Here's an EXTREME example: suppose families can only name
their girls one of: Emily, Grace or Florida. They name them
according to the following probabalistic "weights":

Emily: 90%
Grace: 9%
Florida: 1%

Not just that, any family with two children that are both
girls must give different names by law.

Let B=boy, G=girl and (XY) denote the case where X is the
older child and Y is the younger

Clearly, if there are no girls or one girl in a family with two
children, there isn't anything hard concerning the chances
that a girl is named "Florida":

(BB): no chance of any girl at all (exclude this case)
(BG): older sibling is a boy, younger is a girl so probability
that the girl is named "Florida" is 1% conditioned on the
(1/4) probability of the (BG) combination, i.e., altogether,
the chances are (1/4)%
(GB): similar to (BG) above and chances are (1/4)%

(GG): either the older girl is named "Florida" or the younger
girl is named "Florida".

a) Older girl is named "Florida": 1%
b) Younger girl is named "Florida". If the older girl's name IS
NOT Florida, you can break it down to two possibilities:

1) Emily: now the second girl's name is Florida 1%/(1-0.9)
or 10% of the time because she can't be named Emily
2) Grace: now the second girl's name is Florida 1%/(1-0.09)
or about 1.0989% of the time because she can't be named
Grace

Thus, if the first girl's name IS NOT Florida, altogether, the
probabilities above sum to 90%(10%)+(9%)(1.0989%) or
about 9.0989% of the time (since Emily is the older girl's
name 90% of the time and Grace is the older girl's name 9%
of the time).

Altogether, the probabilities for a) and b) above sum to
about 10.0989%. Then, multiplying by (1/4) gives about
2.525%. Thus, the probabilities of having a girl with the
name "Florida" for each of the cases are:

(BG): 0.25%
(GB): 0.25%
(GG): 2.525%

Then, the probability that both children are girls, GIVEN
that one of the children is a girl named "Florida" is

2.525%/(2.525%+0.25%+0.25%)~83.47%, substantially
higher than 50%.

Of course, it's an artificial example, but it's just to illustrate
the point that the probability can be greater than 1/2.

Other than two quotes? Surely you must know. If you
don't, ask again later.

I never said that I wanted you to. But if you must, go ahead.

Even if somebody wanted you to, why bother asking?

Which airport?

I don't know, but you can always ask around. Do there exist
other families with exactly two children? Are there any with
two children with at least one girl?

It's all imaginary, even the family you referred to.

I didn't hear it. Did you? How does it sound?

The more data the better.

Facts aren't everything unless you're Wittgenstein or a
disciple thereof.


Context.

This is false. Let B(t) be a standard Brownian motion. Let A be the event that B(2)>0. Let C_n(x) be the event that B(1) is between x and x+(1/n). Then P(C_n(x)) --> 0 as n --> infinity, and P(A) = 1/2, but

P(A|C_n(x)) --> P(B(1)+x>0) = Phi(x),

where Phi is the cdf of a standard normal random variable. By adjusting x, we can make this converge to any limit in (0,1) that we please.

Similarly, one can construct examples where the conditional probabilities converge to 0, or 1, or do not converge at all.

This is just silly -- given that the problem explicitly states that there is a girl named Florida. First and second at that point become somewhat meaningless -- and completely meaningless in light of your new law stating that names must be unique.

I would stand on my belief that the answer is between 1/3 and 1/2, approaching 1/2. The exceptions are in cases that we knew all girls are named Florida (which would include 1/3) or where we know that the name Florida is singularly unique (which would include 1/2).

I don't believe that the question, as stated, allows for any probability outside that range.

Obviously, if the first girl's name is NOT named Florida, and
there is a girl of the family named Florida it must be the
second girl. Read the post carefully.

You should understand that "uniqueness of names" was a
reasonable assumption in my first post in this thread. "The
law" was in the EXTREME example to clarify the point.

How can the answer be "approaching 1/2"?

The case of all girls named Florida would be arguably an
even MORE EXTREME example than the one I posted.

