Combining conditional branches

What you will learn

Use an exhaustive source partition to compute an event probability and explain how mixture weights change it.

Prerequisites: Unequal outcome weights · Testing event independence

A conditional rate needs its source weight

Suppose a process first chooses source A or source B and then produces success or failure. A success rate within A answers a question about A cases only. It does not tell you how often the process reaches A. To recover the overall success probability, account for both stages rather than averaging the two conditional rates by habit.

Our model selects one of eight equally likely source tickets: two lead to A and six to B. Within the chosen source, one of four equally likely branch cases is selected. A has three successes; B has one. The source probabilities are 2/8 and 6/8, while conditional success rates are 3/4 and 1/4. These rates are explicit assumptions, not measurements from real players.

Multiply down a path, add across disjoint paths

The complete event has two disjoint ways to occur: A then success, or B then success. Their probabilities are (2/8)(3/4)=6/32 and (6/8)(1/4)=6/32. Adding gives 12/32=3/8. The denominators refer to different stages: four is the number of equally likely cases inside a chosen branch, not a claim that all eight source tickets are split into a single four-case population.

The four complete paths are A-success, A-failure, B-success and B-failure. They partition the experiment, so their probabilities sum to one. An unweighted average of 3/4 and 1/4 would give 1/2. That would match a different experiment that chooses the two sources equally often. It does not match the stated 2/8 versus 6/8 selection.

Check coverage before applying the sum

A source partition must cover every possible source case without overlap. Otherwise adding branch contributions can omit probability or double-count it. For a finite partition C1,…,Ck with positive branch probabilities, P(E)=ΣP(Ci)P(E given Ci). A zero-probability branch contributes zero, but its ordinary conditional ratio is undefined and should not be manufactured by dividing by zero.

In the second activity model, source A receives six tickets and B two, while the conditional rates stay fixed. Success becomes (6/8)(3/4)+(2/8)(1/4)=5/8. No within-source improvement occurred; only the source mixture changed. Poker and quant models also need this separation between how often a situation occurs and what happens conditional on reaching it. The lesson does not supply real-opponent frequencies.

Hands-on: Route source mass through conditional branches

Construct source and within-source populations, select success leaves and reconcile weighted path probabilities before changing the source mixture.

Reading branch workshop…

Common mistakes

• Averaging conditional rates without their source weights.
• Multiplying by the marginal event rate instead of its branch-conditional rate.
• Adding overlapping or incomplete branch categories.
• Treating a mixture change as proof of a within-group improvement.

Keep these ideas

• Partition sources into disjoint exhaustive events.
• Multiply each source weight by its conditional event rate.
• Add the disjoint contributions.
• Keep mixture changes separate from conditional-rate changes.

Further reading

• MIT — Conditioning and Bayes’ rule, lecture 2 (2018) ↗

  John Tsitsiklis and Patrick Jaillet’s annotated lecture slides, especially pages 2, 5 and 8–10: conditioning, partitions, total probability and Bayes’ rule. Our 100-case populations are original assumed models, not observations of real opponents. Checked 2026-10-21.