Zero, because such a family would bring great shame upon the father whose seed is too impure to produce a male heir. He would then be honor-bound to abandon one of the shameful children, and replace her with a stolen gypsy boy, claiming him as his own.

Here is one way of looking at it.

There are N families with exactly two children each. N/4 have two boys, N/2 have a boy and a girl, N/4 have two girls.

Of the N girls, K are named "Florida." They are in three kinds of families:

K1 of them have a sister named "Florida."
K2 of them has a sister with another name.
K - K1 - K2 of them have a brother.

(K1/2 + K2) / (K - K1/2) is the fraction of families with with daughters named Florida that consist of two girls. If we agree to call this the probability that a family with a girl named Florida will have two girls, then we avoid a lot of the technicalities and deviant solutions. The only remaining questions are what are the expected values of K1 and K2 given K.

Suppose I pick K girls at random to be named Florida. I expect K1 = K*(K-1)/[2*(N-1)], K2 = 2*K*(N-K)/[4*(N-1)]. In that case, the probability is:

{K*(K-1)/[4*(N-1)] + 2*K*(N-K)/[4*(N-1)]} / {K - K*(K-1)/[4*(N-1)]}
=[(K-1) + 2*(N-K)] / [4*(N - 1) - (K-1)]
=(2*N - K - 1) / (4*N - K - 3)

This is always between 1/2 and 1/3, and decreasing in K. For K = 1 it is 1/2, for K = N it is 1/3.

Some people want to set K1 = 0. In that case, there are different ways to distribute the extra Floridas. One way is to pick K families with girls at random, then name one of their girls (or their only girl if they only have one) Florida. In that case, K2 is expected to be K/3 and the probability is 1/3 for all K.

There are other ways to pick the Floridas, which can give different answers.

Note that I have modified the problem by moving the supposition of a family having two children outside the first question so that the supposition can be valid for all three questions. I also changed the first instance of the word a to the in in the third question, in order to clarify that the same family was being referenced.


Assumptions:

1. The genders are not independent.

2. A boy or girl is equally likely is equally likely, except in the case of identical twins.

3. Identical twins are equally likely to be either both boys or both girls.



Let P(t)=P(identical twins)

Let the sample space S={BB BG GB GG}

Note that P(BB) = P(GG) and P(BG) = P(GB)



1. The probability that both are girls = P(GG)

= (1/4)*P(children are not identical twins)+(1/2)*P(children are identical twins)

= (1-P(t))/4+P(t)/2 = (1+P(t))/4


2. The probability of having two girls if at least one of the children is a girl
= P(GG|GB U BG U GG) = P(GG)/P(GB U BG U GG)
= P(GG)/(1-P(BB))

= [(1+P(t))/4]/[1-(1+P(t))/4] = (P(t)+1)/(3-P(t))


3. What is the probability of a family having two girls if one of the children is a girl named Florida? This question is intended to be tricky. We are given the name of the given girl, but that is not enough information to be useful. We cannot assume that only one of the children has the name Florida. Consider George Foreman's sons all being named George. A more likely possibility is that Florida is the family name rather than a first name. We simply do not know. Thus, the probability for question 3 is the same as question 2, the probability that both children are girls, given that at least one child is a girl.

P(t) should actually be P(monozygotic multiple birth) in order to include twins, triplets, etc. It's worth noting that, in reality, such births are more likely to result in female children.

D'oh! Never mind about triplets, etc., since there are only two children.

I hate to bring this up, but there is a possibility that one (or
more) of the children died, resulting in exactly two "living"
children, so your idea is valid.

I've rewritten the question to better illustrate why knowing the name and gender of one of the children does not affect the odds:

Two copper Lincoln pennies are flipped sequentially (and neither lands on edge).

1. What is the probability that both pennies landed heads up?

2. Given that one of the pennies landed heads up, what is the probability that both pennies landed heads up?

3. Given that one of the pennies landed heads up and was minted in 1943, what is the probability that both pennies landed heads up?

