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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

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348 answers to this question

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

Mine: They have two kids. Given that one of them is a girl, what is the probability that the other is also a girl?

Are they different?

"One of them is a girl" is a statement. And as thus we don't know why this statement is supplied so a condition can not be made. So in all instances of a statistical analysis we have to consider the statement as a random event and are just as likely to be informed "One of them is a boy".

"Given that one of them is a girl" is a condition. Saying that in all instances of a statistical analysis we will always be supplied with the information that one of them is a girl if their is at least one girl.

The difference is with the wording of this problem. All other instances of this equation that I have read on the Internet give a condition (witch is something I overlooked before in my haste.... so Dr. Math is right!). That is not the case with wording of problem in the OP.

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dahw

Fairly robust response. However:

If you can show that these two sentences have different semantic meanings, then I'll accept your answer, but as it is, I don't see that being the case.

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

Mine: They have two kids. Given that one of them is a girl, what is the probability that the other is also a girl?

Are they different?

Please read one of my previous posts. I believe this sets out the equivocal nature of the question.

Semantics: noun

1. (linguistics) The science of the meaning of words. Semantics is part of linguistics.

2. The study of the relationship between words and their meanings.

3. The individual meanings of words, as opposed to the overall meaning of a passage.

Take a close look at 3.

3. The individual meanings of words, as opposed to the overall meaning of a passage.

We all know that the premise of the OP is to apply logic and come up with the answer 1/3.

However, when we deconstruct the sentence it is possible to to answer 1/2. Here's why:

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

Lets take two specific parts of this passage: "one of them" and "the other".

The children are referred to separately. No reference is made to "BOTH" children. If the OP stated "What is the probability they are both girls" then the best answer is 1/3. But it doesn't. The first and second elements of the statement are unconnected, independent. We are asked to assess the probability that "the other kid" is a girl in isolation because in the second part of the statement no reference is made to the original girl. This is crucial. We are not asked to make a correlation or combination. We are only asked to evaluate the probability of "the other kid" being a girl.

In this instance, "the other kid" has an equal chance of being either a boy or a girl.

You may consider this to be semantic pedantry, but the wording of such problems must avoid ambiguity. If the OP read:

"There is a 2 child family who have at least one girl. What is the probability that they are both girls?"

Then this is an open-and-shut case. 1/3. Now look at our problem again:

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

Not the same is it? Same intention, different wording, different answer.

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Ok I can't resist having another go :)

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.

1. If a family has two kids and the older one is a girl, what is the probability that the younger child is a girl?

2. If a family has two kids, one of them is a girl, what is the probability that the other kid is also a girl?

To be fair, 1/2 is in fact the correct answer to the first question. Let us represent a boy with 'B' and a girl with 'G', with the older child coming first, and assume that the boy:girl ratio is precisely 50:50. This produces four possible combinations: BB, BG, GB and GG. However, in the case of the first question, we are told that the older child is a girl, thus rendering the combinations BG and BB impossible. We are thus left with GB and GG, in which one out of the two equal possibilities contains a girl as the younger child. The probability of the younger child being a girl is thus 1/2.

Now let us look at the second question, which states that at least one of the children is a girl. This means that out of the four possibilities, only BB is impossible owing to the fact that it does not contain a boy. As the second question does not state whether the girl is the older or the younger child, it is possible to have any one of GB, BG or GG. In other words, the girl we know of could have an older brother, a younger brother or a sister.

Note that the last possibility, GG, should only be counted once. This point can be confusing and thus merits a further explaination. First, let us look at GB and BG:

GB = there is an older girl who has a younger brother.

BG = there is a younger girl who has an older brother.

Clearly, these two situations are different, and thus represent two distinct possibilities. However, let us treat the ways in which GG might occur in the same manner:

GG = there is a younger girl who has an older sister.

GG = there is an older girl who has a younger sister.

Unlike the first pair of sentences, the ones for GG both describe the same situation - the words we use to describe GG simply depends on which of the girls we think the question has already referred to. GG is therefore only one possibility out of three, and thus has a 1/3 probability of occurring. On the other hand, having an older girl and a younger brother is different to having an younger girl and a older brother.

Two children- combinations are BB, BG, GB, GG.

At least one child is a girl - combinations are GG, BG, GB.

Probability that the other chid is a girl 1/3.

Now I am going to ask another question. This question differs slightly from question number 2 (as above), but again asks us the probability that the other child is a girl.

