Following the Urn One Patient at a Time
In a play-the-winner design, the next treatment assignment depends on the urn’s current contents. A patient’s assigned treatment determines which outcome is observed; that outcome then changes the urn, which determines the probabilities for the next patient. To understand a realized assignment sequence, keep these three items separate: the treatment drawn, the response to that treatment, and the urn update made after the response.
This tutorial uses the two-arm, one-ball reinforcement rule: the drawn ball is returned, and then exactly one additional ball is added. A success adds a ball for the treatment just assigned; a failure adds a ball for the other treatment. The procedure is an allocation rule, not a claim that one treatment is more effective.
The rule can be written explicitly. If the assigned treatment is A, success changes the state from \((a_t,b_t)\) to \((a_t+1,b_t)\), whereas failure changes it to \((a_t,b_t+1)\). If the assigned treatment is B, success changes the state to \((a_t,b_t+1)\), whereas failure changes it to \((a_t+1,b_t)\). The assigned ball itself is returned, so it is not subtracted from the starting composition.
For a sequence of patients, the state after one response is the state before the next patient. Thus every update increases the total number of balls by one, and the assignment probabilities for the next patient are calculated from that updated total. This simple check catches many arithmetic errors.
Why the Assignment Probabilities Have This Form
Suppose the current urn contains \(a_t+b_t\) balls and each ball is equally likely to be drawn. Exactly \(a_t\) of those balls correspond to A. The event of assigning A is the event that one of those \(a_t\) balls is drawn, so its conditional probability is \(a_t/(a_t+b_t)\). The same reasoning gives the probability for B. Because the two treatments account for all balls, the probabilities add to one.
This calculation is conditional on the current urn state. It describes the randomization for the next patient, not the probability of a success, the probability that a treatment is superior, or the probability of a whole multi-patient sequence.
Worked Sequences
Worked Example: Four Patients Starting From Equal Counts
Consider a hypothetical trial with initial urn state \((3,3)\). The specified sequence is: patient 1 receives A and succeeds; patient 2 receives B and fails; patient 3 receives A and fails; patient 4 receives B and succeeds. For each patient, calculate the assignment probabilities before the draw, then update the urn using that patient’s assigned treatment and response.
Patient 1. The urn begins with 3 A-balls and 3 B-balls, for 6 total. Therefore \(P(A)=3/6=0.5000\) and \(P(B)=3/6=0.5000\). The specified draw assigns A. A success on A adds one A-ball, so the state becomes \((4,3)\), with 7 balls.
Patient 2. Before this assignment, \(P(A)=4/7\approx0.5714\) and \(P(B)=3/7\approx0.4286\). The specified draw assigns B. B fails, so one ball for the other treatment, A, is added. The state becomes \((5,3)\), with 8 balls.
Patient 3. The probabilities are now \(P(A)=5/8=0.6250\) and \(P(B)=3/8=0.3750\). The specified draw assigns A. A fails, so a B-ball is added; the new state is \((5,4)\), with 9 balls.
Patient 4. The assignment probabilities are \(P(A)=5/9\approx0.5556\) and \(P(B)=4/9\approx0.4444\). The specified draw assigns B. B succeeds, so a B-ball is added and the state becomes \((5,5)\), with 10 balls. These are the probabilities for the next assignment, if another patient is enrolled.
The realized assignments need not match the more probable treatment at every step: patient 2 receives B even though A has the larger probability. A probability describes the random mechanism before a draw; it does not dictate the outcome of that draw.
Worked Example: Starting With More B-Balls
Suppose a hypothetical trial starts with state \((2,4)\), and the specified sequence is B-success, A-failure, B-failure. The aim is to calculate the next assignment probabilities after each observed response.
First assignment. There are 6 balls, so \(P(A)=2/6=1/3\approx0.3333\) and \(P(B)=4/6=2/3\approx0.6667\). B is assigned and succeeds. Add one B-ball: the state is now \((2,5)\), total 7.
Second assignment. From \((2,5)\), \(P(A)=2/7\approx0.2857\) and \(P(B)=5/7\approx0.7143\). A is assigned and fails. Failure on A adds a B-ball, giving state \((2,6)\), total 8.
Third assignment. From \((2,6)\), \(P(A)=2/8=0.2500\) and \(P(B)=6/8=0.7500\). B is assigned and fails. Failure on B adds an A-ball, so the new state is \((3,6)\), total 9. The probabilities for a further assignment would be \(P(A)=3/9=0.3333\) and \(P(B)=6/9=0.6667\).
