Why Unequal Allocation Needs Careful Counting
In a two-treatment trial, equal allocation is not the only option. A design may assign twice as many participants to treatment A as to treatment B, for example. Under complete randomization with unequal allocation, the total number assigned to each treatment is fixed in advance, while the identities of the participants assigned to each treatment are selected at random. The unequal counts change both the assignment probabilities and the number of possible assignments.
The key is to separate three quantities: the treatment counts, the probability that a particular participant receives a treatment, and the number of valid assignments of all participants. The target ratio alone does not give the number of assignments; the total sample size and the rule for choosing assignments are also needed.
This definition assumes that the set of eligible participants is fixed, that each participant can receive exactly one of the two treatments, and that the allocation procedure samples uniformly from all valid assignments. It is a fixed-count design: once the treatment totals are set, one participant’s assignment is not independent of everyone else’s. The earlier material on complete randomization establishes the uniform-sampling idea; here the counts need not be equal.
Counts, Individual Probabilities, and Assignment Totals
An allocation ratio specifies relative counts, not necessarily the counts themselves. A 2:1 ratio means \(n_A=2k\) and \(n_B=k\) for some positive integer \(k\). If \(N\) is divisible by 3, these counts use all \(N\) participants exactly. For a 3:2 ratio, \(n_A=3k\), \(n_B=2k\), and the total must be divisible by 5 for the ratio to be exact.
Once the counts are fixed, each participant has the same marginal probability of receiving treatment A. To see this, consider a particular participant \(i\), and let \(Z_i=1\) if that participant is assigned to A and \(Z_i=0\) otherwise. Exactly \(n_A\) of the \(N\) values \(Z_i\) equal 1 in every valid assignment. Under uniform randomization, no participant is favored, so the chance that \(Z_i=1\) is the fraction of treatment-A slots among all slots.
The number of valid assignments is the number of ways to choose which \(n_A\) participants receive A; the remaining participants receive B. Therefore,
Each valid assignment has probability \(1/\binom{N}{n_A}\). These results depend on uniform selection among the valid assignments. Merely specifying treatment counts does not guarantee uniform randomization.
Worked Example: A 2:1 Allocation With 12 Participants
A hypothetical trial will assign 12 participants in a 2:1 ratio, with twice as many assigned to A as to B. Find the treatment counts, an individual’s assignment probabilities, and the number and probability of the valid assignments.
Counts. Write the counts as \(2k\) and \(k\). Since \(2k+k=12\), \(k=4\), so \(n_A=8\) and \(n_B=4\). The counts have the required ratio and sum to 12.
Individual probabilities. For any particular participant \(i\),
Thus each participant has a \(66.7\%\) probability of assignment to A and a \(33.3\%\) probability of assignment to B. These are assignment probabilities, not probabilities of a treatment response.
Number of assignments. Choose the 8 participants assigned to A; the other 4 must receive B:
There are 495 valid assignments. Since each is equally likely, the probability of any particular valid assignment is \(1/495\approx0.00202\). An assignment with 7 participants on A and 5 on B has probability zero under this fixed-count design because it is not valid.
Why Individual Assignments Are Dependent
Equal marginal probabilities do not imply independent assignments. If one participant is assigned to A, there is one fewer A place available for the others. More generally, if \(r\) distinct participants are specified and \(s\) of them are required to receive A, the probability of that particular pattern is
provided the remaining counts are feasible; otherwise the probability is zero. The numerator counts valid assignments of the \(N-r\) unspecified participants after the specified assignments have been made.
For two distinct participants \(i\) and \(j\), this gives
This is generally not equal to \(P(Z_i=1)P(Z_j=1)=(n_A/N)^2\). The distinction matters whenever a question concerns a group of participants rather than one participant at a time.
Worked Example: Joint Probabilities Under a 2:1 Allocation
In a hypothetical study of 9 participants, complete randomization assigns 6 to A and 3 to B. Find the probability that two specified participants both receive A, and the probability that the first specified participant receives A while the second receives B.
There are \(\binom{9}{6}=84\) valid assignments. For the first event, both specified participants are on A, leaving 4 A places among the 7 remaining participants:
The product of their marginal probabilities would be \((6/9)^2=4/9\approx0.444\), which is different. The assignments are dependent because the total number assigned to A is fixed.
For the second event, participant 1 is assigned to A and participant 2 to B. This leaves 5 A places among the 7 remaining participants:
The same probability follows by first assigning participant 1 to A and then calculating the chance that participant 2 is among the 3 B places: \((6/9)(3/8)=18/72=1/4\). Both calculations reflect the fixed totals.
Derivation of the Marginal Probability
A short counting argument verifies the individual probability formula. Out of \(\binom{N}{n_A}\) valid assignments, the assignments that place a specified participant \(i\) on A choose the other \(n_A-1\) A participants from the remaining \(N-1\). Therefore,
The last equality follows because \(n_A!/(n_A-1)!=n_A\) and \(N!=N(N-1)!\). Replacing A with B yields \(P(Z_i=0)=n_B/N\). The same result also follows by symmetry: in every valid assignment there are \(n_A\) A labels distributed across \(N\) participant positions uniformly.
Worked Example: A 3:2 Allocation and the Number of Assignments
A hypothetical study will use a 3:2 ratio with 20 participants. Calculate the counts, the probabilities for one participant, and the number of valid assignments.
