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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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## Recommended Posts

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A lot of the initial confusion in this riddle is that it was worded differently the first few days. Initially, it was implied that a couple was going to have 2 children (future tense), and the question was, if the first child was a girl, what would the probability be for the second to also be a girl.

I do see with the wording now, the actual intent of the OP. It is indeed 1/3 the way it is worded now, just as it is 1/4 the way it was worded originally.

Not only did the OP never change the wording of the riddle, if it was worded the way you claim, the answer to the problem you wrote above is a probability of 1/2, not 1/4.

This is why in roulette black and red are 2:1 payout

1:1 payout.

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It does come down to the wording.

"One of them is a girl"

versus

"One of which is a girl"

The statement in the riddle is not "One of which" though.

"One of them is a girl" means the same exact thing as "One of which is a girl"!

Nor is it "At least one of them is a girl"

You're saying that adding "at least" to "One of them is a girl" changes the meaning? It doesn't! Did you think that "One of them is a girl" didn't mean at least one? Did you think it meant both?

It's a matter of assigning. As soon as you say "one of them" you've assigned, say, the first born, or the second born to be a girl.

Oh really? So which one did the OP assign, the first or the second? The answer is 'neither'. "One of them" means either one of them. Not the first and not the second.

But then again, I may just be "one of them" guys.

Yup!

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okay i've had enough of this.

ANYONE who thinks it is 1/2, PLEASE come forward and state your case. Those who do not we will assume agree with us that it is 1/3. To anyone that still thinks it is 1/2, we will correct the flaw in your arguement.

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You're saying that adding "at least" to "One of them is a girl" changes the meaning? It doesn't! Did you think that "One of them is a girl" didn't mean at least one? Did you think it meant both?

Actually, in Propsguy's defense, these two statements can be interpreted differently. "At least one" clearly allows for more than one. However, "One of them" can be interpreted to mean "one and only one". Of course, if this is the interpretation you take, then the answer to the puzzle is "0%", as some clever individual has already posted (Sorry, I forget who it was).

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Since Propguy's argument wasn't that the answer is 0%, but 50%, it's not him you are defending. He wasn't arguing that "One of them" means "one and only one". He said that it meant a specific one and I agree with Scraff's analysis.

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It does come down to the wording.

"One of them is a girl"

versus

"One of which is a girl"

The statement in the riddle is not "One of which" though.

"One of them is a girl" means the same exact thing as "One of which is a girl"!

Nor is it "At least one of them is a girl"

You're saying that adding "at least" to "One of them is a girl" changes the meaning? It doesn't! Did you think that "One of them is a girl" didn't mean at least one? Did you think it meant both?

It's a matter of assigning. As soon as you say "one of them" you've assigned, say, the first born, or the second born to be a girl.

Oh really? So which one did the OP assign, the first or the second? The answer is 'neither'. "One of them" means either one of them. Not the first and not the second.

But then again, I may just be "one of them" guys.

Yup!

Ok guys, I was willing to let it go, but now I've been forced to reach into my bag of tricks and open up a can of logical whup-a** on you all.

"One of them is a girl. What is the probability the other kid is also a girl?"

"One of them." "The other kid."

and again,

"One of them." "The other kid."

I'm not saying the OP assigned one of the children to be a girl. The question itself assigns one of them.

Seems clear to me:

There is "one", and there is the "other." We are asked about the probability of "The other kid."

I don't care if it's the first or the second born. I don't care how you look at it: "One of them is a girl." As soon as you ask about "the other" you've made that "other" the subject of the probability.

And the probability of "the other one" being a girl is 1/2.

Nuff said.

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I'm not saying the OP assigned one of the children to be a girl. The question itself assigns one of them.

Seems clear to me:

There is "one", and there is the "other." We are asked about the probability of "The other kid."

I don't care if it's the first or the second born. I don't care how you look at it: "One of them is a girl." As soon as you ask about "the other" you've made that "other" the subject of the probability.

