Two Children — A Probability Puzzle

Mr. Jones has two children. At least one is a boy. What is the probability he has two boys?

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Legend

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Pruned — ruled out by condition

Qualifies — not two boys

Two boys — counts toward answer

Answer

1/3

One of three equally-weighted qualifying outcomes is B,B.

13 = ¼¼+¼+¼

Why not 1/2?

There are two ways to have exactly one boy: G,B and B,G. Birth order matters — they are distinct outcomes each with probability ¼.

G,G is ruled out, leaving three equally-likely qualifying outcomes — not two.

Mr. Jones has two children. You knock on the door; one child (chosen at random) opens it. It is a boy. What is the probability he has two boys?

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Legend

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Pruned — boy didn't open door

Boy opened — not two boys

Two boys — counts toward answer

Answer

1/2

B,B families produce two "boy opens door" leaves; mixed families produce one. The weights balance exactly.

P(B,B ∩ boy) = ¼
P(boy opens) = ½
→ ¼ ÷ ½ = 1/2

Why different from problem 1?

Problem 1 conditions on the event "at least one boy exists." This problem conditions on observing a randomly chosen child.

B,B families are twice as likely to show you a boy — so they get twice the weight in the conditional, pushing the answer from 1/3 to 1/2.

Mr. Jones has two children. At least one is a boy born on Tuesday. What is the probability he has two boys?

Click any node to reveal its branches

Legend

Click to expand branches

Pruned — ruled out by condition

Qualifies — not two boys

Two boys — counts toward answer

Answer

13/27

With N=7 days, 13 of the 27 qualifying pairs are two boys.

2N−14N−1 = 1327

Why 13/27?

A B,B family has two chances to produce a Tuesday boy — one per child. So B,B outcomes are over-represented among Tuesday-boy families.

As N → ∞ (rarer trait), P → 1/2. Try the slider:

N = 7 days

Simulation

Generate random two-child families. Keep only those with at least one B-Tue. Watch the fraction of two-boy families converge on 13/27.

Step-by-step — one trial

CHILD 1 —

CHILD 2 —

0 Families generated

0 Kept (≥1 B-Tue)

0 Two boys among kept

          —
          Simulated P(BB | B-Tue)
        

Simulated probability converging on 13/27 ≈ 0.481

Each batch filters to ≥1 B-Tue, then plots running P(BB).

Why filter, not pre-select?

A common mistake: "generate one B-Tue child, then generate the other." That fixes which child is the Tuesday boy, changing the conditioning. Generate both freely and discard families where neither is B-Tue.

Why so many trials?

The filter keeps only ~1 in 7 families. You need many raw families to accumulate enough kept ones. With only 100 kept, 0.481 is close enough to 0.5 that sampling noise swamps the signal.

Weekend Problem

Design a simulation for the door-opening problem.

No filtering needed — the generative process matches the conditioning directly. What changes about how you record outcomes?