Comparing Target Maps While Holding the Coin Rule Fixed

Why Hold the Coin Rule Fixed?

In a doubly adaptive biased coin design (DBCD), the target map and the coin rule play different roles. A target map converts the available response information into a desired allocation-share vector. The coin rule combines that vector with current allocation shares to produce the probabilities used to randomize the next participant. If both parts change in a comparison, the resulting probability difference cannot be attributed to either one alone.

This tutorial isolates the target-map contribution. We compare two maps using the same response estimates, allocation counts, number of participants already assigned, and DBCD tuning parameter. The resulting assignment probabilities are conditional probabilities for the next assignment at that fixed state; they do not by themselves describe the allocation path or long-run performance of either design.

Definition: For $K$ treatment arms, let $N_k(n)$ be the number assigned to arm $k$ among the first $n$ participants, and let $s_k(n)=N_k(n)/n$ be its current allocation share. A target map $M$, evaluated at the response estimates available at that time, supplies target shares $\rho_k^{(M)}$, where $\rho_k^{(M)}>0$ and $\sum_{k=1}^K \rho_k^{(M)}=1$. Under the DBCD coin rule with tuning parameter $\gamma\geq 0$, the next-participant assignment probability is $$ q_k^{(M)}(n+1)= \frac{\rho_k^{(M)} \left\{\rho_k^{(M)}/s_k(n)\right\}^{\gamma}} {\displaystyle\sum_{j=1}^K \rho_j^{(M)} \left\{\rho_j^{(M)}/s_j(n)\right\}^{\gamma}}. $$

The target share $\rho_k^{(M)}$ is a desired allocation proportion, not the probability of assigning the next participant to arm $k$. The latter is $q_k^{(M)}(n+1)$. When $\gamma=0$, the rule gives $q_k^{(M)}=\rho_k^{(M)}$. For positive $\gamma$, the rule adjusts for the current shares: an arm with a target share greater than its current share has its unnormalized weight multiplied by a factor greater than one relative to its target-share weight.

For the direct comparisons below, assume that both maps are evaluated on exactly the same response-estimate vector, both output strictly positive target shares, and the same positive current allocation counts and $\gamma$ are used. There are no arm-specific caps, clipping operations, or other constraints altering the stated rule. These conditions let us compare the maps at one fixed state. If a current share is zero, this formula is not directly defined; initialization and any zero-count handling must be specified separately rather than silently substituted into the calculation.

A Useful Odds Identity

Normalization can obscure which part of the calculation changes. Comparing the odds for two arms removes the common normalizing denominator. For arms $i$ and $j$, the DBCD rule gives

\frac{q_i^{(M)}}{q_j^{(M)}}= \frac{\rho_i^{(M)}}{\rho_j^{(M)}} \left\{ \frac{\rho_i^{(M)}/s_i}{\rho_j^{(M)}/s_j} \right\}^{\gamma} = \left(\frac{\rho_i^{(M)}}{\rho_j^{(M)}}\right)^{\gamma+1} \left(\frac{s_j}{s_i}\right)^\gamma .

The last factor depends only on the shared allocation state and shared $\gamma$. Thus, when maps $A$ and $B$ are compared at that same state,

\frac{q_i^{(A)}/q_j^{(A)}}{q_i^{(B)}/q_j^{(B)}} = \left[ \frac{\rho_i^{(A)}/\rho_j^{(A)}} {\rho_i^{(B)}/\rho_j^{(B)}} \right]^{\gamma+1}.

This identity is a concise justification for the comparison: the change in pairwise assignment odds is determined by the change in the maps’ target odds, amplified by the exponent $\gamma+1$. The state can affect each map’s probabilities (in particular, when $\gamma>0$); when it is held fixed, it cancels from the cross-map ratio of pairwise odds.

Worked Example: Three Arms at a Shared State

Suppose the same hypothetical response estimates, $(0.60,0.45,0.30)$, are passed to two candidate target maps. At these estimates, map $A$ supplies $(0.50,0.30,0.20)$, while map $B$ supplies $(0.40,0.40,0.20)$. These are stipulated outputs for illustration, not claims that either map is preferable. The current allocation counts are $(40,35,25)$, so $n=100$ and the current shares are $(0.40,0.35,0.25)$. Use the same coin-rule parameter $\gamma=1$ for both maps. All target and current shares are positive, and each target vector sums to one.

