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## Question

Orginal text from forum reads:

"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 probability of each gender is 1/2.

Of course its not 1/2 else would make it a lousy puzzle..."

and then given and accepted answer in the puzzle is this:

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

Now my question is, is this solution correct?

So this is what I think, please correct me if I'm wrong because this one really puzzled me, even though it seems so simple.

I think the answer is 1/2 still. Now before you guys all get mad, see my logic:

Now we know one of the children is a girl. And as previously stated, the only possibilities are:

GG

GB

BG

BB.

Now the original poster of the question said since we know one is a girl, so then that eliminates the BB out of the sample space. So since only 1 of the remaining 3 has a girl, so then that means the answer is 1/3. I DISAGREE with this logic.

So here's my logic:

Knowing that one of the children is a girl, there's 2 possible scenarios:

Scenario 1. The given girl is the elder child.

Scenario 2. The given girl is the younger child.

Now assuming that both these possibilities are equal (i.e. 50%). Then we have the following.

Scenario 1: probability that the remaining child is a girl is 50% (because we have {GG and GB})

Scenario 2: probability that the remaining child is a girl is 50% (because we have {GG and BG})

Thus total probability should be 0.5 * 0.5 + 0.5 * 0.5 = 0.5 = 50% (i.e. 0.5 * scenario 1 + 0.5 * scenario 2)

Is my logic correct? Please elaborate.

## 34 answers to this question

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There is a difference in how you choose to interpret the problem. You aren't actually given a girl, you are given the fact that there is a girl among them. That's what makes the difference. If there were in fact a "given girl", your logic would be correct. And I personally think that the problem is ambiguous about that, so I understand your point of view. But when given only the fact that there is at least one girl, the probability is 1/3 for them both to be girls.

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You're counting GG twice. In your explanation, your four possible scenarios are GG, GB, GG, and BG. You're not allowed to count GG twice, because it doesn't matter the order in which the girls are born. If you only count GG once, the probability is 1/3.

Why are we allowed to count BG and GB if the order in which the children are born doesn't matter?

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Argue all you want but the information given to you about the first child helps in no way. The child is either a boy or a girl so it is 50% 1/2 however you put it.

Edited by Plumbstar Tom

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Well, I always see it this way:

since we've to calculate the probability of one situation given the other situation, so we can use Baye's theorem. (though we could've simply said it to be 1/2 since the events are mutually independent)

So, Probability of 2 girls = P(GG) = 1/4

Probability of a girl child = P(G) = 1/2

So, required probability = P(GG) / P(G) = 1/4 / 1/2 = 1/2

The other felony people use is Goat's example, which is just WiFoM. (for more info on WiFoM, check the link in my signature)

Edited by KlueMaster

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Well, I always see it this way:

since we've to calculate the probability of one situation given the other situation, so we can use Baye's theorem. (though we could've simply said it to be 1/2 since the events are mutually independent)

So, Probability of 2 girls = P(GG) = 1/4

Probability of a girl child = P(G) = 1/2

So, required probability = P(GG) / P(G) = 1/4 / 1/2 = 1/2

The other felony people use is Goat's example, which is just WiFoM. (for more info on WiFoM, check the link in my signature)

Actually P(G) is not 1/2, it's 3/4 (there are 4 possible combinations, and 3 of them include a girl). So then the calculation is 1/4 / 3/4 = 1/3.

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Argue all you want but the information given to you about the first child helps in no way. The child is either a boy or a girl so it is 50% 1/2 however you put it.

But Plumbstar, the question is NOT "What is the probability that one of the children is a girl?" It is also not asking, "A mother gives birth to one child, and it is a girl. What is the probability that the second child will be a girl?" The question is "What is the probability that one child is a girl GIVEN THAT the other child is a girl?"

The trick here is to see that both children have already been born. They're both sitting in a room somewhere. The four possible combinations are:

Child 1 boy, Child 2 boy

Child 1 boy, Child 2 girl

Child 1 girl, Child 2 boy

Child 1 girl, Child 2 girl

We KNOW that one of them is a girl, so that immediately eliminates the first scenario. Of the remaining three, only one of them provides ANOTHER girl. That's where the 1/3 probability comes in.

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Stop arguing, the problem is AMBIGUITY. All of you are right, you are just answering different questions. Let's turn this into two different non-ambiguous questions instead:

Question 1. There are two siblings. They are not both boys. What is the probability that they are both girls? Answer: 1/3

Logical explanation: it's twice as likely to have a boy and a girl as it is to have two girls. This relation between the two doesn't change by eliminating the two boys scenario. So the total probability is 1, and since boy and girl is twice as likely, we get P(both are girls)=1/3.

Mathematical explanation: the probability of event A given event B, P(A|B), is defined as P(A and B)/P(B). In this case, event A is the siblings both being girls, event B is the siblings not both being boys. Obviously event A -> event B, so P(A and B)=P(A)=1/4. P(B), the probability of them not both being boys, is 1 minus the probability of them both being boys, i.e. 1 - 1/4 = 3/4.

So P(A|B) = P(A and B)/P(B) = P(A)/P(B) = (1/4)/(3/4) = 1/3.

