For the discussion of math. Duh.

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vatar
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Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?
http://www.codinghorror.com/blog/archiv ... tml?r=1183

I think the author is wrong here. This is not the same as the Monty Hall Problem. I think the original statement can be simplified to "there is one child, who has a sister." 50%.

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### Re: Boy or Girl Paradox

Sorry, but it doesn't reduce to that. If it had said "Someone has two children and the eldest is a girl," then it would reduce in that way, because order matters, so there are only two possibilities: G,B and G,G. However, it says "one is a girl," so there are three possibilities: G,B, B,G, and G,G. Thus, in two of three cases, she has a brother.

So, rather than reducing to "there is a child who has a sister" it must remain as "there are two children and at least one has a sister."
Last edited by quintopia on Fri Jan 02, 2009 4:56 pm UTC, edited 1 time in total.

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### Re: Boy or Girl Paradox

Right off the top of my head, that when the first child mentioned is a girl, if I was forced to guess, I would say the second child would be "girl." Sort of both mathematics and biology.

There are some couples who will produce one gender over another. When you factor this small probability in when you hear "girl" you're better off guessing "another girl." I believe when you hear "boy" you should guess "another boy" as well.

Edit: But with noticing how the question was aksed, I'll have to agree with the article's result. If they asked "Hey, whats the other child?" Then it would be a different story. But instead they ask "whats the probability NOW for 1 boy 1 girl, after I just told you we already have a girl?"

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### Re: Boy or Girl Paradox

For argument's sake...

Choose one of the children. Keep choosing until you get a girl. Send this girl to another room. What is the sex of the remaining child?

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### Re: Boy or Girl Paradox

I get the argument as well. It is 2/3...

Blatm
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### Re: Boy or Girl Paradox

Looking at it this way helped me:

What are the odds that 2 a family of 2 children has a boy/girl pair? What are the odds of a girl/girl pair? What is the ratio of the odds?

The unfinished game at the bottom is a clearer example of the idea. Given 2 heads and 1 tail, what are the chances that on the next throw Harry wins? What are the odds Harry wins after exactly 2 more throws? The difference that makes this case easier for me to understand is the fact that there are no cases that have to be ruled out (like the boy/boy case in the actual question), so I don't have to deal with ratios (which I don't like).

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### Re: Boy or Girl Paradox

A pregnant woman is expecting twins. She tells us that 1 of those twins is a girl. Does that mean that there is a 2/3 chance that the other is a boy?

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### Re: Boy or Girl Paradox

vatar wrote:A pregnant woman is expecting twins. She tells us that 1 of those twins is a girl. Does that mean that there is a 2/3 chance that the other is a boy?

No because we know the girl must be the older of the two, like quintopia said.

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### Re: Boy or Girl Paradox

What? They're unborn twins. How can one be older? And even then, how do we know the girl she means is that one?

The case of twins is complicated by the fact that a significant fraction of twins are identical, in which case they must be the same sex. (Come to think of it, this complicates the original question too, but you usually ignore that in these problems.)

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### Re: Boy or Girl Paradox

Look at the combinatoric problem, not the biological problem. Ignore that the odds of a child being a girl is not exactly 50-50, and that twins are more likely to be the same sex than opposite. Think of them as coin flips.

I agree with Cosmologicon. They are unborn, so age is irrelevant.

andy11235
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### Re: Boy or Girl Paradox

I get the arguement, but it still feels like a 169.

If the person had 1 boy and 1 girl, it makes sense that they would be as likely to say they have 1 boy instead of 1 girl. This change of behavior brings it back to 50%.

Code: Select all

`   G   B   TGG 1/4 0   1/4GB 1/8 1/8 1/4BG 1/8 1/8 1/4BB 0   1/4 1/4T  1/2 1/2 1`

For each combination of GB/BG, the person is half as likely to make this statement which makes the chance of this happening equal to GG.

Otherwise they are probably talking about a specific child (eg the eldest), which also makes it 50%, or they could mean exactly what the words actually mean - they have exactly one girl (and one boy).

Or we can just assume that their interpretation is the only *correct* one.