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

Now let us look at the third question, which states that at least one of the children is a boy. This means that out of the four possibilities, only GG is impossible owing to the fact that it does not contain a boy. As the second question does not state whether the boy is the older or the younger child, it is possible to have any one of GB, BG or BB. In other words, the boy we know of could have an older sister, a younger sister or a brother.

Note that the last possibility, BB, should only be counted once. This point can be confusing and thus merits a further explaination. First, let us look at GB and BG:

GB = there is a younger boy who has an older sister.

BG = there is an older boy who has a younger sister.

Clearly, these two situations are different, and thus represent two distinct possibilities. However, let us treat the ways in which BB might occur in the same manner:

BB = there is a younger boy who has an older brother.

BB = there is an older boy who has a younger brother.

Unlike the first pair of sentences, the ones for BB both describe the same situation - the words we use to describe BB simply depends on which of the boys we think the question has already referred to. BB is therefore only one possibility out of three, and thus has a 1/3 probability of occurring. On the other hand, having an older boy and a younger girl is different to having an older girl and a younger boy, and the probability of the family including a girl is therefore 2/3.

Two children - combinations are BB, BG, GB, GG.

At least one child is a boy - combinations are BB, BG, GB.

Probability of one child being a girl is 2/3.

This question is slightly different from the orginal question, and although it gives us a different answer (2/3) I have posted it to support the argument that there are three possibilities.

It tells us that one of the children is a boy. It asks the probability that the other child is a girl. In this case the boy can have a brother an older sister or a younger sister thus 2/3.

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You say:

-----------------

"There is a 2 child family who have at least one girl. What is the probability that they are both girls?"

Then this is an open-and-shut case. 1/3. Now look at our problem again:

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

Not the same is it? Same intention, different wording, different answer.

-----------------

No it's not a different answer. Consider:

"They have two kids ... "

This means that with equal probability the kids are same or mixed gender.

"... one of them is a girl ..."

This means that the kids are either both girls or mixed gender, the second option having twice the probability of the first.

"... what is the probability that the other kid is also a girl?"

This means they are both [one + the other = both] girls, and that probability [against the twice as likely mixed case] is 1/3.

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Has any one heard of The Monty Hall Problem.

I am posting this is support of the 1/3 camp.

I have taken this from wikipedia. To view the whole Wikipedia page please go to:

http://en.wikipedia.org/wiki/Monty_Hall_problem

Monty Hall problem

From Wikipedia, the free encyclopedia

180px-Monty_open_door.svg.png

The Monty Hall problem is a probability puzzle loosely based on the American television game show Let's Make a Deal. The name comes from the show's host, Monty Hall. The problem is also called the Monty Hall paradox, as it is a veridical paradox in that the solution is counterintuitive.

A well-known statement of the problem was published in Parade magazine:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? (Whitaker 1990)

Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car, from 1/3 to 2/3.

When the problem and the solution appeared in Parade, approximately 10,000 readers, including nearly 1,000 with Ph.D.s, wrote to the magazine claiming the published solution was wrong. Some of the controversy was because the Parade version of the problem is technically ambiguous since it leaves certain aspects of the host's behavior unstated, for example whether the host must open a door and must make the offer to switch. Variants of the problem involving these and other assumptions have been published in the mathematical literature.

The standard Monty Hall problem is mathematically equivalent to the earlier Three Prisoners problem and both are related to the much older Bertrand's box paradox. These and other problems involving unequal distributions of probability are notoriously difficult for people to solve correctly, and have led to numerous psychological studies. Even when given a completely unambiguous statement of the Monty Hall problem, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief.

Now you may say that this is nothing like the original posted question, but I have posted it because the solution for both "paradox's" is quite counter-intuitive.

The Monty Hall Problem demonstrates that you cannot trust your instincts in even fairly simple situations involving chance. You must sit down and work out the details to check that your intuition is correct.

Now apparently most people said there was now a 1/2 (50-50%) chance of choosing the door with the car, but by showing them that door 3 had a goat, it increased the contestants odds from 1/3 to 2/3.

Do you get what I'm trying to say?

If you go to the above link it goes into much more detail with lots of math stuff (equations) that I don't understand so would never post, incase I would have to later explain them (and I wouldn't be able to).

I thought by referring to another probability paradox might help the 1/2 camp understand the 1/3 camps answer.