Arguably, someone could say that in the case that "Florida
is singularly unique", the probability is either 0 or 1, i.e, we
know "Florida" comes from a specific family and we can
know the certainty of the sex of Florida's sibling (there's
only one "Florida" and thus, only one family), so the
required probability can be conceivably 1 if she has a sister.

You mention it's merely a "belief", a belief that is consistent
with the idea that families are capable of giving the same
name to two children.

Here is the probabilist's response, as promised:

Regarding the second problem:

In my opinion, context has little, if anything, to do with this problem. The problem, as stated, contains virtually everything needed to solve it. The only thing the reader must supply is that the sex of different children are independent, and boys are just as likely as girls. These assumptions seem fairly universal and do not depend on the context in which the problem appeared.

The correct answer is 1/3. This line of reasoning:

makes it look like we are answering the same question, but we are not. There is a flaw in the above, which can be corrected as follows:

We have violated one of the principles of applied probability: one should not add to or ignore any of the available information when forming conditional probabilities.

In this case, we added information, and that changed the result. Equally dangerous, and perhaps more common, is for people to selectively ignore information -- sometimes to simplify the math, sometimes because they believe it is irrelevant, or sometimes because it does not support their desired conclusions.

The fact that we added additional information to the problem is the source of error, and one can see this without making reference to any collectives of families, or any hypothetical occurrences of the experiment with different outcomes.

As for the third problem:

The answer is p_1/(p_1 + 2p_2), where p_1 is the prior probability that one of the children is a girl named Florida, given that the family has two girls, and p_2 is the prior probability that one of the children is a girl named Florida, given that the family has one girl and one boy. This answer is based on the assumptions (evidently implied by the problem) that boys are not named Florida and different children in the same family do not have the same name.

There is not enough information in the problem to determine p_1 and p_2, so a numerical answer cannot be given. However, if we add additional information to the problem, then we can get a numerical answer. But of course, if we do this, then we are not solving the original problem, and our answer will depend on the additional information we provide.

For example, suppose we assume that "Florida" is a name that is never given to the first-born girl in a family. In this case, p_2 = 0, and the answer is 1.

Alternatively, suppose we assume that parents only choose the name "Florida" after they have already had at least one son. In this case, p_1 = 0, and the answer is 0.

Evidently, a whole litany of other imaginative assumptions could be concocted to produce any numerical answer between these two extremes. But none of these assumptions are present in the original problem.

We have established the assumption that families will give their children unique names -- without regard to the assumption that those names are common or rare. This is the equivalent of "I flip two coins of different denominations, the quarter landed on heads."

Do I now need to consider whether the quarter was flipped first or second?

As I see it, the "law" invalidated your point.

The more unique the name, the closer 1/2 approximates the correct answer. When the name is absolutely unique, 1/2 becomes the correct answer.

Your weighted example was intended to exceed a probability of 1/2. My universal example intended to prove there was a confidence limit of 1/3.

You're assuming knowledge we don't have. Obviously "someone" knows it, but I can't see how that has any bearing on probabilities.

In the original problem, they may well have been capable of giving the same name to two children. This is why the answer can only approach 1/2.

Ah, but you've overlooked something! [ But since you
prefaced your explanation with "IMO", you can always
change your opinion! ]

Technically, I'd disagree because question 2, in "total
isolation" from question 1, doesn't quite necessarily have
ALL THE NECESSARY INFORMATION nor the meaning for
which you give the "correct" answer of 1/3. Look at
question two very carefully now:


Now what is the probability of having two girls if at least
one of the children is a girl?


In isolation from question 1, arguably it could mean: out
of all families that have at least one girl, how many (or
what proportion) of those families having at least one girl
would have two girls (and here, this would ambiguously
mean two or more girls). Those families could have
precisely one child, a girl, (in which case it is impossible
for there to be "two girls") or even three or more children
(and the more children, the more likely that family has
"two girls"). Without the context of question 1, we
don't have the information that it is in the context of
those families having precisely two children. Then, we
would have to do some dirty work and find the
distribution of the number of children of families that
have at least one girl in the "real world".

Context!

This I'd consider one of the "correct" solutions, likely the
correct solution! But obviously, NOT in the "spirit" of all the
questions that the author intended (but he's an economist,
right?).

