One Girl - One Boy

[Guest]

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

[Guest]

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

The question is not asking you what is the probability of the second child.

It is asking what is the probablity in a two child family that two of the children are girls.

We start with working out what combinations you can have in a two child family. Placing them in birth order.

First - Second

Boy - Boy

Girl - Girl

Boy - Girl

Girl - Boy

We know that one child is a girl so we eliminate the Boy - Boy option.

That leaves us three other options where only one option can be our answer - Girl - Girl.

And that option was one option out of three options (where there is as least one girl)

1/3

[Guest]

Now to me this to is a random event that applies to them alone. So when I work it out I work it out with a series of random events.

And the way you worked it out was wrong. You kept claiming that a GB or BG is thrown out after we find out there is no possibility of a BB couple. That's wrong regardless of how you read the question.

However all the 1/3 people out their and say that "one of them is a girl" is the same as "given any random couple with at least one girl" and take a look at all the couples and find only the one with girls and work the math out to 1/3. However in probability how you get your information is just as important as the information you get. This being that all the 1/3 people out their don't think that the information being supplied is a random event. The scenario being you walk up to a guy and ask him if he "has at least one girl" and he responds "yes" or "no". However I believe that the information being supplied is as random as the couple, and their is nothing that implies otherwise. I walk up to some guy and ask him if he as any kids and he says "I have two, one of them is a girl". With this information the odds of the other kid being a girl is 1/2 because it is just as likely that we could have been told that one of them is a boy.

Nice try. The riddle never stated how we got the information; so we don't start taking into account how someone would answer question that was never asked and isn't part of the riddle. Where are you getting "The scenario being you walk up to a guy..."?

The answer is 1/3. And you're changing your tune on your rationale for 1/2. You're rationale was that either BG or GB should be thrown out. It would be refreshing to see someone just admit they were wrong once in a while instead of pulling imaginary scenarios out of thin air to justify a wrong answer.

How dose

They have two kids, one of them is a girl.

turn into

Given any random couple with at least one girl.

How doesn't it? The probability is the same either way!

Unless what you're now doing is claiming that if someone states they have one girl we'd have to wonder why they'd state this this way. That's irrelevant! No one stated anything! And if you're going to make that claim, how can you justify 1/2? All that can be justified is the answer that there's not much of a chance at all, because if someone states they have two kid and one is a girl, the assumption is that the other is a boy. But again, that statement was not made in the riddle!

I don't see how given information we have, this can be anything but random event. Perhaps you can tell me?

It doesn't matter whether it's a random event, specific couple, hypothetical situation, etc. The probability remains the same.

In the riddle, we're asked about the probability of a specific couple. The probability is the same for them as it is for all couples combined or a random couple.

[Guest]

To all in the 1/3 camp. Sorry to ruin your day, but...

...We know that one child is a girl. She is either older or younger than her sibling. She cannot be both. Now take a list of the possible sibling pairs:

Older: Younger:

1) B,B

2) B,G

3) G,B

4) G,G

If she is older then 2&4 apply. If she is younger then 3&4 apply.

So: p(GG)=p(older&GG)+p(younger&GG)=0.5)*0.5+0.5*0.5= 0.5

Thank you.

[bonanova]

To all in the 1/3 camp. Sorry to ruin your day, but...

...We know that one child is a girl. She is either older or younger than her sibling. She cannot be both. Now take a list of the possible sibling pairs:

Older: Younger:

1) B,B

2) B,G

3) G,B

4) G,G

If she is older then 2&4 apply. If she is younger then 3&4 apply.

So: p(GG)=p(older&GG)+p(younger&GG)=0.5)*0.5+0.5*0.5= 0.5

Thank you.

You're welcome.

So what you claim is that when 2-boy families are eliminated,

half of the remaining families have 2 girls, and the other half

are mixed-gender families.

Given that p[bB]=0. then p[GG] = p[bG or GB] = 1/2.

Do you really believe that 2-girl families and mixed-gender families occur with the same probability?

