A Coin Tossing Puzzle

What this lecture covers

This recitation problem tests intuition about conditional probability. Alice tosses a possibly biased coin twice and wonders whether knowing "the first toss is heads" makes both tosses being heads more likely than only knowing "at least one toss is heads." The video verifies her guess and then generalizes the argument to any two nested events.

It's a compact exercise in translating a verbal claim into symbols and comparing two conditional probabilities using only the definition of conditional probability and basic set relationships.

Key ideas

• Translating words into events: "first toss is heads" and "second toss is heads" become events A and B, with "both heads" as A ∩ B.
• Conditioning on a subset simplifies the numerator: since A ∩ B is a subset of A, the intersection (A ∩ B) ∩ A reduces to just A ∩ B.
• Monotonicity of probability: because A is a subset of A ∪ B, P(A) is never greater than P(A ∪ B).
• Comparing fractions with equal numerators: a larger denominator means a smaller fraction, so P(A∩B|A) ≥ P(A∩B|A∪B).
• The result doesn't depend on the coin's bias: the inequality holds for any probability of heads, fair or not.
• Generalization: for any events C ⊆ D ⊆ E with C ∩ D = C ∩ E, the same argument shows P(C|D) ≥ P(C|E).

Before you watch

• Know the definition of conditional probability and the multiplication rule, covered in the second lecture of this course.
• Be comfortable with basic set relationships such as subset and intersection.

Check your understanding

1. Why does (A ∩ B) ∩ A simplify to A ∩ B?
2. Why is P(A) always less than or equal to P(A ∪ B)?
3. For a fair coin, what are the two conditional probabilities in this problem, and why is one strictly larger than the other?
4. What are the two conditions that events C, D, and E must satisfy for the generalized inequality to hold?