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### Re: Boy or Girl Paradox

You are given two binary digits. 1 represent boys, 0 represent girls. "What are the odds that person has a boy and a girl?" is the equivalent of saying "what are the odds that your two binary digits don't match?" Obviously 50-50.

option 1) 11 - match
option 2) 10 - no match
option 3) 01 - no match
option 4) 00 - match

Now I tell you that one of the binary digits is a 0. This clearly eliminates option 1. So that leaves 2, 3, and 4, so 2/3 are no match, right? Wrong! If I tell you what one of the digits is, you only have to guess the other. This additionally eliminates either option 2 or option 3. "What are the odds that person has a boy and a girl?" is no longer equivalent to "what are the odds that your two binary digits don't match?," instead it now becomes "what are the odds that the child I gave you no information about is a boy?," which is still 50-50.

If the question is "what are the odds that two binary digits (one of which is a 0) match?," the answer is 50-50. I have given you complete information about one of the digits, so it is no longer a 2-digit question, it is a one digit question.

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### Re: Boy or Girl Paradox

vatar wrote:If the question is "what are the odds that two binary digits (one of which is a 0) match?," the answer is 50-50. I have given you complete information about one of the digits, so it is no longer a 2-digit question, it is a one digit question.

No, you haven't given me complete information about one of the digits, because you haven't told me if it's the first or second digit.

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### Re: Boy or Girl Paradox

Sets are not ordered. There is a set containing two children, at least one of which is a girl. The set of two boys is not a subset of this set, which is the problem with the 2/3 solution. It is perfectly valid to say "remove a girl from the set." Use an iterator. Iterate through the set, removing the first one that is a girl. This leaves one child in the set, which has a 1/2 chance of being a boy, not a 2/3 chance.

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### Re: Boy or Girl Paradox

If the sets aren't ordered, why did you list four possibilities above? Clearly options 2 and 3 are the same, then.

Look, this is silly to argue about, because you can do the experiment in real life. Pick at random a large collection of families with two children. About 25% of them will have two boys, 25% two girls, and the other 50% a boy and a girl. Throw out the ones that have no girls. Of the families that remain, which is more common: two girls, or a boy and a girl?

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### Re: Boy or Girl Paradox

vatar wrote:instead it now becomes "what are the odds that the child I gave you no information about is a boy?," which is still 50-50.

Not true. This is the key fallacy lying behind all of your arguments: you are not given information about a single child; you are given information about the family. Specifically, you are told that the family contains a girl. This allows you to eliminate all possibilities where none of the children are girls, without altering the relative likelihoods of the other possibilities, by Bayes theorem. Therefore the probability is 2/3, not 1/2.
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### Re: Boy or Girl Paradox

"There are three men on a train. One of them is an economist and one of them is a logician and one of them is a mathematician. And they have just crossed the border into Scotland (I don’t know why they are going to Scotland) and they see a brown cow standing in a field from the window of the train (and the cow is standing parallel to the train). And the economist says, ‘Look, the cows in Scotland are brown.’ And the logician says, ‘No. There are cows in Scotland of which at least one is brown.’ And the mathematician says, ‘No. There is at least one cow in Scotland, of which one side appears to be brown."

Look, this is silly to argue about, because you can do the experiment in real life. Pick at random a large collection of families with two children.

You are arguing the logician point of view. If you do an experiment in real life, there are brown cows, and black cows, and boy cows... In the problem we are given, there is one family, which has one daughter, and another child, not a Pascal triangle of families in a 1:2:1 proportion.

If you are asked to predict the sex of the girl, and the other, there is only one unknown here. You could use 1:2:1 if there were two unknowns. Order is not one of the unknowns, because it isn't necessary. I can say that other = boy in either case of BG or GB.

If the sets aren't ordered, why did you list four possibilities above? Clearly options 2 and 3 are the same, then.

They are the same. A family that has a daughter and a son is not the same as a family that has a son and a daughter?

Think of it in reverse. There is a family with one child, who is either a boy or a girl. (1/2, right?). Add randomly B1 or G1 to the set. They have an additional child, who we are told is a girl. Add G2 to the set. That doesn't change the odds that the original child was a boy.

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### Re: Boy or Girl Paradox

vatar, tell me which of the following statements (if any) you find objectionable.

Bayes' Theorem wrote:P(A|B) = P(B|A)P(A)/P(B)

Statement 1 wrote:Given that a couple has a boy and a girl, the probability that at least one of their children is female is 100%.

Statement 2 wrote:Given that a couple has two children, the probability that they are of different sexes is 50%.

Statement 3 wrote:Given that a couple has two children, the probability at least one of them is female (i.e. that they aren't both male) is 75%.

