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# One Girl - One Boy

## Question

Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

Assume safely that the porbability of each gender is 1/2.

Ofcourse its not 1/2 else would make it a lousy puzzle...

Ans: 1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.

This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

• Created

## Recommended Posts

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Hmm, you do have a point. And I'm beginning to think that this question cannot be answered by experimentation. The reason I say that is surveying about the genders of people's children is very different from them just coming out and telling you. If you ask people about their children, you have no prior knowledge and thus have to remove those cases where you do not get the results you want. However, if a bunch of people come up and tell you the same condition (that they have two children and one girl), every case has to be considered; no case is just thrown out, meaning there's no dropping of combinations. So now the question I'm pondering is, if a thousand people come up to you and say said condition, how many of the unmentioned children are girls?

In this case, there are only two possible scenarios. Either the unmentioned is a girl or a boy. You can't consider the four cases, because that assumes that we took our sample families out of a larger group of families with any possible combinations, but that is not the case. We have our sample population; we can't compare it to the rest of the population. So I would have to say my answer to the original question is 50%.

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Oh good grief.

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Oh good grief.

Sorry, I'm always doubting myself.

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This is getting ridiculously obtuse. benjer - you were right. Now extend your program to show the percentage of 2 child families who had 2 boys. and you'll find .... well see what you find.

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Ummm... I didn't read this entire thread (waste of time), but the issue here is simply a combination/permutation issue.

Since it doesn't matter whether the child stated as the girl is the first born or second born, there isn't as many scenarios as the "answer" claims.

A child is a girl

Possibilities:

Girl-Girl

Girl-Boy } Permutation of

Boy-Girl } the same thing

Boy-Boy

We can rule out the boy-boy field since there is already the claim that one child is a daughter.

What we are left with is

Girl-Girl

Girl-Boy

That is a 50% chance.

If they claimed that the FIRST child was a girl, then you can include the permutations and assume the 1/3 chance of the second child being a girl.

Problem solved.

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Hmmm, the more I think about it, the more issues there are with it.

It depends on whether you look at it small scale or large scale.

Small scale it appears to be 50% chance because the birth of one child doesn't affect the other.

However, if you place a limitation on it, and look at it as a large scale model, then the answer is 1/3

For a large scale model you have 100 families.

25 G-G

25 G-B

25 B-G

25 B-B

You know one child is a girl, what are the odds the other child is a girl would be 25/75 (since the B-B are eliminated).

However, this is placing limitations and additional information on the question.

Like was said, small scale and large scale are contradicting each other

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Ummm... I didn't read this entire thread (waste of time), but the issue here is simply a combination/permutation issue.

Since it doesn't matter whether the child stated as the girl is the first born or second born, there isn't as many scenarios as the "answer" claims.

A child is a girl

Possibilities:

Girl-Girl

Girl-Boy } Permutation of

Boy-Girl } the same thing

Boy-Boy

We can rule out the boy-boy field since there is already the claim that one child is a daughter.

What we are left with is

Girl-Girl

Girl-Boy

That is a 50% chance.

If they claimed that the FIRST child was a girl, then you can include the permutations and assume the 1/3 chance of the second child being a girl.

Problem solved.

Hmmm, the more I think about it, the more issues there are with it.

It depends on whether you look at it small scale or large scale.

Small scale it appears to be 50% chance because the birth of one child doesn't affect the other.

However, if you place a limitation on it, and look at it as a large scale model, then the answer is 1/3

For a large scale model you have 100 families.

25 G-G

25 G-B

25 B-G

25 B-B

You know one child is a girl, what are the odds the other child is a girl would be 25/75 (since the B-B are eliminated).

However, this is placing limitations and additional information on the question.

Like was said, small scale and large scale are contradicting each other

1st red. You're missing the fact that there is twice as much chance of a GB combination (as BG also comes into this category)

2nd red. Surely Vice-versa?

3rd red. No. As two - 2 child families have, by definition, 4 children, the chances are that there are going to be 2 of each gender. If family A has (at least) 1 girl, the chances of the other child is a girl is one in 3. ie of the 2 boys and 2 girls, 1 girl is accounted for, leaving 1 girl and 2 boys.

