Thursday, November 10, 2011

Girl Named Florida solutions

In The Drunkard's Walk, Leonard Mlodinow presents "The Girl Named Florida Problem":
"In a family with two children, what are the chances, if one of the children is a girl named Florida, that both children are girls?"
I like this problem, and I use it on the first day of my class to introduce the topic of conditional probability.  But I've decided that it's too easy.  To give it a little more punch, I've decided to combine it with the Red-Haired Problem from last week:
In a family with two children, what are the chances, if at least one of the children is a girl with red hair, that both children are girls?
Just like last week, you can make some simplifying assumptions:
About 2% of the world population has red hair.  You can assume that the alleles for red hair are purely recessive.  Also, you can assume that the Red Hair Extinction theory is false, so you can apply the Hardy–Weinberg principle.  And you can ignore the effect of identical twins.
Before I present my solution, I want to sneak up on it with a series of warm-up problems.
  1. P[GG | two children]: if a family has two children, what is the chance that they have two girls?
  2. P[GG | two children, at least one girl]: if we know they have at least one girl, what is the chance that they have two girls?
  3. P[GG | two children, older child is a girl]: if the older child is a girl, what is the chance that they have two girls?
  4. P[GG | two children, at least one is a girl named Florida].
  5. P[GG | two children, at least one is a girl with red hair, and the parents have brown hair].
  6. P[GG | two children, at least one is a girl with red hair].


Problem 1: P[GG | two children]

If we assume that the probability that each child is a girl is 50%, then P[GG | two children] = 1/4.



Problem 2: P[GG | two children, at least one girl]

There are four equally-likely kinds of two child families:  BB, BG, GB and GG.  We know that BB is out, so the conditional probability is

P[GG | at least one girl] = P[GG and at least one girl] / P[at least one girl] = 1/3.



Problem 3: P[GG | two children, older child is a girl]

Now there are only two possible families, GB and GG, so the conditional probability is 1/2.  Informally we can argue that once we know about the older child we can treat the younger child as independent.  But if there's one thing we learn from this problem, it's that our intuition for independence is not reliable.




Problem 4: P[GG | two children, at least one girl named Florida]

Here's the one that makes people's head hurt.  For each child, there are three possibilities, boy, girl not named Florida, and girl named Florida, with these probabilities:

B: 1/2
G: 1/2 - x
GF: x

where x is the unknown percentage of people who are girls named Florida.  Of families with at least one girl named Florida, there are these possible combinations, with these probabilities

B GF: 1/2 x
GF B: 1/2 x
G GF: x (1/2 - x)
GF G: x (1/2 - x)
GF GF: x^2

The highlighted cases have two girls, so the probability we want is the sum of the highlighted cases over the sum of all cases.  With a little algebra, we get:

P(GG | at least one girl named Florida) = (1 - x) / (2 - x)

Assuming that Florida is not a common name, x approaches 0 and the answer approaches 1/2.  So it turns out, surprisingly, that the name of the girl is relevant information.

As x approaches 1/2, the answer converges on 1/3.  For example, if we know that at least one child is a girl with two X chromosomes, x is close to 1/2 and the problem reduces to Problem 2.

If this problem is still making your head hurt, this figure might help:
Here B a boy, Gx is a girl with some property X, and G is a girl who doesn't have that property.  If we select all families with at least one Gx, we get the five blue squares (light and dark).  The families with two girls are the three dark blue squares.

If property X is common, the ratio of dark blue to all blue approaches 1/3.  If X is rare, the same ratio approaches 1/2.


Problem 5: P[GG | two children, at least one girl with red hair, parents have brown hair]


If the parents have brown hair and one of their children has red hair, we know that both parents are heterozygous, so their chance of having a red-haired girl is 1/8.

Using the girl-named-Florida formula, we get

P[GG | two children, at least one girl with red hair, parents have brown hair] = (1 - 1/8) / (2 - 1/8) = 7/15.

And finally:


Problem 6: P[GG | two children, at least one girl with red hair]


In this case we don't know the genotype of the parents.  There are three possibilities: Aa Aa, Aa aa, and aa aa.

We follow these steps:

1) Use the prevalence of red hair to compute the prior probabilities of each parental genotype.

2) Use the evidence (at least one girl with red hair) to compute the posteriors.

3) For each combination, compute the conditional probability.  We have already computed

P(GG | two children, at least one with red hair, Aa Aa) = 7/15

The others are

P(GG | two children, at least one with red hair, Aa aa) = 3/7
P(GG | two children, at least one with red hair, aa aa) = 1/3

4) Apply the law of total probability to get the answer.