Statement 4 wrote:100% * 50% / 75% = 2/3

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### Re: Boy or Girl Paradox

@vatar:

Suppose we have two children. What are their possible sexes? Well, suppose we have no girls, i.e. two boys. The probability of this is 0.5*0.5 = 0.25.

On the other hand, suppose we don't have no girls, i.e., we have a girl. The prob of this is 1 - 0.25 = 0.75. Then the other child is a girl or a boy with equal probability. In particular, the probability that it's a girl is 0.5. So the probability for two girls is 0.75*0.5 = 0.375.

So two girls are more likely than two boys!

See the mistake in the reasoning?
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Blatm
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### Re: Boy or Girl Paradox

I think a large problem in understanding this is that it is counter-intuitive for most people, so they're on the defensive, looking to find problems with explanations so they can justify their intuition. If instead they knew the answer, but didn't know why, understanding the reasoning would be much easier. That said, could someone please write a program and run a few simulations?

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### Re: Boy or Girl Paradox

Running simulations is meaningless; at best, it justifies the Bayesian logic. It looks to me like the problem some people are having is with the Bayesian point of view in the first place, that is, on the meaning of the expression [imath]P(A | B)[/imath] in the first place.

vatar, the question of trying to assign a probability to this event in the first place means that this isn't a question about one family; it's a question about what would happen if we took a random sample of families, looked at only the ones with one girl, and computed the proportion with a boy and a girl.

Perhaps, as I believe someone's already mentioned, the result would make more sense if rephrased in terms of coin flips: I flip two coins without telling you the results, and I tell you that at least one of them flipped heads. What is the probability that I flipped a heads and a tails?

Perhaps the result would make more sense if, like in the Monty Hall problem, I boosted the numbers: I flip a hundred coins and I tell you that at least ninety-nine of them were heads. What is the probability that I flipped ninety-nine heads and a tails? (Answer: 100/101.)

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### Re: Boy or Girl Paradox

To be clear: yes I know running a simulation is in no way a proof. However, if people see that the simulation agrees closely with what people are saying, they might be more motivated to make an effort to understand the theory, despite their intuition.

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### Re: Boy or Girl Paradox

A simulation would look like this: there are four types of two-child families if twins aren't allowed.

Older boy, younger girl: approximately 25%.
Older girl, younger boy: approximately 25%.
Older boy, younger boy: approximately 25%.
Older girl, younger girl: approximately 25%.

We are told that the family has a girl, so that eliminates case three. Of the other three cases, two are a boy-girl family and one is a girl-girl family, which is exactly what the Bayesian logic says.

Here's another point of view: the large-number-of-coin-flips scenario suggests that this phenomenon is entropic in nature. A collection of some boys and some girls has a greater entropy than a collection of nearly all boys or a collection of nearly all girls; in the example I gave, a hundred heads out of a hundred is maximally ordered and consequently occurs with tiny probability whereas ninety-nine heads out of a hundred occurs a hundred times more often because the tails can occupy one of a hundred different positions (that is, microstates). In the original problem, the Bayesian calculation just tells us that boy-girl families are more disordered than girl-girl families since there are two possible birth orders (microstates).

To drive the point home even further, suppose that we are looking at particles traveling randomly along a line and that "boy" particles travel to the left while "girl" particles travel to the right and we observe some pressure in the girl direction. Should we conclude that all the particles are traveling in the girl direction?

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### Re: Boy or Girl Paradox

Why should we assume that girl/girl, boy/girl, girl/boy, and boy/boy are all equally likely? This is true given a random family, but given the fact that the family has at least one girl changes this. Certainly the probability of boy/boy becomes zero, but it seems plausible to me that the probability of girl/girl given this information is higher than that of either boy/girl or girl/boy individually.

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### Re: Boy or Girl Paradox

Logically, a person could argue that while it was not stated that only one child was a girl, it makes sense, assuming that the person you are speaking with is cooperating with you, tht they are not both girls. Most likely, if they were, the statement would have said so.

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### Re: Boy or Girl Paradox

tehtmi wrote:it seems plausible to me that the probability of girl/girl given this information is higher than that of either boy/girl or girl/boy individually.

Plausible doesn't mean anything; that's the same kind of intuition that leads people to the wrong answer to the Monty Hall problem. Can you provide any justification for a mechanism that would increase this probability?

Perhaps what you're getting hung up on is the notion of choosing "a random two-child family with at least one girl." It's plausible that a family with some girls is likely to have more girls, perhaps because intuitively there is some trait that steers the gender ratio one way or another, but in the assumptions we are making for this problem that does not occur. We choose a random two-child family with at least one girl by ignoring the two-child families with two boys (as above) and looking at the rest (as above). If you agree with me here, there's nothing to argue about.