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Tearz,

It was posted here a year ago.

It has a long discussion thread and similar groups of intuitants who absolutely know it's a wash.

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I wasn't trying to start a new post, I was using an example of a paradox to support my argument, just like others have used the coin toss, penny experiment, coloured marbles in a bag etc..... Whether it is now believed to be a wash or not, my use of this paradox was purely in support of my argument how I see it. I will refrain from using it in the future.

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"One of them is a girl" is a statement. And as thus we don't know why this statement is supplied so a condition can not be made.

<snip>

"Given that one of them is a girl" is a condition.

Interesting. "One of them is a girl" is a statement and not a condition, but if that sentence were to start with "Given that" then that statement for some reason becomes a condition?

"One of them is a girl" is a statement. And as thus we don't know why this statement is supplied so a condition can not be made.

So, unless we know the reason the statement was supplied, we're not dealing with conditional probability? This is plain silly. We're dealing with conditional probability.

So in all instances of a statistical analysis we have to consider the statement as a random event and are just as likely to be informed "One of them is a boy".

Why does that matter? It doesn't. What matters is what we were told. We could have been told one of them is a boy, but we weren't. This information isn't useless just because the information could have been different.

The difference is with the wording of this problem. All other instances of this equation that I have read on the Internet give a condition (witch is something I overlooked before in my haste.... so Dr. Math is right!). That is not the case with wording of problem in the OP.

"One of them is a girl" is a condition.

Dr. Math is right? Here's how it's worded on his website:

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

Why is 2/3 the correct answer in that case and 1/3 not in the case of the OP's riddle?

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Interesting. "One of them is a girl" is a statement and not a condition, but if that sentence were to start with "Given that" then that statement for some reason becomes a condition?

Yes. By changing the statement to "Given that one of them is a girl" adds structure to the information. The information is structured in such a way that we will always be given that their is a girl. However if the information being supplied is as random as the couple then half the time when the have a GB/BG mix we will be told that one of them is a boy. Why because their is no structure to it.

So, unless we know the reason the statement was supplied, we're not dealing with conditional probability?

Yes and no. If you walk up to a guy on the street and he says.

"Hi my name is Teanchi and this is my wife Beanchi. We have two kids, one of them is a girl" (aka the riddle)

Then your odds of the other one being a girl is best guess 50%. Why? Because in a GB/BG mix it is just as likely we will be informed that one of them is a boy. Their is no structure to the information.

But if you walk up to a guy and he says.

"Hi my name is Teanchi and this is my wife Beanchi. We have two kids."

Then you ask

"Is at least one of them a girl"

and he reply

"Yes"

Then the odds are 1/3 because their is structure to the information. The fact that he could have answered "no" changes your odds 1/3.

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

Why is 2/3 the correct answer in that case and 1/3 not in the case of the OP's riddle?

Because "In a two-child family, one child is a boy" is a statement (however false) that sets the guidelines for the question. It says that in any given two-child family, one child is a boy. Kind of like saying "In a bathroom, their is toilet paper". Their a plenty of bathrooms out their without toilet paper (I hate it when that happens) but the initial statement sets the condition that all bathrooms have toilet paper. Now I know what you are going to say. "how is the statement in the riddle any different". Well because the statement riddle pertains only to Teanchi and Beanchi ("they have two kids"). So it more like saying "In your bathroom, their is toilet paper. What is the probability that I have toilet paper in mine".

http://www.mathpages.com/home/kmath036.htm

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Why it can't possibly be 1/3:

There are two ways of looking at this problem:

Method 1: You are looking at the first child, and only the first child.

Following your second spoiler:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

You are looking at the first child, and it is a girl, so the third case is invalid as well as the fourth, leaving the first two choices, 50% each boy and girl. This has been mentioned several times, and is at the root of the "Gambler's fallacy" and other people's explanations in which the first event is already decided and therefore the second is independent, and therefore a probability of 1/2.

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Method 2: You are looking at any one of the children in the original spoiler diagram.

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

This is how the author of the problem explained his 1/3 solution, which is incorrect. Take a look at the diagram: by the author's logic, you can pick a girl from the left or the right. This invalidates picking any of the boys, not just the last possibility, although of course that one is completely invalidated, while the middle 2 are only halfway invalidated.

The possibilities now look like this:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Knowing that you can pick the left or the right child, there are 4 girls to pick from! If you pick either of the girls from the first choice, you have another girl child. If you pick either of the girls from the middle two rows, you have a second boy child. 50-50%, 1/2 probability, not 1/3.