Even with the knowledge that 1943 copper pennies are extremely rare, it is easy to recognize that knowing the date of the penny that we know landed heads up does not affect the probability that that penny was flipped first or second or the probability of the other penny landing heads up.


Now, back to children:

Ignoring the possibility of twins and assuming the probability of any particular child being female to be 1/2, given a family with two children, the possibilities of gender that make up our sample space are:

{BB, BG, GB, GG}.

Each possibility is equally likely, i.e. P(BB) = P(BG) = P(GB) = P(GG) = 1/4.


1. What's the probability of both children being girls?

Only GG satisfies that condition, so the probability is 1/4.


2. Given that one of the children is a girl, what is the probability that both children are girls?

Since we are given the information that one of the children is female, eliminating the possibility of BB, our sample space becomes:

{BG, GB, GG}.

Each possibility is equally likely. Since only GG satisfies our condition of both children being girls, the probability is 1/3.

3. Given that one of the children is a girl named Florida, what is the probability that both children are females. Here, our sample space doesn't change from the previous question. (Note, this is where the author of the problem, Leonard Mlodinow, erred in his logic.) To understand why, it might be helpful to consider each possibility in the sample space to be a subset of the sample space:

BG = {BfGf BfGnf BnfGf BnfGnf};
GB = {GfBf GfBnf GnfBf GnfBnf};
GG = {GfGf GfGnf GnfGf GnfGnf},

where f implies the child has the name Florida, and nf implies not named Florida. So,

{BG, GB, GG} = {{BfGf BfGnf BnfGf BnfGnf} {GfBf GfBnf GnfBf GnfBnf} {GfGf GfGnf GnfGf GnfGnf}}.

Each subset of the sample space has a total probability of 1/3. Eliminating the possibilities that do not contain a girl named Florida, we are left with:

{{BfGf BnfGf} {GfBf GfBnf} {GfGf GfGnf GnfGf}}

We can further simplify the sample space by disregarding the name of the possible male child, leaving:

{{BGf} {GfB} {GfGf GfGnf GnfGf}}

Eliminating possibilities within the subsets of the sample space does not affect the total probability of any subset, only the possibilities within that subset. (In fact, we use the same logic to correctly solve the Monty Hall problem.) So, we have:

P(BGf) = 1/3
P(GfB) = 1/3
P(GfGf)+P(GfGnf)+P(GnfGf) = 1/3

Since only the possibilities in the last subset of our sample space contains two girls, the probability of both children being girls, given one child is a girl named Florida, is 1/3.

FYI The official correct answer to the girl named Florida puzzle has been given out by the Wall Street Journal as 1/2.

For more info see question 3 at
http://blogs.wsj.com/numbersguy/prob...d-winners-321/

The OP linked to the quiz in the WSJ. That the official answer is lim(P(GG|one child is a girl named Florida),P(F),0) = 1/2 does not imply that the correct answer is lim(P(GG|one child is a girl named Florida),P(F),0) = 1/2.

However, I have passed the problem w/ official answer along to a couple Statistics professors to get their opinion. The initial response, before having the opportunity to closely examine Dr. Mlodinow's logic, is that knowing the name in the third part of the question does not alter the probability from part two.

And, in fact, the correct answer turns out to be 1/3. (I worked it out on the way to class this morning.)

Mathematically:

Let S = {BB, BG, GB. GG} be the order specific set of possible genders of two children, boy or girl.

P(BB) = P(BG) = P(GB) = P(GG) = 1/4

Let r = {ff, fn, nf, nn} be the possibility of a child being named Florida.

Let R = {F, N} be a subset of r, such that F = {ff, fn, nf} and N = {nn}.

We know the family has two children, one of which is named Florida.

Thus:

P(F|BG) = P(F|GB) = P(F|GG) = 1.

P(BG∩F) = P(BG)*P(F|BG) = 1/4.

P(GB∩F) = P(GB)*P(F|GB) = 1/4.

