300=0.6461\) and \(P(A_{101}=B\mid\mathcal H_{100})=0.3645/1.0300=0.3539\). They sum to 1.0000, up to rounding. The target for A is 55%, but its next-patient probability in this state is about 64.61%; the two quantities answer different questions.
Reading the Result and Checking the Calculation
In each example, the result is conditional on the information available before the assignment. For example, \(P(A_{N+1}=A\mid\mathcal H_N)=0.60\) means that, given the current trial history and the specified rule, the next assignment is to A with probability 0.60. It is a probability for a randomized decision, not a statement that the next patient is partly assigned to each arm, and not a forecast that the final allocation proportion will equal 0.60.
A reliable hand calculation can be organized as a short sequence. Carry out each step using the same state and the same target values throughout; do not update the estimates or counts halfway through the calculation.
Write \(n_A,n_B\), confirm \(N=n_A+n_B\), and calculate \(\pi_A=n_A/N\) and \(\pi_B=n_B/N\).
Use the response estimates with the prespecified target map. Check that the resulting \(\rho_A,\rho_B\) are the intended values and sum to one.
For each arm, evaluate \(\rho_i(\rho_i/\pi_i)^\gamma\). Keep sufficient precision in intermediate values.
Divide each weight by \(w_A+w_B\), check the probabilities sum to one, then state the result as a conditional probability for the next assignment.
Several quick diagnostics follow from these steps. The current shares should sum to one, the targets should sum to one, and the normalized assignment probabilities should sum to one. If an arm is below its target, its ratio \(\rho_i/\pi_i\) is greater than one; if it is above target, that ratio is less than one. These checks can reveal a swapped count, a reversed ratio, or an incorrectly applied exponent. A weight larger than one is not automatically an error, because weights are not probabilities; the normalized result is what must lie between zero and one.
Common Pitfall and Main Alternatives
A common mistake is to calculate one arm’s weight correctly and then miscalculate the other, or to read a weight as if it were already a probability. For instance, with \(\rho_B=0.50\), \(\pi_B=0.60\), and \(\gamma=1\), the B weight is \(0.50(0.50/0.60)=0.4167\), not \(0.8333\). The correction is to evaluate the ratio first, multiply by the target, add both weights, and divide each weight by that sum. In the first worked example, this gives weights \(0.625\) and \(0.4167\), and probabilities \(0.60\) and \(0.40\).
The DBCD update differs from two simple alternatives in what it uses. Assigning with probabilities \((\rho_A,\rho_B)\) uses the target shares but does not directly account for the current allocation shares. Deterministically assigning to whichever arm is below target removes the random draw and is not the DBCD rule. The weighted DBCD rule uses both target and current shares and then randomizes according to the normalized weights. A minimum-probability constraint, if part of a design, would also alter the probabilities and must be applied according to that design’s stated rule rather than silently added to this calculation.
The hand calculation is only as appropriate as its inputs and assumptions. A different target map can give different targets from the same response estimates. An undefined or unsuitable response estimate, a boundary target, or a zero arm count can prevent the direct formula from being used as written. Further, the calculation describes one update, not long-run operating characteristics or a guarantee of a final allocation. Its value is precise and local: given a valid state and a specified rule, it gives the conditional probabilities for the next random assignment.
Check Your Understanding
For each problem, calculate or interpret the update using the stated two-arm DBCD rule. Full solutions follow the questions.
- There are 25 patients on A and 75 on B, with targets \((0.50,0.50)\) and \(\gamma=1\). Find both next-patient probabilities.
- There are 60 patients on A and 40 on B, with targets \((0.70,0.30)\) and \(\gamma=1\). Find both next-patient probabilities.
- If the rule gives \(P(A_{N+1}=A\mid\mathcal H_N)=0.65\), what does this say about the next assignment, and what does it not guarantee?
- What prevents direct use of the stated weight formula if an arm has count zero and \(\gamma>0\)?
Solutions.
1. The current shares are \((0.25,0.75)\). The weights are \(w_A=0.50(0.50/0.25)=1\) and \(w_B=0.50(0.50/0.75)=1/3\). Their sum is \(4/3\), so the probabilities are \(1/(4/3)=0.75\) for A and \((1/3)/(4/3)=0.25\) for B.
2. The current shares are \((0.60,0.40)\). The weights are \(w_A=0.70(0.70/0.60)=49/60\approx0.8167\) and \(w_B=0.30(0.30/0.40)=0.225\). Their sum is approximately \(1.0417\), so the probabilities are \(0.8167/1.0417=0.7840\) for A and \(0.225/1.0417=0.2160\) for B.
3. Conditional on the current history and the stated rule, the next assignment is to A with probability 0.65 and to B with probability 0.35. It does not guarantee that the next patient will be assigned to A, nor that 65% of the eventual trial sample will be assigned to A.
4. A zero count gives current share \(\pi_i=0\), so the ratio \(\rho_i/\pi_i\) is undefined. The direct calculation therefore requires a design-specified initialization or other safeguard before this update can be applied.