I was a bit surprised to see that this puzzle on Marginal Revolution hadn't made its way to 2+2 -- maybe it's something that's been addressed before -- but I'd be interested in comments from more than a few of you.

From Marginal Revolution and WSJ --

1. Suppose that a family has two children. What is the probability that both are girls?

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

3. Ok, now here is the stumper. What is the probability of a family having two girls if one of the children is a girl named Florida?


The why seems to be the most interesting part of the solution.

1) 1 in four BB BG GB GG

2) 1 in 3 GB BG GG

3) 1 in 2 FloridaB FloridaG

1) and 2) are trivial

3)

Interesting.

I'll have a few reasonable assumptions:

1) The parents won't give identical names to their children
2) "Florida" is a rarer name than average
3) In the context of the question, I'll assume that the
parents "name" children based on some finite probability
space

Let the probabilty that a girl will be named "Florida" be p.
Also, if the first child is a girl not named "Florida", and
the second child is a girl, the probability of the second
girl being named "Florida" will be p/(1-delta) where delta
is strictly positive because of 3) above; more specifically,
if one "female name" is chosen, it can't be chosen again
among finitely many choices. Also, delta should be larger
than p because of 2).

There are three cases (with B=boy, G=girl) with at least
one girl:

BG: p
GB: p
GG: p+(1-p)p/(1-delta)

Now (1-p)p/(1-delta)>p since delta>p.
Thus, the probability in the GG case above is strictly
greater than 2p, say 2p+epsilon.

Then, the probability of two girls given one child is a girl
named Florida is (2p+epsilon)/(2p+2p+epsilon)>1/2.

Thus, the probability is strictly greater than 1/2.

I didn't know that the other children were named Delta and Epsilon.

That changes everything.

The thread on Marginal Revoution is now up to 216 comments -- but has resolved itself as a contest between MikeP and J_Thomas -- neither one of who might be right. I suppose I come here because it isn't being resolved there.

Inquiring minds want to know -- who's right? Fat chance the economists will get it.

I haven't looked at the link carefully, but I'd be pretty sure
that nobody that has posted there is "correct" for question 3.

Usually economists are "incorrect" because their models
aren't "accurate" or they don't include the "correct"
assumptions. I wouldn't trust economists to solve any math
problem (perhaps I'd trust physicists).

I'm quite sure that the "correct" answer to question 3 is
"strictly greater than 1/2". Of course if "Emily" replaced
"Florida" in question 3, the answer would be strictly less
than 1/2. ("Emily" was the most popular choice of names for
baby girls in the U.S. for every year in the last decade.)

Then you do consider the popularity of the name as significant. Which means you must somehow discount any assumption that parents would give their children unique names.

I would say 1/4.

This problem does not specify how I came to know that at least one of the children is a girl. So I will make some mild assumptions about this.

Here is what happened. The two children were standing behind a curtain, out of sight to me. Also behind the curtain was a vagina detector. The children flipped a coin to decide who would step into the vagina detector. Moments later, I heard the vagina detector go off. My conclusion: at least one of the children is a girl.

Now, the solution. Let

A = both are girls
B = the detector went off.

I need P(A|B) = P(A and B)/P(B). I get

P(A and B) = (1/4)*1 = 1/4, and
P(B) = (1/4)*0 + (1/4)*(1/2) + (1/4)*(1/2) + (1/4)*1 = 1/2.

So the answer is: (1/4)/(1/2) = 1/2.

Again, this problem does not specify how I came to know that one of the children is a girl named Florida. So I will make some mild assumptions about this.

Here is what happened. My friend, Larry, stole the passports of the two children. I told him I just wanted to know if there was at least one girl, and I wanted a name. So he decided to look at the passports and: if both were boys, he would tell me; if they were opposite sex, he would give me the girl's passport; if both were girls, he would flip a coin to decide which passport to give me. In the end, he gave me a passport. I opened it and it was for a girl named Florida.