And the probability of "the other one" being a girl is 1/2.

Nuff said.

We've gone over and over this.

It is perfectly fine to speak of them as "one" and "the other". With "one" of them being a girl, the probability of the "other" being a girl is 1/3. The distribution still looks like this:

GG

GB

BG

If you are going to say "One of them" means that it can't be two of them, then as already has been brought up, you can say the chances are 0% that the other child is a girl. But the OP has made clear later in this thread that that is not an option and we are to assume the OP means "at least one of them" is a girl.

As soon as you ask about "the other" you've made that "other" the subject of the probability.

That is correct. And as you can see by the distribution above, the other child is a girl 1/3 of the time.

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Yes, we have gone over and over it, but I've been spurred to attempt to make my point again.

I will agree that the answer is supposed to be 1/3;1/3 is the answer the judges are looking for. But as the puzzle is written, the answer is 1/2.

The puzzle is written as:

"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."

The puzzle should have been written as:

Ok, so Teanchi and Beanchi are a married couple (don't ask me who's he who's she)!

They have two kids. At least one of the kids is a girl. What is the probability that both kids are girls?

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

There is a big diffence between the two puzzles above (aside from spelling and grammar mistakes).

The way the puzzle was presented, we are asked the probability of one child. That "other kid". The probability of any "other kid" being a girl is 1/2.

The way the puzzle should have been presented, the answer is 1/3. The way it should have been presented makes the gender of both children a variable.

The way it was presented makes the gender of only one of the kids a variable. Not a particular kid, as I may have previously indicated, but one of them. Take yer pick!

In the simplest terms, the 1/3 camp has come to their answer using the gender of both kids as variable. As a member of the 1/2 camp, I have come to my answer using the gender of only one kid as variable. Just as the question asks: "What is the probability the other kid is also a girl."

I don't care which of the kids is the "one of them" girl. We are asked about the probability of "the other kid."

We are never asked about the probability of both kids being girls.

(By the by, I think this is a great puzzle, and a great forum. Thanks.)

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The puzzle is written as:

"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."

The puzzle should have been written as:

Ok, so Teanchi and Beanchi are a married couple (don't ask me who's he who's she)!

They have two kids. At least one of the kids is a girl. What is the probability that both kids are girls?

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

There is a big diffence between the two puzzles above (aside from spelling and grammar mistakes).

The way the puzzle was presented, we are asked the probability of one child. That "other kid". The probability of any "other kid" being a girl is 1/2.

Please look at the distribution again. You will see that the answer is 1/3 either way the riddle is presented:

GG

GB

BG

"What is the probability that the other kid is also a girl?"

In only one of the three choices is the other kid a girl.

"What is the probability that both kids are girls"?

In only one of the three choices are both kids girls.

For the 'other kid' to be a girl, that must mean the riddle is asking about when 'both kids' are girls. How can it be possible that the other kid is also a girl without them both being girls? It can't. Which is why it doesn't matter which way it is phrased.

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so now I accept that 1/3 is the correct answer. I kept getting hung up on the (hopefully undisputed) FACT that the probability of any one child being a girl is 1/2. The answer, as has been elucidated so frequently, is that the question isn't what's the probability of a child being a girl, but the probabilty of the case where one is a girl, will the other also be a girl. So, as has oft been typed:

BG

GB

GG

If they had 3 children and we know one is a girl:

what is the probability that at least one of the others is a girl?

what is the probability that both of the others are girls?

GBB

GGB

GBG

GGG

at least one of the others is a girl = 3/4

both of the others are girls = 1/4

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If they had 3 children and we know one is a girl:

what is the probability that at least one of the others is a girl?

what is the probability that both of the others are girls?

GBB

GGB

GBG

GGG

at least one of the others is a girl = 3/4

both of the others are girls = 1/4

Nope:

GGG

GGB

GBG

GBB

BGG

BGB

BBG

If they had 3 children and we know one is a girl:

what is the probability that at least one of the others is a girl?