P(GG∩F) = P(GG)*P(F|GG) = 1/4.

P(GG∩F)/(P(GG∩F)+P(GB∩F)+P(BG∩F)) = (1/4)/(3/4) = 1/3.


The trick is recognizing that the conditional probabilities are given in the problem. Dr. Mlodinow set up his mathematics as though the name of both children were unknown, but they are not both unknown. We know one of the names, so no matter how rare the name might be, the probability that one of the two children has it is 1. Setting the problem up as he did would be like calculating the odds that exactly one king is flopped in hold 'em with nCr(50,3)*nCr(4,1), after the flop comes 10-K-A. Once we see exactly one king on the flop, the probability that exactly one king comes out on the flop is 1.

In other words, knowing the name and gender of one of the children doesn't change the probabilities at all compared to only knowing the gender of one of the children.

Thanks to Mr. Sklansky for the helpful insight.

LapsedPythagorean, I will explain this way:
Say we know of two unseen children named "Florida" and "Alabama ". We are told Florida is a girl. What are the chances that both are girls? Alabama has an equal chance of being a girl or boy, right? So chances for both being girls has to be 1/2. There is only one way that a boy-girl could occur and one way for two girls occur. This situation is the equivalent to Part 3 of OP's puzzle.

The same reasoning goes for "Given that one of the pennies landed heads up and was minted in 1943, what is the probability that both pennies landed heads up?"
The correct answer=1/2 not 1/3 The non-1943 coin has p(H) of 1/2 while the 1943 penny is fixed at p=1.0 at H. So the chances of both being H must be 1/2.

Your example differs from my example, and both parts of the problem proposed by Dr. Mlodinow, as referenced by the opening post of this thread. You have provided a way to identify the two kids, Alabama and Florida, relative to each other. That does indeed change the probabilities.

That example is like saying that two coins identified as a dime and a quarter (prior to the toss) are both flipped. What is the probability that both coins landed heads up, given that the quarter landed heads up. In that case, the only unknown factor is how the dime landed. There are two possibilities, Heads, or tails.

The problem I proposed with the pennies did not offer such a way to identify the two coins relative to each other. We know that one of the pennies landed heads up, and that penny was minted n 1943. But, we don't know if that penny was flipped first, nor were we given any way to differentiate between the pennies. For all we know, both could have been minted in 1943. Knowing that the penny that we know landed heads up was minted in 1943 offers us no additional useful information.

If we toss two fair coins, there are four possible results. They can land HH, HT, TH, or TT. Each result is equally likely. Note that there is a 25% chance that both coins land heads up (i.e. P(HH) = 1/4), a 25% chance that both coins land tails up (i.e. P(TT) = 1/4), and a 50% chance that one coin lands heads up and one coin lands tails up (i.e. P(HT)+P(TH) = 1/2).

Now, suppose we tossed the same two fair coins, only this time we know that one of the coins landed heads up. We want to know the probability that both coins landed heads up. Note that we don't know which coin landed heads up. Now, instead of four possibilities, we have three possibilities. We can rule out TT, and are left with HH, HT, and TH as possible outcomes, all equally likely. Note that there are two ways for one head and one tail to appears, but only one way for two heads to appear. Therefore, the probability of both coins landing heads up is 1/3. Formally, we would write P(HH|(HH∪HT∪TH)) = P(HH∩(HH∪HT∪TH))/P(HH∪HT∪TH) = P(HH)/P(HH∪HT∪TH) = (1/4)/(3/4) = 1/3.

Once again, imagine the same two fair coins tossed sequentially. This time, let's say that we know the first coin landed heads up. We want to know the probability that both coins landed heads up. This time, we have enough information to eliminate two possibilities, TT and TH, and are left with two possible outcomes, HH and HT, each equally likely. Therefore, the probability that both coins landed heads up, given that the first coin landed heads up, is 1/2. Formally, we write P(HH|(HH∪HT)) = P(HH∩(HH∪HT))/P(HH∪HT) = P(HH)/P(HH∪HT) = (1/4)/(1/2) = 1/2.

