Why Calculate Allocation Odds?
In a doubly adaptive biased coin design (DBCD), the target shares describe the allocation the design seeks, while the current shares describe the allocation already observed. The next assignment probability depends on both. Calculating the odds makes their separate contributions visible: a target may favor one arm, while the current allocation may call for correction in the other direction.
This tutorial conditions on a particular sequential state. The target shares and current allocation counts are treated as known at that state, so the calculation concerns the probability of the next assignment, not a guarantee about the final allocation. Any response-adaptive procedure used to obtain the target must provide a defined target at the current step; the odds calculation itself takes that target as given.
The calculation assumes both current counts are positive, both target shares are positive, and \(\gamma\) is finite and nonnegative. These conditions make the ratios in the weight formula well-defined. The probabilities are conditional on the information available at step \(n\); they do not require a particular outcome model beyond having a defined target share. In particular, the target shares are not themselves the next-assignment probabilities except in special cases.
Deriving the Two-Arm Odds
For positive assignment probabilities, define the allocation odds for arm 1 relative to arm 2 as \(\pi_{n+1,1}/\pi_{n+1,2}\). Because both probabilities divide their respective weights by the same normalizing sum, that sum cancels in their ratio. Substituting the two DBCD weights gives
Since \(q_{n,2}/q_{n,1}=N_{n,2}/N_{n,1}\), the current-share term can also be calculated directly from the counts. Another useful form separates the target odds from the correction:
The bracket compares the target odds with the current allocation odds. If arm 1 is underrepresented relative to its target, the bracket exceeds 1; if it is overrepresented, the bracket is below 1. The exponent \(\gamma\) controls how strongly that discrepancy contributes to the assignment odds at this state.
To convert odds \(O\) into a probability, use \(O=\pi_{n+1,1}/(1-\pi_{n+1,1})\). Solving gives \(\pi_{n+1,1}=O/(1+O)\), and \(\pi_{n+1,2}=1/(1+O)\). This conversion is essential: odds are not probabilities.
How the Three Inputs Contribute
Taking logarithms gives a compact way to compare target imbalance and current imbalance:
The first term is the target odds on a log scale. The second term is gamma times the difference between the target and current log odds. At a fixed state, increasing gamma moves the log assignment odds in the direction of that difference. This describes the algebraic effect of gamma; it does not by itself establish that every chosen value is appropriate for a particular design.
When current shares equal target shares, the bracket is zero, so the assignment odds equal the target odds for every finite \(\gamma\). When \(\gamma=0\), the correction term disappears and the next-assignment probabilities equal the target shares. These are useful checks on both reasoning and implementation.
Worked Example: Target Imbalance With Equal Current Shares
Suppose a hypothetical two-arm design has target shares \(\rho_{n,1}=0.60\) and \(\rho_{n,2}=0.40\). There have been 10 assignments to each arm, so \(q_{n,1}=q_{n,2}=0.50\). Let \(\gamma=1.5\). The shares are positive and sum to 1, and both current counts are positive, so the odds formula applies.
The target odds are \(0.60/0.40=1.5\), and the current odds are \(0.50/0.50=1\). Therefore,
Convert the odds to a probability:
Thus, at this state the next participant is assigned to arm 1 with probability about 0.7337, or 73.37%. This exceeds the 0.60 target share because the current allocation is equal across arms while the target favors arm 1, and \(\gamma\) amplifies that target-versus-current discrepancy.
Worked Example: Current Overrepresentation and Two Gamma Values
Consider a hypothetical state with target shares 0.60 for arm 1 and 0.40 for arm 2. Of 20 participants already assigned, 15 received arm 1 and 5 received arm 2, so the current shares are \(q_{n,1}=0.75\) and \(q_{n,2}=0.25\). The target and current shares are positive and each pair sums to 1. Compare \(\gamma=0.5\) and \(\gamma=1\).
For \(\gamma=0.5\), target odds are \(0.60/0.40=1.5\), while the current odds are \(0.75/0.25=3\). The correction factor is \((1.5/3)^{0.5}\):
For \(\gamma=1\), the correction is stronger:
The current allocation has more arm 1 assignments than the 0.60 target share would suggest. Accordingly, increasing gamma decreases the next-assignment probability for arm 1 at this fixed state. This comparison illustrates the calculation, not a recommendation for a particular gamma.
