my take: there can’t be 2/3 chance the second child is male. If options (1) female/male and (3) male/female are distinguishable from one another this implies automatically that there is an additional factor which makes the placement of the male and female important. this additional factor cuts the probability of options (1) and (3) occurring by half so that the two together have the same chance as option (2). i wasn’t very clear in the email when i suggested an illustration of such a factor as one of age because such a factor regardless has to exist when it was decided that there is a difference between options (1) and (3). my conclusion is that (1) and (3) don’t really mean 2/3 chance that the other sibling is male because the probability of (1) and (3) are cut in half by the fact that the placement of the female and male matters in the chance of those two options occurring, whereas in option (2) female/female age does not.
Here’s where the flaw is in your logic. If males and females are born with equal probability, then in two-child families, the number of families with one boy and one girl is twice the number of families with two girls. Additional information (such as ordering by age) doesn’t change that.
why should a general trend that doesn’t take into consideration the fact that one gender of a pair of siblings has been ascertained affect a particular case where that of one is already known? such a general trend that does take into consideration the fact that one of the two is already female will transform into the same exact question in consideration.
Start with the general trend. In two-child families, there are four possiblities, which you quoted in your first post.
Now, what is it about the particular case that affects the relative probablities? If you know, for example, that one child is female, it *does[/s] affect the relative probability of option four (male/male), because having one female child excludes that option.
However, knowing that one child is female does not affect the relative probability of any of the other options. If you disagree, then articulate, exactly, what it is about this particular case that affects options 1, 2, or 3.
by knowing one is female, option 4 becomes impossible. options 1 and 3 either can become identical, OR ELSE distinguishable.
if distinguishable then that means it matters where the female lands, either in child A or child B. if the female lands in child B then option 1 becomes impossible. if the female lands in child A then option 3 becomes impossible. the female has 50% chance to land in A and 50% chance to land in B. options 1 and 3 therefore have half the chance applied to it again due to the importance of whether the female is A or B.
you can only make options 1 and 3 the same option, or else make them distinguishable from one another and follow through with the logic in the previous paragraph.
This is the whole point of this thought problem: You don’t know if the female in question is child A or child B. This is the whole point of statistics: in individual cases, there is some information you don’t have, and probablities are adjusted based on what you know or don’t.
Suppose the problem were stated this way: “There is a family with two children. You have been told the eldest child in this family is a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal?” In this case, you can rule out both option 3 and option 4 (because you do know the female in question is child A), and the correct answer is 1/2.
But this isn’t how the problem is stated. It’s stated: “There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal?” You’re given no information on the ordering. This is a different problem than the one in the previous paragraph, because the amount of information you know is less, and the answer is 2/3.
(And, in addition, the question could be worded this way: “There is a family with two children. You have no idea of the sex of either child. What are the odds they have a son, assuming the biological odds of having a male or female child are equal?” This is a different problem yet, with the answer being 3/4.)
Actually, with the information given, the answer is 1:2. Worded differently it would be 2:3.
However, all other things being equal, since we were told there is a daughter, we have to consider that, as well. If they have one of each gender, the odds of being told “we have a daughter” and the odds of being told “we have a son” are 1/2.
Without being told the gender of either child, the odds of there being 1 boy and 1 girl are 2:4 (b-g, b-b, g-b, g-g), which reduces to 1:2.
The odds of being told one is a girl AND that there actually is 1 boy and 1 girl would then be expressed as the odds of being told the gender times the odds of 1boy-1girl divided by the odds of being told the gender. (Bayes formula)
This is expressed then as (1/2 * 1/2)/1/2, which reduces to 1:2.
but if no one cares about ordering, then no one should care about the difference between option (1) female/male and option (3) male/female. the author of the question can decide to say there is no regard for ordering, in which case these two options become one and the same. and when it is one and the same then it is 50% likely vs female/female.
Well, no. You don’t get to do that. Once you’ve been told “we have a daughter,” the odds of being told that are 100%. I would think this is obvious.
As an example, suppose the question were worded: “There is a family with two children. You have been told this family has a son. What are the odds they have a son, assuming the biological odds of having a male or female child are equal?” What is the answer to this question?
Well, no it’s not. Not at all. I suspect you’re being confused by the meaning of probability. Let’s change the problem into a whole-number equivalent.
In a small city, there happens to be 1000 families with two children.
In half these families, the first child is a male. In the other half, the first child is a female. Likewise, the second child’s sex is split evenly.
That leads to three groups of families:
Group A (250 families) has a female first child and a male second child.
Group B (250 families) has a female first child and a female second child.
Group C (250 families) has a male first child and a female second child.
Group D (250 families) has a male first child and a male second child.
OK, with me so far?
Now, some families have a daughter. In fact, 750 have a daughter: groups A, B, and C.
Of those families that have a daughter, how many also have a son? In this case, only groups A and C have a son: 500 families, in other words.
So, you can say, out of those families who have a daughter, 2/3 also have a son.
The fact that, when unordered, groups A and C are indistinguishable doesn’t reduce the number of families in those groups.
it doesn’t reduce the number, but there’s this new additional probability:
for the daughter to be in Group A, the daughter must be first child. the daughter has a 50/50 chance of being either first or second child. if the daughter happens to be the second child, there is no chance it will be in Group A.
simply knowing that the family has a female, brings this added problem of whether the female is first or second to Group A and C.
btw it’s great laughs for me that every single person i’ve talked to disagrees with me but i’m sure it is causing you guys a lot of aggravation so if no one agrees with me i’ll just let this be.