I'm too lazy to do the algebra, so I got Mathematica to do it for me.  Here is the notebook with the answer.

In general, if p is the prevalence of red hair alleles,

P(GG | two children, at least one with red hair) = (p^2 + 2p - 7) / (p^2 + 2p - 15)

If the prevalence of red hair is 0.02, then p = sqrt(0.02) =  0.141, and

P(GG | two children, at least one with red hair) = 45.6%

Congratulations to Professor Ted Bunn at the University of Richmond, the only person who submitted a correct answer before the deadline!

At least, I think it's the right answer.  Maybe we both made the same mistake.

For more fun with probability, see Chapter 5 of my book, Think Stats, which you can read here, or buy here.

22 comments:

  1. Unfortunately, these problems are based on the ambiguous usage of the word "given" in the mathematics of probability. To illustrate, consider this Problem 7:

    On a game show, you are offered the choice of three boxes. One contains $1000, but the other two are empty. You choose Box #1, but before opening it, the host says "GIVEN that Box #2 is empty, would you like to switch to Box #3?" Should you?

    Using the events M1, M2, and M3 to indicate where the money is; and E1, E2, and E3 to indicate empty boxes, the methods you used to answer your two-child questions produce:

    P(M1|E2) = P(M1 and E2)/P(E2) = P(M1)/[P(M1)+P(M3)] = 1/2.
    P(M3|E2) = P(M3 and E2)/P(E2) = P(M3)/[P(M1)+P(M3)] = 1/2.

    So there is no benefit to switching. But if, instead, the host had opened Box #2 to show you it was empty, this is a classic problem in probability. The accepted answer is that switching doubles your probability of winning the $1000.

    The difference in the solutions is whether the "given" information, that Box #2 is empty, is a prior constraint on the problem, or if it is the result of a random choice. Did the host tell you about Box #2 because it was the only one he knew was empty, or did he know about all of them and choose an empty one from the two you didn't select? The former situation produces the answer I found above, but the latter requires modeling his choice. In M3, he had no choice; but in M1 he had the choice of two boxes. So if O2 means he opens Box #2:

    P(M1|O2) = P(M1 and O2)/P(O2) = [P(M1)/2]/[P(M1)/2+P(M3)] = 1/3.
    P(M3|O2) = P(M3 and O2)/P(O2) = P(M3)/[P(M1)/2+P(M3)] = 2/3.

    The answers to your Problems #2, $4, #5, and #6 are correct only if we assume the information included in the statement - like "at least one is a girl with red hair" - was used as a requirement to select the family. That is, that the fact was selected first, and then it was determined that the chosen family fit the fact. If, instead, we assume one fact was chosen from the set of one or two similar facts that apply to the family, all of those answers are 1/2.

    I won't try to tell you which model is more reasonable for your problems. But when Martin Gardner popularized this kind of question (almost identical to Problem #2) in the May, 1959 issue of Scientific American, he retracted his answer of 1/3 in the October issue saying it couldn't be answered as written. The "random choice" model is strictly preferred in the Game Show Problem, and it is the better model for how Leonard Mlodinow actually phrased this question in his book. That was more like "You recall that a family of two has two children, and that one is a girl named Florida. But you don't recall whether both are girls." The answer is 1/2 unless you met this family because you were looking for girls named Florida.

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  2. @JeffJo. It's true that English is ambiguous; nevertheless, when people read these problems, they almost always interpret them as intended.

    If you want to disambiguate, read "probability of A given B" as the conditional probability, P(A|B), which is well defined.

    These problem are just for fun. If you interpret them differently and get the "wrong" answer, I promise not to take off any points.

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  3. But "as intended" isn't always right. Like in the Month Hall Problem; people interpret it the same way, and there that interpretation it's wrong.

    It's wrong here, too. P(BB|you know one is a boy)=1/3 requires that P(GG|you know one is a girl)=1, because it assumes you will always know one is a boy in the three cases where it is true.

    It's a bad assumption to make, and it is the reason why your Florida answer is unintuitive. People won't interpret the extra information the same way.

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  4. Oh, I forgot to comment on "If you want to disambiguate, read 'probability of A given B' as the conditional probability, P(A|B), which is well defined." Yes, it is well defined. But that is not the definition you are using, which is my point. The definition is: "The probability the outcome is an element of the set A, given that the outcome is an element of set B and could be any element set B."

