Why Equal Scores Need an Explicit Rule
Minimization compares the imbalance score that would result from assigning an incoming participant to each treatment arm. Sometimes the candidate scores are equal. Neither arm then has a lower score under the chosen criterion, so an additional rule is needed to select an assignment. That rule is not a minor software detail: it determines the assignment probability at a tie and can change the history on which later decisions depend.
This tutorial focuses on three common choices: random allocation at a tie, alternating between arms across ties, and always assigning a prespecified arm at a tie. These rules resolve only equal-score cases. When one arm has a lower score, the minimization procedure’s usual assignment rule still applies.
To make the calculations concrete, examples below use a common unweighted score: after each candidate assignment, calculate the absolute difference between the arm counts at every level of each covariate, then sum those absolute differences. The tie-breaking logic applies to other prespecified scores as well. The examples assume that all relevant counts are up to date, that the same score definition is applied to both candidate assignments, and that the incoming participant’s covariate values are known before assignment.
Three Rules for a Tie
With random tie-breaking, assign the participant to \(A\) or \(B\) using a fair random choice, so that each arm has probability \(1/2\) conditional on the tied history. The random choice should be generated and recorded as part of the allocation procedure. This rule leaves both arms possible at the tie, though it does not make other, non-tied assignments random.
With alternating tie-breaking, assign successive tied participants to opposite arms. The protocol must specify the initial arm—for example, the first tie goes to \(A\)—and clarify that alternation is over tie events, not every enrolled participant. If an intervening participant is assigned when scores are unequal, that assignment does not ordinarily reset the tie sequence.
With a fixed-arm tie rule, assign every tied participant to a prespecified arm, such as \(A\). This rule is simple to implement, but the arm is fully determined whenever a tie occurs. “Fixed arm” refers only to ties; it does not imply that all participants receive that arm.
| Rule | Assignment at a tie | What must be specified |
|---|---|---|
| Random | Fair choice between \(A\) and \(B\) | Randomization method and record of the choice |
| Alternating | Opposite to the arm assigned at the previous tie | Initial arm and what counts as a tie event |
| Fixed arm | Always the prespecified arm | Which arm is fixed |
These descriptions assume exact equality of the computed scores. With weighted or rounded scores, the implementation must state whether a tie means exact equality or equality within a specified tolerance. Otherwise two sites, or two versions of a program, could classify the same candidate scores differently.
Worked Example: Calculating a Tie From Two Covariates
Consider a hypothetical trial with arms \(A\) and \(B\), and two binary covariates, \(X\) and \(Z\). Before a new participant is assigned, the arm counts are equal at every level: for \(X=1\), both arms have 5 participants; for \(X=0\), both have 4; for \(Z=0\), both have 6; and for \(Z=1\), both have 3. The incoming participant has \(X=1\) and \(Z=0\).
If assigned to \(A\), the absolute count difference becomes 1 at \(X=1\), 0 at \(X=0\), 1 at \(Z=0\), and 0 at \(Z=1\). Thus \(S_A=1+0+1+0=2\).
If assigned to \(B\), the difference is also 1 at \(X=1\) and \(Z=0\), and 0 at the other two levels. Thus \(S_B=1+0+1+0=2\). Since \(S_A=S_B\), this is a tie. The score alone does not select an arm; the specified tie rule does.
How a Tie Choice Changes Later Decisions
Equal candidate scores mean that either assignment has the same value under the current score. They do not mean that the two assignments create identical covariate-count configurations. Those configurations become the starting point for the next participant’s comparison. The distinction is important: a tie rule can leave the current score unchanged across the two choices while changing which arm is favored for someone who arrives next.
Worked Example: Tracing Alternating, Fixed, and Random Ties
Suppose all incoming participants have the same covariate profile and the score is the absolute difference in the arms’ counts for that profile. At the start, there are 10 participants in each arm for this profile. Four participants arrive in sequence. If the arms are currently tied in count, either assignment creates a difference of 1 and the scores tie. If the counts differ by 1, assigning to the smaller arm creates a difference of 0, while assigning to the larger creates a difference of 2.
Alternating rule. Let the first tie go to \(A\). Participant 1 is assigned to \(A\), giving counts 11 and 10. Participant 2 is assigned to \(B\), because that produces difference 0 rather than 2. Counts return to 11 and 11. At the second tie, alternation assigns participant 3 to \(B\). Counts become 11 and 12, so participant 4 goes to \(A\) to reduce the difference to 0. The sequence is \(A,B,B,A\), with final counts 12 and 12.
Fixed-arm rule. If every tie goes to \(A\), participants 1 and 3 receive \(A\), while participants 2 and 4 receive \(B\) to minimize the non-tied scores. The sequence is \(A,B,A,B\), also ending at 12 and 12.
Random rule. Each of the two ties is resolved by a fair random choice. For example, if the tie choices are \(A\) and then \(B\), the resulting sequence is \(A,B,B,A\). If the tie choices are \(A\) and then \(A\), it is \(A,B,A,B\). Each of the four possible ordered pairs of tie choices has probability \(1/4\); in this simple setting, each produces two assignments to each arm overall.
