Why the Rule Uses Both Treatment and Response
In response-adaptive randomization, information from earlier participants can influence the probabilities used to assign later participants. The classical two-arm randomized play-the-winner rule does this with an urn: each ball represents a treatment, and each observed binary outcome determines the colour of one additional ball. The rule changes future assignment probabilities, but it does not guarantee that the treatment with more balls is more effective, nor does it make any assignment deterministic.
The name can be misleading. “Play the winner” does not mean that the next participant must receive whichever treatment most recently produced a success. Instead, a ball is drawn at random from the current urn. A success adds a ball of the same colour as the treatment just given; a failure adds one of the opposite colour. The drawn ball itself is returned in either case.
Definition and Assumptions
The operational rule requires an urn with at least one ball of each colour initially. The draw must be random and uniform over the balls currently in the urn. Each participant must receive the treatment indicated by the draw, and the outcome used by the rule must be recorded before the next participant’s draw. The success criterion must be defined in advance as a binary outcome. A protocol must also specify what happens if that outcome is unavailable; the classical rule alone does not prescribe an update for a missing or ambiguous outcome.
These are assumptions for implementing the allocation procedure. A commonly used additional model for studying its probabilistic behaviour assumes that, conditional on treatment, participant outcomes are independent Bernoulli variables with constant success probabilities \(p_A\) and \(p_B\). Under that model, \(P(\text{success}\mid A)=p_A\) and \(P(\text{success}\mid B)=p_B\) for each participant. The urn rule itself does not require this model: it can be followed using observed binary outcomes without assuming that success probabilities are constant. The model is an extra assumption, not a property guaranteed by randomization.
Let \(a\) and \(b\) denote the current numbers of A-colour and B-colour balls. Since a ball is drawn uniformly, the next assignment probabilities are
This follows because there are \(a+b\) possible balls to draw and exactly \(a\) of them indicate A (and \(b\) indicate B). Thus, the current urn composition sets the allocation probabilities for the next participant. The composition changes only after an outcome has been observed and the extra ball has been added.
The Four Treatment–Outcome Cases
The colour added depends on both the treatment assigned and the outcome. The drawn ball is always returned; the final column describes the one additional ball.
| Assigned treatment | Observed outcome | Ball added | Meaning |
|---|---|---|---|
| A | Success | A-colour | Reinforce A |
| A | Failure | B-colour | Reinforce the alternative |
| B | Success | B-colour | Reinforce B |
| B | Failure | A-colour | Reinforce the alternative |
The table makes clear why the outcome label must be interpreted relative to the assigned treatment. A failure on A does not mean that no ball is added; it means that the opposite-colour ball is added. Likewise, a success on B reinforces B, not A.
Worked Example: Updating After a Success on A
Suppose a hypothetical trial begins with 3 A-colour balls and 2 B-colour balls. A uniform random draw selects an A-colour ball, so the participant is assigned A. The participant has a success. Find the urn composition after the outcome and the assignment probabilities for the next participant.
Before the assignment, the probability of drawing A is \(3/(3+2)=3/5=0.600\), and the probability of drawing B is \(2/5=0.400\). In this hypothetical sequence, an A ball is drawn and returned. Because the outcome is a success on A, a separate A ball is added.
The updated counts are \(a=3+1=4\) and \(b=2\), for a total of \(4+2=6\) balls. The next assignment probabilities are therefore
The success raises the chance of assigning A on the next draw from 0.600 to 0.6667. It does not make A certain: two B-colour balls remain in the urn.
Worked Example: Failure on B and Then Success on A
Consider a hypothetical urn that begins with 2 A-colour balls and 2 B-colour balls. The first draw selects B, and the participant assigned B has a failure. The next draw selects A, and the participant assigned A has a success. Track the counts after both outcomes.
Initially, the chance of assigning either treatment is \(2/(2+2)=0.500\). After the first draw, the B ball is returned. Failure on B calls for an additional A ball, so the counts become \(a=2+1=3\) and \(b=2\). There are now 5 balls, giving next-draw probabilities \(P(\text{assign A})=3/5=0.600\) and \(P(\text{assign B})=2/5=0.400\).
In the stated sequence, an A ball is then drawn and returned. Success on A adds another A ball. The counts become \(a=3+1=4\) and \(b=2\), for a total of 6. The next assignment probabilities are \(4/6=0.6667\) for A and \(2/6=0.3333\) for B.
Both outcomes increased the A count, but for different reasons: failure on B added the opposite colour, while success on A added the same colour. The treatment assigned and the outcome together determine the update.
