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Topic: MEDIUM: TWO-CHILD FAMILY I & II (Read 8469 times) |
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S. Owen
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MEDIUM: TWO-CHILD FAMILY I & II
« on: Aug 12th, 2002, 1:13pm » |
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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.
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Lance
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Re: MEDIUM: TWO-CHILD FAMILY
« Reply #1 on: Aug 23rd, 2002, 7:33am » |
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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.
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AlexH
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #2 on: Aug 23rd, 2002, 10:32am » |
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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
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Rodrick Crider
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #3 on: Aug 23rd, 2002, 10:44am » |
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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)
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Mark Lindsay
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #4 on: Aug 23rd, 2002, 12:31pm » |
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on Aug 23rd, 2002, 10:32am, AlexH wrote: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 |
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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%.
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S. Owen
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #5 on: Aug 23rd, 2002, 1:14pm » |
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on Aug 23rd, 2002, 12:31pm, 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. |
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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."
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Mark Lindsay
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #6 on: Aug 27th, 2002, 2:02pm » |
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on Aug 23rd, 2002, 1:14pm, 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.
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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
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S. Owen
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #7 on: Aug 27th, 2002, 2:25pm » |
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on Aug 27th, 2002, 2:02pm, 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. |
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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).
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Chronos
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #8 on: Aug 27th, 2002, 3:40pm » |
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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%.
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Carl_Cox
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #9 on: Sep 16th, 2002, 5:14pm » |
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on Aug 27th, 2002, 3:40pm, Chronos wrote: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.
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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?
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Chronos
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #10 on: Sep 17th, 2002, 12:56pm » |
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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%.
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Carl_Cox
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #11 on: Sep 18th, 2002, 7:19pm » |
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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?
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Chronos
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #12 on: Sep 19th, 2002, 11:35am » |
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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.
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Mark Gilbert
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #13 on: Dec 19th, 2002, 7:42pm » |
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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.
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Icarus
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #14 on: Dec 20th, 2002, 3:14pm » |
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on Dec 19th, 2002, 7:42pm, 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.
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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".
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Mark Gilbert
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #15 on: Dec 20th, 2002, 11:42pm » |
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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.
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Mark Gilbert
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #16 on: Dec 21st, 2002, 12:11am » |
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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!
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S. Owen
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #17 on: Dec 21st, 2002, 2:17am » |
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on Dec 19th, 2002, 7:42pm, Mark Gilbert wrote: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. |
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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.
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redPEPPER
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #18 on: Dec 21st, 2002, 8:20am » |
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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.
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| « Last Edit: Dec 21st, 2002, 8:26am by redPEPPER » |
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Icarus
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #19 on: Dec 21st, 2002, 6:23pm » |
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Well said, redPepper.
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BNC
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #20 on: Dec 25th, 2002, 10:15am » |
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on Aug 27th, 2002, 3:40pm, 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%. |
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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
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| « Last Edit: Dec 25th, 2002, 10:21am by BNC » |
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Chronos
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #21 on: Jan 17th, 2003, 3:33pm » |
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Quoth Mark Gilbert: Quote:| 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. |
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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.
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rmsgrey
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #22 on: Apr 15th, 2003, 10:52am » |
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on Jan 17th, 2003, 3:33pm, 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. |
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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
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Chronos
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #23 on: Apr 17th, 2003, 5:12pm » |
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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.
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rmsgrey
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Re: MEDIUM: TWO-CHILD FAMILY I & II
« Reply #24 on: Apr 22nd, 2003, 9:37am » |
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on Apr 17th, 2003, 5:12pm, Chronos wrote:| 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. |
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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.
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