From an Outcome to a State Change
In a two-arm randomized play-the-winner design, the urn’s composition records the current allocation mechanism. A patient is assigned by drawing a treatment ball, and the ball is returned before the urn is updated. The outcome then determines which treatment’s count increases. Writing this operation as a state transition makes the rule auditable: it shows exactly how a success or failure changes the composition, without confusing the treatment assigned with the treatment that receives reinforcement.
The earlier material on the two-arm randomized play-the-winner rule introduced the outcome-dependent reinforcement idea. Here we represent that rule precisely as changes in counts. Unless stated otherwise, assume two treatments, A and B; a binary outcome classified as success or failure for each assigned patient; one treatment ball drawn uniformly from the urn for each assignment; return of the drawn ball; and exactly one additional ball placed in the urn after the outcome is known. A success reinforces the treatment the patient received. A failure reinforces the other treatment.
The state is a pair of nonnegative integer counts, not a record of the entire assignment history. The same state can be reached through different histories. For the standard one-ball rule, each transition adds one ball to exactly one treatment count. No existing ball is removed: returning the drawn ball preserves the pre-draw composition, and the reinforcement adds one ball.
The Four One-Patient Transitions
There are four cases because either treatment can be assigned and either outcome can follow. On success, add a ball matching the assigned treatment. On failure, add a ball for the alternative treatment. Thus the treatment assigned determines which outcome counts as “the other treatment” in a failure update.
| Treatment assigned | Outcome | Ball added | Transition from \((a,b)\) |
|---|---|---|---|
| A | Success | A | \((a,b)\to(a+1,b)\) |
| A | Failure | B | \((a,b)\to(a,b+1)\) |
| B | Success | B | \((a,b)\to(a,b+1)\) |
| B | Failure | A | \((a,b)\to(a+1,b)\) |
Two rows have the same count change: an A success and a B failure each add an A ball; an A failure and a B success each add a B ball. The reason differs, so a transition record should retain both the assignment and the outcome rather than recording only which ball was added.
Worked Example: A Success on Treatment A
Suppose, hypothetically, that the urn is in state \(U_{n-1}=(4,3)\), a patient is assigned to A, and the patient has a success. The drawn A ball is returned. Under the success rule, one additional A ball is placed in the urn.
The transition is
The total number of balls changes from \(4+3=7\) to \(5+3=8\). The A count increases by one; the B count does not change. The patient’s success does not change the assignment already made. It only determines the subsequent update.
Worked Example: A Failure on Treatment B
Consider a hypothetical state \(U_{n-1}=(2,5)\). A patient is assigned to B and has a failure. Failure reinforces the alternative treatment, so the added ball is A, not B.
Therefore,
There were seven balls before the update and eight afterward. The B count stays at five even though B was the assigned treatment. This is the case most easily reversed when an update is written from memory: a failure does not always add a ball of the treatment assigned.
Indicator Form and a Short Derivation
A compact formula is useful for checking a sequence or translating the rule into a program. Let \(I_n=1\) if patient \(n\) is assigned to A and \(I_n=0\) if assigned to B. Let \(Y_n=1\) for success and \(Y_n=0\) for failure. The A count increases when A is assigned and succeeds, or when B is assigned and fails. These two cases are disjoint, so their indicators add:
Likewise, the B count increases when A is assigned and fails, or when B is assigned and succeeds:
These expressions reproduce all four transition rows. For example, for assignment to B with failure, \(I_n=0\) and \(Y_n=0\), so the A increment is \(0+1=1\) and the B increment is \(0+0=0\). For assignment to A with success, \(I_n=1\) and \(Y_n=1\), giving an A increment of one and a B increment of zero.
Adding the two increments gives
The equality follows by grouping the terms with \(I_n\) and \(1-I_n\): the first group is \(I_n[Y_n+(1-Y_n)]=I_n\), and the second is \((1-I_n)[(1-Y_n)+Y_n]=1-I_n\). Their sum is one. Consequently, after \(n\) completed updates,
This is a direct check on the update bookkeeping. It relies on exactly one ball being added per completed patient and none being removed. A rule that adds a different number of balls would have a different total-count relation.
Worked Example: Following Several State Transitions
Suppose a hypothetical trial starts a sequence of four completed patients at state \((3,2)\). Their assignments and binary outcomes are: A success, B success, A failure, and B failure. Apply each transition in order.