In other words, while by specifying that the first choice is not a boy you have completely eliminated one outcome, you have also made it half as likely to pick either of the two middle solutions.

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There are two ways of looking at this problem:

Method 1: You are looking at the first child, and only the first child.

You are looking at the first child, and it is a girl

Method 2: You are looking at any one of the children in the original spoiler diagram.

you can pick a girl from the left or the right.

Construct a truth table of the premise "One child is a girl" for the 4 cases BB/BG/GB/GG.

What you find is that "One is a girl" is false for BB and true otherwise.

"One is a girl" is logically equivalent to "they are not both boys".

Included among the things that premise does not do is to assign the gender of a selected child.

Both of your methods scrutinize individual children, which ends up trivializing the OP to:

"The probability that a child is a girl is 1/2. What is the probability that child X is a girl?"

Because "one is a girl" means only "not both are boys" we're told [only] about gender combinations.

And we're asked about combinations. Specifically, the case "both are girls." [one + the other = both]

The expectation for 100 two-child families is: 25 have 2 girls; 25 have 2 boys; 50 are mixed gender.

Removing "both are boys" shrinks the sample space to 75 -- 25 with two girls, and 50 with mixed gender.

These probabilities are 1/3 and 2/3, respectively.

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Construct a truth table of the premise "One child is a girl" for the 4 cases BB/BG/GB/GG.

What you find is that "One is a girl" is false for BB and true otherwise.

"One is a girl" is logically equivalent to "they are not both boys".

Included among the things that premise does not do is to assign the gender of a selected child.

Both of your methods scrutinize individual children, which ends up trivializing the OP to:

"The probability that a child is a girl is 1/2. What is the probability that child X is a girl?"

Because "one is a girl" means only "not both are boys" we're told [only] about gender combinations.

And we're asked about combinations. Specifically, the case "both are girls." [one + the other = both]

The expectation for 100 two-child families is: 25 have 2 girls; 25 have 2 boys; 50 are mixed gender.

Removing "both are boys" shrinks the sample space to 75 -- 25 with two girls, and 50 with mixed gender.

These probabilities are 1/3 and 2/3, respectively.

I see what you are saying. The problem states that "one of them is a girl". You interpret that to mean that if either child is a girl, you are told about it, and left to guess about the other one. With three cases of equal probability, and if you are guaranteed to pick the girl of the family if there is one, then I agree. If, however, you choose a random child and it is a girl, I still maintain that the probability that the second child will also be a girl is 1/2 as shown above.

This reminds me of a two-sided card puzzle which I will post right away.

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Yes. By changing the statement to "Given that one of them is a girl" adds structure to the information. The information is structured in such a way that we will always be given that their is a girl.

You're making no sense. "Given that" does nothing to the statement "One of them is a girl". "We will always be given that their is a girl" either way.

However if the information being supplied is as random as the couple then half the time when the have a GB/BG mix we will be told that one of them is a boy. Why because their is no structure to it.

What? We don't know what we would be told half the time. You've even admitted that we don't know why we were supplied the information. All we know is that there is a particular couple that has two children and that one of them is a girl.

Yes and no. If you walk up to a guy on the street and he says.

"Hi my name is Teanchi and this is my wife Beanchi. We have two kids, one of them is a girl" (aka the riddle)

No, that's not the "aka the riddle". Stop making things up already! No representative from the couple has made any statements.

Then your odds of the other one being a girl is best guess 50%. Why? Because in a GB/BG mix it is just as likely we will be informed that one of them is a boy. Their is no structure to the information.

See above. We know nothing about the likelihood of receiving any information in the riddle. All we know is that the couple has two children and one of them is a girl. We know NOTHING about how the information was obtained or why it was obtained.

But if you walk up to a guy and he says.

"Hi my name is Teanchi and this is my wife Beanchi. We have two kids."

Then you ask

"Is at least one of them a girl"

and he reply

"Yes"

Then the odds are 1/3 because their is structure to the information.

Good. Then you should have to agree that 1/3 is the correct answer in this riddle also as we no nothing about how the information is obtained.

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

Why is 2/3 the correct answer in that case and 1/3 not in the case of the OP's riddle?

Because "In a two-child family, one child is a boy" is a statement (however false) that sets the guidelines for the question. It says that in any given two-child family, one child is a boy.