Yes, I know you are a probabilist, but are you the "probabilist"?

In any case, thanks for your posts!

Oh, you're right! Question 2, in total isolation, is not answerable. I think we should change it to

I think this works in isolation. I will tell the probabilist what a moran he is.

By the way...

No offense, but I seem to get a little headache every time you talk about all those other families. I wish there was some way to do probability without them.

Doh! The jig is up, I guess. Well, it's probably for the best, since I feel guilty about frittering away my time here as it is. At least I didn't get bogged down over at Marginal Revolution! Looking at that site really makes me appreciate the community we have here at 2+2. It is what it is thanks to posters like yourself and other intelligent contributors that give of their time. Keep up the good posts.

More on adding information

1. Introduction

Let us consider a slightly wordier version of Question 2 in the OP:

Now, Joe Blow comes along and wants to take a stab at this puzzle. But it is a little too abstract for Joe's mind. Joe likes to think in concrete terms and real-life circumstances. So Joe invents a story to go along with the puzzle. In the story, Joe is on a bus, he meets a man with two children, they have a conversation, the man tells Joe various things, etc, etc, etc. Joe is very careful to ensure that his story does not contradict anything given in the puzzle. Once his story is crafted, Joe solves the puzzle, gets a number, and is quite satisfied.

Soon after, though, Joe discovers that the entire internet is against him. They say he is wrong, his story has a hundred different flaws, and so on. Joe is utterly confused. So he goes back and invents a new story. This time, Joe is outside the family home, standing on the street, a girl walks out the front door, etc, etc, etc. Joe is very careful to ensure that his story does not contradict anything given in the puzzle. Once his story is crafted, Joe solves the puzzle and, to his surprise, he gets a different number! And people are telling him he is still wrong!

Frustrated, Joe declares that the puzzle is flawed and that it does not have a unique numerical solution. He says it is ambiguous and the answer depends on what additional assumptions you make. In this case, though, Joe is wrong. And if he would just stop making up stories, he would be fine. If he would just deal with the puzzle in the abstract as it is given, he would be okay. So where did Joe get this nasty habit of making up stories?

2. Adding information in classical mathematics

Consider the following puzzle:

Here is one legitimate way to solve the puzzle. We know there is a unique solution, so why not just assume y = 4? We are adding a superfluous bit of information to the problem, and this information does not contradict what we are given, so logically this extra assumption cannot affect the answer.

Now we get x + 4 - |x - 4| = 6, which becomes x - 2 = |x - 4|. There are two cases to consider. If x >= 4, then this becomes x - 2 = x - 4, which has no solutions. Otherwise, if x < 4, then this becomes x - 2 = 4 - x, which has one solution: x = 3. So the solution to the puzzle is x = 3.

Now imagine someone else did the same thing, but with y = 5, and they ended up with a different answer. After double- and triple-checking everyone's math, we were certain that no mistake was made. If that were to happen, then we could legitimately conclude that the puzzle is flawed, and there is not a unique numerical answer.

We do this all the time in math puzzles. As long as the puzzle has a unique answer, we are free to add superfluous assumptions to it, as long as they do not contradict the puzzle. Of course, extra assumptions are not really necessary, but if they help us wrap our mind around the problem, then they can serve a useful purpose. This is a helpful habit that we are all accustomed to.

3. Adding information in formal logic

In fact, this habit runs even deeper. Imagine I have a collection of premises, {P_1,...,P_n}, and a potential conclusion, C, that interests me. I would like to know whether

(a) {P_1,...,P_n} implies C, or
(b) {P_1,...,P_n} implies not-C.

Suppose I somehow know that one of (a) or (b) is true. In other words, I somehow know that {P_1,...,P_n} are sufficient to logically determine the truth value of C. In that case, I am free to add any new premise, P, to my collection, as long it is consistent with the others. If I discover that {P_1,...,P_n,P} implies C, then it must also be the case that {P_1,...,P_n} alone implies C. This technique could be quite helpful, if the new premise P makes it easier for me to see the logical connections between the premises.

On the other hand, if I discover two premises, P and P', each consistent with {P_1,...,P_n}, such that {P_1,...,P_n,P} implies C, but {P_1,...,P_n,P'} implies not-C, then I will conclude that my original premises were not actually sufficient to determine the truth value of C.

