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        riddles >> medium >> MEDIUM: TWO-CHILD FAMILY I & II
        
(Message started by: S. Owen on Aug 12^th, 2002, 1:13pm)

Title: MEDIUM: TWO-CHILD FAMILY I & II
Post by S. Owen on Aug 12^th, 2002, 1:13pm

I like this pair because the different wordings (and different answers) highlight the subtleties of both. This will clear up any confusion about the very similar (and strangely controversial) "I-flip-a-coin-and-one-is-heads" problem here too.

I think we are to assume that boys and girls are equally probable.

I: In a two-child family, one child is a boy. What is the probability that the other child is a girl?

If one is a boy, then we know there are not two girls. We are left with the equally probable boy-girl, girl-boy, boy-boy scenarios. In 2 of the 3 the other child is a girl: 2/3.

II: In a two-child family, the older child is a boy. What is the probability that the other child is a girl?

The crucial difference is that we are given information about exactly one child. This gives us no information about the other; the other child has a of 1/2 chance of being a girl. Or, you could say that this leaves us with a boy-girl, boy-boy situation, each of which is equally probable.

Title: Re: MEDIUM: TWO-CHILD FAMILY
Post by Lance on Aug 23^rd, 2002, 7:33am

I fail to see the difference between I and II ? Why are they not both 2/3 or both 1/2 probability ? The information that the boy is older seems irrelevant. All that matters is that one of them is a boy. Which is exactly like the first one.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by AlexH on Aug 23^rd, 2002, 10:32am

Telling you which child the boy is narrows down the situation more than does simply telling you one is a boy.

Initially 4 possible states (listing older child first): BB,BG,GB,GG

Given statement one is a boy we have
3 possible states: BB, BG, GB

Given statement older child is a boy we have
2 possible states: BB, BG

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Rodrick Crider on Aug 23^rd, 2002, 10:44am

For the first scenario...

In a two-child family, one child is a boy. What is the probability that the other child is a girl?

... I propose that the answer is 100%. The question states that "one", not "at least one", child is a boy. That only leaves us with two possible combinations: BG and GB. In both of those cases the "other" child is a girl.

(assuming that boys and girls exhaust the set of possible children)

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Mark Lindsay on Aug 23^rd, 2002, 12:31pm

on 08/23/02 at 10:32:11, AlexH wrote:

But BG and GB are the same thing. You are counting the same thing twice. If you count BG and GB as distinct, you have to consider B1B2 and B2B1 as distinct.

Say there is a girl and a boy. The boy is Mike and the girl is Judy.

Now say next door there are two boys. One is named Tom and the other is named Sam.

Now, if you consider Mike + Judy to be different from Judy + Mike, you have to consider Tom + Sam to be different from Sam + Tom.

So, you have four scenarios possible:

G1 + B1
B1 + G1

B1 + B2
B2 + B1

Or, you can just simplify it, by not paying attention to the order:

B + G
B + B

Either way, it is 50%.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by S. Owen on Aug 23^rd, 2002, 1:14pm

on 08/23/02 at 12:31:05, Mark Lindsay wrote:

But BG and GB are the same thing. You are counting the same thing twice. If you count BG and GB as distinct, you have to consider B1B2 and B2B1 as distinct.

By your same logic, flipping two coins should yield a heads and tails 1/3 of the time, which it doesn't.

AlexH is right, since girls are distinct from boys, but boys are not distinct from other boys, at least in the context of this problem. Your analogy doesn't quite line up since you start talking about distinct individuals.

BG and GB are not the same, since it is not the same to have "older boy, younger girl" as to have "older girl, younger boy."

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Mark Lindsay on Aug 27^th, 2002, 2:02pm

on 08/23/02 at 13:14:50, S. Owen wrote:

By your same logic, flipping two coins should yield a heads and tails 1/3 of the time, which it doesn't.

Yeah, I see how that doesn't work. I guess I still can't make myself understand how knowing that the 1st coin flip was heads makes the second coin flip anything other than 50%, which is what it seems like is going on here.

mark

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by S. Owen on Aug 27^th, 2002, 2:25pm

on 08/27/02 at 14:02:41, Mark Lindsay wrote:

I guess I still can't make myself understand how knowing that the 1st coin flip was heads makes the second coin flip anything other than 50%, which is what it seems like is going on here.