For map $A$, the unnormalized weights are target share times target-to-current-share ratio:

w^{(A)}= \left( 0.50\frac{0.50}{0.40}, 0.30\frac{0.30}{0.35}, 0.20\frac{0.20}{0.25} \right) = (0.6250,\,0.2571429,\,0.1600).

Their sum is $1.0421429$. Normalizing with the unrounded weights gives

q^{(A)}= \left( \frac{0.6250}{1.0421429}, \frac{0.2571429}{1.0421429}, \frac{0.1600}{1.0421429} \right) \approx(0.5997,\,0.2467,\,0.1535).

For map $B$, the weights are

w^{(B)}= \left( 0.40\frac{0.40}{0.40}, 0.40\frac{0.40}{0.35}, 0.20\frac{0.20}{0.25} \right) =(0.4000,\,0.4571429,\,0.1600).

Their sum is $1.0171429$, giving

q^{(B)}= \left( \frac{0.4000}{1.0171429}, \frac{0.4571429}{1.0171429}, \frac{0.1600}{1.0171429} \right) \approx(0.3933,\,0.4494,\,0.1573).

With map $A$, arm 1 has the largest next-assignment probability; with map $B$, arm 2 does. The approximate changes $q^{(B)}-q^{(A)}$ are $(-0.2065,\,0.2027,\,0.0038)$. Small discrepancies when adding displayed rounded probabilities are due to rounding; the unrounded probabilities sum to one. Since the response estimates, counts, and coin rule were identical, these differences are attributable to the alternative target outputs at this state.

Comparisons at Other Fixed States

A comparison at one state is a conditional calculation, not a general ranking of maps. The following examples use new hypothetical inputs to show how to make that calculation transparent. Within each example, the estimates, allocation state, and coin rule remain the same for both maps.

Worked Example: Two Arms With Unequal Current Shares

Suppose response estimates $(0.80,0.60)$ are supplied to two candidate maps. At those estimates, map $A$ returns target shares $(0.55,0.45)$, and map $B$ returns $(0.35,0.65)$. The current counts are $(70,30)$, so the current shares are $(0.70,0.30)$. Let $\gamma=1$. Both maps have positive targets, and both use the same state and rule.

For map $A$, the weights are $0.55^2/0.70=0.4321429$ and $0.45^2/0.30=0.6750$. Their sum is $1.1071429$, so

q^{(A)}= \left(\frac{0.4321429}{1.1071429}, \frac{0.6750}{1.1071429}\right) \approx(0.3903,\,0.6097).

For map $B$, the weights are $0.35^2/0.70=0.1750$ and $0.65^2/0.30=1.4083333$, summing to $1.5833333$. Therefore

q^{(B)}= \left(\frac{0.1750}{1.5833333}, \frac{1.4083333}{1.5833333}\right) \approx(0.1105,\,0.8895).

Both assignment probabilities differ from their respective targets because the current state is unequal and $\gamma=1$. In particular, map $A$'s target for arm 1 is $0.55$, but its next-assignment probability is about $0.3903$: arm 1 is already above its target share. Comparing $0.55$ directly with $0.1105$ would confuse map $A$'s target share with map $B$'s assignment probability.

Worked Example: Distinguishing a Small Target Change From a Small Probability Change

Return to a three-arm setting with the same response estimates $(0.60,0.45,0.30)$ for both maps, current shares $(0.50,0.30,0.20)$, and $\gamma=1$. At those estimates, map $C$ outputs $(0.60,0.25,0.15)$, while map $D$ outputs $(0.55,0.30,0.15)$. The current shares are positive and both target vectors are positive and sum to one.

Map $C$'s weights are $0.60^2/0.50=0.7200$, $0.25^2/0.30=0.2083333$, and $0.15^2/0.20=0.1125$. They sum to $1.0408333$, so its assignment vector is approximately $(0.6918,0.2002,0.1081)$. Map $D$'s weights are $0.55^2/0.50=0.6050$, $0.30^2/0.30=0.3000$, and $0.15^2/0.20=0.1125$, summing to $1.0175$. Its assignment vector is approximately $(0.5946,0.2948,0.1106)$.

