Allocating sample across control and treatment

In a randomised control trial, should your treatment arm(s) and control be allocated the same number of observations? What are the efficiency gains?

Summary

1. In a simple T vs C comparison, the answer is almost always yes. Various other considerations do not have large efficiency gains.
2. If you have an RCT with K treatment arms, in most cases you should consider scaling up the control arm by sqrt(K). If you care about comparing treatments to each other, do not inflate the control arm.
3. If testing many highly similar treatments, you could increase size of control even further AND consider borrowing strength across treatment arms for large efficiency gains.
4. Gains can be large for large $K$. For K = 5 you reduce sample size by ~1/9 by sizing up control arms. For K = 5 and highly correlated treatments you can reduce sample size by even 1/3 by borrowing information.

Single treatment vs control

In a randomised controlled trial with two arms, T and C, variance for difference of means is

\[Var(\delta) = \sigma_T^2/n_T + \sigma_C^2/n_C\]

where $\sigma^2$ is variance in each group and $n$ is sample size. In inferring treatment effects it is optimal to set $n_T = n_C$, a 1:1 randomisation, if (1) we assume that treatment does not change variance, and (2) cost per control or treatment subject is the same.

If either (1) or (2) is violated, it is very easy to solve: put relatively more sample where outcomes are noisier or where subjects are cheaper. (1) usually has little practical impact.¹

In some cases it is practical to set the treatment arm larger because we learn additional information. In medicine, for example, we use the treatment arm to learn about both efficacy and safety, so a treated patient is more valuable than a control patient.

Multi-arm treatment

If we have $k$ treatment arms, it can be optimal to set sample sizes so that

\[n_C = \sqrt{k} \cdot n_T.\]

This follows from the same total variance formula. Each additional control observation is used in every treatment-versus-control comparison, so control observations have higher value. Setting $\sqrt{k}$ scaling minimises total variance across $k$ comparisons.

However, there are other optimal allocations for a multi-arm trial. They depend on what the goal is.

• If the goal is to also estimate an average treatment effect across all $k$ arms compared to control, you would set $n_C = k \cdot n_T$, i.e. set $n_C$ to the same size as all treatment arms together.
• This can be improved on further if you assume that different treatment arms are similar but not the same. See below.
• If the main value of the trial is comparison between treatment arms, or ranking treatments against each other, one may instead want more sample in treatment arms.

How large are the gains?

If the goal is maximising power in $k$ individual comparisons against a shared control, the implied total sample savings add up as $k$ increases.

$k$	Optimal control:treatment ratio	Total sample saving
1	1:1	0.0%
2	1.4:1	2.9%
3	1.7:1	6.7%
4	2:1	10.0%
5	2.2:1	12.7%

To give this a visual flavour: for $k = 5$, if equal ($n_T = n_C$) allocation gives about 80% power for some treatment effect, the optimal allocation raises that to about 85% at the same total sample size. (This plot was adjusted to a case where MDE is targetting 0.1 SD increase, but the gains from optimal allocation would scale accordingly. That is, the x axis is not relevant here.)

Borrowing of information across treatment arms

The efficiency gains can be much larger if the active treatments are closely related and we account for that in the model. Simple examples include testing different doses of the same drug or different delivery methods for a behavioural intervention. If we see an intervention work at low intensity of delivery, we should be highly confident (but not certain) it will work at high intensity.

Even without modifying allocation, the gains from this procedure can be very large when correlations are high. Continuing with the $k=5$ example and $\sqrt{k}$:1 allocation, at 90% correlation the required sample would be about 30% smaller if using a partially pooled model—if the goal is estimating significant arm-specific effects.

In these cases it is also optimal to further increase size of control arm. As noted, if $k$ treatment arms are pooled into a single analysis, then it is optimal to set $n_C = k \cdot n_T$. Therefore usually the optimal size of $n_C$ will be somewhere between $\sqrt{k}$ times (“no pooling”) to $k$ times (“full pooling”) larger than $n_T$.

Moving to 5:1 allocation in this highly correlated example would reduce sample size by further 7%. Taken together, that is about a one-third saving relative to no pooling under the classical $\sqrt{k}$ allocation.

To be clear: these gains are purely illustrative and in practice we do not know how closely correlated treatments will be. Therefore this choice should be used cautiously both in design and analysis of experiments.

Practical implications

There are four practical conclusions.

• J-PAL already has a short, accessible guide to power calculations. People should make use of it, or revise it slightly to incorporate the points on allocation and stratification (J-PAL 2025).
• In a multi-arm trial with $k$ separate comparisons against a common control, control arm should be scaled up by $\sqrt{k}$. This can shrink samples by 10% if you have 4 treatment arms.
• The solution is different when concerned with average treatment effect across all arms (larger control arm) or comparisons across arms (larger treatment arms).
• If treatment arms are closely related, partial pooling should be done as an additional step. It offers larger efficiency gains than the optimal sample allocation alone.

References

Kondylis and Loesler, 2021. You’re probably doing it right: Experimental design with heterogeneous treatment effects. Development Impact Blog. URL J-PAL. 2025. Power Calculations. Abdul Latif Jameel Poverty Action Lab. URL

1. The ratio $\sigma_T / \sigma_C$ matters only when it is far from 1. Even in the extreme case where $\sigma_T / \sigma_C = 2$, optimal allocation shrinks sample size by about 10%. If $\sigma_T / \sigma_C = 1.25$ the gain is about 1%. Generally 1:1 split is a good choice (see Kondylis and Loesler 2021 for more thoughts). ↩