We know that in the general population p[bB] = p[GG] so that means when p[bB] not = 0,

p[bB] = p[GG] = p[mixed] = 1/3.

Thanks for news !

Most of us thought p[bB]=p[GG]= 1/4 and p[mixed] = 1/2.

Now we know better.

Wait.

We know that she is older or younger, so 2, 3 and 4 all apply.

p[GG] = p[older or younger]*p[GG] = 1 * 1/3 = 1/3.

Whew!

You had us believing you were serious for a moment.

[Guest]

You're welcome.

So what you claim is that when 2-boy families are eliminated,

half of the remaining families have 2 girls, and the other half

are mixed-gender families.

Given that p[bB]=0. then p[GG] = p[bG or GB] = 1/2.

Do you really believe that 2-girl families and mixed-gender families occur with the same probability?

We know that in the general population p[bB] = p[GG] so that means when p[bB] not = 0,

p[bB] = p[GG] = p[mixed] = 1/3.

Thanks for news !

Most of us thought p[bB]=p[GG]= 1/4 and p[mixed] = 1/2.

Now we know better.

Wait.

We know that she is older or younger, so 2, 3 and 4 all apply.

p[GG] = p[older or younger]*p[GG] = 1 * 1/3 = 1/3.

Whew!

You had us believing you were serious for a moment.

Thank you for your edifying brand of sarcasm.

Lets start with four combinations of siblings (let’s assume all

sitting in a room). Those combinations can be written as follows:

(1)bb

(2)gb

(3)bg

(4)gg

Then one child was randomly selected and it turned out to be a boy.

Consequently, we know that selected sibling did not come from sibling-

set #4.

We now know that the three possible combinations of siblings, given

the impossibility of ‘gg’ due to a boy having been already randomly

selected, are as follows:

(1)bb

(2)gb

(3)bg

The randomly selected boy is one of the boys found within the three

sets listed above.

To simplify matters we may rewrite the three possible sets as follows:

(1) Bob, Bill

(2) Gayle, Brian

(3) Bart, Gabriella

NOTE: With the same probably of 25%, the randomly selected boy can

either be: Bob, Bill, Brian, or Bart. In other words, given the

already randomly selected boy (who must be either: Bob, Bill, Brian,

Bart), there are four equally possible sibling outcomes that can

obtain:

(A) Given Bob, then 100% chance that sibling Boy-Bill obtains

(B) Given Bill, then 100% chance that sibling Boy-Bob obtains

© Given Brian, then 100% chance that sibling Girl-Gayle obtains

(D) Given Bart, then 100% chance that sibling Girl-Gabriella obtains

There is a 25% chance that any one of those four outcomes obtains.

Consequently, there is a 25% chance of the second sibling being one

of either: Bill, Bob, Sally or Jane – which reduces to a 50% chance

that given a selected-boy, the remaining sibling would too be a boy.

If the above reasoning is sound, then your solution must be false

because it contradicts the above solution. But rather than leaving

you to find the error in your position, please let me try to perform

an internal critique of it. You noted that 1 minus the probability of

two girls in a family of two children is 75%. Of course that is true.

Accordingly, as you noted, the probability of having at least one boy

within a family of two children is 75%. That would be relevant if the

question were: “Given two children, what is the probability of having

at least one boy?” which is 75%, of course. However, the question

that we are to be concerning ourselves with is: “Given a randomly

selected boy with a sibling, what is the probability that his sibling

is also a boy?” – which is different question entirely. With respect

to the problem we are to be solving, the “given” is the randomly

selected boy - which can be the first *or* second child of the two

children! The sample space that you concerned yourself with, (which

gets you off on the wrong track I’m afraid), ignored the relevant

statistic that from a two-boy family the randomly selected boy can be

either the first or second boy.

From a sample space consideration, the problem is to be viewed thusly:

Girl sibling, randomly selected boy

Randomly selected boy, girl sibling

Randomly selected boy, boy sibling

Boy sibling, randomly selected boy

We know that the randomly selected boy

is either older or younger than his sibling. Accordingly, if he has a

girl sibling, then there are two, not one, ways in which he might

fall within that order. That statistic *is* implied in your

reasoning. To be consistent, however, we must also recognize that

there are two, not one, ways in which the *already* randomly selected

boy with a boy sibling can fall within the order of his family. That

statistic is *not* implied by the probability of there being at least

one boy in a family of two children!