No, it doesn't. "In a two-child family, one child is a boy" is talking about one family. It's not a statement "however false" about all two child families.

Even in your made up example where Teanchi makes a statement, we have no reason to assume all two-child families have one boy.

Kind of like saying "In a bathroom, their is toilet paper". Their a plenty of bathrooms out their without toilet paper (I hate it when that happens) but the initial statement sets the condition that all bathrooms have toilet paper. Now I know what you are going to say. "how is the statement in the riddle any different". Well because the statement riddle pertains only to Teanchi and Beanchi ("they have two kids"). So it more like saying "In your bathroom, their is toilet paper. What is the probability that I have toilet paper in mine".

Teanchi and Beanchi have toilet paper in their bathroom. What is the probability that they have toilet paper in their bathroom?

A two child family has a boy. What is the probability that they have a boy?

It's the same. It doesn't make any difference whatsoever that the couple has been named.

Your example above is equivalent to saying the following.:

Kind of like saying "In a family with two children, one is a boy". Their a plenty of two-child families out their without a boy (I hate it when that happens) but the initial statement sets the condition that all two-children families have one boy. Now I know what you are going to say. "how is the statement in the riddle any different". Well because the statement riddle pertains only to Teanchi and Beanchi ("they have two kids"). So it more like saying "in your two-child family there is a boy. What is the probability that my two-child family has a boy".

If that were how the riddle were framed, the answer would be 1/1 because your version already stated "the condition that all two-children families have one boy". But even if the riddle did state that all two-child families have one boy, the answer would still be 1/3 because the riddle asks what is the probability the other child is a boy (actually, we're dealing with girls in the riddle). It doesn't ask what the chance of having a boy is. Again, if it did and it were framed for us to believe that all two-child couples have one boy (and the riddle is not framed that way) and asked what the probability of having one boy were, the answer would be 1/1.

I see what you are saying. The problem states that "one of them is a girl". You interpret that to mean that if either child is a girl, you are told about it, and left to guess about the other one.

There's no interpreting how anything was told to us; we just deal with what we were told. With the information that was give, the answer is 1/3.

With three cases of equal probability, and if you are guaranteed to pick the girl of the family if there is one, then I agree. If, however, you choose a random child and it is a girl, I still maintain that the probability that the second child will also be a girl is 1/2 as shown above.

We've all agreed with this throughout the thread. But the riddle is not about "the second child".

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I see what you are saying. The problem states that "one of them is a girl". You interpret that to mean that if either child is a girl, you are told about it, and left to guess about the other one. With three cases of equal probability, and if you are guaranteed to pick the girl of the family if there is one, then I agree. If, however, you choose a random child and it is a girl, I still maintain that the probability that the second child will also be a girl is 1/2 as shown above.

This reminds me of a two-sided card puzzle which I will post right away.

Well put. B))

Is this your card puzzle?

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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

Your spolier logic regarding sample space is faulty. The sample space you present are the combinations possible before any children are born. Once one child is known to be a girl, the sample space for a family of two becomes:

Girl - Girl

Girl - Boy

And the chance that the second child is a boy is 50%. Probability only applies to unknowns, not things already known. To argue otherwise, you would have to explain how the birth of one childs affects the gender of the next.

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ShawnInToronto, please read more of the posts in this thread. What you've described has been explained ad nauseum. The riddle does not ask about the probability of "the second child". We don't know if a boy was born first or not, so there are three equal possibilities:

GG

GB

BG

See the following links for further information.

http://mathforum.org/dr.math/faq/faq.boy.girl.html

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

http://www.mathpages.com/home/kmath036.htm

The rationale behind the OP's answer has been explained more than enough.

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Let mine be one more vote for itsclueless's correct (also, rational :P ) answer. The probability of the "other child" being a girl is 1/2.

The answer offered by brainden.com is incorrect, and even the cusory consideration that observing one sibling changes the probability of the other's gender should set off red flags in people's minds.

The proof that I give to people is as follows:

The two facts we know are:

two children have been born (fact A)

at least child is female (fact B)

Fact A gives us the four possible populations (male/female configurations):

M-M

M-F

F-M

F-F

It then also seems logical to eliminate any combination where at least one female is not present (i.e. M-M), leaving,

M-F

F-M

F-F

and then concluding a 67% probability of 2M and 1F. The problem is that these three remaining combinations are not equally likely. That is, we've botched the application of Bayes' rule.