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Hmm, I think I've had a breakthrough. Let me put it in terms of the OP:

A parent comes up to you and says, "I have two children and one of them is a girl." They then proceed to tell you the gender of the other child. Being the analytic people that we are, we wonder what the odds that the second child was a girl were. While we ponder this, another parent happens to come up and says, "I have two children and one of them is a boy." A hundred parents then proceed to do the same, with the gender of the first child varying. (It also just so happens that, if each parent has at least one girl, they will say, "One of them is a girl.")

Now I ask, can we just discard the cases where the first child was a boy? I propose not, for in doing so we would be selectively taking members out of a sample group that needs to be complete, not to mention skewing the results by doing so selectively. So what are we to do? What if we changed our question, while keeping the same general idea intact: "Given the gender of one in two children, what are the odds that the other is a girl?" This question encapsulates the original question, as well as allowing for other scenarios. But now the answer is obvious, isn't it. In this question, it is obvious that the gender of the "other child" is completely independent of the gender of the first. And thus its answer can be nothing other than 50%.

So think what you will. I did change the question, but as I said, I believe the question is a perfect substitute for the original as it asks the exact same thing but allows for a complete sample group.

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Apart from 1 thing...........If one of the 2 children is a boy, the chance of the other being a girl is 2/3

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On the contrary. Look at this program (and let me know if it's wrong ):

class Program

{

static void Main(string[] args)

{

Random random = new Random();

int numberOfFamilies = 1000;

int numberSecondGirl = 0;

int numberGirlGirl = 0;

int numberBoyGirl = 0;

for (int count = 0; count < numberOfFamilies; ++count)

{

Family family = new Family();

int childIndex = random.Next(2);

bool firstChild = family.Children[childIndex];

bool otherChild = family.Children[(childIndex == 0) ? 1 : 0];

if (otherChild)

{

++numberSecondGirl;

if (firstChild)

{

++numberGirlGirl;

}

else

{

++numberBoyGirl;

}

}

}

Console.WriteLine("Out of {0} families, {1} of the \"other children\" were girls. Out of those, {2} of the first children were girls ({3}%), while {4} of them were boys ({5}%)",

numberOfFamilies, numberSecondGirl, numberGirlGirl, ((double)numberGirlGirl / (double)numberSecondGirl) * 100D,

numberBoyGirl, ((double)numberBoyGirl / (double)numberSecondGirl) * 100D);

}

}

public class Family

{

private static readonly Random random = new Random();

public bool[] Children = new bool[2];

public Family()

{

for (int index = 0; index < Children.Length; ++index)

{

Children[index] = random.Next(2) == 1;

}

}

}

For one run, this resulted in "Out of 1000 families, 515 of the 'other children' were girls. Out of those, 271 of the first children were girls (52.6%), while 244 of them were boys (47.4%)." The results are kinda backwards from what you stated, but I think you get the idea.

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so what were the other 485? If you're claiming that they're boys, then that makes 485*2 boys = 970 + (244 out of your 515 families) which leaves 786 children of whom 271 are girls.and the other 515 are ???.

Or are you taking it that the girl you've been introduced to is the eldest child?

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so what were the other 485? If you're claiming that they're boys, then that makes 485*2 boys = 970 + (244 out of your 515 families) which leaves 786 children of whom 271 are girls.and the other 515 are ???.

Or are you taking it that the girl you've been introduced to is the eldest child?

In the other 485, all of the "other children" were boys, and the first children could be boys or girls. I could modify the program to show that the first children were equally distributed among boys and girls in this group as well. That leaves 515 "other" girls, 485 "other" boys, (271 + ~0.5 * 485) "first" girls, and (244 + ~0.5 * 485) "first" boys, giving a near perfect distribution. And if you look at the program, I randomly select from the first and second child of the family (though it wouldn't make a difference either way; I just did that to make it more realistic), so I'm not assuming anything about the children I've been introduced to.