Now, the solution. I will make all the same assumptions as bigpooch. Let

A = both are girls,
B = Larry handed me a passport for a girl named Florida.

I want P(A|B) = P(A and B)/P(B)

I get

P(A and B) = (1/4)[p + (1 - p)p/(1 - delta)](1/2)
= (1/4)[2p + epsilon](1/2),

and

P(B) = (1/4)*0 + (1/4)p + (1/4)p + (1/4)[2p + epsilon](1/2).

So the answer is:

[2p + epsilon]/[2p + 2p + 2p + epsilon],

which is strictly bigger than 1/3, and tends to 1/3 as epsilon goes to 0.

I looked at the link and now believe that the form of the
question below is "less ambiguous" (from a WSJ "quiz"):

http://blogs.wsj.com/numbersguy/prob...-321/#comments


3. 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?


Then, obviously the solution can only be 50%. On the other
hand, it isn't as interesting as question 3 in the OP and the
resulting "correct" solution.

When I was working out the problem, the necessity of
the name being rare would result in a probablity strictly
greater than 1/2. I don't see how your last sentence
makes any sense whatsoever.

The only reason that I could see parents naming a
second child the same name as the first is if the first
child died in early infancy; if the first child's life was
long enough to make any lasting impression on the
parents, it is reasonable to assume that they would
give the second child a different name. The OP's
question in its original form doesn't imply much of
anything about the children other than obvious
assumptions:

1) A child can be either a girl with probability 1/2
or a boy with probability 1/2.
2) Parents name their children

Your answer to question 2 isn't "correct" since "at least" is
in the original question.

I found it useful to abstract this to the following question:

1. You have a bag of 1,000,000 marbles, half of which are red and half of which are blue.
2. The blue marbles have a finer distinction: 50 of the 500,000 are Aqua colored.

Q1. What is the probability that a random, independent draw of 2 marbles are both blue? A1: 1 in 4 because it's equally likely to be RR, RB, BR or BB

Q2. What is the probability that a random, independent draw of 2 marbles, at least one of which is blue, are both blue? A2: 1 in 3, because we know it can't be RR, it is equally likely to be RB, BR or BB

Q3. What is the probability that a random, independent draw of 2 marbles, at least one of which is aqua, are both blue?

A3: This one is tougher, but solvable:

So,

It's not very intuitive that way, though. However, if we express it algebraically, we get:

Where x the percent chance of the "rare" event. Now if we reduce that down as much as we can, we get:

So, we can see from that that as the percent chance of the rare event (x) approaches 0, the function will approach .5, and as x approaches 1, it will approach .25/.75 or .3333

It looks as though this has some ramifications on the Bayes theorom: As the probability of the "Given" side of the statement approaches 0, the probability of the "Dependent" result approaches the probability of it independent of the "Given".

Independence isn't really the case in Q3, so I would argue
that P(BA)<>P(AB) because the names given to the children
are not independent (where BA denotes that "Florida" is the
younger girl). Also, because the names of the children
aren't very likely to be identical (except maybe in the case
where the first died in early infancy), it's not unreasonable
to set P(AA)=0.

Your analysis is good in the sense that it conveys the right
approach to the general question where you know at least
something about one of the girls. The answer to the
general question is likely to be in the range [1/3,1/2] but
not always.

.

Let me clarify. Your detection scheme doesn't show the
equivalence of the two statements:

a) if at least one of the children is a girl
b) the detector went off

but rather only shows half of the cases when exactly one of
the children is a girl.

Almost always in these problems, when a statement that is
made such as "if such and such", we consider the universe
of all such cases. In mathematics, when we say "if such
and such", we can only conclude "then..." if it is true for all
cases, such as "if p is a prime number, p is either 2 or an odd
number".

The answer you gave to question 2 only answers in the
case of b) but a) includes cases where b) isn't true.

I'm certain I don't follow your thinking here ... or in your first response where you answered "strictly greater than 1/2". What conditions would lead you to a conclusion > 1/2?

There are a number of "very reasonable" assumptions
that most people will agree with in my analysis.