4/7

what is the probability that both of the others are girls?

1/7

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Nope:

drats - that's what I get for thinking too fast.... :-)

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they want the probability that the 2nd kid is a girl

so stop makin a fuss it makes no diff if the 1st is a girl of boy!!

the ansa is a 1/2 thats the way it was worded, thats the way u gotta ansa it, simple math!

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they want the probability that the 2nd kid is a girl

Oh, really?

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they want the probability that the 2nd kid is a girl

I believe the original question was about the probability that both kids were girls.

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For a more definitive answer, I thought I would check with Wikipedia. Their puzzle is slightly different and worded better, but, as expected, their solution to the puzzle matched mine.

Note that the article starts off with a different puzzle, but then moves on to this one.

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The chance of the second child being a girl is a 50/50% chance. That chance will never change. If I put a penny and a dime into a bag, close my eyes and pick out a coin, what are the chances of it being a penny or a dime? 50/50%. If I put back the coin that I picked out and placed it back into the bag, what are the chances of me picking up a dime? 50/50%.

No matter how many times the wife bears a child, the chances of her child being a girl or a boy will always be 50/50%.

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The chance of the second child being a girl is a 50/50% chance. That chance will never change. If I put a penny and a dime into a bag, close my eyes and pick out a coin, what are the chances of it being a penny or a dime? 50/50%. If I put back the coin that I picked out and placed it back into the bag, what are the chances of me picking up a dime? 50/50%.

No matter how many times the wife bears a child, the chances of her child being a girl or a boy will always be 50/50%.

sajow4, read this thread in its entirety and the links I provided explaining the answer. If you're still convinced the answer is a probability of 1/2, then come back and you can explain where you think I and everyone who thought the way you do yet eventually came around to believing that 1/3 is the correct answer is wrong. But please don't post until you do the required reading and make sure you understand what the riddle is asking.

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Martini, I apologize if I sounded rude or impulsive, I just think that my answer is correct (this being obvious, as I would not have written that answer had I not thought it correct), because no matter what, the chances should always be 50/50%.

And yes, I did read the Wikipedia article.

This is what it states:

"For a single birth, there are two possibilities (a boy or a girl) with equal probability.

Therefore, for two births, there are four possibilities: 1) two boys, 2) two girls, 3) first a boy, then a girl, and 4) first a girl, then a boy; all of them have equal probability.

We are given that one of the children is a boy. Thus, only one of the four possibilities -- two daughters -- is eliminated. Three possibilities with equal probabilities (1/3) remain.

Out of those three, only one -- two sons -- is what we are looking for. Hence, the answer is 1/3."

This is being that there is 4 children - In this case there is only 2. Even in the beginning of the article, it states that the child being a girl or a boy is in equal probablility. Since the problem stated in Wikipedia is with 4 children, and the current one in this forum is 2, that changes everything. This is saying that if a couple have 4 children, and we know that the first one is a boy, what is the probablility of the other 3 children being boys or girls.

This problem differs from the Wikipedia article.

Had this problem be the same one in the article stated above, I would agree with you, as it is the corect answer. But being as it is a different question with 2 children instead of 2, I must differ.

I apologize if I have irked you in any way.

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This is being that there is 4 children - In this case there is only 2.

No, the Wikipedia article contains the same riddle. There are only two children.

Even in the beginning of the article, it states that the child being a girl or a boy is in equal probablility.

"For a single birth, there are two possibilities (a boy or a girl) with equal probability."

No one is disputing that. However it doesn't state, "the child being a girl or a boy is in equal probablility".

The child's probability of being a boy is 1/3.

Since the problem stated in Wikipedia is with 4 children, and the current one in this forum is 2, that changes everything.

Again, take another look at the Wikipedia entry. It's with two children and is essentially the same riddle as given by the OP.