The difference between the last example and the one previous is the in the last example, we were able to identify which coin we knew landed heads up, while in the previous example, we only knew that one of the coins landed heads up, but we didn't know which one. That bit of information is relevant when calculating probabilities.

There's no need to take my word for it, though. You can toss two coins a couple hundred times (or more), and record the results. Better yet, write a simple coin tossing simulation program in your favorite programming language. Count the number of times both coins land heads up. Then count the number of tosses in which either coin lands heads up. Divide the first number by the second number. The result should be near 1/3. Now count the number of times in which the first coin landed heads up. Divide the number of HH results by your new number. The result should be near 1/2.

A priori, we have these matchups:

[BB, BG, GB, GG]

thus making it 1/4 that both are girls.

If the father says "One of them is a girl", we can exclude [BB], leaving us with

[BG, GB, GG]

If he says "This one is a girl", pointing at the first one, we can exclude [BB, BG], leaving us with

[GB, GG]

Many people seem to forget this point - there is a difference between identifying a specific one of them that is a girl, and just talking about them as a pair where at least one of them is a girl. In the latter case, it could be any one of them, or both.

So, when we're down to [BG, GB, GG], the probability that both are girls is 1/3, but when we know that a specific one is a girl, and thus are down to [GB, GG], the probability is 1/2.

Now, let's throw the name Florida into the mix.

Still, a priori, we have these matchups:

[BB, BG, GB, GG]

Then, the father tells us that one of them is a girl named Florida. First of all, that allows us to exclude [BB]. The remaining set [BG, GB, GG] can be further specified to:

[[BGf], [GfB], [GfG, GGf]]

that is, the members of the initial set are specified into subsets. This seems to make it 2/4 = 1/2.

I'm not sure about the next part of the solution, but I can imagine that, in fact, each subset is equally likely to be the case, no matter the number of members in it. That is:

P(BGf) = P(GfB) = P([GfG, GGf]) = 1/3
P(GfG) = P(GGf) = P([GfG, GGf]) / 2 = 1/6
P(GG) = P(GfG) + P(GGf) = 1/6 + 1/6 = 1/3

Can't chose sides here. Anyone? Does this bring anything to the discussion?

You seem to have left out [GfGf] as a possibility, but the assumption that only one child can be named Florida won't change the end result of the problem.

As you suggest, each subset is equally likely. The members of each subset are not all equally likely. P(BGf) = P(GfB) = P(GfG) = P(GGf) is not a true statement, but P(BGf) = P(GfB) = P(GfG)+P(GGf) is a true statement.

That assumption gives us the Florida-Not Florida (or Florida-Alabama) way of distinguishing girls and so the answer is 1/2

?? then it follows that the p(GG)=1/2

What's the likelihood of two siblings in the US have different first names? Would .99 or more be about right? Do you think using .99 instead of 1.0 is going to make much difference here? I say No, Hope you agree.

Saying "Forida is a girl" is pointing out a girl and is specifc enough so that p=1/2. Doesn't matter what characteristic someone uses to point to the girl as long as it is unique or close to unique. It could be the "taller one is a girl", the "older one is a girl", "the one whose birthday is Jan 3,2000 is a girl", "the one on the left is a girl" each make p=1/2. The fact that there is a tiny chance (<.001 in the US, I would think) that the names are the same won't affect p much.

If one of the siblings was randomly put behind a curtain and we are asked to guess the sex , we can ask if it is Florida and get a useful answer. Our chances of guessing the sex correctly would be elevated compared to not having any identifier.

Incorrect. That assumption only changes the possibilities from:

{BG={BGf}, GB={GfB}, GG={GfG, GGf, GfGf}}, with P(BG) = P(GB) = P(GG) = 1/3,

to:

{BG={BGf}, GB={GfB}, GG={GfG, GGf}}, with P(BG) = P(GB) = P(GG) = 1/3.

