From Pairwise Results to a Count of Successful Comparisons
A multi-arm trial can be designed to answer a question about the number of treatments that succeed, rather than requiring success from every comparison or counting any single success as sufficient. For example, a decision rule might require at least two of three treatment-control comparisons to meet a prespecified criterion. The relevant operating characteristic is then the probability that the number of successful comparisons reaches that threshold.
The success rule must be specified before calculating its power. It includes which comparisons are counted and what qualifies as success for each comparison, such as rejection under a chosen testing procedure. Power is evaluated under a specified configuration of treatment effects; it is not the probability that a treatment is effective or that the null hypothesis is true.
This definition depends on the joint behaviour of the success events. If the events are independent, their probabilities can be combined using an independence calculation. If they are dependent, the joint distribution must be accounted for. In a shared-control trial, for instance, comparisons may have related test statistics because they use the same control group. Dependence does not change the definition; it changes how its probability is calculated.
The success events should also match the planned analysis. If success means a comparison rejects using a multiplicity-adjusted procedure, \(S_i\) must refer to rejection under that procedure, not under an unadjusted test substituted for convenience. The event probabilities used below are hypothetical values assumed to come from the specified procedure under the stated effect configuration.
Calculation When Comparisons Are Independent
Suppose the \(m\) success indicators are independent under the treatment-effect configuration of interest, and write \(p_i=P(S_i)\) for the power of comparison \(i\). The number of successes \(N\) is then a sum of independent Bernoulli random variables, which need not have the same success probability. Its probability-generating function is
To see why, \(z^N=z^{I_1+\cdots+I_m}=\prod_i z^{I_i}\). Independence gives \(E(z^N)=\prod_i E(z^{I_i})\), and \(E(z^{I_i})=(1-p_i)+p_i z\) because \(I_i\) is either 0 or 1. The coefficient of \(z^r\) in \(G_N(z)\) is \(P(N=r)\): each factor contributes either a failure term or a success term, and the product’s \(z^r\) terms correspond to exactly \(r\) successes. Therefore,
where \([z^r]G_N(z)\) means the coefficient of \(z^r\). This gives a direct calculation for small numbers of comparisons and a convenient computational method for larger \(m\). When every comparison has the same power \(p\), the number of successes has a binomial distribution and \(P(N\geq k)=\sum_{r=k}^{m}\binom{m}{r}p^r(1-p)^{m-r}\). Equal marginal powers alone do not imply this binomial model; independence is also required.
Worked Example: At Least Two Successes in Three Independent Comparisons
Suppose a hypothetical trial has three prespecified comparisons. Under a specified treatment-effect configuration, their powers are \(p_1=0.80\), \(p_2=0.70\), and \(p_3=0.60\). Assume that the three success events are independent and that success is defined by the same planned analysis procedure used to obtain these powers. Find the at-least-two power.
There are two ways to meet the rule: exactly two successes or exactly three. The probability of exactly two is the sum of the probabilities for the three possible patterns:
The probability of exactly three successes is
Thus the at-least-two power is \(0.452+0.336=0.788\), or 78.8%. Under the stated effect configuration and independence assumption, the trial has a 78.8% probability of meeting its requirement that at least two comparisons succeed.
Calculation When Success Events Are Dependent
Without independence, the product of marginal powers does not generally give the probability of a pattern of successes. A general expression sums the joint probabilities of every success pattern containing at least \(k\) successes. For a subset \(A\) of \(\{1,\ldots,m\}\), let the event that exactly the comparisons in \(A\) succeed be
The patterns are mutually exclusive and exhaustive. Therefore, at-least-\(k\) power is the sum of the probabilities of those patterns for which \(|A|\geq k\). This is exact if the required joint probabilities are known, though enumeration involves \(2^m\) patterns. For a small number of comparisons, listing the patterns is transparent; for larger problems, a joint model or simulation may be more practical.
Worked Example: At Least Two Successes From a Joint Distribution
Consider three hypothetical comparison success events with the following joint distribution under one specified effect configuration. A pattern such as 101 means that comparisons 1 and 3 succeed and comparison 2 does not. The probabilities are stipulated to be the probabilities under the planned analysis.
| Pattern | Probability | Number of successes |
|---|---|---|
| 000 | 0.20 | 0 |
| 100 | 0.10 | 1 |
| 010 | 0.10 | 1 |
| 001 | 0.10 | 1 |
| 110 | 0.15 | 2 |
| 101 | 0.15 | 2 |
| 011 | 0.10 | 2 |
| 111 | 0.10 | 3 |
The entries are nonnegative and sum to \(0.20+0.10+0.10+0.10+0.15+0.15+0.10+0.10=1.00\), as a joint distribution must. At least two successes occur in patterns 110, 101, 011, and 111. Hence
The at-least-two power is 50.0%. This calculation uses the joint pattern probabilities directly; it does not require independence. The example also illustrates why the independence assumption should not be inserted just because separate marginal powers are available.