So the concept of adding superfluous information to well-posed problems in order to aid us in thinking about a problem, is a concept that lies at the very heart of logic itself.

4. Adding information in probability

The example in Section 3 above has an exact analogue in probability theory: if P(C|P_1,...,P_n) = 1, then P(C|P_1,...,P_n,P) = 1; and if P(C|P_1,...,P_n) = 0, then P(C|P_1,...,P_n,P) = 0. In other words, if we are dealing with certainties, then we can feel free to add any superfluous conditions we like, and we will not change those certainties.

However, with a little thought, nearly anyone can recognize that P(C|P_1,...,P_n) = p does not imply P(C|P_1,...,P_n,P) = p when 0 < p < 1. In fact, adding a new condition P will very rarely leave the original probability unchanged. New conditions typically change the probability, and this is why we need formulas such as Bayes' Theorem. So if we are dealing with probabilities that are not 0 or 1, then we cannot add superfluous conditions as we see fit. In fact, we should not add anything whatsoever.

So that fantastic technique that comes from logic itself, and which we use almost instinctively to aid our thought process, fails miserably in a probabilistic setting. This is where Joe Blow fails. His instincts -- the same instincts that are extremely effective in classical math puzzles and problems of formal logic -- drive him to make up a superfluous story to help him wrap his mind around the abstract. But it is exactly that superfluous story which is screwing everything up. The puzzle is well-posed with a unique answer, but to arrive at that answer, one must resist the urge to add concreteness to the problem. One must deal with the abstract nature of it as it is. This seems to be easier said than done for many people.

Thought I'd take a stab at Grunching this:

Since x+y-|x-y| must equal

(a) x+y-(x-y) if (x-y)>=0 OR
(b) x+y+(x-y) if(x-y)<0

When solving (a) we find that y must be equal to 3, and since we know that y>3, we can dismiss (a), leaving us with only (b). If we solve (b), we find that x+y+x-y=6 which implies that 2x + 0y = 6, and that x=3.

Interestingly this solution means that x=3 independent of y. y can be any value here (as long as y>3) since x+y-|x-y| effectively "devolves" into 2x + 0y = 6.

Good post!

Technically, the "analogue" in probability theory doesn't even
necessarily hold in probability spaces, i.e., the statements

if P(C|P_1,...,P_n) = 1, then P(C|P_1,...,P_n,P) = 1; and
if P(C|P_1,...,P_n) = 0, then P(C|P_1,...,P_n,P) = 0

aren't even necessarily true.

For example, if the probability space is isomorphic to the
interval (0,1] and the set S=intersection(P_1,...,P_n) has
positive measure, but S-C={r}, a singleton (with zero
measure), then clearly, P(C|P_1,...,P_n) = 1 (everything
is "C" except one point), but with P={r}, clearly,
P(C|P_1,...,P_n,P) = 0.

In fact, for the above example, instead, S-C could be
any set of measure zero and P could be any subset of
S-C. If the restriction is such that P has positive
measure in S=intersection(P_1,...,P_n), then the first
statement is an appropriate "analogue". Most of the time,
in poker, or "everyday" probability, the spaces are finite
or discrete and not continuous, so usually those two
statements could apply.

Obviously, that doesn't change the main purpose of your
writing section 4; if anything, it even makes your point
that much stronger: i.e, the "miserable" failing that
results from adding "incorrect" assumptions.


Thus, for question 3 in the OP, if there is specifically
one number in the solution, it could be considered as 1/2.
On the other hand, to arrive at that "solution", there is an
"interpretation" about what it means that a girl is "named
Florida" in the context of a probability space. Was question
3 in the OP considered an "unambiguous" question that
would necessarily result in a specific numeric answer?


Now if question two were formulated:

Now what is the probability of having two girls if you know at
least one of the children is a girl from a conversation you had
with one of the parents (and the family has [exactly]
two children)?

this would imply something about how you "know" that is
the case. If you know that is the case, you might think
the answer should be 1/3, but here's some food for thought
(from Wikipedia link:

http://en.wikipedia.org/wiki/Boy_or_Girl
):

An ambiguous real-life version
-----------------------------

Two old classmates, Anna and Brian, meet in the street, not having seen each other since they left school.