To simplify this even more, we might consider that there is a

randomly selected boy sitting in a room who has only one sibling:

That boy may have:

(a) An older brother

(b) A younger brother

© An older sister

(d) A younger sister

If the randomly selected boy is older than his sibling, then the

probability of (a) and © can be eliminated, leaving a 50% chance of

his having brother. If the randomly selected boy is younger than his

sibling, then the probability of (b) and (d) can be eliminated,

leaving also a 50% chance of his having a brother. Added to that, the

randomly selected boy has a 50% chance of being older than his

sibling, and a 50 chance of being younger than his sibling.

Consequently, it the statistically irrelevant whether the randomly

selected boy is older or younger than his sibling because the

probability of his sibling being a boy is the same 50% no matter

whether the randomly selected boy is older or younger, and the

randomly selected boy has the same chance of being older than his

sibling as he does being younger than his sibling. Therefore, we need

not know the age of the randomly selected boy to know that the

probability of his sibling being a boy is 50%.

Finally, we can demonstrate this empirically with playing cards. Let

red cards stand for girl and black cards for boys. If a boy is

randomly selected, then we can discard the two red 2s.

Red 2, Red 2

Black 3, Red 3

Red 4, Black 4

Black 5, Black 5

The remaining cards are now:

Black 3, Red 3

Red 4, Black 4

Black 5, Black 5

Now place the two red cards, which represent sibling girls, aside and

keep the four black cards in your hand. Shuffle the black cards and

draw a card at random. If you draw either of the black 5s, then the

corresponding sibling would be a boy. If you draw a black 3 or black

4, then the corresponding sibling would be a red card, which

represents a sibling-girl.

Probability of drawing a black 3 is ¼ or 25%, corresponding to girl-

sibling

Probability of drawing a black 4 is ¼ or 25%, corresponding to a girl-

sibling

Probability of drawing a black 5 is 2/4 or 50%, corresponding to a

boy-sibling.

Therefore 50% or 1/2, not 1/3.

Gazza

[Guest]

Of course that is true.

Accordingly, as you noted, the probability of having at least one boy

within a family of two children is 75%. That would be relevant if the

question were: “Given two children, what is the probability of having

at least one boy?” which is 75%, of course. However, the question

that we are to be concerning ourselves with is: “Given a randomly

selected boy with a sibling, what is the probability that his sibling

is also a boy?” – which is different question entirely.

No, that's not the question we're concerning ourselves with. You've randomly selected a boy with one sibling. Now we have a specific child. You're right that in this case the probability of his sibling be a girl is 1/2. But no boy has been randomly selected in the riddle.

"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."

These are our possibilities:

GG

GB

BG

Check out the following links:

http://www.learninghaven.com/articles/ridd...in-teasers.html

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

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

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

[Guest]

The answer is obivously 1/2. This is like grade 1 math. The probability of the second baby being a girl has nothing to do with the 1st baby. Since we know that first baby is a girl. So both baby can only make one of the two possbilities:

Girl + boy

Girl + girl

note that the first baby cannot be a boy. So all the people what think it is 1/3 is wrong. Boy-girl is not an option.

now we calculate the probability of each of the 2 cases.

girl-boy = 1/1 * 1/2 = 1/2

girl-girl = 1/1 * 1/2 = 1/2

We could just completely ignore the fact the first baby is even there. I only did this so all the people that think otherwise would understand the problem better.

Now if we do boy-girl it would be : 0/1 * 1/2 = 0

So the answer is plain and obivously 1/2. What a lame question

[bonanova]

Pardon the sarcasm.

I just found it really improbable that you believe girl-girl and mixed gender families occur with equal likelihood.

And that certainly follows from your answer: that removing the boy-boy case leaves the girl-girl case with 50% likelihood.

