Talk:Boy or girl paradox

As this is linked from MontyHallProblem we need to be very careful (as described there) as to how the question is phrased:

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

The 'one child is a boy' is ambiguous because it doesn't explicity explain that the other case 'one child is a girl' is being excluded.

For instance if a parent of a two-child family walks into a room accompanied with a boy (one of their children) is the probability that the other child is a boy or girl anything but 50/50? The answer is NO.

If I have a room full of parents of 2 children families (randomly selected) and I ask all those with a boy (implying 1 or more) to step forward - THEN and only then have I skewed the odds that for the parents that have stepped forward to 1/3 2/3.

Again this is a very subtle point, and worth making explicitly. The fact to are making a decision is very important in this problem.

This example from rec.puzzles.faq makes the 'question step' explicit http://www.faqs.org/ftp/faqs/puzzles/faq

2.3. ==> oldest.girl <== [probability] If a person has two children, and truthfully answers yes to the question "Is at least one of your children a girl?", what is the probability that both children are girls?

The answer is 1/3, assuming that it is equally likely that a child will be a boy or a girl. Assume that the children are named Pat and Chris: the three cases are that Pat is a girl and Chris is a boy, Chris is a girl and Pat is a boy, or both are girls. Since one of those three equally likely possibilities have two girls, the probability is 1/3.

You're welcome to clarify the article, but please keep in mind that the "ambiguity" here is just part of a very general phenomenon affecting all probability problems: giving yourself too much information skews the answer from the textbook result. For example, if I see a two-child family with two boys, then they certainly have at least one boy; nonetheless the probability that they have a girl is 0, not 1/2 or 1/3. I don't get 1/3 because I have more information than is in the statement of the question.

Likewise, in your example, if the parent walks into the room accompanied by a boy, I know that "the child here is a boy", which gives me strictly more information than "one of their children is a boy". With the extra information, it is no surprise that I arrive a correct probability different from the textbook result.

My point is that the abstract question "In a two-child family, one child is a boy. What is the probability that the other child is a girl?" is unambiguous, even if it can be confusing. When translating it into a real-world situation, we must be careful not to introduce the wrong information, and the "question step" you identify is a good way of doing that. But strictly speaking, it is not necessary to analyze the problem. Melchoir 19:09, 5 April 2006 (UTC)[reply]

Questioning

Well, this page is busy today. The recent addition is mistaken; the intent of the question is not to identify any boy at all. I can get a reference on that, but for now I'll revert and clean up the language. Melchoir 19:46, 5 April 2006 (UTC)[reply]

Small Change

I made a small change to the article to try and clear up a common problem. (One I had myself once.)

This doesn't seem to work

Given two variables, U and K, each of these being a children, one (U) whose gender is unknown to us, and one (K) whose gender is known to us. This is the only difference between U and K ; U may be older than or younger than K without any relevance to the case.

U can be either a boy, or a girl. K can be either a boy, or a girl.

The following combinations are possible.

U is a boy, K is a boy. U is a boy, K is a girl. U is a girl, K is a boy. U is a girl, K is a girl.

Since we know the gender of K (in this case, let's assume K is a girl), two possibilities are eliminated : the two possibilities where K is a boy. That leaves us two possibilities, one of which has U as a girl, and one of which has K has a boy. Note again that ages is of absolutely no relevance here : U can be either older than or younger than K without changing a thing.

Essentialy the problem appears to be the mistaken notion that, because you don't know the ages of the children, both G-B and B-G are possible. However, this is false : if both G-B and B-G are possible, then G-G need to appear twice : once for G-G where the girl whose gender is known to us is the first G, and one for G-G where the girl whose gender is known to us is the second G.

Just my two cents.--Damian Silverblade 17:44, 13 April 2006 (UTC)[reply]

Agreed - as I pointed out earlier how the question is phrased makes all the difference, there is a very strong argument that as the question is *currently* phrased the correct answer is 50/50.

Also agreed (or phrased alternately, the three possibilities listed in the sample space are not equally likely). This article is ridiculous. Can we make a motion for deletion? Topher0128 02:58, 12 July 2006 (UTC)[reply]

The article may be phrased badly, but it is indeed a well-known example, and I've seen it discussed in a book describing probability problems from a cognitive perspective. For now, I'll simply add the {{unreferenced}} tag. But it's not a joke. Melchoir 03:04, 12 July 2006 (UTC)[reply]

The phrasing of the first formulation of the question says nothing about the set of Boy/Girl being ordered, and yet the sample evaluation assumes that it is important. The maths are correct just for a different formulation. The second formalization does explicitly order the children(older/younger). The first should read something like

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

and the samples would be oldest then youngest {BB, BG, GB, GG} GG is not possible, since one is a boy so three possibilitys remain, {BB, BG, GB} The girl is the oldest in only 1/3 prob.