    The last part is what you overlooked. The problem statement gave us a necessary condition to define set B, not set B itself. In order to satisfy the definition of a conditional probability, we need both the necessary conditions and sufficient conditions that define it. Saying the family has at least one girl does not mean it could be just any family that has at least one girl.

    If you had asked "Given that a family was selected from all two-child families that have at least one red-headed, left-handed, soccer-playing, girl named Florida," then it is no longer unintuitive why the probability changes as the seemingly unimportant information is added. The two events that partition this well-defined set B - B&{1Girl} and B&{2Girl} - change at different rates because all qualifying families are included and the second one is nearly twice as likely to fit the new information. But if, as readers almost universally interpret at first, you interpret the added information as something that is determined about the girl after the family is selected, and not before, it is not a sufficient condition and the answer can’t change.

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  5. Hi Allen,

    You've made an independence assumption concerning both daughters being named Florida. That's clearly incorrect.

    Neil

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    1. Tell that to George Foreman. He has five sons, all named George.

      More seriously, since Florida is a rare name, this simplification has a very small effect.

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    2. Five sons called George calls more attention to the fact it isn't independent.

      I agree the effect is small (apart from where you take x->1/2), but you also seem interested in small effects in your example.

      To be honest, I think such examples make probability sound a lot more complex than it really is.

      The lesson I'd like students to take from classes is that the first thing they should do is write down all the assumptions they're making in their modelling and then the probabilistic calculus should follow directly from there.

      Delete
    3. The effect is bigger than you suppose, using your requirement that there be a "Florida" in the family before selection. Let a "G" in the family ordering event indicate any girl, and an "X" indicates any child. Assume the gender of Florida's sibling is independent; that is, P(GG|FX)=P(GG|XF)=1/2. If the event F means there is a Florida in the family, Bayes' Rule says:

      P(GG&FX|F) = P(GG|FX)*P(FX|F) = P(FX|F)/2
      P(GG&XF|F) = P(GG|XF)*P(XF|F) = P(XF|F)/2.

      Since FX and XF are the only two ways for a family to have a girl named Florida, and they are independent, their probabilities must sum to 1, so:

      P(GG|F) = P(GG&FX|F) + P(GG&XF|F)
      = P(FX|F)/2 + P(XF|F)/2
      = 1/2.

      The only flaw in this derivation is that, while P(GG|FX) is 1/2, P(GG|XF) is not. If we similarly disallow any other name to be duplicated, the chances a younger-child Florida will have a sister are greater than 1/2. This is because more common names are removed from the name pool, than rare names, when the older child is a girl. Taking that into account, it's not too hard to show that P(GG|F) = (2+C-F)/(4+C-F), where F is the probability the first girl (whether or not she is the first child) will be named Florida, and C a sort of "average probability" over all girl's names.

      But then, for Mlodinow's actual question, there is no requirement to recall "Florida" just because there is one in a family. The answer is always 1/2 because you are recalling a specific child.

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    4. I still think that the assumption of independence has little or no effect on the results. As an example, let's look at one possible kind of dependence:

      Let's say that the probability that the first child is GF is x, and that if the first child is GF, then the probability that the second child is GF is ax for some value of a.

      In that case:

      B GF: 1/2 x (same as before)
      GF B: x 1/2 (same as before)
      G GF: (1/2 - x) x (same as before)
      GF G: x (1/2 - ax)
      GF GF: ax^2

      So the probability of two girls, given that at least one child is a girl named Florida is (1-x) / (2-x), which is the same as what we got before, regardless of the value of a.

      Can you give an example of a dependence that makes a difference?

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  6. I did give examples. If you assume there can't be two girls named Florida in the same family (this is the approximation we are debating the effect of), but ignore the duplication of other names, then the probability of two girls, given a Florida, is exactly 1/2. Regardless of how common or rare "Florida" is as a name.

    If you disallow any duplicated names, the probability of two girls, given a Florida, is greater than 1/2 by an amount that depends more on the average name's probability than Florida's probability. It is (2+C-F)/(4+C-F) = 1/2 ~= 1/2+(C-F)/12. (Note that your x is my F/2). I'm omitting the derivation because it is detailed, not because it is difficult.

    You did a couple of things wrong. You want to adjust the probability the second *girl* is named Florida, not the second child, and you allowed two Florida's. Using your notation, you should get:

    B GF: x/2
    GF B: x/2
    GF G: x/2 (Because the second girl can't be named Florida)
    G GF: ax/2
    GF GF: 0 (Because you can't have two girls named Florida)

    P(two girls|Florida) = (1+a)/(3+a)~= 1/2+(a-1)/12. And note that a>1, so this is greater than 1/2.

