From Dose-Level Decisions to Trial-Level Probabilities
A 3+3 trial follows a sequence of decisions. At each dose, the observed number of dose-limiting toxicities (DLTs) determines whether the trial escalates, expands the cohort, or stops. The probability of selecting a dose therefore depends not only on its own DLT probability, but also on the probability that the trial reaches it and on what happens at any higher dose tested.
This tutorial uses path enumeration: calculate the probability of each relevant sequence of dose-level decisions, then add the probabilities of paths that lead to the same selection. It considers a fixed ladder of two or three dose levels. The specified DLT probabilities are assumed known for the calculation; they are hypothetical inputs, not estimates inferred from trial data.
The conventional rule used here is as follows. Treat three participants at a dose. With zero DLTs, escalate. With one DLT, treat three additional participants at the same dose; escalate if there is still only one DLT among all six, and stop for toxicity if there are two or more. With two or three DLTs among the first three, stop for toxicity without expansion. A dose that meets the rule has at most one DLT among six, or zero among three.
The calculation assumes a fixed dose-specific DLT probability, independent participant outcomes conditional on dose, complete and timely DLT observation before each decision, and strict application of the rule. The dose ladder is fixed, cohorts are enrolled sequentially, and there is no dose skipping, de-escalation, replacement of unevaluable participants, or other modification. These assumptions make the paths below well-defined. If outcomes are dependent, observation is incomplete, or the decision rule changes, the probabilities must be recalculated under the changed model.
Probability of Advancing at One Dose
At dose \(i\), the number of DLTs in the first three participants has a binomial distribution with three trials and probability \(p_i\). Zero DLTs occur with probability \((1-p_i)^3\). Exactly one occurs with probability \(3p_i(1-p_i)^2\). If exactly one occurs, the trial expands to six participants, and advancement requires all three additional participants to have no DLTs. Under the independence assumption, that conditional probability is \((1-p_i)^3\).
The stop probability \(b_i\) includes both stopping after at least two DLTs in the first three and stopping after expansion when the initial cohort had one DLT but at least one of the next three also has a DLT. The two possible outcomes at a dose are exhaustive and mutually exclusive: the trial either passes the rule or stops for toxicity. Thus \(a_i+b_i=1\).
Worked Example: Calculating Advancement at One Dose
Suppose a hypothetical dose has DLT probability \(p_i=0.20\). Calculate the probability that the rule permits escalation beyond this dose.
With three participants, the probability of zero DLTs is \((1-0.20)^3=0.80^3=0.512\). The probability of exactly one DLT is \(3(0.20)(0.80)^2=0.384\). Given exactly one initial DLT, advancement requires zero DLTs among the three additional participants, with probability \(0.80^3=0.512\).
Therefore,
The probability of stopping at this dose for toxicity is \(b_i=1-0.708608=0.291392\). As a check, stopping can occur with at least two DLTs in the initial three, probability \(1-0.512-0.384=0.104\), or after an initial one DLT followed by at least one DLT in the next three, probability \((0.384)(1-0.512)=0.187392\). These add to \(0.291392\).
Enumerating a Two-Dose Ladder
For a two-dose ladder, the trial first encounters dose 1. If it stops there for toxicity, no dose is selected under the rule. If it passes dose 1, it reaches dose 2. Stopping for toxicity at dose 2 selects dose 1 as the MTD; passing the rule at dose 2 selects dose 2, the highest planned dose. These paths are mutually exclusive and account for every possible outcome.
The multiplication follows from the sequential path: reaching dose 2 requires passing dose 1, with probability \(a_1\). Conditional on reaching dose 2, stopping there has probability \(b_2\), while passing it has probability \(a_2\). The total is one because
Worked Example: Selection Probabilities for Two Doses
Consider a hypothetical two-dose trial with \(p_1=0.10\) and \(p_2=0.25\). Calculate the probability of selecting each dose and the probability that no dose is selected.
At dose 1,
At dose 2,
The probability of selecting dose 1 is the probability of passing dose 1 and then stopping for toxicity at dose 2:
The probability of selecting dose 2 is the probability of passing both doses:
The probability of selecting no dose is \(b_1=0.09385\). The three probabilities sum to approximately \(0.3626+0.5436+0.09385=1.000\), with the displayed rounding. Thus, in this hypothetical setting, dose 2 is selected most often, but the trial also has a nonzero chance of stopping at dose 1 without selecting an MTD.
Extending the Paths to Three Doses
For three doses, each selected-dose event is a path through the ladder. Selecting dose 1 requires passing dose 1 and then stopping for toxicity at dose 2. Selecting dose 2 requires passing doses 1 and 2 and then stopping for toxicity at dose 3. Selecting dose 3 requires passing all three doses. If dose 1 itself fails, no dose is selected. The same reasoning extends to longer fixed ladders, but here the formulas are written explicitly for three.