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

This one is described correctly. Domhail 04:03, 12 May 2006 (UTC)[reply]

Rules of Conditional Probability?

Precisely. For example in a tree diagram: the way the question is phrased, "In a two-child family, one child is a boy. What is the probability that the other child is a girl?" , we are surely not only excluding the G-G branch from the conditional tree, we are also assuming that the G-B and B-G trees are identical, as the order in this question is irrelevant; therefore these branches should be merged and the answer remains 1/2. —The preceding unsigned comment was added by 15:34, 5 August 2006 (talk • contribs) 89.145.196.3.

That is not a valid line of reasoning. You might as well say that the probability that a two-child family has two boys is 1/3, since there are three possibilities and the question does not care about order. In probability, there are no rules that deal with such notions as irrelevance and "merging branches". Melchoir 16:59, 5 August 2006 (UTC)[reply]

Coin Examples

(A) If I throw 2 coins and let you see one, have I given you any information about the 2nd (hidden) coin? - Obviously not, its probability of heads or tails remains .5/.5

(B) If I throw 2 coins, and I look at them, you ask me is there at least one head, and I answer truthfuly 'Yes' and show you that coin.

We've now arrived at the subtle (and counter-intuitive) case where the probability that the other coin is a tail is now 2/3. The reasoning is explained in the article page and can be verified using a simple computer program (or indeed throwing coins yourself)

But again the atual 'questioning step' is critical in differentiating (A) & (B) and Missing from the article page.

--Pajh 21:33, 21 April 2006 (UTC)[reply]

THIS ARTICLE IS LIES!!!!!

i dont understand how it can be 2/3. its 1/2!!! i swear this article is completely wrong someone delete it pls.

just because there are 3 possibilities doesnt mean a 1 in 3 possibility. someone driving by my house can either shoot me or not, therefore thjeres a 50:50 chance the next car will shoot me. K. I honestly wouldnt be surprized if everyone whos contributed to this article is completely wrong. The intellectual talent of this place is about a 7th grade level. Seriously ive had to compeltely rewrite some engineering articles coz of the morons here.

oh wait i get it. its like the monty hall dealie.

YES BUT NO

The 2-child family may be either : 2 boys (p = 1/4), 2 girls (p = 1/4), 1 boy and 1 girl (p = 1/2).

The rule is :

    p (A / B) = p (A and B) / p (B)
    Probability that A is true if B is true = Probability that A and B are true at the same time / Probability that B is true

The statement for A is clear :

    A = "One child is a girl"

There are 2 different statements for B :

    Case 1.     B = "(I know one of them,) he is a boy" -> p = 1/2
                        (probability for him to be a boy)
    Case 2.     B = "(I know that) at least one of the two of them is a boy" -> p = 3/4
                        (probability that a 2-child family has at least one boy)

Thus, 2 different statements for (A and B) :

    Case 1.     (A and B) = "The one I know is a boy, the one I don't know is a girl" -> p = (1/2).(1/2) = 1/4
    Case 2.     (A and B) = "One is a boy, one is girl" -> p = 1/2

And 2 different results :

    Case 1.     p (A / B) = (1/4) / (1/2) = 1/2  (= p (A) actually, B has no influence)
    Case 2.     p (A / B) = (1/2) / (3/4) = 2/3

Case 1 sounds ok to me. In case 1, we don't need to know that it's a 2-child family. I still find Case 2 very disturbing. The thing is : a simple, almost automatic, deduction leads from a "Case 1" B to a "Case 2" one. I wonder, is it good to know too much about something ?

--[Strahd] 5:35, 12 August 2006 (Orléans, France)

I guess the simple answer is that you can't use deduction in these problems. Melchoir 17:29, 12 August 2006 (UTC)[reply]

I agree --[Strahd] 19:57, 12 August 2006 (France)

YES BUT NO (II)

Trees !