    As an example that shows a>1, consider a world with only three girl's names: Mary, Ann, and Beth. All else being equal, "Mary" is twice as popular as either "Ann" or "Beth," which are equally popular. The first girl in a family has a 1/2 chance to be named "Mary," and 1/4 chance to be named "Ann" or "Beth." But for the second girl, half of the time she can't be named "Mary." If she can, there is a 2/3 chance she will, but that means the overall probability a second girl will be named "Mary" is (1/2)*(0)+(1/2)*(2/3)=1/3. By symmetry, the chances a second girl will be named "Ann" or "Beth" is 1/3 as well. That makes the "a" for the common name Mary (1/3)/(1/2)=2/3, and the "a" for a rare name like Ann or Beth (1/3)/(1/4)=4/3.

    But all this still assumes that "knowing" one child is a girl is equivalent to selecting the family from all families with a girl, and it is not. Look up Bertrand's Box Paradox - it's the same thing, only with three family/box types instead of four. Or google for an article by Professors Marks and Smith from Pomona College, Department of Economics.

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    1. First, you can't prove that a>1. a is an arbitrary constant I defined to describe a kind of dependence. For any value of a, the probability of two girls, given that at least one child is a girl named Florida is (1-x) / (2-x), which is the same as what you get if you assume there is no dependence.

      If you disallow two girls named Florida, that's a special case of the analysis I did, with a=0.

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  7. I'm sorry, I was assuming you were trying to address what I had said. Youy have to be careful when you start assuming dependence, because you have to include it everywhere. Your mistake is "G GF: (1/2 - x) x (same as before)". Allowing two girls in the same family to have the same name means that you have to do it for every name, including the common ones. And for that to happen, you need to change the probability that this second child can be named Florida, so that second-girls can duplicate their older sister's common name!

    That might be easier to see if you calculate P(two girls|younger child is a Florida) by your values. It's (1/2+ax-x)/(1+ax-x), which is less than 1/2. Why should the second child's name affect the older child's gender? In fact, it can't. This probability must be 1/2, and it is the similarly dependent term for "G GF" that you need to add to both the numerator and the denominator, that will make it 1/2.


    Also note that "the same as what we got before, regardless of the value of a" can't be true, because my much simpler - and so easier to see that it is true - derivations when your a=0 shows that tehanswer is not the same. It is 1/2.

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    1. With a whiteboard and higher-bandwidth communication, I'm sure we could straighten this out quickly, but for now I have to stop. Thanks very much for the comments!

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  8. Well, it would take less bandwidth if you would allow for a two-way discussion, instead of dismissing the problems I find with your solution and refusing to address my solution which has no similar flaws. But I'm stubborn, so I'll try once again.

    Consider a set of girls names N(i) with probability P(i), where the P's sum to 1/2 (so we can keep your notation, which really is more cumbersome than mine). And then construct a square matrix of all possibilities, where we use case i for your x and the sequence of all the other cases j's for your G. The equivalences are:

    1) "B GF: x/2" becomes "B N(i): P(i)/2".

    2) "GF B: x/2" becomes "N(i) B: P(i)/2".

    3) "G GF: (1/2 - x) x" becomes "N(j) N(i): P(j)*P(i)". Note that the sum of P(j), over all j except i, is (1/2-P(i)), so this matches your result.

    4) "GF G: x (1/2-a*x)" becomes "N(i) N(j): ???"; wait a minute. We already have a term for this. It's P(i)*P(j). And it doesn’t involve an "a".

    5) "GF GF: a*x^2" becomes "N(i) N(i): ???"; we have a required result here, too. If we keep steps 3 and 4 - which must give the same result when you reverse i and j - as they are, "a" has to be 1 for the set of all combinations to sum to a probability of 1. If you want "a" to be near zero, you must adjust steps 3 and 4 to account for the dependencies you are assuming. Essentially, the sample space you have constructed is not part of a consistent probability space, so any further probabilities you derive from it are invalid.

    However, if we assume N(i) N(j) is the same whether we start with N(i) as the first child, or with N(j) as the second - and I don’t see how we can't - then we get different results. I've shown you two results. Well, I've shown one and stated the other. I admit that there is an assumption related to dependence that is buried in my analysis which shows the answer is 1/2, but it is a more reasonable one. It is essentially that parents plan for all possible name combinations before any children are born. This means that the probability for any girl to be named Florida is the same regardless of her position in the family. Under that assumption, P(2girls|a Florida is required to be in the family) is 1/2. If we try to assign names sequentially, this probability is greater than 1/2.

