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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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The probability of both children being girls when we know at least one is a girl is 33%.

The normal possibilities for a set of two children are:

Boy/Boy

Boy/Girl

Girl/Boy

Girl/Girl

Since we know one of the children is a girl, we can drop the Boy/Boy possibility.

This leaves only three possibilities, one of which is two girls.

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It's a lousy puzzle.

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its 1/2, so its a lousy puzzle...

lol but the title says "One Girl - One Boy" so maybe there's a 0% chance the second kid is a girl because its a boy?

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It's a lousy puzzle.

Its not, wait for the reasoning till someone gets the right answer!

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are these people or animals, because you never said child

you said kid, which can refer to a goat

and you just said married couple

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are these people or animals, because you never said child

you said kid, which can refer to a goat

and you just said married couple

Yes, we are talking about human-beings here, this is no lateral thinking puzzle.

Its a counter-intutive yet simple conditional probability puzzle!

Good job rookie, these spoiler tabs are excellent!

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the other child has to be a boy,

first you have one girl then the next needs to be a guy

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oops i forgot

probability is 1/2

or 1 out of 2

they have 2 children

1 out of two is a girl then the other 1 out of two has to be a boy

(edit: I suck at spelling so I fixed it)

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oops i forgot

probability is 1/2

or 1 out of 2

they have 2 children

1 out of two is a girl then the other 1 out of two has to be a boy

(edit: I suck at spelling so I fixed it)

The probability of having a girl or a boy is same and is 1/2. i.e. same as saying if a fair coin is tossed then odds of getting a head is 50% and getting a tail is also 50%. Hope this makes is more clear, take another digg @ the question!

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The answer is 25%. Before they had children the probability for a girl first is 50%. The probability for a girl second is 50%. BUT, to have 2 girls the probability moves to half of half, or 25%.

Comparing a coin toss is a common mislead. A single coin toss is of course 50/50. But 2 consecutive tosses is 50/50 for the first, and 25/75 for the second. Taken as a set (as you must in a family), you cannot discount the total probability by looking at the single toss.

Much like this:

I have 3 boxes (a - b - c).

2 have nothing in them.

1 has a million dollars.

If you guess the correct one, you get the money.

You pick one.

Out of the remaining 2, I show you that one is of course empty.

Do you have better, worse, or the same odds if you DO change?

Better. The first guess you have a 1 in 3 chance of being right. Once I remove one, your odds move up to 2 in 3 by selecting the OTHER box, because it would have been like selecting the 2 boxes you didn't pick.

If I had offered for you to trade your ONE pick for the TWO you didn't pick, before showing one was empty would you?

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The answer is 25%. Before they had children the probability for a girl first is 50%. The probability for a girl second is 50%. BUT, to have 2 girls the probability moves to half of half, or 25%.

You might be on the right track but thats not the correct answer.

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imtcb, you have fallen into something known as "gambler's fallacy"...

once a first girl is born, THAT DOES NOT CHANGE THE PROBABILITY OF ANOTHER GIRL BEING BORN.

Yes if you look at the whole thing total, it's 25% for two children to be girls

But we're looking at the next child only... and the fact that a girl was born already does not change the 50/50 probability.

One single coin flip is ALWAYS 50/50 (assuming the coin is standard and all that). It doesnt matter what has come before it- will be 50/50 always. But if you're looking at things as a whole, for example if I flip a coin three times the chances of three heads is 1/2 * 1/2 * 1/2 = 1/8. But that doesnt change the fact that the next coin I flip will either be heads or tails, 50/50.

so if we're just looking at whether the next child is a boy or girl, its 1/2 each.

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imtcb, you have fallen into something known as "gambler's fallacy"...

once a first girl is born, THAT DOES NOT CHANGE THE PROBABILITY OF ANOTHER GIRL BEING BORN.

That is correct BUT,

so if we're just looking at whether the next child is a boy or girl, its 1/2 each.

knowing one of the 2 kids is a girl does alter something else, now am giving away too many hints.. it gives you useful information!

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knowing one of the 2 kids is a girl does alter something else, now am giving away too many hints.. it gives you useful information!

It's 1/2 using conditional probability, so why don't you share your answer already so we can pick it apart.

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The probability of both children being girls when we know at least one is a girl is (1/3) - 33.33%.

The normal possibilities for a set of two children are:

Boy/Boy

Boy/Girl

Girl/Boy

Girl/Girl

Since we know one of the children is a girl, we can drop the Boy/Boy possibility.

This leaves only three possibilities, one of which is two girls.

That sum's it all.. good job Riddari!

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ah so we are looking at the thing as a whole... good job riddari!

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imtcb - they are imdapendant events: the answer is 1/2

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If these are the possible combinations and you eliminated the boy-boy combo that means that you are solving this combo problem backwards. What I am trying to say is that since the untouched chances are 50% 50% because:

1st child / 2nd child

Girl / Girl

Girl / Boy

Boy / Girl

Boy / Boy

so for 2nd child the set of possibilities is {girl, boy, girl, boy} which gives 2 out of 4 for girls and gives you 50%.

YOU HAVE TO eliminate the 1st child boy, 2nd child girl because the 1st child is already known to be a girl.

THEREFORE: your only possible outcomes are girl girl and girl boy which gives you 50% 50%.

There is a puzzle about eliminating doors behind which prizes are hidden. I think it is "the price is right" puzzle; where a bit of knowledge helps you increase your chances of picking the right door with the prize but the puzzle you presented isn't it.