Counts. One ratio unit contains \(3+2=5\) participants. Since \(20/5=4\), there are 4 units: \(n_A=3(4)=12\) and \(n_B=2(4)=8\).
Individual probabilities.
Number of valid assignments. Choose 12 of the 20 participants for A:
Thus there are 125,970 equally likely valid assignments, each with probability \(1/125{,}970\approx0.00000794\). The individual probability of A is \(0.600\), but that does not mean the whole assignment has probability \(0.600\); the two probabilities describe different events.
When the Target Ratio Does Not Fit
An exact ratio may be incompatible with the planned sample size. For example, 14 participants cannot be divided into an exact 2:1 ratio: the ratio requires a total of \(3k\), and 14 is not divisible by 3. A design must then specify integer treatment counts, such as 9 on A and 5 on B, or 10 on A and 4 on B. Those choices give different realized ratios and different individual assignment probabilities.
For \(N=14\), counts of 9 and 5 give \(P(Z_i=1)=9/14\approx0.643\), while counts of 10 and 4 give \(P(Z_i=1)=10/14\approx0.714\). Neither is exactly 2:1. A protocol should state how such rounding is handled rather than leaving the allocation rule ambiguous. Once counts have been specified, the assignment count is still \(\binom{14}{n_A}\), and uniform randomization among those assignments gives each specified participant the same marginal probability.
Express the target as integer counts, such as \(n_A=2k\) and \(n_B=k\) for a 2:1 ratio.
Confirm that the counts are nonnegative integers and add to \(N\). If not, state the integer-count rule being used.
Use \(n_A/N\) and \(n_B/N\) for individual probabilities, and \(\binom{N}{n_A}\) for the number of valid assignments.
Comparison, Diagnostics, and Limitations
With equal allocation and an even sample size, complete randomization fixes \(n_A=n_B=N/2\), so each participant’s probability of A is \(1/2\). Unequal complete randomization uses the same uniform fixed-count principle but changes the counts, marginal probabilities, and number of valid assignments. In contrast, independent assignment with probability \(q\) assigns each participant to A separately, so the total number on A is random rather than fixed. These are distinct designs: using independent coin flips does not implement complete randomization with fixed unequal counts.
| Feature | Unequal complete randomization | Independent assignment |
|---|---|---|
| Treatment totals | Fixed at \(n_A\) and \(n_B\) | Random from assignment to assignment |
| Probability for a participant | \(n_A/N\) for A under uniform fixed counts | The assigned probability \(q\) for A |
| Dependence between assignments | Assignments are dependent when totals are fixed | Assignments are independent by design |
| Number of possible assignments | \(\binom{N}{n_A}\) valid assignments | Potentially \(2^N\) assignment patterns |
Useful checks are straightforward. Confirm that \(n_A+n_B=N\), that the counts match the stated ratio or its explicitly described rounding rule, that the individual probabilities sum to 1, and that the count of assignments uses the correct treatment count. If each of several participants is assigned using the marginal probability \(n_A/N\) as though the assignments were independent, the calculation will generally be wrong for a fixed-count design.
Unequal allocation specifies how participants are distributed, not whether one treatment is better or whether the design is ethically preferable. Its practical consequences for information and patient benefit depend on further design considerations. It also does not by itself prevent chance covariate imbalance or control the order in which treatment labels appear. The probability calculations here apply to the stated fixed-count uniform design; a different assignment mechanism requires its own probability calculation.
Common Mistake
Wrong reasoning: “In a 2:1 design, every participant independently has probability \(2/3\) of receiving A, so I can multiply \(2/3\) for each A assignment and \(1/3\) for each B assignment to get the probability of an allocation.”
Correction: Under unequal complete randomization the counts are fixed and assignments are dependent. For 12 participants with 8 assigned to A, a particular valid assignment has probability \(1/\binom{12}{8}=1/495\), not \((2/3)^8(1/3)^4\). The latter is the probability of that particular pattern under independent assignment with \(q=2/3\), a different design that does not fix the totals.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A hypothetical study has 18 participants and uses a 2:1 allocation. Find \(n_A\), \(n_B\), and each participant’s probability of receiving A.
- For the study in question 1, how many valid assignments are there?
- With 10 participants and fixed counts \(n_A=6\), \(n_B=4\), find the probability that two specified participants both receive A.
- Can 17 participants be divided in an exact 3:2 ratio? If not, what must be specified before assignment probabilities can be calculated?
- Explain why multiplying the individual marginal probabilities is generally not a valid way to calculate the probability of a full assignment under fixed-count complete randomization.
Solutions. (1) A 2:1 ratio has three total parts, so \(k=18/3=6\). Thus \(n_A=12\), \(n_B=6\), and \(P(Z_i=1)=12/18=2/3\approx0.667\). (2) There are \(\binom{18}{12}=\binom{18}{6}=18{,}564\) valid assignments. (3) The probability is \(6/10\cdot5/9=30/90=1/3\approx0.333\); equivalently, \(\binom{8}{4}/\binom{10}{6}=70/210=1/3\). (4) No. An exact 3:2 ratio requires a total divisible by 5, and 17 is not. Integer treatment counts and the rule for choosing them, such as 10 on A and 7 on B, must be specified. (5) Under fixed counts, assigning one participant to A changes the remaining available A places. The assignments are therefore dependent, so a full-assignment probability must be calculated from the uniform set of valid assignments, not by multiplying marginal probabilities as if they were independent.