Anna asks Brian: "Have you got any children?"
Brian answers: "Yes, I've got two."
Anna: "Do you have a boy?"
Brian: "Yes, I do!"
Here, for some reason, the conversation is cut short.

Formally, this corresponds to the second version as Brian only has told Anna that at least one child is a boy. Accordingly, the probability that Brian has a girl should be 2/3. However, in real conversation, if Brian had two boys, he would be more likely to answer, e.g., "Yes, they are both boys," (Grice's maxim of quantity). The fact that he does not answer like that could reasonably be taken by Anna as a clue increasing her posterior probability of one child being a girl above 2/3. This highlights the need for precision when stating such problems in probability.

The idea is "independence". In context of jason1990's post
about "More on adding information", the main reason one
doesn't even have to find the solution the way you did is
that the "hint" indicates there is a unique solution, so you
can happily pick any y>3 that you want, i.e, the solution is
independent of the valid choices of y.

In probability, the notion of "independence" is important.
If we add a "reasonable" assumption, it better be in such a
way that it is "independent". Sometimes even if we do, it
doesn't really answer the question in the original form. In
some cases, we truly can't give a specific answer. In some
cases, adding a truly "independent" assumption in a
"meaningful" way (i.e., the "restricted" space still has some
positive probability) and solving this new problem can
result in giving a specific answer, but it may not necessarily
be the correct one.

Question 3 in the OP obviously assumes that girls can be
named "Florida". If it is impossible for that to be the case,
then the question has no meaning; technically, one could
even argue that since the conditional event of a girl being
named "Florida" is the empty set, the probability isn't even
defined. So to make progress, it's natural to assume a
probability space wherein there is some positive probability
of a child being named "Florida"; however, such a space is
completely different than that space that is referred to in
questions 1 and 2 in that it has "more structure".

One of the problems with question 3 boils down to how
children are "named". It's not hard to see that a second
child's name is not necessarily independent of how the
first child is named; e.g., if the names are distinct. But
the real problem is what is the "interpretation" of the
probability in terms of the probability space underlying
questions 1 and 2? Unfortunately, the probability space
that we refer to in questions 1 and 2 has nothing to say
about "naming" that question 3 asks about.

In other words, there is really no "correct" answer (one
that gives a specific answer) that is correct (as in correct
in some absolute sense) answer to question 3, even in the
context of questions 1 and 2; however, we can surmise the
intention of the question and "fabricate" some reasonable
assumptions that answer question 3 in the "spirit" of the
question. Arguably, one "correct" answer could be 1/2,
but in no way is it THE correct answer.

Well, they are true whenever those conditional probabilities are well-defined. Remember, P(A|B) is undefined whenever P(B) = 0.

As you surmised, the important thing is to ensure that P(P_1,...,P_n,P) > 0. This is the probabilistic analogue of the requirement in Section 3 that P is consistent with {P_1,...,P_n}, or the requirement in Section 2 that the extra assumption does not contradict any of the givens in the puzzle. So, in fact, this little technical exception actually strengthens the analogy.

Obviously the puzzle-makers thought so. They expected everyone to assume that one child's name is independent of both the other child's name and the other child's sex. I suppose they thought this assumption was so natural and obvious (like the assumption in Questions 1 and 2 that the sexes of the children are independent), that they did not need to say anything about it. But it is far from obvious and, in fact, probably not even true in any reasonable, real-life situation.

This actually illustrates another common pitfall in applied probability that is worth highlighting: the unwarranted assumption of independence. Independence seems to be everywhere in probability, and we are very accustomed to working with it. Independence makes the math easier and, without it, we often cannot even begin to calculate the probabilities we are interested in. So we are quite used to assuming independence, and building it right into our models -- so much so that it can become a habit to assume it is there.

This is a very nice example. I would say it also highlights the need for caution and critical thinking when solving such problems in probability.

Didn't read any responses and didn't want to do ANY reasoning and math and therefore not adding ANYTHING to this post.