If you'd like, it's possible to compute the change in probability if we are also given that there is, say, a 1% chance that a given family has a predisposition to daughters (and will, perhaps, birth twice as many daughters as sons on average). I'm a little too lazy to perform the calculation, but it is likely that the change in the answer to the problem is small.

tehtmi
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### Re: Boy or Girl Paradox

What I had in mind was something like this. The parent has two children, and you are told the gender of one of the children. Given only this information, I would say its plausible that the child whose gender you were told was chosen randomly, and that being told the gender of the younger child was equally as likely as being told the gender of the older child. If the older child is a daughter, then there is a 50/50 chance of the other child being a boy. If the younger child is a girl then there is also a 50/50 chance of the other child being a boy. This leads 50% as the answer to the question, and in regard to the effect I was suggesting earlier, a 25% chance of boy/girl, a 25% chance of girl/boy, and a 50% chance of girl/girl.

I am however, not completely sure that my assumption about the parent choosing a child is a valid interpretation.

In the Monty Hall problem, I believe that the trickiness arises from the game show host choosing a door to open with information about the door's contents, which can be used to the contestant's advantage.

In this question I can imagine two models:
In one model, random families with two children at least one of which is a girl are chosen. Two thirds of these families will have a boy.
In the other model, which I suggest is a better interpretation, random families with two children are chosen. The parent chooses one of the two children at random. If that child is a girl, there is a 50% chance that the other child is a boy.

In the first model, I believe the "Monty Hall Effect" comes from whoever is choosing the family to have at least one girl in the first place.

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### Re: Boy or Girl Paradox

tehtmi wrote:What I had in mind was something like this. The parent has two children, and you are told the gender of one of the children. Given only this information, I would say its plausible that the child whose gender you were told was chosen randomly, and that being told the gender of the younger child was equally as likely as being told the gender of the older child.

This point of view distorts the probabilities. After you ignore the cases where the child "chosen" is male, a family with two girls is twice as likely to be "reported" as a family with one girl because a parent has twice the likelihood of "choosing" a female. In effect, the girl-girl families are being counted twice, which accounts for the probabilities you give. But that is not the way the problem is set up: we have no information that a choice or a report is being made and we are counting families, not reports.

I'd like to make this extremely clear. A la Schrodinger, suppose you are given a box that you are told contains two kittens, each of which is equally likely to be dead or alive. You hear meowing. What is the probability that the box contains one live kitten and one dead kitten?

Here nobody is making a choice about which kitten gets to be "reported" and we assume that you cannot distinguish between the sound of one kitten meowing and two kittens meowing.

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### Re: Boy or Girl Paradox

I don't actually know how related they are, but this problem reminds me of one of my favorite riddles.

I have two U.S. coins from recent circulation, whose monetary values total to 35 cents. One of the coins is not a dime. What are the two coins?
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
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tehtmi
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### Re: Boy or Girl Paradox

This point of view distorts the probabilities. After you ignore the cases where the child "chosen" is male, a family with two girls is twice as likely to be "reported" as a family with one girl because a parent has twice the likelihood of "choosing" a female. In effect, the girl-girl families are being counted twice, which accounts for the probabilities you give. But that is not the way the problem is set up: we have no information that a choice or a report is being made and we are counting families, not reports.

Okay, I agree that your analysis is entirely correct.

Maybe what I'm trying to say is that the result of the question is not really applicable in the way it would seem to be given the way the problem is stated. If a parent of two children came up to me and said they had a daughter, I would assume that they were thinking of a particular one of their children, which they had chosen to tell me about for some reason other than just gender. I would assume that if that parent had a son and a daughter, I could have just as well have been told about the son. I am inherently assuming a kind of symmetry between boys and girls. These would be reasonable assumptions, and would change, or "distort" the probabilities. However, they are more applicable to the real life situation suggested (though not actually stated) by the problem, and may be why people find the correct solution frustrating.

Your kitten example (didn't Schrodinger use fully grown cats? ) helped me see this, because alive and dead are inherently different--alive kittens report themselves, but dead kittens don't.

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### Re: Boy or Girl Paradox

That's reasonable. Bayesian logic is a particular interpretation of what probability means that not everybody agrees with. This problem is stated mathematically and is intended to be solved within this mathematical framework, but what to make of the interpretation of that result is less clear.