Care to comment [specifically] on that point?

[Guest]

Pardon the sarcasm.

I just found it really improbable that you believe girl-girl and mixed gender families occur with equal likelihood.

And that certainly follows from your answer: that removing the boy-boy case leaves the girl-girl case with 50% likelihood.

Care to comment [specifically] on that point?

The thing is the first baby is already a girl.

The boy-boy case is just like the boy-girl case = 0/1 * 1/2 =0

The question is only asking for the second girl. The first girl is just there to confuse you.

Edited August 6, 2008 by noobsauce

[bonanova]

The answer is obivously 1/2. This is like grade 1 math. The probability of the second baby being a girl has nothing to do with the 1st baby. Since we know that first baby is a girl. So both baby can only make one of the two possbilities:

Girl + boy

Girl + girl

note that the first baby cannot be a boy. So all the people what think it is 1/3 is wrong. Boy-girl is not an option.

now we calculate the probability of each of the 2 cases.

girl-boy = 1/1 * 1/2 = 1/2

girl-girl = 1/1 * 1/2 = 1/2

We could just completely ignore the fact the first baby is even there. I only did this so all the people that think otherwise would understand the problem better.

Now if we do boy-girl it would be : 0/1 * 1/2 = 0

So the answer is plain and obivously 1/2. What a lame question

I think we all agree that the question My first child is a girl, what is the probability my second child is a girl is lame.

Please read the OP and some of the other posts in this thread.

Then use your 1st grade math to see that the question actually being asked is not lame.

[bonanova]

The thing is the first baby is already a girl.

The boy-boy case is just like the boy-girl case = 0/1 * 1/2 =0

The question is only asking for the second girl. The first girl is just there to confuse you.

No, it's not. The question is not that trivial.

You are only told that the two children are not both boys.

"One of the children is a girl" does not make the first child a girl.

[Guest]

No, it's not. The question is not that trivial.

You are only told that the two children are not both boys.

"One of the children is a girl" does not make the first child a girl.

You just proved i am right. If one of the baby is a girl, then the chance of the baby being a girl is 100%. So the other would still be 1/2. I soppose i didn't make my solution clear by saying let's suppose the first baby is the baby that is the girl

Nothing changed. It doesn't matter if the first girl is a girl or the second one is a girl. It's the same. If you count 1 apple and 1 orange. You can count it two ways, but it doesn't change that one is apple and one is orange.

I am not even trying to say how 1/3 is illogical and not even close to being a possible answer.

Edited August 6, 2008 by noobsauce

[bonanova]

You just proved i am right. If one of the baby is a girl, then the chance of the baby being a girl is 100%. So the other would still be 1/2. I soppose i didn't make my solution clear by saying let's suppose the first baby is the baby that is the girl

Nothing changed. It doesn't matter if the first girl is a girl or the second one is a girl. It's the same. If you count 1 apple and 1 orange. You can count it two ways, but it doesn't change that one is apple and one is orange

That's your mistake.

There are three cases, each with a probability of 1/3.

1. The sibling is an older boy
   
2. The sibling is a younger boy
   
3. The sibling is an older or younger girl. [in the gg case, it doesn't matter which child "a girl" refers to - there is only one gg case - it may not be double counted.]
   

The conditional chances that the sibling is a girl, respectively, in the three cases are

1. 0
   
2. 0
   
3. 1
   

So the probability of the other child being a girl is 0*1/3 + 0*1/3 + 1*1/3 = 1/3.

[Guest]

That's your mistake.

There are three cases, each with a probability of 1/3.

1. The other sibling is an older boy
   
2. The other sibling is a younger boy
   
3. The other sibling is an older or younger girl. [in the gg case, it doesn't matter which one is "a girl" - there is only one gg case - it may not be double counted.]
   

The conditional chances of the other sibling being a girl, respectively, in the three cases are

1. 0
   
2. 0
   
3. 1
   

So the probability of the other child being a girl is 0*1/3 + 0*1/3 + 1*1/3 = 1/3.