The main reason is that if the older girl is given a name
other than "Florida", the younger girl's chance of being
given the name "Florida" is on average higher than the
chance of the older girl being given the name "Florida",
so much so that (1-p)q>p where p is the probability
that a oldest girl is given the name "Florida" and q is
the probability that a second girl is given the name
"Florida" given the oldest girl isn't named "Florida".
The other assumption (a very reasonable one) is that
the parents will not give identical names to the two
children.

For example, if the parents gave the name "Emily" to the
older girl, the choice of names for the second girl is now
restricted to names excluding "Emily" so now the
probability of choosing "Florida" for the second girl's
name should on average be higher than that for the first
girl. This effect is large enough to make it MORE likely
that the second girl is named "Florida" rather than the
oldest girl is named "Florida", i.e, "Florida" is more likely to
be the younger girl's name.

Of course, "Emily" is an extreme choice, being the most
popular name given to female babies for every year in
the last tens years in the U.S. according to government
information. On average, the name of the first girl is
going to be more "popular" than "Florida" and from my
first post in this thread, this leads to the conclusion
that the probability is >1/2. Actually, the result doesn't
technically depend on "Florida" being a rare name, but
assuming this results in the situation described in the
second paragraph above.

This seems to contradict logic. Consider the comment by Hunterotd --

"So, we can see from that that as the percent chance of the rare event (x) approaches 0, the function will approach .5, and as x approaches 1, it will approach .25/.75 or .3333"

Let's look at the case where x approaches 1 -- or better yet is 1. All girls are named Florida. The answer now is 1/3, because knowing her name is Florida provides no more information than knowing she has one head.

Next let's look at the extreme case where we knnow that there is only one girl in the world named Florida -- that should be the only case where the probability is actually 1/2. Knowing that there are any more than 1 girl named Florida should move us closer to x approaching 1.

All other cases must lie between.

Hunterotd hasn't considered the aspects of "naming"
children, i.e., his assumptions aren't as complete. Simply
read and understand my analysis. It's clear from the
analysis that the probability is strictly greater than 1/2.

But your analysis requires interpreting the ordering and probabilities that a name will be given -- when in the original question, the name is a given.

In context, there must be a probability space for the
question to be even "meaningful". The OP is mentioning a
question in the context of a book having something to do
with probabilty; therefore, it is very reasonable to consider
a probabilty space for each of the questions.

Question three is preceded by two questions that are to be
interpreted with the idea of a probability space in mind, so
even though one interpretation is to consider all families
that "in reality" exist (which is a difficult enough gathering
of data similar to that of a census) which I would argue is a
"reasonable" interpretation, it isn't a "correct" interpretation
in context.

For example, one could collect all the data of all families
with precisely two children at a specific time to answer the
first two questions, but that would be simply against the
"spirit" of the question and would be a waste of time if we
have a good enough "model" of the situation that would be
able to answer the question with enough generality if the
"model" has very reasonable assumptions.

Of course a name is a given. But how are names given to
children in the context of a probability space? It's not hard
to assume that the names given to two children of a family
will be different. More subtly, if parents are to choose
names from a list of choices, how do they really do so? Of
course, "in reality", they don't choose a name "randomly"
from the entire universe of names with different weights
assigned to each name; in reality, many families only
consider a very small subset of the universe of names. In
fact, it shouldn't surprise anyone that in many families, this
very small subset doesn't even include "Florida".

In much the same way, we could only consider all of those
families that have precisely two children, one of them
being a girl named "Florida" in all of the world today. Then,
we could give a "real probability" that is actually the case
now, but obviously that would be against the intention of
the question.

Consider all of the solutions that have been examined thus
far; they all assume some probability space so that there
is a "meaningful" answer. They really do assume that the
"model" is such that parents give the name "Florida" to a
girl is with some probability x; however, in my "model" if
a family has precisely two children and the first girl is
given a name other than "Florida", then there is exactly
one fewer choice of names to name the second girl. Since
the first choice was likely to be a much more common name
than "Florida", it's not hard to see that it is more likely that
"Florida" will be a choice for the younger girl when compared
with the older girl if you analyze carefully.