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### Re: Boy or Girl Paradox

I think it has been shown that the probability of a boy-girl pair in this situation is clearly 2/3.

But the question asks for the odds, therefore I believe the answer is 2:1.

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### Re: Boy or Girl Paradox

Vatar, perhaps you are thinking about the situation differently than we are, so I'll ask it a different way:

Suppose I poll all the women in the world, and ask them if they have exactly two children. Those who say no I eliminate from the poll. Then I ask all remaining women if they have at least one daughter. Those who say no I eliminate. I finally ask each remaining woman if she has a son.

Approximately what proportion of yes answers should I expect to receive to the final question?
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### Re: Boy or Girl Paradox

Cauchy wrote:I have two U.S. coins from recent circulation, whose monetary values total to 35 cents. One of the coins is not a dime. What are the two coins?

I'd say the coins are a dime and a quarter, totalling 35 cents and one of them (the quarter) is not a dime! Although I must admit I have heard this puzzle before in the book '100 Ways to Win a Tenner' by Paul Zenon, who recommends doing it as a bar bet.

Re the original question, I'd agree with most here that as the two children must be one of the older child / younger child pairings of GG, GB, BG but not BB, the likelihood of the children being a boy and a girl is 2 out of 3.

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### Re: Boy or Girl Paradox

Obviously the answer is 50/50, the paradox is generated by misinterpretation of data. You have to count the GG pair twice, because there are two girls.

Consider this: to make the problem simpler, there is an older sibling and a younger sibling. The older sibling is first, so in the case of GB the girl is older than her brother.

Obviously, since one of the children is a girl, there is no BB combination. Let's say that the older sibling is female. The only two combinations that allow this are GB and GG. They are both equally likely, so there is a 50/50 chance of getting either boy or girl if the oldest child is female. The same applies when the younger child is female. There are two possibilities (GG and BG) and they're both equally likely. 50/50 again. Notice that GG was counted twice and counts for 50% of the possibilities, while BG and GB only account for 25% each.

Writing the list of all the girl/boy possibilities like this:

GG, BG, GB, BB.

Is fine, except that you must remember that you have explicitly stated that both siblings are distinguishable from each other, because otherwise BG and GB are the same thing. By saying that one is older and one is younger, I have included a clear difference in children, and therefore must count GG twice, because the oldest or the youngest child may be the girl in question.

Alternatively, if you want to prove this without adding a distinguishing factor between siblings, it is still possible. Because the sibling are indistinguishable from each other, eg: their order does not matter, then GB and BG are actually the same thing. Therefore you have three possibilities (BB, BG, GG). Eliminating the BB pair there are two combinations equally likely: BG and GG. Either way, there is a 50/50 chance of having your second child being a girl.

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### Re: Boy or Girl Paradox

andymac wrote:Consider this: to make the problem simpler, there is an older sibling and a younger sibling. The older sibling is first, so in the case of GB the girl is older than her brother.

That changes the question completely. Skep's answer a few posts above this spells out what is going on perfectly. You'd have to change the questionaire to get the 50/50 answer.
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### Re: Boy or Girl Paradox

andymac wrote:Obviously the answer is 50/50, the paradox is generated by misinterpretation of data. You have to count the GG pair twice, because there are two girls.

Yes, but no. Even taking that into account, using the selection method as detailed in the problem each girl in that pair has a lower probability of selection than the girls with brothers, which happens to cancel that effect out completely.

In the land of Statistica, the 2008 Census showed there to be exactly four families, with children as such:

Family A: Anna and Alice
Family B: Barry and Betty
Family C: Colin and Cathy
Family D: David and Dan

First, let's consider a method of selection that is different to that posed in the paradox: Select one of the eight children with equal probability 1/8 of selection. You end up selecting one of the four girls, and you ask her if she has a brother or a sister. What is the probability of a brother? Well, of the four girls, two (Betty and Cathy) have brothers and two (Anna and Alice) have sisters, so 50% is the answer.

Now, let's look at the selection method actually described: Select one family at random, and ask if they have at least one daughter - to which the answer turns out to be yes. Thus, while initially the probability of selecting any particular family was 1/4, this new knowledge means that we can say that the probability that we have selected Family A, B or C is 1/3 each, and Family D is 0. This is the posterior probability, i.e. the probability after the fact that is adjusted to include extra information. So, with this new situation, what's the probability that the family has at least one son? Well, obviously that's only the case if we've picked Family B or C, meaning that there is a 2/3 chance.