Checking Assumptions and Choosing a Calculation
The calculation method is only as credible as the probabilities and assumptions it uses. Before reporting at-least-\(k\) power, check the following:
- Counted comparisons: verify that \(m\) and the comparisons included in \(N\) match the protocol’s success rule.
- Success definition: specify the decision rule for each comparison, including any multiplicity adjustment. Do not substitute a different test’s rejection probability.
- Effect configuration: state the treatment effects under which power is evaluated. Pairwise powers can change across configurations.
- Dependence: use independence only when justified by the design and analysis model. If comparisons share data or otherwise have related outcomes, use their joint distribution or an appropriate joint model.
- Probability calibration: ensure the marginal or joint probabilities are computed under the same analysis rule, allocation, and assumptions as the planned trial.
The probability-generating-function product is an efficient alternative to listing every pattern, but it rests on independence. Exact pattern enumeration is a direct alternative when the joint distribution is available and \(m\) is small. For dependent tests whose joint distribution is not available in closed form, one can simulate trial outcomes under the specified model, apply the full analysis rule to each simulated trial, count the proportion with \(N\geq k\), and report the Monte Carlo uncertainty as well. Simulation does not remove the need to specify a credible joint model.
A basic diagnostic is to check that the calculated probability lies between 0 and 1 and that all probabilities in a stated joint distribution sum to 1. For an independent calculation, check that the \(p_i\) values lie between 0 and 1 and that the polynomial’s coefficients correspond to the possible success counts. A further check is monotonicity: for a fixed joint model, lowering the threshold \(k\) cannot decrease \(P(N\geq k)\). If the calculation violates these checks, revisit the event definitions, arithmetic, or implementation.
Worked Example: At Least Three Successes in Four Independent Comparisons
Suppose a hypothetical four-comparison design has independent success events with powers \(0.90\), \(0.80\), \(0.70\), and \(0.40\) under a stated treatment-effect configuration. Calculate the probability of at least three successes. Independence is assumed here; the values are illustrative, not results from a real trial.
At least three successes means exactly three or exactly four. The probability of four successes is
For exactly three successes, sum the four patterns in which one comparison fails:
Therefore, \(P(N\geq 3)=0.2016+0.4616=0.6632\), or 66.3%. This is the probability, under the assumed independent model and specified effects, that at least three of the four comparisons meet their success criteria.
Common Pitfalls and Practical Limits
A frequent error is to report an individual comparison’s power as though it were the probability of meeting a multi-comparison count rule. Another is to calculate the independent formula from marginal powers even though the success events are dependent. In the joint-distribution example, the at-least-two probability was 0.50. Multiplying marginal probabilities as if independent would answer a different model’s question, not the one specified by that joint distribution.
There is also a distinction between the event counted and the scientific claim. An at-least-\(k\) testing success rule counts comparisons meeting the statistical criterion. It does not, by itself, establish that exactly \(k\) treatments have clinically meaningful effects, nor does its power state the probability that a treatment is truly effective. Such interpretations require the relevant scientific definitions and the assumed effect configuration.
The approach is flexible, but it does not select \(k\), define a clinically meaningful effect, or determine a valid multiplicity procedure. Those are design choices that must precede the power calculation. Its main value is to make the operating target explicit: the probability that the chosen analysis produces at least the required number of successes.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- In a design with five comparisons, what does \(N\) count, and what event defines at-least-three power?
- Three independent comparisons have powers 0.50, 0.60, and 0.70. Find the probability of at least one success.
- Why can’t marginal powers alone determine at-least-two power when the comparison success events are dependent?
- A proposed joint distribution for all success patterns has probabilities summing to 1.08. What must be checked before using it to calculate power?
Solutions. (1) \(N\) counts how many of the five prespecified comparisons meet their success criteria; at-least-three power is \(P(N\geq 3)\) under the specified effect configuration. (2) By independence, \(P(N\geq1)=1-P(N=0)=1-(0.50)(0.40)(0.30)=0.94\), or 94.0%. (3) Marginal powers give individual event probabilities but not the probabilities of joint success patterns; those depend on the joint distribution. (4) The pattern probabilities must form a valid joint distribution, so they must be nonnegative and sum to 1. The stated total of 1.08 signals an error or an incomplete specification that must be resolved first.