A Chess Tournament Problem

What this lecture covers

This recitation problem uses a multi-round chess tournament to practice building probability trees and computing both unconditional and conditional probabilities. Challengers Bo and Chi play a first round, and the winner faces the reigning champion Al in a second round, with fixed win probabilities given for each matchup.

After watching, you should be able to build a sequential tree diagram for a multi-stage random experiment, read off disjoint outcomes for a given event, and apply the definition of conditional probability when the condition itself is an intersection of events.

Key ideas

• Tree diagrams for sequential experiments: each branch is labeled with the probability of that step, and multiplying probabilities along a path from root to leaf gives the probability of that full outcome.
• Disjoint outcomes sum: because the leaves of a correctly built tree are mutually exclusive, the probability of an event is the sum of the probabilities of the leaves that belong to it.
• Complement trick: computing the probability of the complement of an event can be much faster when the complement has fewer outcomes.
• Conditional probability by definition: P(A|B) = P(A and B) / P(B), computed directly from the tree by identifying which leaves satisfy the numerator and denominator events.
• Conditioning on an intersection: when the condition is itself an intersection of two events, it can be treated as a single combined event and the ordinary definition of conditional probability still applies.

Walkthrough

Setting up the tournament and the probability tree (0:00)

The problem is introduced: Bo and Chi play a first round with Bo winning each game with probability 0.6, and the survivor faces Al in a second round. A tree diagram is built for the first round's two games.

Part a: probability a second round is required (1:00)

The lecture identifies the tree leaves where a second round is needed, multiplies along branches, and sums the disjoint outcomes to get the probability, then also computes the probability that Bo specifically wins the first round.

Part a continued: probability Al retains the championship (5:01)

A fuller tree including the second round is used. Rather than summing the many outcomes where Al wins, the lecture computes the complement (the few outcomes where Al loses) and subtracts from 1.

Part b: conditional probabilities given a second round (8:06)

Two conditional probabilities are computed using P(A|B) = P(A and B) / P(B): the probability that Bo (rather than Chi) is the surviving challenger given a second round occurred, and the probability that Al retains his championship given a second round occurred.

Part c: conditioning on an intersection of events (12:17)

The question asks for the probability that Bo won the first round given both that a second round was required and that it consisted of only one game. The lecture shows that treating the intersection of the two conditions as a single event lets the same conditional probability definition apply, then works out the numerator and denominator from the tree.

Before you watch

• Review the definition of conditional probability and the multiplication rule from earlier lectures in this course.
• Be comfortable computing probabilities of disjoint events by addition.

Check your understanding

1. Why can the probability of an event be found by summing the probabilities of the tree leaves that belong to it?
2. When is it more efficient to compute the complement of an event rather than the event itself?
3. How does conditioning on an intersection of two events reduce to the ordinary definition of conditional probability?
4. In part b, why is the denominator the same value in both conditional probability calculations?