Reading a Transition Table
An urn transition table summarizes how an urn’s composition may change after an assignment and response. Reading it correctly requires keeping two probabilities separate: the probability of assigning a treatment from the current composition, and the probability of arriving at a particular next composition. The first describes the next treatment assignment given the state; the second describes how the state itself changes over one step.
We will use a state vector \(S_t=(Y_{1t},\ldots,Y_{Kt})\), where \(Y_{it}\) is the number of balls of treatment color \(i\) in the urn at time \(t\), and \(K\) is the number of treatments. When the urn contains \(Y_{+t}=\sum_{i=1}^K Y_{it}>0\) balls and assignment is proportional to the color counts, the probability of assigning treatment \(i\) in state \(s\) is \(Y_i/Y_+\). A transition table instead records probabilities of the form \(P(S_{t+1}=s'\mid S_t=s)\).
The interpretation depends on the model being used. For the examples below, assume that the current state is known, a treatment is drawn according to the state’s allocation probabilities, a response is generated according to the specified treatment-specific response probabilities, and the replacement rule determines the next composition. Conditional on the current state and assigned treatment, the response probabilities are as stated; the replacement rule is applied as specified. These assumptions make the transition probabilities conditional on the current state well-defined. If the rule depends on additional history not recorded in the state, the composition alone may not be enough to determine a transition row.
Each possible path within a step has probability equal to the assignment probability multiplied by the conditional response probability. If two or more paths lead to the same next composition, their probabilities must be added. The table reports probabilities of destination states, not a separate row for every path.
Worked Example: Two Treatments and Two Routes to Each State
Worked Example: Two Treatments and Two Routes to Each State
Consider a hypothetical binary-treatment urn with current state \(S_t=(2,1)\): two A balls and one B ball. Assign treatment in proportion to the ball counts. Suppose the probability of success is 0.60 on A and 0.50 on B. After a success, add one ball of the assigned treatment’s color; after a failure, add one ball of the other color. There are no removals.
The current urn has three balls, so the allocation probabilities are \(P(A\mid S_t=(2,1))=2/3\) and \(P(B\mid S_t=(2,1))=1/3\). The four treatment-response paths are:
| Assignment and response | Path probability | Next state |
|---|---|---|
| A, success | \((2/3)(0.60)=0.4000\) | \((3,1)\) |
| A, failure | \((2/3)(0.40)=0.2667\) | \((2,2)\) |
| B, success | \((1/3)(0.50)=0.1667\) | \((2,2)\) |
| B, failure | \((1/3)(0.50)=0.1667\) | \((3,1)\) |
The destination \((3,1)\) is reached by A success or B failure. Adding their unrounded path probabilities gives
The destination \((2,2)\) is reached by A failure or B success:
Thus the transition row has probabilities 0.5667 and 0.4333, which sum to 1. The allocation probabilities at the possible next states are different from those transition probabilities: in state \((3,1)\), the next assignment probabilities are \(3/4\) for A and \(1/4\) for B; in state \((2,2)\), they are \(1/2\) for each treatment.
Why the Row Total Is One
A complete transition row must include every possible next state under the stated model. Its probabilities sum to one because the assignment and response paths partition all outcomes of the next step. This fact is a useful check, but it does not by itself prove that every listed entry is correct: an omitted path and a compensating arithmetic error could still produce a total of one.
For a current state \(s\), let \(q_i(s)\) be the probability of assigning treatment \(i\), and let \(r\) index the possible responses under that treatment. Write \(\rho(r\mid i,s)\) for the conditional probability of response \(r\), and let \(g(s,i,r)\) be the next state produced by the replacement rule. Then a destination \(s'\) has probability
where the indicator is 1 when that path leads to \(s'\), and 0 otherwise. Summing over all possible destinations removes the indicator by counting each path once. Since the treatment-assignment probabilities sum to one and the response probabilities sum to one within each treatment,
This also explains why distinct paths that share a destination are added: they are disjoint ways to reach that destination. Multiplying their probabilities would describe the chance that both paths occur, which is not the event represented by the table entry.
Worked Example: Three Treatments and Aggregated Destinations
Worked Example: Three Treatments and Aggregated Destinations
Consider a hypothetical three-treatment urn in state \((1,2,1)\), with one A ball, two B balls, and one C ball. Assignment is proportional to the counts. The success probabilities are 0.80 for A, 0.50 for B, and 0.60 for C. After success, add a ball of the assigned color. After failure, add a ball of the next color in the cycle A to B, B to C, C to A. The urn starts with four balls and grows by one ball each step; all response probabilities lie between zero and one.
The assignment probabilities are \(1/4\) for A, \(1/2\) for B, and \(1/4\) for C. Enumerating each path gives:
| Path | Probability | Next state |
|---|---|---|
| A success | \((1/4)(0.80)=0.2000\) | \((2,2,1)\) |
| A failure | \((1/4)(0.20)=0.0500\) | \((1,3,1)\) |
| B success | \((1/2)(0.50)=0.2500\) | \((1,3,1)\) |
| B failure | \((1/2)(0.50)=0.2500\) | \((1,2,2)\) |
| C success | \((1/4)(0.60)=0.1500\) | \((1,2,2)\) |
| C failure | \((1/4)(0.40)=0.1000\) | \((2,2,1)\) |
Aggregating by destination, the transition row is:
The row total is \(0.3000+0.3000+0.4000=1.000\). Each destination contains five balls, as required by the stated rule of adding one ball. The treatment allocation probabilities at these next states are found from their compositions: \((2/5,2/5,1/5)\) at \((2,2,1)\), \((1/5,3/5,1/5)\) at \((1,3,1)\), and \((1/5,2/5,2/5)\) at \((1,2,2)\).