Case 1 :
First, the child we know, then the other child : BB, BG, GB, GG.
The child we know is a boy : BB, BG.
The other one is a girl : BG.
Probability : 1/2

Case 2 :
First, the elder, then the younger : BB, BG, GB, GG.
One child is a boy : BB, BG, GB.
The other one is a girl : BG, GB.
Probability : 2/3.

In Case 2, I'd use the word "frequency" rather than "probability".

--[Strahd] 6:17, 13 August 2006 (Orléans, France)

Similar question

My parents only got two children. I'm a man. What is the probability that I have a brother? In other words, what is the probability that my sibling has the same sex that I have? INic 10:32, 25 August 2006 (UTC)[reply]

Let's look at the sample space, denoted by {You, Sibling}. Originally, the sample space is {B,B}, {B,G}, {G,B}, and {G,G}. Now that we condition on the fact that you (emphasis intended) are a man, the space has two elements: {B,B} and {B,G}. Thus 50%

The key as to why the 1/3 did not work is that you specified which ("I") was the man. If you had said "I note that (at least) one of my parents' two children is a man", then the 1/3 is correct.

But note if you said "One of my parents' two children picked at random is a man", then we're back to 50% again. Bayes' theorem can show this, but the heuristic here is to note the four possibilities above, and to pick one of the four "B"s at random. You are twice as likely to pick one from the {B,B} group as either from the other two groups (50%, 25%, 25% ~~respectievly~~ respectively). Baccyak4H 20:08, 5 September 2006 (UTC)[reply]

An error is in your first statement. It should be: Now that we condition on the fact that you (emphasis intended) are a man, the space has 3 elements: {B,B} and {B,G} and {G,B}. Thus still 33%. [I do not know whether you are the younger or the older]. --Tauʻolunga 23:37, 5 September 2006 (UTC)[reply]

No, there is no error. The notation used was {You (INic), Sibling (of iNic)}. Since INic is a man, the {G,B} outcome is indeed ruled out. One is not only limited to age order to distinguish the two. The 50% situation could occur if you said "the taller one is a boy", or "the one whose first name comes first alphabetically is a boy", etc., or in this case, "the one with the Wikiname INic is a man."

The way to distinguish the two cases is: The 50% scenario occurs whenever the original statement about being a boy allows that if one knew only that and then were to meet the two children, and if the two children were indeed both boys, then it would be possible in principle to tell which of the two boys was referred to originally. Thus the older one, or "Aaron", or the taller one, or INic, or... (of course, if one was a girl the distinction is trivial). You might need some more info (ages, names, etc.), but in principle you can determine which one was mentioned. If this distinction is impossible in principle ("one is a boy"), then we are in the 1/3 case. Note the potential for ambiguity: when one says "one is a boy", it is easy to picture the problem poser looking at one particular child at random and making such an observation. This is not supposed to be the case, and is the reason for much of the language discussion/disambiguation in the article. Which is why I added "at least" to "one [child is a boy]."

But don't worry; there is a reason this paradox topic is included in WP: there is some very important subtlety that is not always easy to grasp. Baccyak4H 03:21, 6 September 2006 (UTC)[reply]

OK, let's say that when I say I'm a man, I only by that mean that "I note that (at least) one of my parents' two children is a man." Then the probability that I have a brother is 1/3 you say, right? But say that I by I'm a man only mean "My name is INic and that's a male name". Then the probability that I have a brother is 1/2 according to you, correct? Is the probability really dependent on how I say that I'm a man? INic 01:28, 7 September 2006 (UTC)[reply]

Of course it is. If you say "both of my parents' children are men" then the probability that you have a brother is 1. Melchoir 01:32, 7 September 2006 (UTC)[reply]

"Is the probability really dependent on how I say that I'm a man?"

Your rephrasing of the your original statement does not say "you" are a man at all. "I note that (at least) one of my parents' two children is a man." allows for you being a woman and having a brother.

So long as one particular individual is identified somehow as being the man, in this case, the writer here at 01:28, 7 September 2006 (UTC) we are in the 1/2 case. In the case where you had a brother, I could tell in principle which man was you because you would be the one posting here at that time. But if you had said "I note that (at least) one of my parents' two children is a man.", there is no information there even in principle to distinguish whether you were referring to yourself or your brother, in that case.

The probability is dependent solely on whether one particular individual is described as "the man". If so, the prob. is 1/2, since we now can speak of the other particular individual. The chance that any particular individual (and the "other" particular individual in particular ;-)), is a man, is 1/2. The description can be of any sort, but it has to identify (somehow) a particular individual.