The correct approach is to start with as our four equally-likely populations:

M-M

M-F

F-M

F-F

We now sample randomly from one of these four possible populations. We sample a 'F'. (This sampling process is completely equivalent to the statement "there is at least one female in the population". No more and no less information is gained.)

Since we know the a priori probabilities of each of the four configurations (populations) and the probability of sampling an F given each configuration, Bayes' rule tells us what the (conditional) probability of each configuration is after we've observed an 'F'.

I turns out that prob( M-M ) = 0, as expected (i.e. you can't sample a male from female-only population), but for the remaining three,

prob( M-F ) = 1/4

prob( F-M ) = 1/4

prob( F-F ) = 1/2

Or stated more intuitively: the fact that we sampled a female member gives us a higher probability that we sampled it from a female-rich population. In this case, the probability of the F-F config is exactly the same as the combined probabilities of the F-M and M-F configs, giving 50%/50% as expected.

Hence, to summarize, we started out with four populations of equal a priori probability:

M-M : probability 1/4

M-F : probability 1/4

F-M : probability 1/4

F-F : probability 1/4

We observed that at least one of the two children is female, which, through a straightforward application of Bayes' rule, yields our post-observation probabilities:

M-M : probability 0

M-F : probability 1/4

F-M : probability 1/4

F-F : probability 1/2

In configurations 2 and 3, the "other child" is always male. In configuration 4 (which is twice as likely given our observation), the "other child" is always female. Hence:

probability( second child is male ) = 1/4 + 1/4 = 1/2

probability( second child is female ) = 1/2 = 1/2

Q.E.D.

Edited by syonidv
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Wow, I was a doubter, but I have to say I'm impressed. The question I will ask all of you, before all of your bickering and mathematics, is have you tried it out? I wrote a simple program to chuck out the numbers, and the results are pretty astounding. (If someone has already done this, forgive me, but I didn't feel like going through 32 pages checking for it.) I also have to say that in writing the program, I see why it gets the results it gets.

Here's the code (in C#):



static void Main(string[] args)

    {

        int secondGirls = 0, secondBoys = 0;


        for (int count = 0; count < 1000; ++count)

        {

            Family family = new Family();


            if (family.Child1 || family.Child2)

            {

                Console.Write("Family1 has a girl. ");

                if (family.Child1 && family.Child2)

                {

                    Console.Write("The other child is a girl.\n");

                    ++secondGirls;

                }

                else

                {

                    Console.Write("The other child is a boy.\n");

                    ++secondBoys;

                }

            }

        }


        Console.WriteLine("Of the second children, {0} are girls and {1} are boys", secondGirls, secondBoys);


        Console.ReadLine();

    }

}


public class Family

{

    private static readonly Random random = new Random();


    public bool Child1 = random.Next(2) == 1;

    public bool Child2 = random.Next(2) == 1;

}

"Of the second children, 249 are girls and 492 are boys."

This means that for every family where one child was a girl, there was 1/3 of a chance that the other child was a girl.

Think of it as you will, but this program is short, sweet, and to the point, and I think full-proof.

I can also see where some people will argue with my methods. The main point they will make is that I intentionally dropped out the 1/4 of the families with both boys, thus conforming to what the 1/3-ists say. However, I repeat that I was a doubter at first. I thought that this experiment would show that the probability is 1/2.

The main point of the puzzle is that there are already two children, not one child with another being born. I set up my program to set up each family before testing it. If both were boys, I simply discarded the family as it didn't apply to the puzzle. If I had set it up so that the second child was determined only after the first was determined to be a girl (as in a birth scenario), I would obviously get a 50-50 result.

That being said, I see both sides very clearly now, and I imagine the bickering will continue, as my code clearly "leans toward" the 1/3-ist's perspective. All I'll say is the results of this experiment turned my thoughts around. Maybe, just maybe, both sides are right, and it really all depends on how you look at the problem.

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This teaser is about 4 years old now. It was resurrected about a year ago as: http://brainden.com/...8--/&#39; . On the last couple of pages, you'll see there are programs which come to the same conclusion as benjer3's. (including a rare excursion into programming by one Mr F.Pig :P )

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This teaser is about 4 years old now. It was resurrected about a year ago as: http://brainden.com/...8--/&#39; . On the last couple of pages, you'll see there are programs which come to the same conclusion as benjer3's. (including a rare excursion into programming by one Mr F.Pig :P )

***************Sorry, the above link seems to have got corrupted between copying and pasting. Try

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

In my view the word 'ALSO' in OP has restricted the probability to only either the other kid (the kid other than the girl kid) is ALSO a girl, or it is Boy. So answer should be 1/2.