The example shows that an allocation probability can move in either direction over a sequence. It depends jointly on the starting composition and the observed outcomes, not only on the most recent assigned treatment.
Worked Example: Auditing a Recorded Sequence
A hypothetical sequence starts at \((1,3)\): B is assigned and fails; then A is assigned and succeeds; then A is assigned and fails. Reconstruct the state before each assignment and determine the probabilities for one more assignment.
Initially, the total is \(1+3=4\), so \(P(A)=1/4=0.2500\) and \(P(B)=3/4=0.7500\). B fails, which adds an A-ball. The state before the second patient is \((2,3)\), total 5.
Before the second patient, \(P(A)=2/5=0.4000\) and \(P(B)=3/5=0.6000\). A succeeds, adding an A-ball. The state before the third patient is \((3,3)\), total 6.
Before the third patient, both probabilities are \(3/6=0.5000\). A fails, adding a B-ball. The resulting state is \((3,4)\), total 7, so the next assignment probabilities are \(P(A)=3/7\approx0.4286\) and \(P(B)=4/7\approx0.5714\).
An audit can check both the arm counts and the total: three patients have produced three added balls, so the total increased from 4 to 7. If a reconstruction instead ends with 6 or 8 balls, at least one update has been omitted or counted twice.
Assumptions, Checks, and Common Pitfalls
The sequence calculations rely on a small set of operational assumptions. The initial counts are known and contain at least one ball. Each draw is uniform over the balls currently in the urn. The drawn ball is returned before the response-based ball is added. The response for the current patient is available and classified before the next assignment is made. Finally, exactly one additional ball is added under the stated success/failure rule. If the implementation uses a different update, these calculations no longer describe its assignment probabilities.
A practical audit can proceed in order: record the state before each draw; calculate both probabilities from that state; record the treatment actually drawn; record the response; apply one update; and verify that the new total is one greater. A probability check is also useful: both probabilities must be between 0 and 1 and sum to 1. A zero count for one treatment gives that treatment probability zero; if both counts were zero, the next draw would be undefined, which is why a nonempty initial urn is required.
- Confusing assignment probability with certainty. If \(P(A)=0.70\), A is more likely on that draw, but B can still be assigned with probability 0.30.
- Subtracting the drawn ball. The drawn ball is returned. Subtracting it before calculating the next state contradicts the replacement rule.
- Updating for the treatment rather than the response. The assigned arm alone does not determine the reinforcement; success and failure lead to different added balls.
- Forgetting the timing. The probability for patient \(t\) is calculated from the state before patient \(t\)'s draw. The response to patient \(t\) changes the probabilities for patient \(t+1\), not retrospectively for patient \(t\).
A play-the-winner sequence is adaptive, unlike a fixed equal-probability allocation in which each patient’s treatment probability is set in advance and does not change in response to outcomes. It also differs from permuted-block randomization, whose design feature is constrained balance within blocks. An urn sequence can temporarily or persistently favor one arm; it does not guarantee equal final group sizes. Its probabilities are determined by the urn state, so the displayed sequence alone should not be interpreted as evidence that a treatment works better. The rule is also sensitive to its starting composition and to the accuracy and timing of response classification.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- A trial starts at \((4,2)\). What are the assignment probabilities for the first patient?
- From state \((4,2)\), A is assigned and fails. What is the next urn state and the next assignment probability for A?
- Starting from \((2,3)\), B is assigned and succeeds, then A is assigned and fails. Find the state and assignment probabilities after these two patients.
- After five patients, an audit reports that the urn total rose from 6 to 10. Is this consistent with the one-ball update rule? Explain.
Solutions. 1. The total is \(4+2=6\), so \(P(A)=4/6=0.6667\) and \(P(B)=2/6=0.3333\). 2. A failure adds a B-ball, giving state \((4,3)\). The next probability for A is \(4/7\approx0.5714\). 3. From \((2,3)\), B succeeds, so the state becomes \((2,4)\). A then fails, adding a B-ball; the final state is \((2,5)\), total 7, and the next probabilities are \(P(A)=2/7\approx0.2857\) and \(P(B)=5/7\approx0.7143\). 4. No. Five updates, each adding one ball, must increase the total by five. Starting at 6, the total should be 11, not 10.