You're right that knowing that the first coin flip ("first" is crucial) is heads says nothing about the second. That's analogous to part II here, and yeah, the answer is nothing other than 50%.

Part I is analogous to "at least one of two coin flips is heads" - in that case the chance that the other is heads is 33 1/3%. The difference is that this statement does not concern exactly one of the coins (and likewise, part I says something about the state of both children, whereas part II reports only on one child).

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Chronos on Aug 27^th, 2002, 3:40pm

The answer to both questions depends on how you come by the information. To illustrate:

Scenario 1: My mother is at an appreciation dinner hosted by my old high school: Only parents of alumni are invited. During dinner, Mom strikes up a conversation with a man there who she did not previously know. He mentions that he has two children, but says nothing about their genders. Since my high school was all male, my mom knows that at least one of this gentleman's children is male. What's the probability that the other child is male? Here, the answer is 1/3.

Scenario 2: Later in the conversation, my mom asks "So, your son that went here... Was he the older or the younger sibling?". The man answers "He's the older one". Now, my mom knows that this fellow has two children, and the oldest is a boy. But the probability that the other child is a boy is still 1/3.

Scenario 3: As scenario 1, but then the man says "Sorry I'm so tired, but my six-month-old cried all night and kept me awake". Now, we know that the child who went to the high school must be the older one, since the younger one is too young. What's the probability that the baby is male? In this case, it's 50%.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Carl_Cox on Sep 16^th, 2002, 5:14pm

on 08/27/02 at 15:40:39, Chronos wrote:

Whoa. I was cool with logic from 1/3 and 1/2 and all that until we get to scenario 2.

At this point, your mother already knows that one kid was male, because otherwise he couldn't go to your old high school. You mother, being apparently bored, figures out the chances that the other child is a boy is 1/3. This is illustrating the possibilities:
older boy and younger girl (older went to this school)
older girl and younger boy (younger went to this school)
older boy and younger boy (but no knowledge of which went to school).

Now, scenario two roles around. Mom decides to maintain this incredibly interesting conversation after her mathematical solitaire, and learns that the older one went to school. Now she has new information. Doesn't that make the cases
older boy and younger girl
or
older boy and younger boy (now knowing the older went to school)

since we now know that the child can't BOTH be older AND female? Thus now the possibility that the other child (the younger child) is a girl is 50%.

I seem to recall a puzzle about a game show and 3 doors that tried to do your kind of math, too. How did that work?

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Chronos on Sep 17^th, 2002, 12:56pm

After the man answers the question, we are, indeed, reduced to two possibilities (Older child male, younger female, or both male). However, we do not have an equal probability of both. Think of it this way: Suppose that the man had answered "the youngest", instead of "the oldest". The exact same reasoning would apply, would it not? If the answer in Scenario 2 is 50%, then the answer in this alternate scenario would also be 50%.

But then, my mom could reason to herself before she even asked the question: "Suppose that he answers 'The older one'. Then the probability that the other child is male is 50%. But suppose that he answers 'the younger one'. Then the probability is also 50%. But he must answer older or younger, and either way, the probability is 50%.". So, then, Mom can figure the probability at 50% even without asking the question, and we're back to scenario 1. But the probability in scenario 1 isn't 50%, it's 33%. So our premise must be wrong, and the probability in Scenario 2 isn't 50%.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Carl_Cox on Sep 18^th, 2002, 7:19pm

I believe I see. The chance has to do with what you know when you guess. Revealing a differerent incorrect answer later (saying "he's the oldest" when you already guess the other is a boy implies that the oldeest cannot be a girl) doesn't change your chance, because you could still be wrong. I can follow that.

Here's a related question; if you guess that the other child is a girl, would your chances decrease, because one of the possibilities that it was true has disappeared?

ie, scenario 1: reveal that he has at least one boy; chances other child is a girl = 2/3
scenario 2: reveal that the older child is a boy

In this scenario 2, is the guess that the other child a girl still 2/3, or does it decrease, since one of the counted possibilities has disappeared?

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Chronos on Sep 19^th, 2002, 11:35am

Barring flukes of genetics, the other child must be either a boy or a girl. So the probability of "boy" and the probability of "girl" must sum to 1.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Mark Gilbert on Dec 19^th, 2002, 7:42pm

S.Owen, you math is good, but your interpretation of the English language of the question is poor. When a person talks about "one child" versus "the other child" they are specifying particular children just as certainly as if they say, "the older child" and "the other child."