Worked Example: Current Shares Already Match the Target
Suppose a hypothetical trial has target shares 0.60 and 0.40, with 18 of 30 current assignments to arm 1 and 12 to arm 2. The current shares are \(18/30=0.60\) and \(12/30=0.40\), so they match the target. Both counts are positive. Take \(\gamma=2\).
The target odds are \(0.60/0.40=1.5\), and the current odds are also \(0.60/0.40=1.5\). Thus the correction ratio is 1:
The next-assignment probability for arm 1 equals its target share despite \(\gamma=2\). The reason is not that gamma has no role in the design generally; at this particular state, the target-to-current ratio is exactly 1, so its power remains 1.
Comparisons, Diagnostics, and Limitations
The odds expression distinguishes DBCD from simpler alternatives. With \(\gamma=0\), the rule assigns directly according to the target shares, regardless of current allocation. With positive gamma, it also accounts for the current shares. A deterministic procedure that simply assigns the next participant to the arm most behind its target is different: DBCD specifies probabilities, so even a state with one arm currently underrepresented need not force the next assignment to that arm.
| Feature | Target-only assignment | DBCD at a fixed state |
|---|---|---|
| Arm 1 probability | \(\rho_{n,1}\) | \(O/(1+O)\), using target and current shares |
| Role of current shares | None | Enter through the correction ratio |
| When probabilities equal target shares | Always | When \(\gamma=0\), or when current shares match targets |
| Assignment mechanism | Random if implemented as a probability rule | Random with the state-dependent probabilities \(\pi_{n+1,k}\) |
Several checks help detect mistakes. First, verify that each pair of target shares and current shares sums to 1. Second, calculate the correction ratio in the form \((\rho_{n,1}/\rho_{n,2})/(q_{n,1}/q_{n,2})\); this makes the direction of correction easier to inspect. Third, confirm that the two normalized probabilities sum to 1. Finally, check the special cases \(\gamma=0\) and \(q_{n,k}=\rho_{n,k}\).
A common error is to use \((q_{n,1}/q_{n,2})^\gamma\) as the current-allocation factor in the two-arm odds. That reverses the correction: when arm 1 is already overrepresented, it would increase rather than decrease the odds of assigning to arm 1. The correct factor in the expanded formula is \((q_{n,2}/q_{n,1})^\gamma\); equivalently, use the target-to-current odds ratio inside the correction bracket.
A second pitfall is to report the odds as the assignment probability. For example, odds of 0.75 correspond to probability \(0.75/(1+0.75)\), not probability 0.75. Also, the formula is a conditional next-step calculation. It does not say that the realized cumulative allocation will equal the target at every step, nor does it remove the need to define valid target shares and handle states where a ratio is undefined.
Check Your Understanding
Work through each question using the target and current shares at the stated step. Solutions follow the questions.
- Target shares are 0.55 and 0.45, current shares are 0.50 and 0.50, and \(\gamma=1\). Find the odds and the next-assignment probability for arm 1.
- Target shares are 0.60 and 0.40, current shares are 0.75 and 0.25, and \(\gamma=1\). Find the next-assignment probability for arm 1.
- If target and current shares are both 0.35 and 0.65, what is the next-assignment probability for arm 1 for any finite nonnegative \(\gamma\)? Explain.
- A calculation uses \((q_{n,1}/q_{n,2})^\gamma\) as the correction factor multiplying target odds. Explain the error and give the correct factor.
Solutions. (1) The target odds are \(0.55/0.45=11/9\), and the current odds are 1. The allocation odds are \((11/9)^2=121/81\approx1.4938\). Therefore, \(\pi_{n+1,1}=(121/81)/(1+121/81)=121/202\approx0.5990\).
(2) The target odds are 1.5 and the current odds are 3. Thus \(O=1.5(1.5/3)=0.75\), and \(\pi_{n+1,1}=0.75/1.75\approx0.4286\).
(3) The target-to-current odds ratio is 1, so raising it to any finite \(\gamma\) gives 1. The assignment odds equal the target odds \(0.35/0.65\), and the resulting probability for arm 1 is 0.35.
(4) The proposed factor increases the odds for arm 1 when arm 1 is overrepresented, which is the opposite of the correction. The correct factor is \((q_{n,2}/q_{n,1})^\gamma\), or equivalently \(\left[(\rho_{n,1}/\rho_{n,2})/(q_{n,1}/q_{n,2})\right]^\gamma\).