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class Program

{

static void Main(string[] args)

{

Random random = new Random();

int numberOfFamilies = 1000;

int numberGirlGirl = 0;

int numberBoyGirl = 0;

int numberBoyBoy = 0;

int numberGirlBoy = 0;

for (int count = 0; count < numberOfFamilies; ++count)

{

Family family = new Family();

int childIndex = random.Next(2);

bool firstChild = family.Children[childIndex];

bool otherChild = family.Children[(childIndex == 0) ? 1 : 0];

if (firstChild)

{

if (otherChild)

{

++numberGirlGirl;

}

else

{

++numberGirlBoy;

}

}

else

{

if (otherChild)

{

++numberBoyGirl;

}

else

{

++numberBoyBoy;

}

}

}

Console.WriteLine("Out of {0} families, there were {1} girl-girls ({2}%), {3} girl-boys ({4}%), {5} boy-girls ({6}%), {7} boy-boys ({8}%).",

numberOfFamilies,

numberGirlGirl, (double)numberGirlGirl / (double)numberOfFamilies * 100D,

numberGirlBoy, (double)numberGirlBoy / (double)numberOfFamilies * 100D,

numberBoyGirl, (double)numberBoyGirl / (double)numberOfFamilies * 100D,

numberBoyBoy, (double)numberBoyBoy / (double)numberOfFamilies * 100D);

}

}

public class Family

{

private static readonly Random random = new Random();

public bool[] Children = new bool[2];

public Family()

{

for (int index = 0; index < Children.Length; ++index)

{

Children[index] = random.Next(2) == 1;

}

}

}

This yields "Out of 1000 families, there were 239 girl-girls (23.9%), 247 girl-boys (24.7%), 280 boy-girls (28%), and 234 boy-boys (23.4%)." Near-perfect distribution.

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If you have 2 children, both girls, and 1 of them is a girl, then the 'other' child is also a girl. I really don't see your point.

I have 2 daughters. If someone says to me "Do you have at least 1 daughter?", I say "Yes". If they then ask what the gender of my 'other' child is, I don't say "what other child?"

Indeed, sir, but to the question "What is the gender and age of your other child?", the parents would answer, "What 'other' child?"

The problem is unique in that even if there is no specific 'other' child, we have the liberty of assuming 'female' since this is the only possibility. My point is that the problem statement specifically asks for the gender of the 'other child', indicating that the information we have pertains to one child, and not to two collectively. To wit, we are asked for the gender of the 'other child', not "What is the probability both children will be girls?"

I also hold fast to the 1/2-ist view and the hard semantics of 'other' (rather than Mr. Ben's reasonable "both can be correct" view) because applying conditionals to groups (e.g. "at least one of X or Y is") can lead to inconsistencies for even slight variations of the representative simulation.

Consider a dentist's office that serves 500 2-child families. Each child has a file. Suppose the dentist asks the receptionist to place a star on the file of any child whose family has at least one girl. Later, the dentist asks the receptionist to look through all the files with stars on them and for each one write down the gender of the patient's sibling.

In this case, the list of siblings would be roughly 2/3 female and 1/3 male.

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"Other" answers to "One." it does not answer to "At least one."

But be careful not to identify "One" with a specific one.

That leads to 1/2 unambiguously, but the OP does not support that interpretation

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That leads to 1/2 unambiguously, but the OP does not support that interpretation

I believe the mere presence of "other" in the OP implies that our given information pertains specifically to one individual (not two collectively) and the question asks us to infer information about the "other" individual.

At its heart, the 1/3-ist interpretation is that the given information pertains to the joint distribution of two individuals, and the solution is an inference that can be made about the conditional joint distribution (i.e. the fact that in the FF case, we may conveniently neglect the issue of selecting an "other" and conclude "female" since both siblings are female).

But perhaps this speaks to the point of many posts in this forum, which is that the problem is poorly stated. The answer depends on what specifically is meant by "two kids; one girl"--how that information is obtained.

Thanks to all for the discussion, at any rate.

Edited by syonidv
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I believe the mere presence of "other" in the OP implies that our given information pertains specifically to one individual (not two collectively) and the question asks us to infer information about the "other" individual.

At its heart, the 1/3-ist interpretation is that the given information pertains to the joint distribution of two individuals, and the solution is an inference that can be made about the conditional joint distribution (i.e. the fact that in the FF case, we may conveniently neglect the issue of selecting an "other" and conclude "female" since both siblings are female).

But perhaps this speaks to the point of many posts in this forum, which is that the problem is poorly stated. The answer depends on what specifically is meant by "two kids; one girl"--how that information is obtained.

Thanks to all for the discussion, at any rate.

Well said. If the OP had said "It's not the case that they are both boys. What is the probability that they are both girls?" this forum would have been much shorter.

Thanks to you, for posting your thoughts clearly and cogently.