If the question was, "one child is a boy, what is the probability that both children are boys?" Then the answer would be 1/3. But, that is a different question.

Admittedly, "one child" lacks specificity and can be taken generally to mean either child. But, once I say "the other one," that generality is resolved, and I am clearly talking about this one here and that other one over there. Any other interpretation of the English would be strained. Try to imagine any human speaker saying "one child is X" and "is the other child Y" without having particular children in mind. People just don't talk like that.

I have to say I dislike the riddles because they depend on this obscure reading of the English.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Icarus on Dec 20^th, 2002, 3:14pm

on 12/19/02 at 19:42:03, Mark Gilbert wrote:

Admittedly, "one child" lacks specificity and can be taken generally to mean either child. But, once I say "the other one," that generality is resolved, and I am clearly talking about this one here and that other one over there. Any other interpretation of the English would be strained. Try to imagine any human speaker saying "one child is X" and "is the other child Y" without having particular children in mind. People just don't talk like that.

I have to say I dislike the riddles because they depend on this obscure reading of the English.

Sorry, Mark, but I have to disagree. "The other child" does not convey any sense of position or other means of differentiating the two children which would change the probabilities. "The other child" simply means the child not yet refered to. It does not specify in any way which of the two children was refered to by the original remark. This is not an obscure interpretation. It is the common interpretation of the word "other".

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Mark Gilbert on Dec 20^th, 2002, 11:42pm

Hi Icarus,

Whether I have any information about the children that allows me to distinguish them is irrelevant. What matters is whether the questioner is refering to specific children, and clearly he is. He has chosen to refer to one child as "one child" and the other child as "the other child". The sex of the child whom he calls "the other child" is independent of the sex of the child whom he calls "one child."

The word "one" in the first sentence is not telling us how many children are boys. If it meant that, then Rodrick Crider would be correct - it would mean "exactly one". But, the use of "the other" in the second sentence rules out this interpretation.

My New Oxford American Dictionary gives the relevant definition of "one" as, "denoting a particular item of a pair or number of items." As long as the questioner's "one" refers to a particular child, the answer is 1/2.

My interpretation is the obvious one, and if the questioner wished to express my meaning, the given riddle would be a natural way to do it. If the questioner really wished to ask, "In a two-child family, at least one child is a boy. What is the probability that the children are a boy and a girl." Then he chose an obtuse and easily misconstrued way of expressing it.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Mark Gilbert on Dec 21^st, 2002, 12:11am

By the way, Chronos is also wrong. In all three scenarios, the probability that the other child is a girl is 50%. In all three scenarios, we are talking about particular children. Rather than distinguishing them as the older one and the younger one, we are distinguishing them as the one who went to the school and the one who didn't.

The riddle, Two-Child Family II, could have been written, "In a two-child family, the child that went to school X is a boy. What is the probability that the other child is a girl?" Or, "the child whose name has the higher numerological sum," or "the child that likes broccoli," or whatever. In any case, the answer is 1/2. There is nothing magic about the ages of the children or the order in which they arrived. The thing that (supposedly) distinguishes riddle II from riddle I is that riddle II talks about a particular child and asks about the other child. (I argue that both riddles are best interpreted as referring to particular children, but that is another argument.)

Cheers!

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by S. Owen on Dec 21^st, 2002, 2:17am

on 12/19/02 at 19:42:03, Mark Gilbert wrote:

I'm inclined to agree. I can accept that the first statement wants to have the sense of:

In a two-child family, there is a boy. What is the probability that there is a girl as well?

... but it is not clear that this is exactly what the speaker is conveying. It'd be clearer if it were put this way, but then it wouldn't look as clever as a riddle.

I think that "one child is a boy" is a plausible colloquial alternative to "there is a boy" (at least in U.S. English I'd say, right or wrong), so I don't think the conventional interpretation put forth above is wrong. Certainly though, there is an ambiguity, and that makes the riddle fairly questionable.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by redPEPPER on Dec 21^st, 2002, 8:20am

TCF II has a pretty accurate wording. One specific child is designated as the older one with no possible mistake. We learn something about that specific child but nothing about his sibling, leading to a probability of 1/2 for that one to be a boy too.

