Girl Named Florida

As I mentioned in the previous post, by adding the condition that a family with two children has a girl named Florida the odds go from 1:3 to 1:2 that the other child is a girl.

Florida was one of the top 1000 female names between about 1900-1930 according to Mlodinow and the Social Security office. But now let’s say it’s a 1:1,000,000 name for girls. The possibilities for families include (assuming they won’t have two girls named Florida): (b,b), (b,n), (b,F), (n,b),(n,F), (n,n), (F,b), (F,n), where b=boy, n=girl not named Florida, F=girl named Florida.

Since we know the family has a girl named Florida we can throw out (b,b), (b,n), (n,b), and (n,n). That means there are 4 ways to have two children families with a girl named Florida, (b,F), (n,F), (F,b), and (F,n), two ways with boys and two ways without.

For more nuanced analysis of this problem check out:
There once was a girl named Florida (a.k.a Evil problems in probability)
Two-Child Paradox Reborn?

3 thoughts on “Girl Named Florida

  1. This problem is usually asked ambiguously – there are two possible interpretations, with two different answers. It was first popularized by Martin Gardner in the May, 1959 issue of Scientific American, and in the October issue he withdrew the answer 1/3 because of this ambiguity. But Mlodinow ignored it in his book, even though the question, as he asked it, leans more toward an answer of 1/2.

    Does the problem mean: (1) A family is selected at random from all two-child families with at least one girl; or (2) A family is selected at random, and it is observed only after this selection that they have two children and at least one girl.

    The solution Mlodinow uses, of counting the cases as they exist, applies only to #1. In #2, you have to consider the possibility that you would observe “at least one is a boy” when there is also a girl. The proper count is that 1/4 of the time there are two boys and you observe a boy, 1/4 of the time there are a boy and a girl and you observe a boy, 1/4 of the time there are a boy and a girl and you observe a girl, and 1/4 of the time there are two girls you observe a girl. The answer is 1/2, not 1/3.

    If you add in the bit about the girl being named Florida, and ignore the absurdity of two girls in the same family being named Florida (which is not as insignificant as Mlodinow claims), the answer to #1 changes to (2-F)/(4-F), where F is the proportion of girls who get that name. But it means that you choose the name “Florida” first, and then sought a family meeting this additional requirement. This is not a reasonable interpretation of the question Mlodinow presented. The answer changes because a family with two girls is essentially twice as likely to have one named Florida, as compared to a family with only one girl. But the answer to #2 stays the same, which is what one should expect.

    Incidentally, if you disallow duplicate names in a family? Then you have to do it for all possible names. The answer can be shown to depend on the proportion of girls with an “average” name. I call it C, since more probable names can be called “common” and less probable names “uncommon.” The answer, to either question since you are now identifying a specific girl, turns out to be (2+C-F)/(4+C-F), which is greater than 1/2 by a non-negligible amount.

    • Hi Jeff,
      Thanks so much for the reply! To post this I wanted to do the problem myself, self-assigned homework if you will. I was running into some of these same issues you brought up and from the links I posted I found out there are all sorts of questions about the problem and the results.

      What I love, and what I think is really the point of all this is, we all have to be very careful. What appears straight forward is often not. Way different results can sometimes appear with the turn of a phrase in the question. Is this chaotic behavior in the official chaos theory way, i.e. the butterfly effect? And often our intuition is just wrong.

      I’ll have to post the Monty Hall problem someday as that really twists the mind. –Bob

  2. Actually, the Monty Hall Problem and the Two Child Problem are both members of the same family of problems, just with different numbers of cases. The underlying issue was first noticed in 1889 by a French mathematician named Joseph Bertrand, and it is called Bertrand’s Box Paradox.

    The characteristic that defines the family of problems is that the actual case you have – whether it is Bertrand’s box with two coins inside it, a game show decision, or a family of two children – must have at least one of two properties, but can have both. Bertrand’s coins could be gold or silver. There are two doors you didn’t pick that could have a goat (yes, it is critical to look at goats, not cars). A family of two could include boys, or girls.

    Bertrand pointed out that when you learn of *A* property your case has, and you want to find the probability your case has just that property, that (1) the answer can’t be different based one which property you learned about, so (2) it must be the same answer you would give if you hadn’t learned anything. But the answer you get by counting how many cases exist with that property is different, hence the paradox.

    The paradox is resolved by not counting *ALL* of the cases where what you learned is true, but only the cases where that is what you would learn. Count only half of Bertrand’s boxes that have one of each kind of coin. Count only half of the possible games where you originally picked the right door. Count only half of the families with a boy and a girl. But authors like Mlodinow don’t like this answer, even if they compare the Monty Hall Problem to Bertrand’s Box Paradox because the numbers are the same. The reason it is better to switch doors is because Monty Hall could have opened a different door if your original choice was correct, but not if it was incorrect. Which makes “incorrect” more likely.

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