The assumption that GfGf is not a possible result is an invalid assumption. We don't have sufficient information to justify it. However, in Klyka's calculation, that assumption did not affect the total probability of the subset GG, only the probabilities of the individual members of that subset. Since we are only interested in the total probability of that subset, the assumption did not affect Klyka's end result.

Let P(GfG|GG) = P(GGf|GG) = x_1.
Let P(GfGf|GG) = y.

P(GfG|GG)+P(GGf|GG)+P(GfGf|GG) = P(GG|GG) = 2*x_1+y = 1.

If we ignore the possibility of GfGf, and let

P(GfG|GG) = P(GGf|GG) = x_2,

We have:

P(GfG|GG)+P(GGf|GG) = P(GG|GG) = 2*x_2 = 1.

In our first equation, 2*x_1+y = 1, we have one equation with two variables. Solving for x_1, we find x_1 = (1/2)-(y/2)

In our second equation, 2*x_2 = 1, we have one equation with one variable. Solving for x_2, we find x_2 = 1/2.

All this tells us is that if Florida has a sister, the probability that that sister is named Florida is y. We weren't given any other information, and we cannot apply information that we don't have.

Consider the mathematical solution given by Dr. Mlodinow (emphasis mine --LP):

Since there seems to be some confusion over the girl-named-Florida problem, here is a more detailed mathematical solution:
We ask is the probability that a family with two children has two girls if we know that they have a girl named Florida.
Suppose x = probability that a girl is named Florida.
Write B for boy, GF for girl named Florida, and GN for girl not named Florida.
Then before we are told anything about the family we can write the sample space and probabilities as:
B-B p=.25
GF-B p=.25x
GN-B p=.25* (1-x)
B-GF p=.25x
B-GN p=.25 (1-x)
GF-GN p=.25x(1-x)
GN-GF p=.25x(1-x)
GF-GF p=.25 x*x
GN-GN p=.25 (1-x)*(1-x)
Now, given that one of the children is a girl named Florida, we can rule out many of these. We have left:
GF-B p=.25x
B-GF p=.25x
GF-GN p=.25x(1-x)
GN-GF p=.25x(1-x)
GF-GF p=.25 x*x
According to the rules of conditional probability the chances a family has two girls if it has a girl named Florida are thus:
(Total Probability of GF-GN, GN-GF, and GF-GF) / (Total Probability of all 5 of the above events) =
= .25x*x + 2*.25x*(1-x) / [.25x*x + 2*.25x*(1-x) +2*.25x]
= [2x-x*x] / [4x – x*x]
[= [2-x] / [4-x]]
note that this approaches ½ as x approaches 0 (i.e., the odds approach ½ if the name is extremely rare, which is what the problem assumed).
note also that the answer approaches 1/3 as x approaches 1 (i.e., the odds would be 1/3 – the same as the 2-daughter problem - if all girls were named Florida).
You might think of it this way: if you know the family has a child with a rare girl’s name, then you might suspect that they have two girls because that doubles the odds that one of their children will have the rare name.

With respect to maths, Dr. Mlodinow's model works well. Where he makes his mistake is in not recognizing that we know x = 1, since it was defined as the probability of something that we already know to be true (it was given in the problem).

We have to be able to positively differentiate one from the other. Knowing one is named Florida does not imply the other is not named Florida. Knowing that one was born on January 3rd, 2000 does not imply the other has a different birthday. The probability of both children being girls remains 1/3.

On the other hand, If we are asked the probability of both of two children being girls if we know the taller child is a girl, we have enough information to rule out two of four possibilities. Let T indicate the taller child, and S indicate the shorter child.

Before knowing the taller child is female, we have the following possibilities:

{T:B&S:B, T:B&S:G, T:G&S:B, T:G&S:G}

Knowing that the taller child is a girl leaves us with the following possibilities:

{T:G&S:B, T:G&S:G}

The probability that the shorter child is a girl is 1/2. Likewise, knowing that the older of two children is a girl, or the leftmost of two children is a girl also offers sufficient information to rule out two possibilities instead of only one.