TCF I, on the other hand, has poor wording. By contrast with TCF II we could be led to think that it must mean something different and favor the probability of 1/3, but if we examine it on its own, it's not accurate enough.

Some argued that "one child" means "only one child", and that you'd have to say "at least one child" if you mean otherwise. This reasoning leads to a 0 probability.

Others argued that "one child" means "one specific child" and also propose alternate wordings such as "one of the children" if you mean otherwise. That leads to a 1/2 probability.

Others argued that in the absence of additional info, "one child" can't be understood as "only one child" or "one specific child" and that the only fair meaning is "one random child", especially by contrast with TCF II. The probability here is 1/3.

The fact is, in these three situations you could fairly say "one child is a boy" without lying or making a grammatical blunder. At worst you wouldn't be accurate enough, and that's what makes this riddle a bad one. But I don't think the English language allows us to say that one is more correct than the others.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Icarus on Dec 21^st, 2002, 6:23pm

Well said, redPepper.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by BNC on Dec 25^th, 2002, 10:15am

on 08/27/02 at 15:40:39, Chronos wrote:

The answer to both questions depends on how you come by the information. To illustrate:

...

Scenario 2: Later in the conversation, my mom asks "So, your son that went here... Was he the older or the younger sibling?". The man answers "He's the older one". Now, my mom knows that this fellow has two children, and the oldest is a boy. But the probability that the other child is a boy is still 1/3.

Scenario 3: As scenario 1, but then the man says "Sorry I'm so tired, but my six-month-old cried all night and kept me awake". Now, we know that the child who went to the high school must be the older one, since the younger one is too young. What's the probability that the baby is male? In this case, it's 50%.

Sorry, but I can't seem to understand the difference between scenario 2 and scenario 3. In both cases you learn that the older one is a boy (I don't see any additional relevant information in the "6 months old"). So, why the different answers ? ???

BNC

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Chronos on Jan 17^th, 2003, 3:33pm

Quoth Mark Gilbert:
Quote:

No, that doesn't work, because some families don't have any child who went to my school, and some families have two children who went to my school.

Consider all two-child families. One fourth of them have two boys, one fourth have two girls, and half of them have one of each.

Now consider all of the two child families where one of the children might have gone to my school. None of those families has two girls, so we can say that of all two-child families at my school, one third are both boys, and two thirds are a boy and a girl. At the time of scenario 1, we know that the gentleman's family is one of those families, so the cases are 1/3 and 2/3

As for the difference between scenarios 2 and 3: In scenario 2, we know that at least one kid went to the school. Maybe only one did, in which case "The one who went to school here" is a unique specifier. But on the other hand, maybe they both went to school there. In scenario 3, we know that exactly one went to my school, so "The one who went here" is a unique specifier.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by rmsgrey on Apr 15^th, 2003, 10:52am

on 01/17/03 at 15:33:44, Chronos wrote:

Quoth Mark Gilbert:
No, that doesn't work, because some families don't have any child who went to my school, and some families have two children who went to my school.

Consider all two-child families. One fourth of them have two boys, one fourth have two girls, and half of them have one of each.

Now consider all of the two child families where one of the children might have gone to my school. None of those families has two girls, so we can say that of all two-child families at my school, one third are both boys, and two thirds are a boy and a girl. At the time of scenario 1, we know that the gentleman's family is one of those families, so the cases are 1/3 and 2/3

As for the difference between scenarios 2 and 3: In scenario 2, we know that at least one kid went to the school. Maybe only one did, in which case "The one who went to school here" is a unique specifier. But on the other hand, maybe they both went to school there. In scenario 3, we know that exactly one went to my school, so "The one who went here" is a unique specifier.

In scenario 2, when asked "So, your son that went here... Was he the older or the younger sibling?" the fact the man picks either alternative implies that "your son that went here" is a unique specifier. If it was ambiguous as you suggest, the question as posed would be unanswerable. Therefore either answer (older or younger) gives a 50% chance by excluding the possibility of two sons both going.

