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Boy or girl paradox
You will learn: Use observation likelihoods to distinguish family filters from random child reports.
Start with: Conditioning and independence
A family has two children. You learn that at least one is a boy. Is the probability that both are boys one half or one third? Before counting families, specify how a family entered your sample and how its information was reported. The same sentence can be produced by different experiments.
Our simplified model gives each child an independent binary label B or G, with probability p of B. These are model categories, not a claim that real demographic outcomes follow this exact model. Order distinguishes the first and second child. At p = 1/2, GG, GB, BG, and BB each have probability 1/4.
Predict, then change the protocol
Evidence probability under the original sampling scheme: 0.7500000.
Filled left bars: original probability. Open right bars: probability after the stated observation. The same underlying outcomes remain in the table, including excluded ones.
| Outcome | Prior | P(observation | outcome) | Joint mass | Posterior |
|---|---|---|---|---|
| GG | 0.2500000 | 0.000000 | 0.000000 | 0.000000 |
| GB | 0.2500000 | 1.000000 | 0.2500000 | 0.3333333 |
| BG | 0.2500000 | 1.000000 | 0.2500000 | 0.3333333 |
| BB | 0.2500000 | 1.000000 | 0.2500000 | 0.3333333 |
Both children B, given this observation: 0.3333333. Order identifies first and second child. B/G are two categories in this simplified independent model, not a demographic claim. A tag is independent and uniform over 7 categories.
With At least one B, retain every family satisfying that condition. GG disappears; GB, BG, and BB retain equal weight. Exactly one of the three retained outcomes is BB, so the answer is 1/3. The open bars describe the retained population, not a simulation run.
Now select Random child reported as B. Choose a family without using its labels, then select one of its children uniformly. Retain the observation only if that child is B. BB always produces the report, whereas GB and BG produce it only half the time. Their selected masses are 1/4, 1/8, and 1/8. BB therefore accounts for half the retained observations. The unreported child’s label remains independent of the reported child’s label under these assumptions.
Derive the two answers
The first protocol has evidence probability 1 − (1 − p)². Every BB family supplies the evidence, so divide p² by that denominator. For p greater than zero this simplifies to p/(2 − p). The random-report protocol instead has evidence probability p and joint probability p², giving p. At p = 0 neither experiment can produce the evidence: the conditional probability is undefined, not zero.
The table separates prior, observation likelihood, selected mass, and posterior. This is Bayes’ rule on four outcomes. A family twice as likely to produce the report receives twice the relative weight after observing it.
Why a weekday changes the answer
Choose At least one tagged B, with seven equally likely, independent tags per child. A tag can stand for a weekday in the puzzle; uniformity and independence are explicit assumptions. A mixed pair qualifies with probability 1/7. A BB pair qualifies with probability 1 − (6/7)² = 13/49: either child can supply the tagged B, but the case where both do must be counted only once.
At p = 1/2 the selected masses, with denominator 196, are 0, 7, 7, and 13. The answer is therefore 13/27, not exactly one half. In general it is p(2 − q)/(2 − pq), where q is the probability of the particular tag. The control sets q = 1/d for d possible tags. At d = 1 the tag adds no information and the experiment reduces to “at least one B.”
Make a prediction
A parent chooses a child at random and reports that the child is B with a particular tag. Does 13/27 apply?
Explore the answer
No. That reports one randomly selected child, rather than filtering for any family with a tagged B. Under independence, the other child’s B probability stays p, or one half at the default. The mechanism producing the information is part of the evidence.
What an ordinary conversation leaves unspecified
A person may volunteer an interesting fact rather than report a random child. The probability that each family produces that sentence then needs its own model. There is no universal answer from the sentence alone. Assign likelihoods to the four prior outcomes; the calculation becomes unambiguous once those likelihoods are given.
Reference and next step
Grinstead and Snell, Introduction to Probability, chapter 4 develops finite conditional probability. The reporting and tag calculations above follow the displayed model. Continue to Bertrand’s box for the same weighting with physical coins.