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    1. I am confused about what we are disagreeing about. Here's what I think we have so far:

      1) If we make the simplifying assumption that there is no dependence between the names of siblings, the answer is (1-x) / (2-x), which is 1/2 for small x.

      2) Neil said that the independence assumption is wrong.

      3) I said I don't think the independence assumption makes much difference. Then I proposed a model of one possible kind of dependence and showed that the answer is still 1/2. I asked if there is some other kind of dependence that yields a different answer.

      4) If I understand your most recent reply, you propose a different model of dependence, and show that the answer is still 1/2.

      So your model supports my assertion that the independence assumption has little or no effect on the answer.

      Are we both disagreeing with Neil, and agreeing with each other, or am I missing something?

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  9. 1) Does the problem mean "A family was selected at random from all families of two that include a girl named Florida," or " A family was selected at random from all families of two, and it was observed that it includes a girl named Florida." The answer to the former is some variation of (1-x)/(2-x). The answer to the latter is 1/2. And the reason the variations on (1-x)/(2/x) seem unintuitive, is because the latter is the more reasonable interpretation. This was first discussed by Joseph Bertrand in 1889, in his famous "Box Paradox," and repeated by Martin Gardner when he popularized the Two Children Problem in 1959.

    2) It is not reasonable at all to model for duplicated names. The one example you have is the exception that proves the rule.

    3) But if you must use the former interpretation, and allow some duplicated names, there are two effects of reducing the assumption of independence, not just the one you are modeling. That one is small but not negligible if you do it right. The other is much larger.

    To see it, repeat your analysis with two "fudge" terms. I'll use B to represent any boy, G for any girl (Florida or not), F for Florida, and N for non-Florida. This lets me represent a more useful set of families. Let x1 be the probability of a Florida entering a family without a G already, x2 when there is an N already, and y when there is an F already. (You essentially used x1=x2=x, and y=a*x.) Now we get:

    P(FB) = x1/2 (same as your solution)
    P(BF) = x1/2 (same as your solution)
    P(FG) = x1/2 (sum of two terms in your solution)
    P(FF) = x1*y (the second term for the above, but not needed in the answer)
    P(NF) = x2/2-x1*x2 (essentially the same as your solution)
    Answer = (x1+x2-2*x1*x2)/(3*x1+x2-2*x1*x2)
    = (1-x)/(2-x) if x1=x2=x.

    You are right that y=a*x does not factor in this answer. But the difference between x1 and x2 does, it is also dependent on what you assume for a, and it is greater than the effect of x itself.

    As an approximation, x2=x1/(1-2*c)~=x1*(1+2*c), where c is the probability for a girl with a representative name. It increases because a name must be removed, at least partially, from the possible names for the second girl; yet the remaining ones still must sum to 1/2. Putting these into the answer should give (1+c-x)/(2+c-x), which is greater than 1/2 when c>x. And c is much bigger than x. (In fact, I can write a definition for c that makes this the exact answer.)

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  10. Part 1:

    Here's an "urn" problem that I call the Generic Bertrand's Box Problem. Suppose an urn contains many identical boxes. The contents of each box can always be described as either "gold" or "silver," and some can be both. These properties are symmetric - the probability a box's contents is just gold is the same as the probability a box's contents are silver. Call this probability X. Then, the probability a box has just one property is 2X, and the probability it has both is 1-2X.

    Suppose I pick a box at random and, upon examining its contents, I tell you it is gold. What is the probability it has just one color? Many people are tempted to say that, since X boxes are eliminated this way, the probability is X/(1-X). But if this is true, the probability would also have to be X/(1-X) if I had told you it was silver. And if it is X/(1/-X) regardless of what color I tell you it has, it is X/(1-x) even if I don’t tell you a color. Yet we know this probability is 2X.

    This apparent paradox is resolved by recognizing that the condition in the problem is not that the box is gold, but that *I* *told* *you* *it* *was* *gold.* Since I have to choose what color to tell you about when a box has two colors, some boxes that are gold may be eliminated as well. The probability is actually X/(X+P(1-2X)), where P is the probability I would tell you a such a box is gold (which means the probability saying it is silver is 1-P). And if P=1-P=1/2, the answer reduces to 2X.