Question 2. You meet a child. The child is a girl. What is the probability that her only sibling is a girl as well? Answer: 1/2

Logical explanation not necessary, everyone should understand this.

Mathematical explanation: same as above, only event B here is that the person you meet is a girl, which makes P(B)=1/2.

So P(A|B) = (1/4)/(1/2) = 1/2

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But Plumbstar, the question is NOT "What is the probability that one of the children is a girl?" It is also not asking, "A mother gives birth to one child, and it is a girl. What is the probability that the second child will be a girl?" The question is "What is the probability that one child is a girl GIVEN THAT the other child is a girl?"

The trick here is to see that both children have already been born. They're both sitting in a room somewhere. The four possible combinations are:

Child 1 boy, Child 2 boy

Child 1 boy, Child 2 girl

Child 1 girl, Child 2 boy

Child 1 girl, Child 2 girl

We KNOW that one of them is a girl, so that immediately eliminates the first scenario. Of the remaining three, only one of them provides ANOTHER girl. That's where the 1/3 probability comes in.

Okay, see, the trouble I'm having is that when you say there's a girl, the girl could either be Child 1 or Child 2 right? So surely the given girl could be Child 1 of the girl girl scenario OR Child 2 of the girl girl scenario. Help explain?

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this has been argued to death aleady. the answer is: it depends on the question being asked.

everyone is correct so far based on the reasoning they used.

shakingdavid has the right idea.

someone somewhere flips two coins. they tell you that both coins did not come up heads. what's the probability they both came up tails? here its 1/3. (order matters, HT != TH.)

someone somewhere flipped two coins. they tell you one of the two coins is tails. what's the probability the other coin came up tails? here it's 1/2. (order doesnt matter here, getting HT is equivalent to TH)

based on the way the orginal question is worded, i'd say 1/2.

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But Plumbstar, the question is NOT "What is the probability that one of the children is a girl?" It is also not asking, "A mother gives birth to one child, and it is a girl. What is the probability that the second child will be a girl?" The question is "What is the probability that one child is a girl GIVEN THAT the other child is a girl?"

The trick here is to see that both children have already been born. They're both sitting in a room somewhere. The four possible combinations are:

Child 1 boy, Child 2 boy

Child 1 boy, Child 2 girl

Child 1 girl, Child 2 boy

Child 1 girl, Child 2 girl

We KNOW that one of them is a girl, so that immediately eliminates the first scenario. Of the remaining three, only one of them provides ANOTHER girl. That's where the 1/3 probability comes in.

Nope still dont see it

Question states one is a girl whats is the probability that the other is a girl?

Well there are two options it could be a boy or a girl, 1/2

Agreed that this could go on and on though

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I hate to beat a dead horse here, but I have not seen a correct answer.

The space of solutions for having two kids is (B, b), (B,G), (G,g). There is no ordering of the kids.

We are told that at least one is a girl which removes the (B,b) case.

But (and this is the tricky bit) we have:

(B,G) with the "named" girl being G --> here the other child will be the boy

(G,g) with the "named" girl being G --> here the other child will be the girl "g"

(G,g) with the "named" girl being g --> here the other child will be the girl "G"

Out of possible three choices there two possible ways for the other child to be a girl. This makes the probably 2/3!

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I second what Shakingdavid said. I heard this problem a while back on a forum that was discussing interview questions for jobs in finance. Both answers are correct depending on the interpretation of the wording.

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I disagree. The puzzle's original logic is very sound.

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It may help to think of the problem in terms of how this could arise in the real world.

Imagine that I am looking at census records. In this census there were two questions

of interest to us: "Do you have two children?" and "Do you have a daughter?". So, we

imagine that we make a pile of records for which both of these questions have a "yes"

answer. Call this pile X. Suppose that another question on the census is "Do you have

a son?" What portion of the records in pile X would you expect to have a "no" answer

to the question about a son?

We can use a coin flip to make fake census records of this type. Each record would

require three coin tosses to make -- one for each of the three questions. Then we can

actually make pile X and count what portion has a "no" response to the question about

a son. I suggest Twinhelix try this and see what happens!

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I hate to beat a dead horse here, but I have not seen a correct answer.

The space of solutions for having two kids is (B, b), (B,G), (G,g). There is no ordering of the kids.

We are told that at least one is a girl which removes the (B,b) case.

But (and this is the tricky bit) we have:

(B,G) with the "named" girl being G --> here the other child will be the boy

(G,g) with the "named" girl being G --> here the other child will be the girl "g"

(G,g) with the "named" girl being g --> here the other child will be the girl "G"

Out of possible three choices there two possible ways for the other child to be a girl. This makes the probably 2/3!

Hehe not to be rude but this one made me laugh, cuz at first you say order doesn't matter, the when you get to the (G,g), you made order matter again... I'm sure 2/3 is incorrect?

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The answer is obvious. The other child is a boy, there is no probability that the child could be a girl because the question clearly states there are two children and ONE is a girl. If there are two children and one girl the other must be a boy.

Then again, if this has to be a probability question (as the question does ask for a probability) 1/3 seems right to me if the order of birth matters, if it does not 50/50 is right because you only have 2 options.

