### Tip of the Week #9
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# The Problem with Girls

Teaser Problem: **Bob is the new man in your office. You hear that he
and his wife have two children. Later, you learn that at least one is a
girl. What is the probability that both children are girls?**

This question is non-trivial. For most people, intuition gives a wrong answer. That's
why working with probability is fun and rewarding.

Interested persons may want to devise a spreadsheet Monte Carlo simulation to test: **What
is the probability that both children are girls given that you know that there is at least
one girl?**

Answer: 1/3.

The separate, ordered births of two children can be: {BB, BG, GB, and GG}.
Assuming that births are equally-weighted by sex and independent, each outcome in the set
is equally likely, 1/4. If we learn that Mr. and Mrs. Smith have at least one girl,
that rules out only the possibility of BB. Any of outcomes {BG, GB, and GG} are
possible and equally likely.

The problem appears in "Mathematical Recreations: The Interrogator's Fallacy"
by Ian Stewart, *Scientific American*, Sept. 1996, pp 172-5.

The featured problem of the article is the Interrogator's Fallacy. Most jurors in a
murder trial would tend to believe the confession of an accused. We would be more
confident of guilt with a confession than without. In probability algebra notation:

P(guilty | confession) > P(guilty) > P(guilty | no confession)

The problem is that we presume that a guilty person is more likely to confess than an
innocent person. This is not necessarily true as found in witchcraft trials (e.g., time of
Spanish Inquisition). More recently, terrorists have been found to be fortified against
interrogation. Some innocent persons are actually *more* likely to confess (to end
the interrogation) than a guilty, hardened terrorist. The reasoning is attributed to
Robert A. J. Matthews. The "Problem with Girls" example is used to explain the
mathematical basis for the Interrogator's Fallacy.

--John Schuyler, August 1996, revised Dec. 2004