For the first patient, A success adds A: \((3,2)\to(4,2)\). For the second, B success adds B: \((4,2)\to(4,3)\). For the third, A failure adds B: \((4,3)\to(4,4)\). For the fourth, B failure adds A: \((4,4)\to(5,4)\).
The full sequence is
The initial total is five, and the final total is \(5+4=9\), as required after four updates. The sequence also makes clear that changes depend on both assignment and outcome: the third and fourth patients fail, but their failures reinforce different treatments because they were assigned to different arms.
Checks, Diagnostics, and Common Mistakes
A transition sequence can be checked locally, one patient at a time, and globally by checking the final counts. For every completed patient, verify that the assigned arm and outcome identify exactly one of the four cases; that exactly one count increases by one; and that neither count decreases. Then verify that the final total equals the initial total plus the number of updates. If any check fails, the written sequence does not implement the stated one-ball rule.
- Failure treated as “add to the assigned arm.” That reverses the failure rule. First name the assigned treatment; then add the opposite treatment’s ball.
- Forgetting that the drawn ball is returned. A draw is not a removal. In the stipulated rule, the net change is one added ball, not a subtraction from the assigned arm followed by an addition.
- Updating before the outcome is known. Assignment selects the treatment for the current patient. The outcome determines the reinforcement after that assignment; it does not retroactively alter it.
- Using an unclassified or missing outcome as a failure. The update rule requires a defined success/failure classification. Missing or unresolved outcome information cannot be silently coded as failure without changing the procedure.
- Passing the total-count check but mislabeling treatment counts. A sequence can add one ball each time and still put it in the wrong arm. Check each transition against the assignment-and-outcome table as well as checking the total.
A practical diagnostic is to write a temporary audit row for each patient: prior state, assignment, outcome, ball added, and new state. If the recorded new state cannot be obtained from the prior state by adding the indicated ball, the update is inconsistent. This audit is especially useful when treatment labels in a data file are coded numerically or when “success” is defined using an endpoint with a specific direction.
How This Update Differs From Alternatives
The transition notation describes the rule’s composition update, not every possible way of assigning participants. The key distinction is whether assignments remain randomized and whether outcomes change the urn. The following comparison keeps those features separate.
| Procedure | What happens to the state after a patient? | Role of the outcome |
|---|---|---|
| Two-arm randomized play-the-winner, one-ball rule | One ball is added to exactly one treatment count | Success reinforces the assigned arm; failure reinforces the other |
| Random allocation without reinforcement | The urn composition stays unchanged | Outcome does not alter future urn counts |
| Deterministic play-the-winner | Assignment follows a success/failure rule rather than a random urn draw | Outcome can directly determine the next assignment |
In randomized play-the-winner, the next allocation is still made by a random draw from the current urn. The transition changes the composition that governs later draws; it does not make the next assignment deterministic. By contrast, ordinary random allocation without reinforcement has no outcome-driven state transition. These distinctions matter when describing or reproducing a design: a state update rule alone should not be mistaken for the assignment rule, and a deterministic response rule should not be called randomized simply because it responds to outcomes.
The one-ball formulas assume a binary outcome and one reinforcement ball. If the endpoint has more categories, if a different number of balls is added, or if an outcome is not yet available, the transition definition must be adjusted explicitly. The four rows above should not be applied unchanged to such a design. For the stated rule, however, the transition table and invariant fully specify how each completed patient changes the two treatment counts.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- From state \((6,2)\), a patient assigned to A has a failure. What is the new state, and what is the total number of balls afterward?
- From state \((1,4)\), a patient assigned to B has a success. Which count increases, and what is the new state?
- A sequence begins with five balls and includes three completed updates. What must the final total be under the one-ball rule?
- A student says, “The patient was assigned to B and failed, so I remove one B ball and add one A ball.” Identify the error and give the correct update for a prior state \((3,5)\).
Solutions. (1) A failure adds B, so \((6,2)\to(6,3)\); the total is \(9\). (2) B success adds B, so the B count increases and \((1,4)\to(1,5)\). (3) The final total must be \(5+3=8\). (4) The drawn ball is returned, so no B ball is removed. The failure adds one A ball, giving \((3,5)\to(4,5)\).