Why the Ball Count Changes by One
A frequent bookkeeping question is whether a patient’s draw removes a ball permanently. It does not. If there are \(a\) A balls and \(b\) B balls before a draw, returning the selected ball restores those counts before reinforcement. The outcome then adds exactly one ball, so the total number of balls increases by one per completed participant. For example, starting with \(a_0+b_0\) balls, after \(n\) completed updates the total is \(a_0+b_0+n\).
This count check is a useful justification and an implementation diagnostic. If a ledger shows that a draw permanently removed a ball, or that two extra balls were added after one outcome, it does not implement the stated rule. The return and the reinforcement are two distinct actions: put back the ball that was drawn, then add one new ball.
Worked Example: Auditing a Short Hypothetical Ledger
A hypothetical urn starts with 3 A-colour balls and 1 B-colour ball. Four successive draws and outcomes are recorded as follows: A-success, A-failure, B-failure, A-success. Assume the listed ball colours are the actual results of the uniform draws. Audit the updates and find the final composition.
Initially, \(a=3\), \(b=1\), and the total is 4. After A-success, return the A ball and add A: \(a=3+1=4\), \(b=1\), total 5.
After A-failure, return the drawn A ball and add the opposite colour, B: \(a=4\), \(b=1+1=2\), total 6.
After B-failure, return the B ball and add the opposite colour, A: \(a=4+1=5\), \(b=2\), total 7.
After A-success, return the A ball and add A: \(a=5+1=6\), \(b=2\), total 8. The final urn has 6 A-colour balls and 2 B-colour balls. This agrees with the total-count check: the initial 4 balls plus 4 added balls gives 8. The next assignment probabilities would be \(6/8=0.750\) for A and \(2/8=0.250\) for B.
Diagnostics, Comparisons, and Limitations
An implementation can be checked without estimating any treatment effect. After each participant, verify that the drawn ball was returned, exactly one additional ball was added, and its colour matches the assigned-treatment/outcome rule. Recalculate the total as the initial number plus the number of completed updates. At every assignment, the two probabilities should be nonnegative and sum to one; with both colours present, each probability is strictly between zero and one.
Common failure modes include reversing the failure rule, treating a success as a reason to remove a ball, applying an outcome before it is observed, and confusing the probability of drawing a colour with the observed outcome rate on that treatment. These are different quantities: urn counts determine the next assignment probability, while outcomes are recorded for participants already assigned. In the classical rule, both colours remain present if both were initially present, because no ball is removed.
Unlike a fixed allocation ratio, the play-the-winner rule allows allocation probabilities to depend on past responses. Complete randomization and permuted-block randomization use their own allocation mechanisms to control assignment counts; they do not use the success/failure reinforcement described here. The urn rule can therefore adapt allocation, but it does not promise a predetermined balance between treatments. Nor does a higher urn probability by itself establish that a treatment is more effective: it reflects the accumulated history of assignments and outcomes under the rule.
The rule also has limitations. Its behaviour depends on the initial composition and on the definition and timing of the binary outcome. Delayed, missing, or inconsistently coded outcomes require an additional protocol decision before the procedure can be implemented reliably. Finally, adaptation of assignment probabilities is not, by itself, a complete analysis plan: clinical conclusions require an analysis suited to the design and the data actually collected.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- An urn contains 4 A-colour balls and 3 B-colour balls. A participant is assigned B and has a success. What are the new counts and the next assignment probabilities?
- An urn contains 2 A-colour balls and 5 B-colour balls. A participant is assigned A and has a failure. Which colour is added, and what are the new counts?
- Why is it incorrect to say that a success on B adds an A-colour ball because the outcome was “good”?
- An urn begins with 5 balls and 6 participants have completed updates. How many balls should it contain if every update followed the classical rule?
- State one way in which this rule differs from permuted-block randomization, and one limitation of the rule.
Solutions. (1) Return the B ball and add a B ball, so the counts are \(a=4\), \(b=3+1=4\), total 8. The next probabilities are \(4/8=0.500\) for each treatment. (2) Failure on A adds the opposite colour, B. The counts become \(a=2\), \(b=5+1=6\), total 8. (3) A success adds a ball matching the assigned treatment; therefore success on B adds B, regardless of the informal description “good.” (4) Each completed update adds exactly one ball, so the urn contains \(5+6=11\) balls. (5) Permuted-block randomization uses a block-based allocation mechanism rather than reinforcement from observed responses. One limitation of the urn rule is that missing or delayed outcomes need an additional protocol decision before the next update can be made.