    In 1889, French Mathematician Joseph Bertrand formulated the first example of this problem, and the argument about paradox, as a cautionary tale about how not to solve conditional probability problems. You must always take into account how the information was obtained. He used exactly three boxes; one with two gold coins, one with two silver, and one with one of each. This way, X=1/3. He asked for the probability the both coins were the color of a single coin that was withdrawn and its color observed. Since most people view the two boxes that remain possibilities as equivalent, they answer (1/3)/(1-1/3)=1/2. Bertrand showed that the answer is 2X=2/3, eschewing the pedantic possibility that the selection of a coin in the box was biased.

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  11. Part 2:

    The famous Monty Hall Problem is another example; in fact, it is numerically equivalent, although the logical equivalence is not as intuitive as it could be. Textbooks that discuss such probability problems often compare the two. "Gold" is "the door to the right of the door you choose, utilizing wraparound if necessary, has a goat." "Silver" is "the door to the left has a goat." Every game has one of these properties, and one in three has each possible combination, so X=1/3.

    Due to the laws governing game shows in the United States, and everywhere else I hope, P=1/2. So even after Monty hall reveals a goat, the chances are still 1/3 that your door has the car, 0 that the opened door has the car, and so 2/3 that the remaining door has the car. Given the choice to switch, you should. (Some people will make the point that no matter what P is, you cannot decrease your chances by switching. Since P~=1/2 is not a practical possibility, and since the problem statement usually does not allow for knowing what it is even if it is not 1/2, there is no reason to suppose P is not 1/2 except to claim that you don’t need to assume any value. But I doubt anybody would suggest assuming P=1 was reasonable.)

    The original Two-Child Problem is also an example. "Gold" is "There is at least one girl," and "Silver" (maybe it should be "Bronze"?) is "There is at least one boy." With four possibilities, X=1/4. The incorrect answer is (1/4)/(1-1/4)=1/3. The correct answer is (1/4)/((1/4)+P/2) = 1/(1+2P). And given no reason to suppose P and 1-P can be different, one should assume P=1/2 and the answer is 1/2. Certainly, there is no reason to assume P=1.

    But when textbooks present all three problems, they usually do fall into the trap Bertrand warned against for the Two Child Problem. The authors of such texts have not tried to defend their answers by defending their implicit choice of P=1; in fact, they do not recognize the need for choosing a P, even when they do recognize I for the Monty Hall Problem. Instead, they perpetuate the myth that they provided a correct answer by adding an auxiliary condition like "Girl named Florida" or "boy born on a Tuesday." Continuing to implicitly assume P=1, they show the answer is (2-A)/(4-A), where A is the probability a (properly-gendered) child will meet the auxiliary condition. This answer meets with almost universal rejection, because it requires a "P=1"-like bias not only toward the gender, but the auxiliary condition as well. Which is intuitively absurd. Often they include other faulty logic designed to justify their answer.

    Yet until these authors recognize the assumptions they are implicitly making, and heed Joseph Bertrand's 120-year-old warning, they will continue to misguide their students.

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    1. This is interesting stuff, and I don't mind if you post it here, but you might want to consider creating your own blog.

      Delete
  12. The girl named Florida part is almost certainly wrong. The question is, given a two children with one a girl named Florida, what is the probability of two girls?

    But you define the mutually exclusive and exhaustive events: B, G, GF and ask for P(G, G| GF) which must be zero. It's even easier to see if the events are B, GNF, and GF in which case P(G)= P(GNF) + P(GF). There's a linguistic error between the problem and solution.

    The solution also fails common sense. Any attribute of girls with probability x must be true of some and not of others: girls born on Tuesdays, girls born on rainy days...but what about girls whose best friends are named Courtney? Girls who ate chocolate last week but not last month in the car going to school. this information, however slight, affects the probability of there being two girls?

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    1. I withdraw my objections. The blog is right. Sigh.

      I finally had to simulate the sucker to believe it and I think I have a convincing story.

      I think we severely underestimate size of the ratio of (GF, GNF) + (GNF, GF) + (GNF,GNF) to (GF, GNF) + (GNF, GF) + (GNF,GNF) + (GF, B) + (B, GF) when GF is rare. It really does approach simply the probability of the other child being a girl or boy.

      At the other extreme, if the characteristic provides no information whatsoever, then P(GF) approaches P(G) and we're back to P(GG|given 1 G).

      Cheers.

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    2. Hi Steve, I'm glad you got a chance to work it out. Running simulations is a good idea -- maybe I will write a follow-up with some code so people can download it and run it themselves.

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