But It could also be like this: You are blind folded before two kids. you are told one of the kids is a Girl. NOW you are asked to touch one kid. THEN you are asked what is the probility that the OTHER kid is a girl. In this case there are three possibilities:

You touched may be a BOY, Other is a GIRL. OR

You touched may be a GIRL, Other is a GIRL. OR

You touched may be a GIRL, Other is a BOY.

But this answer will suit only when 'ALSO' word is deleted from OP.

Edited by bhramarraj
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Wow, I was a doubter, but I have to say I'm impressed. The question I will ask all of you, before all of your bickering and mathematics, is have you tried it out? I wrote a simple program to chuck out the numbers, and the results are pretty astounding.

Your simulation is flawed, specifically at the point where you query

if( family.Child1 || family.Child2 )
The 'or' condition encapsulates the condition that the family has at least one girl, but it does not encapsulate the restriction that once a girl has been observed, we need to definitively assign this girl as "sibling #1" or "sibling #2". Consider the corrected code, which honours the constraint that the observed child must be either child 1 or child 2 (I wrote it in javascript, and you can run it by simply pasting it into a site such as this one:
(function() {

    var BOY = 0, GIRL = 1;

    var FIRST = 0, SECOND = 1;


    var second_girls = 0, second_boys = 0;


    for( var i = 0; i < 100000; i++ ) {

        var family = {

            child1: Math.random() < 0.5 ? BOY : GIRL,

            child2: Math.random() < 0.5 ? BOY : GIRL

        };


        var child_we_happen_to_see = Math.random() < 0.5 ? FIRST : SECOND;

        var gender_of_child_we_see = (child_we_happen_to_see == FIRST) ? family.child1 : family.child2;


        if( gender_of_child_we_see == BOY )

            continue; /* this is our conditioning; omit any trial where a boy is observed */


        var gender_of_other_child = (child_we_happen_to_see == FIRST) ? family.child2 : family.child1;

        if( gender_of_other_child == BOY )

            second_boys++;

        else second_girls++;

    }


    alert( "Number of boys is " + second_boys + ".\n" +

        "Number of girls is " + second_girls + ". This is " +

        (new Number( 100*second_girls/(second_girls + second_boys))).toFixed( 2 ) + '%.' );

})();

In my run, I get "Number of boys is 25020. Number of girls is 24799. This is 49.78%."

This makes intuitive sense. The number of included trials is approx. 50,000--all those cases where we happened upon a family and happened to observe a girl child with them. And of those families, approximately half have a boy as the second child.

I believe somebody mentioned this earlier, but an excellent analog that helps make sense of "why you need to constrain the observed child to either child 1 or child 2" is called the "Gambler's Ruin" problem.

Consider I have two boxes, A and B. I tell you that there is some non-zero amount of money in the boxes, and that one box contains exactly twice the amount of money as the other one. But I don't tell you which one is which.

Suppose you pick box A. It has some unknown amount of money, X, in it. But then you think: wait a minute, there's a 50% chance that box B has twice as much money, or 2X, and a 50% chance that it has half as much money, or X/2. So if we calculate the expected value for the amount of money in box B we get

E(B) = 0.5*(2X) + 0.5*(X/2) = 1.25X

Or, roughly 25% more money than in box A. So we decide to swap. But now we let our (slightly larger) amount of money be X and the same logic applies to swapping back to box A.

Hence, by simply swapping A with B again and again, we can make our amount of money grow to infinity! We're rich! :D

Of course, we can disprove this fallacy by honouring the "our box must be only one of either A or B" constraint. That is, the calculation

E(B) = 0.5*(2X) + 0.5*(X/2)

assumes that the box we're holding is in a fuzzy state of simultaneously being both the greater AND the lesser money box (X/2 is only valid if the former is true, and 2X is only valid if the latter is true).

In the same way, "if( family.Child1 || family.Child2 )" assumes the child we've observed is in a fuzzy state of simultaneously being both the first AND second child. But it has to be one or the other.