In general, it is possible to generate the 1/3 case but it is extremely fragile, which probably explains why its counterintuitivity. Attaching the "known" boy to any means of distinguishing between the two children (being able to say "the other one") collapses the state to the more familiar 1/2 case. "Knowing at least one of my two children is a boy, what is the probability that my eldest is a boy?" definitely works (I think) because there's no suggestion of attaching the "boy" property to any non-shared property of the two children

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Chronos on Apr 17^th, 2003, 5:12pm

That's true... I *think* that I can fix that by changing the question to "Did your oldest go to school here?", because that way, we don't have to worry about the possibility that they both went to my school.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by rmsgrey on Apr 22^nd, 2003, 9:37am

on 04/17/03 at 17:12:41, Chronos wrote:

Hmmm, have to run the numbers on that one - assuming all boys in the family went to the school, you're twice as likely to get the answer 'yes' as 'no' - looking at families: 1/3 BG, "yes"; 1/3 BB, "yes"; 1/3 GB, "no". So if he answers "yes", you're back to 1/2 each way.

If you know that the school only accepts one child per family (the reverse of normal school policy in real life) but that in any given family, all sons are equally likely to have gone to the school regardless of age, then, following through from scenario 1): 1/3 eldest only boy, "yes"; 1/6 both boys, "yes"; 1/6 both boys, "no"; 1/3 youngest only boy, "no". Given he says "yes", probabilities are 2/3 BG, 1/3 BB.

Depending on the (known) probability of both boys from a given family going to the school (given at least one of them does) you can get any value for the probabilities between the two extreme cases above.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by Chronos on Apr 22^nd, 2003, 9:38pm

But what if we assume that the probability for each boy to go there is independent? It seems to me that this would be a better model than "all or none" or "exactly one".

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by rmsgrey on Apr 23^rd, 2003, 1:04pm

If each boy goes to the school with independent probability p, then the probability of both boys going to the school in a BB family with at least one attending the school is p²/(p²+p(1-p)+p(1-p)) or p/(2-p). So the probability of "yes" given a BB family is p/(2-p)+(1-(p/(2-p)))/2=1/(2-p)

cases as previously give probabilities:

BG,y - 1/3
BB,y - 1/3(2-p)
BB,n - (1-p)/3(2-p)
GB,n - 1/3

So P(both boys|"yes")=(1/3(2-p))/(1/3+1/3(2-p))=1/(3-p)

Which gives any probability between 1/2 and 1/3 according to the value of p. When p=1 you get the all or none case, but as p tends to 0 you converge on the exactly one case.

Besides, you're only really interested in the probability, q, that the second boy in a BB family attends the school given that the first does. You have no interest in the probability that the first boy in a family attends as you're already restricted to those cases where just that has happened.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by javacodeman on Jul 7^th, 2007, 9:01pm

Okay, I read this Friday afternoon, wasn't settle with the answer that I found here, but didn't give it much thought. I was lying in bed tonight and my mind wondered to it. Here is the solution: It doesn't matter if the order is known or not, it is still a 50/50 chance.

For the case that the older one is known to be a boy, everyone agrees that it is 50/50, so I'll move on to the case where at least one child is a boy.

We are told that at least one child is a boy, but we don't know which. Let us chose one and see what the possibilities are if we are right:

I. If child1, then these are the possibilities:
A) child1 = boy and child2 = boy
B) child1 = boy and child2 = girl

But we could have chosen wrong (a 50/50 shot). If we chose incorrectly, these are the possibilities:

II. If child2, then these are the possibilities:
C) child1 = boy and child2 = boy
D) child1 = girl and child2 = boy

Since there is a 50/50 shot at I. or II. and a 50/50 shot at the choices below them, it is easy to see that A), B), C), and D) all have equal likelihood.

Since half of the possibilities have a boy/girl combo, there is a 50/50 shot that the other child is a girl.

java

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by javacodeman on Jul 7^th, 2007, 9:26pm

Just a follow-up to some arguments that might be posted against my previous post.

Either order matters or it does not. You cannot put an order constraint on the b/g and g/b situation and not also put it on the b/b and b/b situation.

I'd like to say that order doesn't matter: one is a boy and the other has a 50/50 chance at being a girl, so there you go. But some people like to make it more formal and talk about order--just be sure order is applied uniformly.

If b1/g2 and b2/g1 are unique, then so are b1/b2 and b2/b1.

If you argue differently, it is paramount saying to this:

Quote:

I have two coins. One I will place on heads, the other I flip. In one scenario, I'll tell you the order of the coins (either place, then flip or flip, then place) while in the other, I will not tell you the order.