Reading a Table Across Several States
A transition table may contain one row for each current state. Each row is conditional on its own state; rows are not probabilities of the current states themselves. To find an unconditional distribution of the next state, the current-state distribution must also be known. This distinction matters when interpreting a table that includes both transition entries and allocation probabilities.
Worked Example: Using a Transition Table with an Initial State Distribution
Suppose a hypothetical urn model has two states, \(s_1=(2,1)\) and \(s_2=(1,2)\). The table below gives the probability of the next state conditional on the current state. Assume both states have three balls and assignment is proportional to the counts. Initially, the state is \(s_1\) with probability 0.25 and \(s_2\) with probability 0.75.
| Current state | Next state \(s_1\) | Next state \(s_2\) | Row total |
|---|---|---|---|
| \(s_1=(2,1)\) | 0.70 | 0.30 | 1.00 |
| \(s_2=(1,2)\) | 0.20 | 0.80 | 1.00 |
The probability of being in \(s_1\) after one step is the sum of the two ways to arrive there, weighted by the initial state probabilities:
Therefore \(P(S_{t+1}=s_2)=1-0.325=0.675\). In state \(s_1\), the allocation probabilities are \(2/3\) for A and \(1/3\) for B; in \(s_2\), they are \(1/3\) for A and \(2/3\) for B. Thus the probability of assigning A on the next assignment, after this transition, is
This is not the initial A allocation probability, which was \((0.25)(2/3)+(0.75)(1/3)=0.4167\). The transition changes the distribution over states, and the allocation probability is then averaged over the resulting state distribution.
Checks, Alternatives, and Limitations
A reliable reading proceeds from paths to destinations and only then to allocation probabilities. A path tree is often the clearest alternative when a row has only a few possible assignments and responses: list each path, multiply its conditional probabilities, and combine paths by destination. A transition table is more compact when many rows or repeated calculations are needed. Both represent the same conditional probabilities when constructed from the same model.
Use the following checks when examining a table:
- Check the conditioning state. Every row should identify the current composition to which its probabilities apply.
- Check the assignment probabilities. Under proportional allocation, the current color counts divided by the total must sum to one.
- Check destination completeness. Every assignment-response path with positive probability should appear in exactly one destination entry.
- Check the updates. For each path, apply the stated replacement rule to the current counts and verify the listed next composition.
- Check row totals and units of interpretation. Transition probabilities are dimensionless conditional probabilities, not urn counts or treatment proportions.
- Check state sufficiency. If response probabilities or the update rule depend on earlier history, that history must be included in the state for a row to be valid.
A row total different from one usually signals an omitted path, duplicated path, incorrect conditional probability, or arithmetic error. A row total of one is necessary but not sufficient: verify the destinations and the aggregation as well. Rounded entries may sum to a value slightly different from one; when that happens, check the unrounded values before diagnosing a model error.
A transition table is limited to the state space and rule it represents. It does not, by itself, establish that a response model is scientifically appropriate, nor does a finite table automatically describe every state that could occur under a different update rule. In particular, do not infer that a treatment is more likely to be assigned merely because a destination state has a large transition probability. That probability concerns where the urn goes; allocation probabilities are calculated from the composition at the state where an assignment is made.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- In state \((3,1)\), what are the proportional allocation probabilities for treatments A and B?
- Two disjoint paths to the same next state have probabilities 0.18 and 0.27. What is the probability of that destination, and why?
- A complete transition row lists probabilities 0.22, 0.31, and 0.47. What check can you make, and what does passing it not guarantee?
- A state \(s_1\) has probability 0.60, and \(s_2\) has probability 0.40. The conditional probabilities of next state \(u\) are 0.25 from \(s_1\) and 0.50 from \(s_2\). Find the unconditional probability of \(u\) next.
- Explain why a transition probability of 0.70 from state \(s\) to state \(u\) is not necessarily the treatment allocation probability at \(u\).
Solutions. (1) There are four balls, so \(P(A\mid(3,1))=3/4=0.75\) and \(P(B\mid(3,1))=1/4=0.25\). (2) The paths are disjoint, so add: \(0.18+0.27=0.45\). (3) The row total is \(0.22+0.31+0.47=1.00\), as required for a complete row. This does not guarantee that the destinations or individual calculations are correct. (4) Weight each conditional transition probability by its current-state probability: \(P(u)=(0.60)(0.25)+(0.40)(0.50)=0.15+0.20=0.35\). (5) The 0.70 is the conditional probability of reaching \(u\) from \(s\); treatment allocation at \(u\) is determined by the urn composition in \(u\), such as its color counts divided by its total under proportional allocation.