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Did you get to the end of my post? I came to a very important conclusion, I think. Your program assumes that the parents randomly told us the gender of one of their children, which is perfectly logical. On the other had, mine assumes that if a girl is present, the parents would mention that, as that is what we're testing for. I believe that that is the whole teaser in a nutshell. It all depends on how you test it. The problem is, the 1/3-ists can't prove the 1/2-ists wrong as any 1/3-ist proof doesn't apply to the 1/2-ists, and vice versa.

Let's do a little mind experiment. Say we went to test this out for real. You go around asking parents with two children what the gender of one of their children are, while go around asking if one of their children are a girl. You then (if they say a girl) ask the gender of the other child, while I (if they respond yes) do the same. Our results will be obvious: about 1/2 of your second children will be girls, while about 1/3 of mine will be girls. Did we do something wrong? Well yes and no. Yes because we tested for two completely different things. I can't say to you, nor you to me, "You should have gotten my results," because we each tested for something different. However, there is nothing wrong with either of our methods of testing.

The big question now is which is right? I propose that they are both right. It just depends on how you read the question. So if anyone ever points a gun at your head and asks you, "I have two children. One of them is a girl. What's the gender of the other one?" you should first ask him, "If you had a boy and a girl, would you tell me one was a girl, or would you pick arbitrarily between them?"... and that would probably leave him stumped enough for you to get away. :P

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I see where you're coming from, Ben, but let me play the contrarian.

The specific "condition" statement offered by the problem is: "They have two kids, one of them is a girl."

The question to be answered (Q2BA) is: "What is the probability that the other kid is also a girl?"

I agree with your assessment on the difference between the 1/3-ist question and the 1/2-ist question, but the Q2BA makes it clear that the 1/3-ist question is not well-defined in all contexts.

Suppose I ask the 1/3-ist question: "Do you have at least one girl?" and the parents reply "Yes." We'll call this a "yes" family.

Then I ask: "What is the gender of the other child?"

This question is well-defined if the parents have one boy and one girl, but for the approx. 1/3 of the time that the "Yes" families have two girls, they'll ask what 'other' child you're talking about. You didn't specify a 'first' child such that there can be an 'other' child in this case.

Or to make the point clearer, suppose the second question is: "What is the gender and age of the other child?"

If you asked this question, the parents' response would appropriately be "Uh... actually.. we have two daughters. Which one do you mean?", which is ostensibly them telling you "there is no 'other' child; you didn't specify one to begin with".

The 1/2-ist perspective doesn't suffer this drawback. For the 1/2-ist question: "Think of which of your kids did <totally random action>. Is this kid a girl?" (e.g. "Think of which of your kids you saw chewing bubblegum last. It is a girl?", or "Which kid do you have with you? Is it a girl?", etc.) and they answer 'yes'.

Now you have a well-defined 'other' in all contexts, as well as the condition statement being satisfied.

And since families with two girls will be "Yes" families twice as often than 1-boy 1-girl families (per Bayes' rule) under this scheme, our 1/2-ist philosophy comes by its name rightfully. ;)

My $0.02, at any rate. :)

Edited by syonidv
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.

Suppose I ask the 1/3-ist question: "Do you have at least one girl?" and the parents reply "Yes." We'll call this a "yes" family.

Then I ask: "What is the gender of the other child?"

This question is well-defined if the parents have one boy and one girl, but for the approx. 1/3 of the time that the "Yes" families have two girls, they'll ask what 'other' child you're talking about. You didn't specify a 'first' child such that there can be an 'other' child in this case.

Or to make the point clearer, suppose the second question is: "What is the gender and age of the other child?"

If you asked this question, the parents' response would appropriately be "Uh... actually.. we have two daughters. Which one do you mean?", which is ostensibly them telling you "there is no 'other' child; you didn't specify one to begin with".

The 1/2-ist perspective doesn't suffer this drawback. For the 1/2-ist question: "Think of which of your kids did <totally random action>. Is this kid a girl?" (e.g. "Think of which of your kids you saw chewing bubblegum last. It is a girl?", or "Which kid do you have with you? Is it a girl?", etc.) and they answer 'yes'.

Now you have a well-defined 'other' in all contexts, as well as the condition statement being satisfied.

And since families with two girls will be "Yes" families twice as often than 1-boy 1-girl families (per Bayes' rule) under this scheme, our 1/2-ist philosophy comes by its name rightfully. ;)

My $0.02, at any rate. :)

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?"

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