The final arm counts happen to be balanced under all three rules in this example. The order of assignments is not the same, however, and neither are the probabilities of individual tie assignments. Balanced totals in this particular sequence do not establish that the rules are interchangeable in other covariate configurations.
Worked Example: Equal Total Scores, Different Next Assignment
Suppose the current marginal counts are balanced at every level of binary covariates \(X\) and \(Z\). An incoming participant has \(X=1,Z=0\), and the score sums the absolute differences over all levels of both covariates. Assigning this participant to either arm gives score 2: one count difference of 1 for \(X=1\), and one of 1 for \(Z=0\).
If the tie rule assigns this participant to \(A\), the differences become \(+1\) at \(X=1\) and \(+1\) at \(Z=0\), where a positive sign means more participants in \(A\). All other level differences remain 0. Now suppose the next participant has \(X=1,Z=1\). Assigning this person to \(A\) gives absolute differences \(2\) at \(X=1\), \(1\) at \(Z=0\), and \(1\) at \(Z=1\), for a total score of 4. Assigning to \(B\) gives differences \(0\) at \(X=1\), \(1\) at \(Z=0\), and \(1\) at \(Z=1\), for a total of 2. The lower score therefore favors \(B\).
If the first tied participant had instead gone to \(B\), the signs at \(X=1\) and \(Z=0\) would both be negative. For the next participant with \(X=1,Z=1\), assigning to \(A\) would give a total score of 2, while assigning to \(B\) would give 4. The next assignment would favor \(A\). Thus, the tied choices had the same immediate total score, but they led to opposite preferred assignments for the next profile.
A Short Justification of the Path Dependence
The previous examples illustrate a general feature of sequential allocation: an assignment changes the state on which the next decision is based. In the one-profile example, let \(d=n_A-n_B\) be the current count difference. The imbalance score is \(|d|\). At a tie, \(d=0\), and either assignment produces a score of 1:
The two resulting states are \(d=1\) and \(d=-1\), not the same state. For the next participant, assigning to the arm that reduces the difference produces score 0, while assigning to the other produces score 2:
The score comparison is symmetric, but the favored arm reverses with the sign of \(d\). This is why a tie rule can affect the sequence even when either tied assignment is equally good by the current scalar score.
Assignment Probabilities, Diagnostics, and Limitations
At a tied history, the conditional probability of assignment to \(A\) is \(1/2\) under a fair random tie rule. Under a fixed-\(A\) rule it is 1, and under a fixed-\(B\) rule it is 0. Under alternation it is 1 or 0 according to the previously specified tie history and initial arm. These probabilities describe the tie decision conditional on the observed allocation history; they do not describe the probability of all possible trial assignments.
The practical distinction is consequential. Random tie-breaking preserves a random choice at ties, but does not by itself guarantee allocation concealment: if the rest of the procedure is predictable, future assignments may still be anticipated. Alternating and fixed-arm rules are deterministic conditional on the history. If the history and rule are known, the tied assignment is predictable. The relative importance of that predictability depends on how allocation is implemented and whether recruiters can learn the upcoming assignment before a participant is irrevocably enrolled.
Useful implementation checks include:
- Recalculate both candidate scores for recorded tie cases and confirm that they are equal under the documented score definition.
- For random ties, verify that the random choice is generated only when a tie occurs and that the chosen arm is recorded.
- For alternation, verify the initial arm and confirm that the sequence changes only at tie events.
- For a fixed-arm rule, check that the same prespecified arm is used at every tie.
- Check how software handles rounded or near-equal scores; inconsistent tolerances can silently change the rule.
A common mistake is to say, “The two candidate scores are equal, so it does not matter which arm is assigned.” The immediate score is equal, but the resulting count configurations can differ, and later preferred assignments can change, as the worked examples show. Another mistake is to call alternation random because the arm varies over time. Alternation is deterministic once the starting arm and tie history are known; variation is not the same as randomization.
No single tie rule is best in every setting. A fair random tie avoids always favoring the same arm at equal scores, but requires reliable random-number generation and documentation. Alternation can distribute tied assignments across arms over successive ties, but is predictable and depends on an initial convention. A fixed-arm rule is operationally simple but systematically favors its chosen arm at tie events. Whichever rule is used, it should be prespecified and included in the description of the allocation procedure. Analyses that rely on the randomization mechanism must reflect the actual rule, including the probabilities used at ties.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- At a tied history, what is the conditional probability of assignment to arm \(A\) under a fair random tie rule and under a fixed-\(A\) rule?
- Under an alternating rule, the first tied participant is assigned to \(B\). A later participant is assigned when scores are unequal. Which arm should receive the next participant who encounters a tie?
- In the one-profile example, current counts are equal. What are the two possible imbalance scores immediately after assigning the next participant?
- Why is it incorrect to conclude that a tie choice cannot affect subsequent allocations merely because the candidate scores are equal?
Solutions. (1) The probabilities are \(1/2\) and 1, respectively. (2) If alternation is over tie events, the next tied participant goes to \(A\); the intervening non-tied assignment does not change the tie sequence. (3) Each candidate assignment creates a count difference of 1, so both scores are 1. (4) The assignments can create different covariate-count configurations even when their current scalar scores match; the next participant’s candidate scores are calculated from that changed state and may favor opposite arms.