Off of pure gut (and my gut has gotten super huge studying for stupid bar), i say

1) 1/4
2) 1/3
3) 1/3

Lots of interesting mathematics posted in this thread (and several interesting formulations that would be correct if you believe their assumptions - for instance the red, blue, and aqua marbles.)

My own first impression, given the ill-defined question as originally posed, is that there is no difference between "one is a girl" and "one is a girl named X" except that in this case, X gives us some information that the family likes to give their children weird names. I don't have any basis for determining if X is more likely to be a first girl's name or a second girl's name.

And please don't mention this to Sarah Palin. (She's the governor of Alaska, and her five kids are named, in alphabetical order, Bristol, Piper, Track, Trig, and Willow. Heaven help us if she has another.)

why ask why?

Answer is 1/2. When informed 1 child is a girl we are left with gg, bg, gb (3 combos). When informed that 1 child is called Florida we still have 3 combos but we know either the boy or girl is named Florida, so we have bf-g , gf-b combos (f stands for Florida). When informed that the girl is the one named Florida it eliminates bf-g leaving us with only gf-b and gg (2 combos). Thus gg is one of two equal choices remaining.

This works just same if we were informed that that the girl was "younger" (instead of named Florida) or was taller or many other characteristics.

Why are they equal?

Because the chances of having a b or g are considered equal so bb gg bg gb are considered equally likely. Its like tossing a fair coin twice. HH TT HT TH are equally likely i hope you agree. If told the second toss resulted in an H then only HH and TH were possible and there is no reason their relative likelihoods have changed. The chances of HH must be 1/2.

Sure, you mentioned bb, gg, bg and gb are equally likely and
we should assume that from the context of questions 1 and 2
because question 1 is implying a probability space for which
there are these four "fundamental events" (nonempty and
whose union is Omega, the entire space) and these each
have equal probability assuming independence and the
probability of a boy = probability of a girl = 1/2.

But look at what you posted:

Question 3 actually states there is a girl named Florida, not
that there is a child named "Florida" and at least one of them
is a girl. Technically, someone could say that bf-gf is also
possible, and you haven't included that possibility. If you
think names should be distinct, then there could be a
difference between gf-g(not f) and g(not f)-gf where the
"symbols" before the "-" refer to the first child. Also, if the
names are necessarily distinct, but the first child is a boy
named "Florida", then the probability of b(any)-gf is strictly
less than gf-b since b(any) includes bf (in which case gf is
not possible).

As you stated, that leaves you with gf-b and gg, but since
you should be consistent with your notation (you mentioned
bb, bg, gb, gg), it is actually gf-b, b-gf and gg. But what
do you mean by gg? It is really gf-g(not f), g(not f)-gf and
gf-gf. There is no reason to believe gf-g(not f) has the same
probability as g(not f)-gf, not to mention that (gf-b union
b-gf) has the same probability as gg = union (gf-g(not f),
g(not f)-gf, gf-gf).

first off question 2 and 3 are the same except with a name but that means nothing u could call someone what ever u want the fact is u have 1 girl already so u have a 50% chance. remember people dont produce boys and girls equal in ratio. so there for u can only calculate the chances of the one childs gender. so i believe the answer is that there is 50% chance the kids will both be girls.

i also noticed that Q's 2&3 say what is the probability of the family having to girls in it they dont say the girls have to be children. so in that case there is 100 percent there is 2 girls in the family.because it does say ONE of the children is a girl. the mother is the other girl in the family. there is no evidence that the mother is dead or away so there for she must account for the second female. there for there is 100%.
in conclusion 1 of those to answers should be right i believe it is number 2. well in my humble opinion anyway.

The puzzle writer meant you to assume that the children have different names even though it is technically possible that they are the same. In coin tossing puzzles it is technically possible that a coin will land on its edge but we tend to ignore that rare event since it is brain teaser for fun.

You can split gg into two parts by name but their sum of probabilties is still equal to gg. It does not matter to the solution of the problem. We are not asked to specify which girl has which name.

Trust me its 100%
read the question carefully
only question 1 ask the probability of having 2 female children
questions 2 & 3 only ask how many girls in the family.