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Devoted
Donor
Location: New England
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Strange, once you understand the Monty Hall problem, try the Girl Named Florida problem.
Quote:
You know that a certain family has two children, and that at least one is a girl. But you can’t recall whether both are girls. What is the probability that the family has two girls — to the nearest percentage point?
Answer: Assume that there are an equal number of boys and girls, and that the gender of each child in the family is independent of the other’s (more on that below). Then there are four, equally likely possibilities for a two-child family’s history of procreation: Either a girl was born first, and then a boy; or a girl was born first, and then another girl; or boy then boy; or boy then girl. But you know that there is at least one girl in the family, so you can eliminate the (boy, boy) possibility. That leaves three scenarios, and only in one are there two girls. So the probability of two girls is one in three — 33%, to the nearest percentage point.
Comments: There’s an assumption I should have made explicit for this problem and the next problem: that there are an equal number of boys and girls (although in 2005, there were nearly 5% more boys than girls 14 and under, according to the Census Bureau). Kudos to Messrs. Newcombe and Plourde, and to Glenn Tippy and Gregg Skinner, for noting the importance of gender ratio at birth.
Results: 43% got this right. The range of answers was 25% to 80%, the median and mode were 50%, and the mean was 43.4%.
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That wasn't so bad, right? Just wait...
Quote:
You know that a certain family has two children, and you remember that at least one is a girl with a very unusual name (that, say, one in a million females share), but you can’t recall whether both children are girls. What is the probability that the family has two girls — to the nearest percentage point?
Answer: Use the same logic as above, only this time there are three possibilities for each child: Boy, girl with the specific unusual name (let’s for the sake of argument make it Florida, the one used in Mr. Mlodinow’s book), and girl with a different name. Then there are five possibilities for a family with two children, one of them named Florida:
(boy, girl-F)
(girl-NF, girl-F)
(girl-F, boy)
(girl-F, girl-NF)
(girl-F, girl-F)
Unlike in question No. 2, these are not all equally likely. The last scenario is particularly unlikely, assuming the two children’s names are independent, because Florida is such an unusual name. So for the sake of this calculation, we can ignore it. The other four scenarios are, approximately, equally likely, because we’ve assumed that there are the same number of boys as girls, and nearly all girls have names other than Florida. In two of those four scenarios, the family has two girls. So the probability of two girls is about two in four — 50%, to the nearest percentage point. (There’s ample discussion of this question, and a more-detailed explanation from me, in the comments.)
Comments: It seems paradoxical that the girl’s name would make a difference, and in fact 75% of readers thought the answers to No. 2 and No. 3 were the same, including 68% of those who got No. 3 right. Mr. Mlodinow suggested I reward points for No. 3 only to those who also correctly answered No. 2. I disagreed, pointing out that his book is, after all, about the role of randomness in our lives. The final chapter makes a convincing case that much of what society defines as success is due to luck.
Results: 70% got this right. The range of answers was 15% to 100%, the median and mode were 50%, and the mean was 47.5%. Just 21% of readers got both this and No. 2 right.
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I still can't wrap my head around this one.
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I can't read your signature. Sorry.
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