I apologize if I have irked you in any way.

Not at all. Sorry if I came across as irked, but I just wanted you to go over it all again because various explanations for the correct answer have been given several times and I didn't want to have more posts explaining the same thing unless necessary.

Please, feel free to discuss any part of the solution given that you believe is incorrect; that's what forum is all about.

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Sajow4, I do not understand how you get that the problem on the Wikipedia page is concerning 4 children. The problem is clearly stated as:

If we encounter someone with two children, given that at least one of them is a son, what is the probability that the other is also a son?

There are only two children, but four different possible combinations of gender. The only differences between that question and the one here is that it is more clearly stated and asks abouts sons instead of girls.

Neither questions designates a specific child as a specific gender. They simply set a condition that at least one of the children be a specific gender. This is how we eliminate one of the four possibilities, leaving the other three.

To put your penny/dime example into the same conditions, your question would need to be something like this:

A person has a bag in which they keep one penny and one dime. They decide to make two selections from this bag, returning the selected coin after each selections. Providing they select a dime at least once, what is the probability that they select a dime twice?

For the sake of the puzzle, each coin has an equal chance of being selected.

As stated for the original question, there would be three possible outcomes. The person could select a penny then a dime, a dime then a penny or a dime both times. They could not select a penny both times because that would violate the condition. So, there is a one in three chance that the person would select a dime twice.

I think that makes three different alternative questions, but I still like the bear one the most.

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Of all the couples in the world with 2 children, statistically speaking,

what fraction have a 1 boy and 1 girl?

and what fraction have 2 boys and 0 girls?

So, just taking these couples,

what fraction have 2 boys and 0 girls?

He answered: One third; and then: Oh.... OK I get it now.

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No, the Wikipedia article contains the same riddle. There are only two

I am so sorry! I apologize - I looked over the Wikipedia page again, and I realized that I had made a mistake. You are correct. I had not paid attention and was rushed, and thus I overlooked it.

I still do not understand your reasoning.

No matter what, the chances of a child being a boy or a girl will always be 50%.

I believe what you are thinking is that the chances of the second child completing either "boy, boy", "boy, girl" or "girl, girl" is 33% chance. You ARE correct in this sense.

But in the question of anything, absolutely any problem concerning the gender of a child, the chances will always be 50% - No matter what.

Think about it this way: A couple is having a child - But they wish not to know the gender of the baby. What are the chances of the child being a boy/girl? 50%

I believe what you are thinking is: A couple is having twins - And they do not wish to know the gender of the babies. There is a 33% chance for each that the babies will be either: A, boy, boy, B, girl, girl, C, boy, girl. I believe this is what you are thinking.

It is a 33% chance that the children, together, will be either A, B, or C. There is an indefinate 50% chance that in any circumstance that a child will be a boy or girl. That is the way that the Lord created humans to be born - An absolute equal chance of either gender.

Please also think about this: How could the chances be 33% chance of the second child being a boy or girl - When there is only 2 possible possibilities, a boy or a girl. Does that make it more understandable?

I believe I am correct, but I would also like to know if I have made an error anywhere.

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I still do not understand your reasoning.

No matter what, the chances of a child being a boy or a girl will always be 50%.

The probability of a child being a boy or a girl will always be 50%,

regardless of the gender of any previously born siblings. Agreed.

But in the question being asked the two children are already born,

and their family falls into a certain class.

We are told that one of the children is a girl.

Some families with two children don't have any girls, much less two girls.

So we're not talking about just any family.

Then think about the probability of the child that is said to be a girl being either a boy or a girl.

Clearly, it's not 50%.

So, "50% must be the answer" only applies when the scales are not tipped.

Let's see why and how the scales actually are tipped in this case.

To see this, let's consider a case where the scales are not tipped.

"A couple has a girl and then they have a 2nd child.

What is the probability the 2nd child will be a girl?"