I believe my "model" is more accurate and could be tested
empirically (if there is a big enough sample! - good luck!)
for the collection of "rare girl's names" among families with
precisely two children both of whom are girls, i.e., it really
shouldn't surprise anyone who analyzed this very carefully
that a "rare name" could be slightly more likely given to the
second girl rather than the first. But that would be truly a
"waste of time and resources" since it would be much better
to see that the analysis is "correct".

The reason I thought that question 3 was "interesting" is
that since the girl's name "Florida" is rare, that leads to a
rather interesting result if the analysis is conducted very
carefully: namely, that the probability that both children are
girls is strictly greater than 1/2.

It's not a "scheme." It's what happened. I was there. And after that detector went off, I said to myself, "Okay, so what do I know now? Well, I can rule out boy-boy. Can I rule out anything else? No. Okay, so all I really know is that at least one of the children is a girl." And then I asked myself, "Now what is the probability of having two girls if at least one of the children is a girl?"

First -- let me say, if there was some intent you had when inserting the random carriage return and line feed, then please forgive me for encroaching on your form -- but I truly found it difficult to read -- especially when transposed into a monospaced format.

You state --

"Question three is preceded by two questions that are to be interpreted with the idea of a probability space in mind, so even though one interpretation is to consider all families that 'in reality' exist (which is a difficult enough gathering of data similar to that of a census) which I would argue is a 'reasonable' interpretation, it isn't a 'correct' interpretation in context."

In context, you argue that the interpretation (givens) are reasonable but not correct.

"For example, one could collect all the data of all families with precisely two children at a specific time to answer the first two questions, but that would be simply against the 'spirit' of the question and would be a waste of time if we have a good enough 'model' of the situation that would be able to answer the question with enough generality if the 'model' has very reasonable assumptions."

There are no questions to be answered. The proposition is given to us.

I'm tiring from reading your ramblings.

If you would, please respond with situations (real life or not) that would extend the bounds between 1/3 and 1/2.

But obviously your "detection method" is flawed. Not every
family with exactly two children, at least one being a girl,
will be detected. The question assumes something "absolute"
just like almost any math question: "If such and such is
the case...", we need to consider ALL cases. Your method
of detection will NOT detect every family that has two
children for which at least one is a girl in question two
and would be considered "incorrect".

If you want to subject yourself to the scrutiny of "authority",
take your answer to any sample of tenured professors of
mathematics specializing in probability and do you really
believe they would consider your answer "correct"? Hardly!
Not that "authority" would necessarily be "correct", but in
the discipline of mathematics, and more specifically the
theory of probability, answers to questions like this
(specifically question two) are pretty much black and white.

First, I would like to ask, what is the difference between "correct" and correct (without the quotes)?

Second, why do you want me to detect every family with such-and-such children? I don't even own my own vagina detector, so how could I do that? I just happened to be at the airport when that one family was doing their business. And then I had that train of thought which I related to you above. What do other families and other children have to do with it?

And third, why do I need to think about these so-called imaginary "cases" where the detector doesn't go off? I mean, the detector did go off. Should I also consider imaginary "cases" where unicorns flew down and pulled back the curtain and I saw exactly how many girls there were? What's wrong with just considering the facts as they are?

If my answer is wrong, then I would like to know explicitly what is wrong with the simple train of thought I presented to you above.

Anyway, I only know one professional mathematician that specializes in probability theory, and that is also within a 100 mile radius of me right now. So I will ask him later, as you suggested, and let you know what he says. But I wonder: if it's so black-and-white, then why are all the explanations of why I am wrong so convoluted, involving all these other families and imaginary cases?

Think what question 2 asks, in context , given that
it is in a writing about some book about probability. It's
not asking how we get the knowledge at all. The context
is in a probability space and your "detection method" fails
to find part of the probability space that question 2 refers
to.