Edited by Shay-d

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When you look at statistics, you'll note that the world population is basically 50/50 men and women, So one can argue that if the first child is a girl, then the second is most likely (100%?) a boy, depending of course on the family structure of all the other families in the world...

When you look at genetics, you have a 50% chance of getting a boy, and a 50% chance of getting a girl. This is always the same, irrespective of your previous child's gender. This means that a couple that has 15 girls (even a thousand for that matter) will still have 50% chance of getting another girl. (Although in such a case one can do genetic testing to see if there might not be something wrong for them to always get girls, but that is beyond the spectrum of this quiz).

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I hate to beat a dead horse here, but I have not seen a correct answer.

The space of solutions for having two kids is (B, b), (B,G), (G,g). There is no ordering of the kids.

We are told that at least one is a girl which removes the (B,b) case.

But (and this is the tricky bit) we have:

(B,G) with the "named" girl being G --> here the other child will be the boy

(G,g) with the "named" girl being G --> here the other child will be the girl "g"

(G,g) with the "named" girl being g --> here the other child will be the girl "G"

Out of possible three choices there two possible ways for the other child to be a girl. This makes the probably 2/3!

Not quite right. There is a similar problem for which your solution will work: suppose we have two bowls. One contains a white and a black marble, the other one contains two white marbles. You pick a marble at random, turns out you picked a white marble, what is the probability of the other marble in the same bowl also being white? The answer is 2/3 for this problem as shown by your solution.

But in this case of children, it is in fact twice as likely to have a boy and a girl as it is to have two girls. Consider that no matter what sex your first child is, you have a 1/2 chance to get a boy and a girl (your second child has to be of the other sex). But to get two girls, you must first have a first child girl (probability 1/2), then a second child girl (probability 1/2). The probability for this is 1/2 x 1/2 = 1/4. So if we translate to the marble case, for this to be properly translated, you need two bowls with white+black marbles, and one bowl with white+white. The probability will now be 1/2 for the other marble to be white. Which is the same answer as you get to the boy girl problem with a "named" girl.

So to keep it all straight: 1/3 for the "given that both aren't boys" case, 1/2 for the "given one child who's definitely a girl" case, 2/3 for the marble case which is different but similar.

Edited by shakingdavid

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to futher comment on super's resoponce, we are obviouly only interested in the case where a person answers yes to the first two questions. this essentially limits the possibilies to being only yes to the third question or no, with 50/50 chances. but let's say i changed the questions to be... do you have two children? are both childern boys? are both children girls? we know the answer is yes to the first question and no to the second question. we can have one coin toss per question, but count an answer of yes, yes, yes to be a miss-toss. now if you ran the simulation, it would be 1/3. the questions however are very similar.

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to futher comment on super's resoponce, we are obviouly only interested in the case where a person answers yes to the first two questions. this essentially limits the possibilies to being only yes to the third question or no, with 50/50 chances. but let's say i changed the questions to be... do you have two children? are both childern boys? are both children girls? we know the answer is yes to the first question and no to the second question. we can have one coin toss per question, but count an answer of yes, yes, yes to be a miss-toss. now if you ran the simulation, it would be 1/3. the questions however are very similar.

two questions would be are both boys? is one a boy and one a girl?

all question answered leaves a a definitve answer

if one is a boy and one is a girl then the child is a boy, if both are girls the child is a girl

two outcomes of even probability = 1/2

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According to my theory....

If Teanchi and Beanchi love each other more than before,they will have a(boy).

but if they stay to this level of (love),they will have another(girl).

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Argue all you want but the information given to you about the first child helps in no way. The child is either a boy or a girl so it is 50% 1/2 however you put it.

Absolutely right. The question is, what is the chance of being a girl.

options are:

a. Boy

b. Girl

What is the third option to make it 1/3? It has to be 1/2.

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Absolutely right. The question is, what is the chance of being a girl.

options are:

a. Boy

b. Girl

What is the third option to make it 1/3? It has to be 1/2.

You are over simplifying the problem....

Think of a punnet square like this

* Sorry the graph didn't turn out well but set up a punnet square with one boy and one girl on each axis and you will see what I mean

_______Boy______________Girl_____

| BB | BG |

Boy| | |

_________________________________

Girl| BG | GG |

| | |

So, yes there are only two options, but the odds of one of those options is greater; as there are three cases with a girl, and one where it is girl-girl, the odds are 1/3

Edited by harpuzzler

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You are over simplifying the problem....

Think of a punnet square like this

* Sorry the graph didn't turn out well but set up a punnet square with one boy and one girl on each axis and you will see what I mean

_______Boy______________Girl_____

| BB | BG |

Boy| | |

_________________________________

Girl| BG | GG |

| | |

So, yes there are only two options, but the odds of one of those options is greater; as there are three cases with a girl, and one where it is girl-girl, the odds are 1/3

But this is assuming the question is implying the girl does not need to have been born first.

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Where in the puzzle does it say anything about the girl being born first, there is no reason to assume it and this is the main reason as to why people get this wrong. There is a case for the answer being 0.5 but this is based on the method used for selecting the family which is unclear.

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