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The probability that they are both girls is 1/2. We know that one child is a girl. This does not affect the probability of what the other child is. If you are equally likely to have a boy or a girl, having another child does not make any change to that. In fact, there are only 3 possible combinations of gender, two boys, two girls, and one of each. Since two boys are not possible, there is a 1/2 probability of the other child being a girl.

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At first, I thought this 35 page long argument was ridiculous, but after reading bona/syon's posts above, I can understand some of the trouble surrounding the ambiguity of the question. This is the way I see it:

There are four possibilities for boy/girl combinations, as mentioned several times by now (I imagine):

BB

BG

GB

GG

Note that this categorization does take into account the order of the children. We are shown that one of the children is a girl, which of course eliminates the first combination.This is the point where people jump to the answer of 1/3, since there only seem to be three combinations remaining. However, as I stated before, order does matter if you look at the problem this way. Therefore, there are still four possibilities:

G1B2

G2B1

G1G2

G2G1

Where the number indicates order. The girl we see can be either the elder (G1) or younger (G2) sister, so we must look at both possibilities equally. Thus, the probability stands at 1/2, regardless of the order of birth.

Another way to look at this is to completely disregard order in all of the cases. The resulting combinations are:

BB

GB (same as BG!)

GG

When BB is eliminated, the probability of GG remains at 1/2.

Edited by ParaLogic
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The probability that they are both girls is 1/2. We know that one child is a girl. This does not affect the probability of what the other child is. If you are equally likely to have a boy or a girl, having another child does not make any change to that. In fact, there are only 3 possible combinations of gender, two boys, two girls, and one of each. Since two boys are not possible, there is a 1/2 probability of the other child being a girl.

• If OP is taken to identify "one" of the two, and to say that "one" is a girl, the probability is indeed 1/2 that the "other" is, too.

• If OP is taken to say it is not the case that they are both boys, then one can deduce a probability of 1/3, although it is not necessary to conclude 1/3 since our fact-reporter's algorithm has not been disclosed.

• It is incorrect to say there are 3 combinations and one is removed so p=1/2. That reasoning ignores the equal-likelihood requirement: I buy a lottery ticket, and there are only two possibilities: winning and losing. Therefore my chances of winning are ...

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I think discussion so far might have not satisfied many learned friends. May be my explanation would clarify some doubts.

OP says that the couple has two kids, so obviously one will be younger and other elder.

Say couple sends elder kid to attend a function. Now what is the probability of this kid being a girl, if one kid is definitely a girl...?

I think you got it it now....!

Say younger is a girl, then there are two possibilities: elder may be a boy or a girl.

As bonanova stated correctly, OP does not specifies which kid is a girl. so younger kid could be a boy, so the third possibility is that elder could be a girl; here fourth possibility of elder being a boy, when younger is a boy, is ruled out because OP says that one kid is definitely a girl.

Now out of above three possibilities, there is only one possibility of both kids being girls.

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Shouldn't it be a 1/2 as one child is a girl and that another child can be a boy or a girl.

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Shouldn't it be a 1/2 as one child is a girl and that another child can be a boy or a girl.

Not necessarily. What you say is correct: there are only two possibilities. But are they equally likely? Try thinking in terms of "they are not both boys." It's logically equivalent. Do you still get 1/2?
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Old thread, and masses of discussion, but to simplify - There are 4 equally likely combinations of 2 children: Boy/Boy; Girl/Girl; Boy/Girl; Girl/Boy. The odds of the second child being a girlÂ depend upon how the OP is interpreted and the possible combinations (green) each interpretation allows.

Â

1.Â They have two kids, [only]Â one of them is a girl.

Boy/BoyÂ  Girl/GirlÂ Â Boy/GirlÂ  Girl/BoyÂ  -Â  Trick question. In the 2 possible combinations the other child must be a boy. Probability 0.

Â

2.Â They have two kids, [the first]Â one of them is a girl.

Boy/BoyÂ Â Girl/GirlÂ Â Boy/GirlÂ Â Girl/BoyÂ  -Â  There are 2 possible combinations where the first child is a girl. In 1 the other child is too. Probability 1/2.

Â

3.Â They have two kids, [at least]Â one of them is a girl.

Boy/BoyÂ Â Girl/GirlÂ  Boy/GirlÂ  Girl/BoyÂ  -Â  Without the Boy/Boy combination 3 remain, in 1 of which the other child is a girl. Probability 1/3.

Edited by Jellonz
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Geez! ... How do these "fossil" questions keep surfacing?

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