Well, you said it did, above. But you’re right, it doesn’t.
The point of this problem is that you don’t know whether the daughter you’re told about is the first or second child. If you knew whether she was first or second, it would be a different problem. But since you don’t know, the family could be in either group A or group C (or B, of course).
i’m giving this one last go. the chief difference between A and C is that the female comes first in A and second in C.
let’s say we discovered that the “other” (not next) child is male. now there is a 50% chance the male is older or younger. if the male is younger, A is right and C is wrong. if the female is younger C is right and A is wrong. but FF is still true if female 1 is older or female 2 is older.
What are the chances that the family we are considering fits in Group A? we know one is female, there is 50% the other is male, and 50% again the male is younger.
For Group C? we know one is female, there is 50% the other is male, and 50% again the male is older.
For Group B? we know one is female, there is 50% the other is female, and who cares who is older.
personally, i dont mind a half-dozen people disagreeing with me but im on the verge of tears of laughter at the possibility i might be totally bonkers.
also the best analogy cointoss wise i think is this
i ignore all flips until i get T (i know this implies ordering but bare with me) then retain that and the next flip, and i get either H or T. it will then be TH or TT, even chance for either. then i flip once more to randomize TH into TH or HT, and T1T2 into T1T2 or T2T1. there is a 50% chance i get TH, and 50% again it stays TH, and or else 50% it becomes HT. T1T2 can become T2T1 or stay T1T2, in which case it is still TT.
HTchance and THchance=TTchance, or 1/2 chance of other flip being H.
OK ENOUGH if im wrong i’ll stop it for your sake and mine
How about trying it this way, instead of “A” and “B” use youngest and oldest. This stuff about “order” is being misunderstood. Of course there’s an order, if a family has two children, there’s a younger and older. (I will use “youngest” and “oldest” and avoid grammarians.)
(1) Oldest child is female and youngest child is male.
(2) Oldest child is female and youngest child is female.
(3) Oldest child is male and youngest child is female.
(4) Oldest child is male and youngest child is male.
OK, so if you know that one of the children is female, but you don’t know whether it’s the oldest or the youngest, then the only situation that cannot arise is situation (4), and thus the odds are 2/3 that the other child is male.
The question of “order” changes the scenario. If you know that the oldest child if female, then you’ve eliminated cases (2) and (4) and you’re left with 1/2 that the other child is male.
In a nutshell, that’s the essential difference between the two scenarios. It’s not whether there IS an order (of course there’s an order), it’s whether YOU KNOW the order.
And right there is another error. You’re begging the question: in other words, in order to answer the question “what are the odds they also have a son?” you assume (incorrectly) that there’s a 50% chance that the family has a son.
And more begging the question.
You’re fundamentally misunderstanding the problem (as evidenced by the erroneous coin-toss analogy in your next post). The only information you have (that the family has a daughter) allows you to exclude group D (the MM group).
And, one more time, the whole point of this problem is that you don’t have any other additional information. Because you don’t have any other information, you have no means of determining whether the family is in group A, B, or C, or whether one group is more probable than another.
Telecomm, maybe look at this alternate problem: There is a family with five children. You have been told this family has four daughters. What are the odds they also have a son?
Families with 5 children as all daughters are less common than families with 5 children as 4 daughters and one son. In fact, they are 1/6th as common. (For every family of the first type, there are five of the second — assuming a large enough population.) Therefore, when presented with a randomly picked five-child family known to have four (or more) daughters, you know it’s much more likely they have a son than not. The probability of that is 5/6, not 1/2.
For the same reason, though with different numbers of course, a randomly picked two-child family having one or more daughters is more likely to have a son than not — and the probability of that is 2/3.
I think the monkey wrench in the problem lies with the sentence, “You are told the family has a daughter.” In the imagination, this wording suggests a specific child is being pointed to and revealed as a girl. But, there is no specific girl at all. What you’ve really been told, only, is that the girl-count for this family is 1 or more.
No, those are two separate statements. We know for certain they have at least one daughter. However, if the outcome is either older daughter/younger son or older son/younger daughter, then there was a 1:2 chance of them telling us whether they have a daughter and a 1:2 chance of them telling us they have a son. This is wholly different from “They have a son, what are the odds they have a son” (paraphrased) as you put it.
In that case, what was the probability of their telling us both children’s sexes outright? What was the probability of their saying nothing at all?
I’m not sure it makes sense to talk about the probability of any information being told — only that whatever information is given is also true. (There are puzzles out there about liars and unreliable information, but this isn’t one of them.)
Put it this way. You’re at a party and you bump into a stranger. You ask him how many children he has; he says, “two”. You ask him if he has any daughters; he says, “Yes”. Assuming this guy is an honest sort, he had no choices about how to answer these questions. You’re now in the exact same situation, with the same information, as in the original problem.
No, not at all. The chance of learning that this family has a daughter is 100%, because you already know that. You don’t have any knowledge of how this information was acquired, so you can’t make any conclusions based on how you think the information was obtained.
Well, of course it is. The fact they have a son is something you already know, yes?