Yes, but again, that scenario offers more information than is given in the problem. Knowing which of two children is Florida (combined with knowing Florida is a girl) changes the probability that both children are girls, increasing it from 1/3 to 1/2.

D'oh! I missed the edit window.

Dr. Mlodinow's model isn't accurate for the individual elements of the GG subset, as it implies that if both children are girls, they must both be named Florida. With respect to the subset GG without Gn-Gn, it works fine, since 2x-x^2 = 1.

No time for a longer comment, will be back. =)

Saying "one is a girl named Florida" is not the kind of identification that I talked about. It's just as good an "identification" as saying "at least one of them is a girl".

An identification would have to let us know which one is subject to the identification, so we can put that one aside and not look at them as a pair anymore, but at the other one as an individual. Now, instead, we know that one is a girl named Florida, but we have no idea as to which one of them, so we can't put any one of them aside. That is the point of identification.

Bump.

I may need to reconsider this. However, it's late here (actually, early morning), so I won't be very decisive..

"One of them is a girl" leaves us with [BG, GB, GG]. Earlier, I said that "One of them is a girl named Florida" leaves us with the same set, but, the statement may actually be divided into two statements:

"One of them is named Florida", leaving us with [BfB, BfG, GfB, GfG], followed by:

"Florida is a girl", eliminating [BfB, BfG] and leaving us with [GfB, GfG].

Just a thought, not really a belief. I still haven't picked sides.

1 and 2 are easily done by math

3. 50/50.. either the girl is called Florida or not.

Reword the problem to make it easier:

Two balanced one hundred-sided dice are thrown sequentially.

1. What is the probability that both dice came up even (each, not total)?

2. Given that one die came up even, what is the probability that both dice came up even?

3. Given that one die came up "42", what is the probability that both dice came up even?

The answer to number three should be the same as the third part of the "Girl Named Florida" problem if the probability of someone being named Florida is 0.02.

This problem is very confusing. I read Mlodinow's book and I thought I understood the answer - HOWEVER, when I thought that what would happen if there are a million girl's names and all of them are equally probable? Would knowing the name Florida or whatever make any difference - the are all rare? If I were to follow Mr Mlodinow's logic or any of the ones shown here, it should - because the same math holds true. But logic demands then that since all the girl's names are equally probable and that each girl has to have a name, the probability of 2 girls given a girl is 1/2 which we know isn't the case. What gives?

Number 2 is an interesting problem in that it is so easy to get it intuitively wrong. I quickly came to the conclusion of 0.5 when I first saw the problem and then had to do some thinking as to why I intuitively jumped to the wrong conclusion. And it appears that 85% of MBA students in one study also came to the same conclusion (according to wiki). After some further thought I came to the conclusion that the way the problem is presented is a little deceiving in that it does not mimic well the way humans communicate with each other, so therefore it is easy to make the wrong set of assumptions.

For example, if I have two twins and I ask my doctor what their sex is and he/she says: at least one of them is a girl (this is how I heard the problem at first). I will not make the assumption that he/she looked at both kids then determined both their sexes and then just to make it interesting for me and to hide information then said: “one of them is a girl”. If they say “at least one of them is a girl” I will assume they looked at one kid (and one kid only), determined their sex, and then said “at least one of them is a girl”. In that case the answer to the problem becomes 0.5.

The same argument can be made if you meet someone on the street and “Mr. Smith says: ‘I have two children and at least one of them is a boy.’” 99% of people would not communicate this in the manner where they literally meant what they were saying. They would most likely be thinking about one of their children when they were saying it, therefore once again reducing the answer to 0.5. Only very clever statisticians would actually say this statement literally and therefore produce the answer of 1 / 3. This problem is as much psychological as it is mathematical and as a matter of fact to take this problem literally in real life would often lead to use LESS information and therefore to come to less complete conclusions.