So, to see how our answer might be different from 50%,

let's see how the this question differs from the one actually being posed.

Consider:

Families with two children fall into the 4 equally probable classes of

1 boy then 1 boy [25%]

1 boy then 1 girl [25%]

1 girl then 1 boy [25%]

1 girl then 1 girl [25%]

As we just saw, the probability that the 2nd-birthed child is a girl is 50%.

Those classes are 1 boy then 1 girl [25%] and 1 girl then 1 girl [25%].

But the question does not ask the probability for the 2nd-birthed child.

It asks the probability for the other child, given that one of the two children is a girl.

Given that one of the children is a girl eliminates one of the four classes of families above.

We're now looking at the set of families that have two children, one of which is a girl.

We can no longer consider the case of 1 boy then 1 boy.

And that tips the scales.

We now are considering only three equally likely classes of families:

1 boy then 1 boy? Nope, [0%]. These families have been asked to leave the building, so to speak.

1 boy then 1 girl? Yes. [33%]

1 girl then 1 boy? Yes. [33%]

1 girl then 1 girl? Yes. [33%].

Now look at these classes, and ask: given that one of the children is a girl,

what are the chances that the other one is a girl?

Take a minute to be sure of the conditions of the question,

and of the equal probabilities of the classes,

then find the classes that provide a correct answer.

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In the literature, scholars say the answer can be 50%.

The way they arrive at 50% involves how the couple in question is chosen from the set of all couples with two children.

And Yes, Virginia, it really makes a difference.

The "1/3" proponents in this thread [i'm one of them, given the wording skate gave us] interpret the question

in a manner that permits us to simply exclude the 2-boy families from the population of 2-children families

and then look at the fraction of 2-girl families in that smaller set. With that approach, the answer is 1/3.

The 50% proponents argue that the problem might be interpreted in a way that allows us to conclude

that we have been given knowledge of one particular child. That seems a very subtle difference,

but it does change the sample space, and it does change the answer.

And the endless confusion in the debate on these pages is that it is not recognized that the debate

is about the information we feel that we are permitted to assume, and not about the logic.

OK, in some posts the logic was really silly.

But that aside,

Here's how the 50% proponents rephrase the riddle:

More fairly, how the 50% proponents claim the riddle can validly be rephrased:

You meet a woman on the street who is with her daughter.

She says, one.

She doesn't give you the sibling's gender.

OK, they say, here is a random parent with two children, and you know only that [this particular] one of them is a girl.

What are the chances the other is a girl? In this case, the answer is 50%.

That subtle difference in choosing the sample space changes the answer. Note that it does not

include mothers for whom it is known only that the gender of the child not present is female.

Now for all of you 50% people who have jumped up on the table and begun to dance, please note:

I don't disagree, and [gulp] speaking for Martini and others [always dangerous to do]

none of the 1/3 proponents would disagree with this, either.

We only disagree that the problem as stated by skate cannot validly be restated this way.

To summarize ...

The serious debate on this question concerns [only] how the sample space is selected.

You have to look at the precise wording of the posed question in order to set up the categories.

Here's how one scholar distinguishes the two cases:

You meet a woman and ask how many children she has, and she replies "two." You ask if she has any girls, and she replies "yes."

[this is the case of the general population minus the 2-boy families.]

After this brief conversation, you know that the woman has exactly two children, at least one of whom is a girl.

When the question is interpreted this way, the probability that both of her children are girls is 1/3.

You meet a woman and her daughter. You ask the woman how many children she has, and she replies "two."

[This is the case where the gender of one specific child is known to be female.]

So now you know that this woman has exactly two children, at least one of whom is a girl.

When the question is interpreted this way, the probability that both of her children are girls is 1/2.

If I may speak for the "1/3" contingent in this thread, we hold the 1/3 answer because ...

we assert that the problem as stated by skate precludes being recast in this latter form.

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