Looking at it another way, suppose we picked a single family, then picked a child in the family at random and found it to be a girl. Again, obviously, the probability that we picked Barry, Colin, David or Dan is 0. However, the probabilities for the girls are a bit trickier. Since we know that we've obviously picked a family with a girl in it, the probability that the girl is from Family A, B or C is 1/3. The probability that we picked either Anna or Alice, given that we picked a girl from Family A, is 1/2, while the probability that we picked either Betty or Cathy, given that we picked a girl from Family B or C respectively, is 1. So the probability we picked Anna (or Alice) is 1/3 * 1/2 = 1/6, while the probability that we picked Betty (or Cathy) is 1/3 * 1 = 1/3. Hence, the probability that the girl we picked has a brother is equal to the sum of the probabilities of us picking Betty or Cathy, hence 2/3.

Yes it's counterintuitive, but that's the point of the paradox (and while it isn't 100% identical to Monty Hall, it rests on the same premise - adjusting the initial probabilities based on later evidence).
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### Re: Boy or Girl Paradox

andymac wrote:Obviously the answer is 50/50, the paradox is generated by misinterpretation of data. You have to count the GG pair twice, because there are two girls.

It should only be counted once, as it simply means a girl was born first, followed by another girl. This and the BB pair are two of the four possible pairs in two-child families, the other two being boy first followed by girl, and girl first followed by boy. Unless I've misunderstood why GG should be counted twice?

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### Re: Boy or Girl Paradox

Yeah, you're not counting children, you're counting families. The reason you *don't* count GG twice is because that's still only one family. So removing the BB possibility, we have BG, GB, and GG, 2/3 of which have a boy and a girl.

Note that, this being the math forum, we are ignoring psychological factors involved in *why* the parent might have chosen to tell you this, or biological factors involved in why the sex of children might not be independent events or equally likely between male and female.

The problem, as stated, and from a purely mathematical perspective, is the same as the following:
I have two fair coins under my hand, which I have just flipped and looked at. I tell you that one of them is heads, and ask you the probability that under my hand are one heads-up and one tails-up coin.

Note that I did *not* say I told you the up-side of one of the coins. I specifically have told you that one of the coins is heads-up. This information tells you that, of the four equally likely possibilities HT, HH, TH, TT, I definitely did not get TT. But HT, HH, and TH are still equally probable. (The only way they aren't is if they weren't to begin with. If the coins were unfair from the get-go, for instance.) Which means, since probability distributions must add to one, that each of these events must have probability 1/3. And since we have two one-heads-one-tails events, then there is a 2/3 chance that I have one heads-up coin and one tails-up coin.

Anyone who disagrees that it's 2/3 need not go through a long detailed explanation about children and parents and survey questions. They need only explain either why they are rejecting Bayes' theorem, or why they think Bayes' theorem doesn't apply here.
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### Re: Boy or Girl Paradox

t0rajir0u wrote:
tehtmi wrote:it seems plausible to me that the probability of girl/girl given this information is higher than that of either boy/girl or girl/boy individually.

Plausible doesn't mean anything; that's the same kind of intuition that leads people to the wrong answer to the Monty Hall problem. Can you provide any justification for a mechanism that would increase this probability?

A problem (as gmalvik notes in the post before me, but... I go into more detail so whatever) is that in the real world, acts of utterances contain information other than their literal meaning. People are not magic sentence generating machines, they say things for reasons, and thus that changes things. A person may be more or less likely to say "I have at least one daughter" if they have two daughters relative to if they have one daughter. Thus, Bayesian probability should take this into account, and in so far as it isn't, common-sense answers might be more realistic in the real world. (In the Monty Hall problem, this situation is avoided (although people still complain) since it is explicitly stated that Monty always opens a door for you, and that this door always is empty, so the door-opening doesn't contain any "meta-content" in this sense.

Although I don't think this additional information would neccesarily lead to the assumption that BG GB and GG are equiprobable. In order for this to happen, the act-of-utterance likelihood would have to be so arranged that people are more likely to say "I have two children, and one of them is a girl" when they have two girls. If anything, I think the opposite is true, since the English phrase "one of them is a girl" has the connotations that exactly one is a girl rather than just at least one. Thus, the probability of having a boy would be slightly higher than 2/3. (Although I think that 2/3 is as close to a correct answer as you can get by the constraints of this being a puzzle and not actual children, since that additional information is somewhat indeterminate.)