This classic paradox explores how subtly different phrasing in probability problems can lead to distinct outcomes. We’ll examine two versions involving two-child families.

Part A: “At Least One Son”

Given: Ms. Jackson has two children. You are told that at least one is a boy.

Question: What is the probability that both children are boys?

Analysis

There are four equally likely combinations of two children:

  • Boy–Boy (BB)
  • Boy–Girl (BG)
  • Girl–Boy (GB)
  • Girl–Girl (GG)

The “at least one is a boy” condition eliminates GG, leaving:

  • BB
  • BG
  • GB

These remaining cases are equally likely, so:

  • Only 1 out of 3 has both children as boys.

Answer:

\[ P(\text{both are boys} \mid \text{at least one is a boy}) = \boxed{\frac{1}{3}} \]

Part B: “You See a Boy”

Given: Ms. Parker has two children, and you see one of them, a boy.

Question: What is the probability that both children are boys?

Analysis

This is a subtler form of conditioning. You’re not told abstractly that “at least one is a boy.” Instead, you directly observe a boy—without knowing if he’s the older or younger child.

Each family combination (BB, BG, GB, GG) is equally likely to begin with. Now consider the chance that you see a boy under each scenario:

  • BB: Seeing a boy is guaranteed (either child is a boy).
  • BG: 50% chance (you might see the boy or the girl).
  • GB: 50% chance.
  • GG: 0% chance (no boys to see).

Weight the likelihoods based on visibility:

  • BB: 1
  • BG: 0.5
  • GB: 0.5
  • GG: 0

Normalize:

  • Total weight: \( 1 + 0.5 + 0.5 = 2 \)
  • BB has weight 1 out of 2.

Answer:

\[ P(\text{both are boys} \mid \text{you see a boy}) = \boxed{\frac{1}{2}} \]

Conclusion

These two problems differ in how the information is obtained:

  • Part A conditions on a fact about the family.
  • Part B conditions on a specific observation.

This small shift leads to very different probabilities: \( \frac{1}{3} \) versus \( \frac{1}{2} \).

Reference