These expressions use the probabilities of successive decisions along each path. For example, selecting dose 2 requires passing dose 1, passing dose 2, and stopping for toxicity at dose 3. Under the stated model and rule, those sequential conditional probabilities are \(a_1\), \(a_2\), and \(b_3\), respectively. The exhaustive-path check is
Worked Example: Selection Probabilities for Three Doses
Suppose a hypothetical three-dose ladder has \(p_1=0.05\), \(p_2=0.15\), and \(p_3=0.30\). Calculate the probability of each possible selection.
Applying the dose-level formula gives:
The probability of no selected dose is \(b_1=0.02656\). The probability of selecting dose 1 is
The probability of selecting dose 2 is
Finally, the probability of selecting dose 3 is
The rounded probabilities sum to \(0.02656+0.1813+0.4006+0.3915\approx1.000\). This check is important: if the paths are enumerated correctly, all possible trial endings, including stopping at the lowest dose without a selection, must be represented.
How to Check an Enumeration
A practical way to avoid omissions is to draw the dose ladder as a sequence of pass and stop branches. At dose 1, the stop branch has probability \(b_1\) and ends with no selection. The pass branch has probability \(a_1\) and leads to dose 2. At each later dose, its stop branch selects the preceding dose, while its pass branch continues or selects the highest planned dose. This is a compact form of path enumeration.
A second check is to verify that each dose has \(a_i+b_i=1\), and then verify that the probabilities of all terminal paths sum to one. If they do not, check whether a path was omitted, counted twice, or assigned the wrong dose-level probability. Keep the distinction between “stops at dose \(i\)” and “selects dose \(i\)” explicit: toxicity stopping at dose \(i\) selects dose \(i-1\), except at the lowest dose, where it yields no selection.
For every dose \(i\), use \(a_i=(1-p_i)^3+3p_i(1-p_i)^5\), then set \(b_i=1-a_i\).
Include stopping at dose 1, stopping at each later dose, and passing the highest planned dose.
Use a factor \(a_i\) for every passed dose and \(b_i\) for the dose where toxicity stops escalation.
Add paths with the same outcome if needed, and confirm that all terminal probabilities sum to one.
Interpretation, Alternatives, and Limitations
Path enumeration is exact under its assumptions and is often easier to audit than simulation for a short dose ladder. Another exact approach is a recursive calculation: at each dose, multiply the probability of reaching it by its local pass or stop probability, then carry the pass probability forward. Recursion and direct path enumeration give the same answer; recursion is convenient when many dose levels make a written list of paths unwieldy. Simulation can approximate these probabilities by repeatedly generating participant outcomes and applying the decision rule, but it adds Monte Carlo error and is not necessary for the simple fixed ladders considered here.
The calculated quantities are probabilities of the rule’s final selection under specified \(p_i\) values. They do not by themselves establish that a selected dose is safe, effective, or biologically optimal. In particular, the calculation does not account for uncertainty about the \(p_i\), changing risks over time, incomplete follow-up, participant dependence, or deviations from the conventional rule. Nor does it explain all features of trial performance; those require additional operating-characteristic summaries.
A common error is to treat \(1-a_i\) as the probability of selecting dose \(i\). It is instead the probability that escalation stops for toxicity at dose \(i\). If \(i>1\), that event selects the preceding dose, provided the trial reached dose \(i\). The probability of selecting that preceding dose must also include the probability of passing all earlier doses. For example, in a three-dose ladder, selection of dose 2 has probability \(a_1a_2b_3\), not just \(b_3\). The factor \(a_1a_2\) accounts for reaching dose 3.
Check Your Understanding
Answer each question using the fixed-ladder 3+3 rules in this tutorial. Solutions follow the questions.
- If a dose has DLT probability \(p=0.10\), calculate its probability of passing the 3+3 rule.
- In a two-dose ladder, suppose \(a_1=0.80\) and \(b_2=0.30\). What are the probabilities of selecting dose 1 and selecting no dose?
- For a three-dose ladder with \(a_1=0.90\), \(a_2=0.70\), and \(a_3=0.60\), calculate the probabilities of selecting each dose and of selecting no dose.
- Why is \(b_3\) alone not the probability of selecting dose 2 in a three-dose ladder?
Solutions. 1. \(a=(0.90)^3+3(0.10)(0.90)^5=0.729+0.177147=0.906147\), or about \(0.9061\). 2. Selection of dose 1 has probability \(a_1b_2=(0.80)(0.30)=0.24\). No dose is selected if dose 1 fails, with probability \(b_1=1-a_1=0.20\). 3. No dose is selected with probability \(1-0.90=0.10\). Dose 1 is selected with probability \((0.90)(1-0.70)=0.27\); dose 2 with probability \((0.90)(0.70)(1-0.60)=0.252\); dose 3 with probability \((0.90)(0.70)(0.60)=0.378\). These sum to \(1.000\). 4. Stopping at dose 3 selects dose 2 only if the trial first passes doses 1 and 2. The required probability is \(a_1a_2b_3\), which includes the probability of reaching dose 3.