In the situation that I tell you the place/flip order, there will be a 50% chance of a heads/tails combo. In the situation that I do not tell you the order, there will be a 66% chance of a heads/tails combo.

If that is your position, then there is not much hope for you--for you cannot see the forest for the trees.

java

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by rmsgrey on Jul 8^th, 2007, 4:34am

There are 4 possible families:

b1b2
g1b2
b1g2
g1g2

If, for whatever gender-independent order you've picked, child 1 is a boy, then you have 2 possibilities:

b1b2
b1g2

If you pick a child at random, and it turns out to be a boy, you have 4 possibilities (using an underline to indicate which child was chosen):

b1b2
b1g2
b1b2
g1b2

However, if you order the children so boys are listed before girls (keeping the old order if they're the same gender), and pick child 1, then there are three equal possibilities:

b1b2
b1g2
b2g1

The key to the 1/3 answer is to find a way of eliminating the 2 girls possibility without affecting the relative probabilities of the remaining possibilities - if you find a boy, it has to be by a method that has no chance of finding a girl in either of the cases where there's one of each.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by javacodeman on Jul 8^th, 2007, 11:35am

It is true that if one were to randomly fill two cells with either b or g, you would find the following:

1/4 are bb
1/4 are bg
1/4 are gb
1/4 are gg.

In our scenario, the gg possibility has been eliminated, leaving only the first three possibilities. This fact has hoodwinked some of the contributors here. Because this is **not** the same scenario. In the scenario presented, you are not filling both cells randomly and cutting out the gg possibilities. You are guaranteeing one cell to be "b" and randomly filling the other. It doesn't matter which cell you choose to fill with the known "b", either way (or both ways done in a probability tree) lead to half of the remaining possibilities having a girl.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by rmsgrey on Jul 9^th, 2007, 5:56am

on 07/08/07 at 11:35:07, javacodeman wrote:

So, consider a population of 100 2-children families. 25 families are bb, 25 bg, 25 gb and 25 gg.

For scenario I, "(at least) one child is a boy":

75 families fit, of which 25 are bb.

For scenario II, "the elder child is a boy":

50 families fit, of which 25 are bb.

The key point (in scenario I) is that, while you guarantee that there is a boy, you don't specify which child is a boy. Yes, if you look at child 1 being a boy, then at child 2 being a boy, in each case you get a 50% chance of the other being a boy. However, in so doing, you count the bb families twice, once for each boy, while the bg/gb families only get counted once, so when you attempt to combine the two events, you need to take the overlap into account.

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by javacodeman on Jul 9^th, 2007, 10:56am

I see more clearly the divide in our thinking. It depends on whether you are looking at distribution across all families with two children and at least one boy or a specific family with two children and at least one is a boy.

In the normal distribution, it is correct the b/gs(regardless of order) out number the b/bs 2-to-1. But for a specific case where one chance event is already determined, the second chance event is independent of the first.

So if you are looking through a lense that says, "If I go interview 1000 two child families where at least one child is a boy, what are the odds that the other is a girl", then I believe the it is 66% based on distribution.

But for any **one** family, the odds would be 50/50 based on previous arguments. (Just like a coin flip following a heads result is still 50/50 to be heads or tails. But over double flips heads/tails combos will be more prevelent that heads/heads or tails/tails.)
java

edited for mathematical sanity. ;D

Title: Re: MEDIUM: TWO-CHILD FAMILY I & II
Post by rmsgrey on Jul 10^th, 2007, 5:50pm

on 07/09/07 at 10:56:02, javacodeman wrote:

In the normal distribution, it is correct the b/gs(regardless of order) out number the b/bs 2-to-1. But for a specific case where one chance event is already determined, the second chance event is independent of the first.

But you haven't determined which of the two chance events is predetermined.

If you pick any one specific two-child family, knowing nothing about the children's genders, and ask "is one of the children a boy?" then a "yes" answer gives a 1/3 chance of both being boys.

If you pick any one specific two-child family, knowing nothing about the children's genders, specify one of the two children somehow and ask "is that one of the children a boy?" then a "yes" answer gives a 1/2 chance of both being boys.

The key is under what circumstances you would get an answer of "no": in the former case, only if both children were girls; in the latter, as long as the chosen child were a girl, whether the other were male or female.