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the point is:

if whether you had a boy or girl fist does not affect what you have later

the choices therefore are

?/boy(50%)

?/girl (50%)

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If these are the possible combinations and you eliminated the boy-boy combo that means that you are solving this combo problem backwards. What I am trying to say is that since the untouched chances are 50% 50% because:

1st child / 2nd child

Girl / Girl

Girl / Boy

Boy / Girl

Boy / Boy

so for 2nd child the set of possibilities is {girl, boy, girl, boy} which gives 2 out of 4 for girls and gives you 50%.

YOU HAVE TO eliminate the 1st child boy, 2nd child girl because the 1st child is already known to be a girl.

THEREFORE: your only possible outcomes are girl girl and girl boy which gives you 50% 50%.

No, the question says:

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

Let me explain it with the coin example, there are 4 combinations flipping 2 coins:

1) HH

2) HT

3) TH

4) TT

Now what is the probability of one HEAD and one TAIL, clearly it is 2/4. Now let us say that you know one of the toss is a head (and not that the first toss is head). What is the possibility of one head and one tail now? Clearly it is 1/3. If the question said that given the 1st toss is head what is the possibility of one head and one tail then the answer would be 1/2.

Hope this makes it clearer!

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the point is:

if whether you had a boy or girl fist does not affect what you have later

the choices therefore are

?/boy(50%)

?/girl (50%)

Yes and No.

Yes

if whether you had a boy or girl fist does not affect what you have later

No:

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all you ppl that say its 50/50 cuz previous events dont effect the current event are just repeating what i said, but then i saw skale meant "as a whole, looking over the whole thing after two children are born" which it is indeed 1/3 not 1/2, even though the way the puzzle was worded was bad and sugested 1/2

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you meet a family with one child. what are the odds that child's a girl?

you meet a family with 3 children and are informed that 2 of them are girls. what are the odds the last one's a girl?

you meet a family with 20 children and are informed that 19 of them are girls. what are the odds the last one's a girl?

common sense tells us the answer is 50-50 to all of these. statistical analysis tells us they aren't all equal. i think common sense is the right answer here. all of these say exactly the same thing as my first example. there is one child... odds are 50-50.

i'm familiar with skale's analysis and can confirm that most statisticians would agree. but there is even debate among mathematicians as to what the correct answer is. i personally reject the accepted answer and content this problem is too vague to be solved even by math standards.

the problem is a question of the sampling. does the fact that you know the sex of the first child reduce your sample to just families with one daughter or is the original sample of a totally random population still hold. statistics says to use the original sample giving the 1/3 answer but i say that's wrong. as soon as you know the first child is a girl, your sample changes to only families with at least one girl leaving only 2 possibilities. girl/girl and girl/boy and the 1/2 answer. so you see that even in math circles, this question is still debatable. as i said... i'm going with common sense answer. but technically the answer is 1/3... too bad when math doesn't agree with common sense, isn't it?

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The distribution of families with 2 kinders is

25% boy-boy

25% girl-girl and the remaining

50% boy-girl

that's understood.

I got confused because I wrongly assumed that we are prognosing what the sex of the second kinder is going to be knowing that the first child is a girl. But, that's not what you were saying at all.

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you meet a family with one child. what are the odds that child's a girl?

you meet a family with 3 children and are informed that 2 of them are girls. what are the odds the last one's a girl?

you meet a family with 20 children and are informed that 19 of them are girls. what are the odds the last one's a girl?

common sense tells us the answer is 50-50 to all of these. statistical analysis tells us they aren't all equal. i think common sense is the right answer here. all of these say exactly the same thing as my first example. there is one child... odds are 50-50.

i'm familiar with skale's analysis and can confirm that most statisticians would agree. but there is even debate among mathematicians as to what the correct answer is. i personally reject the accepted answer and content this problem is too vague to be solved even by math standards.

the problem is a question of the sampling. does the fact that you know the sex of the first child reduce your sample to just families with one daughter or is the original sample of a totally random population still hold. statistics says to use the original sample giving the 1/3 answer but i say that's wrong. as soon as you know the first child is a girl, your sample changes to only families with at least one girl leaving only 2 possibilities. girl/girl and girl/boy and the 1/2 answer. so you see that even in math circles, this question is still debatable. as i said... i'm going with common sense answer. but technically the answer is 1/3... too bad when math doesn't agree with common sense, isn't it?

You guys are reading the question posed by the puzzle incorrectly. The puzzle does not state that the oldest child is a girl and then ask what the probability is for the youngest child to be a girl. If that were the case, 50/50 would be correct. Instead the puzzle states that one of the children is a girl and then asks what the probability is for the other child to also be a girl. Since there are two children with two possible genders, there are four possible combinations. One of those combinations requires that both children be boys, which we know is not possible. So, there are three possible combinations. Only one of these three possibilities includes two girls. Therefore, the probability that the unspecified child is a girl is one in three.

If my math is correct, a family with twenty children would have 2^20 or 1,048,576 possible combinations, which I am not willing to detail out. Knowing that nineteen of those children are girls would eliminate a lot of the possibilities. In fact, there would be only twenty-one remaining possibilities. One where they are all girls and twenty representing a boy in each position. So, there would be a one in twenty-one chance of the unknown child being a girl.

To put it in terms of sampling, your population would change to families that have an oldest girl or a youngest girl, not exclusively. So, it would include all families who have boy/girl, girl/boy and girl/girl, where the oldest child is listed first. It should be obvious, both statistically and intuitively, that the girl/girl families would constitute approximately a third of this population.

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