The third target share is identical under the two maps, but the third assignment probability is not: changing the first two weights also changes the normalizing denominator. This is why each map must be normalized across all arms before comparing its assignment probabilities.

How to Make the Comparison Diagnostic

A careful comparison is easiest to audit when it follows a fixed sequence. First record the shared inputs; then evaluate each target map; finally apply the coin rule separately and normalize. This prevents a change in response estimates or allocation state from being mistaken for a target-map effect.

Freeze the comparison state.
Record the response-estimate vector, allocation counts and shares, and the value of $\gamma$. State any constraints that are active.

Evaluate both maps.
Write each map's target vector at those same response estimates. Check that its components are positive and sum to one.

Calculate and normalize weights.
For each arm and map, calculate $\rho_k(\rho_k/s_k)^\gamma$; divide each weight by that map's own sum.

Compare at the right level.
Compare target shares with target shares, assignment probabilities with assignment probabilities, and, when useful, pairwise assignment odds using the odds identity.

Useful checks include verifying that all weights are positive, every normalized probability lies between zero and one, and the unrounded probabilities sum to one. Repeating the calculation from the odds identity for a selected arm pair is an independent arithmetic check. A discrepancy can signal a reversed allocation-ratio correction, a wrong current share, or premature rounding. If targets are extremely small, probabilities may also be extremely small; report adequate precision and do not round a positive probability to zero before drawing assignments.

What This Comparison Does—and Does Not—Show

The main alternative comparisons answer different questions. Holding the target map fixed while changing the coin rule investigates the contribution of the randomization mechanism; holding the coin rule fixed, as here, investigates the contribution of target-map outputs. A comparison of complete simulated trajectories can describe how designs evolve over many assignments, but it does not isolate the map effect as cleanly unless other inputs and randomization conditions are controlled. Conversely, the calculation here is deliberately local: it compares probabilities at one shared state and does not establish which map yields better estimation, safety, balance, or long-run allocation.

A target map may be motivated by response estimates, scientific objectives, or allocation constraints. The arithmetic comparison does not validate those choices or assess uncertainty in the response estimates. If the maps would ordinarily receive different estimates because earlier assignments produced different data, that is a separate pathway by which their eventual trajectories may diverge. The fixed-state comparison instead asks: given the same information and allocation history, what changes solely because the target map changes?

Common Mistake

A frequent error is to compare the two target vectors and call their difference the change in next-assignment probabilities. For example, in the first worked example, the target share for arm 1 changes from $0.50$ to $0.40$, but the assignment probability changes from about $0.5997$ to $0.3933$, not from $0.50$ to $0.40$. The coin rule also uses current shares, and normalization couples all arms. The correction is to calculate the full weight vector and normalize it separately for each map. Another error is to change the response estimates or allocation state between calculations: that no longer isolates the target-map choice.

Key takeaway: To isolate target-map choice in a DBCD, hold the response estimates, current allocation shares, and $\gamma$ fixed. Apply the same coin rule to each supplied positive target vector, normalize each result, and interpret the output as a conditional next-assignment probability—not as a target share or a long-run guarantee.

Check Your Understanding

Use the DBCD rule and the distinction between target shares and assignment probabilities to answer each question. Solutions follow.

With $\gamma=0$, two maps output different positive target vectors at the same state. What are the assignment vectors, and what is the effect of the current shares?
For two arms, the same map outputs targets $(0.60,0.40)$ under two calculations, but the current shares differ. Can the assignment probabilities differ when $\gamma=1$?
At a fixed state and fixed $\gamma$, what is needed to attribute a difference in assignment probabilities to target-map choice?
Why can an arm with an unchanged target share have a changed assignment probability when another arm's target changes?

Solutions. (1) When $\gamma=0$, the correction factors equal one, so each assignment vector equals its map's target vector; current shares have no effect on the probabilities. (2) Yes. With $\gamma=1$, the weights include division by current shares, so changing the state can change the normalized probabilities even when the target is unchanged. (3) Use the same response estimates, current allocation shares, $\gamma$, and other active design conditions, and change only the target map. (4) Normalization divides every arm's weight by the sum of all weights; changing another arm's target can change that denominator, and therefore change this arm's probability too.