I have no idea what you did here. The question simply asked if it is a girl or a boy for the baby that is not already a girl. I like how you did this though:

[*]The other sibling is an older boy

[*]The other sibling is a younger boy

[*]The other sibling is an older or younger girl.

Why did you use the 2 cases for boy and combined the 2 cases for a single case for girl. Let me show you how to count 2 apples. you name them apple # 1 and apple #2 , you can count it 2 ways, #2-#1 or #1-#2

Let's now prove how 1/3 is wrong. We know that a baby is a girl, so you can't have boy boy.

Ok, so that leaves boy-girl, girl-boy and girl-girl.

Since one girl has to be a girl, that means that boy-girl and girl-boy represent the samething. What you are actually saying here is that there are 2/3 chance of the baby being a boy, you mixed up combination and premutation togeather with the boy using premutation and girls with combination ( fancy words for whether order matters or not). But it does not matter because if it were premutation there would be 4 possilbilites. boy-girl, girl-boy and girl-girl and girl-girl, you flip the girl-girl so you get it 2 times, just like what you did for boy-girl. The answer would be 2/4 = 1/2 = 50%

Edited August 6, 2008 by noobsauce

[Guest]

But it does not matter because if it were premutation there would be 4 possilbilites. boy-girl, girl-boy and girl-girl and girl-girl, you flip the girl-girl so you get it 2 times, just like what you did for boy-girl. The answer would be 2/4 = 1/2 = 50%

You didn't read all the posts in this thread, did you? You don't get twice as many GG possibilities. Did you see the penny experiment posts in this thread? If you lay out 100 pair of pennies you will get on average an equal amount of the following possibilities.:

HH

TT

HT

TH

Let's say someone asks you to pick a pair at random and remember it. If you ask, "is at least one of the pennies in your pair on heads", and the answer you get is "yes", there is a 1/3 probability that the other penny is also on heads because these are the possibilities left:

HH

HT

TH

You don't get twice as many HH pairs; you get the same amount as HT and TH. If you don't believe it, try the experiment yourself.

[Guest]

Lol, you don't understand the problem. Normally the possibility of a twin or 2 babies both being a girl would be 1/4 or 25% but this case is special.

Let me show you again, since you guys just don't get it.

Suposing that the first baby is the girl:

the possibilities are :

girl-boy = 1 * 1/2 = 1/2

girl-girl = 1 * 1/2 = 1/2

Now, we suppose the the secong baby is a girl:

cases:

boy - girl= 1/2 * 1 = 1/2

girl - girl = 1/2 * 1 =1/2

See where my apple counting skillz come in?

1/2 * 1 and 1 * 1/2 are 2 different cases. They appear to be the same just like the 2 apples.

Total possiblity (1+1)/4 (one case for 1st baby being a girl and 1 for second baby being a girl) = 1/2 = 50%

Like what I've said.

Why do you guys have to over complicate this question. This question is lousy and it is asking what's the chance of a baby being a girl. The fact that 1 has to be the girl make one 100% and the other one has 50% to be a girl so 1/2.

I've read the question at least 4 times before posting. I don't like making mistakes. Stoping accusing me of not reading the problem, you just have to understand the problem. The simple solution is 1/2 the complicated one is 1/2 as well.

Edited August 6, 2008 by noobsauce

[Guest]

Lol, you don't understand the problem.

I get the problem just fine. I've actually done a lot of research on this riddle. You can pick any of the sources below and find that my understanding is just fine and yours isn't.

How is the penny analogy not a good representation of the riddle?

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

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

http://www.learninghaven.com/articles/ridd...in-teasers.html

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

Suposing that the first baby is the girl:

the possibilities are :

girl-boy = 1 * 1/2 = 1/2

girl-girl = 1 * 1/2 = 1/2

Now, we suppose the the secong baby is a girl:

cases:

boy - girl= 1/2 * 1 = 1/2

girl - girl = 1/2 * 1 =1/2

See where my apple counting skillz come in?

We don't know which child is the girl. Noting what the probabilities are in two separate situations is not how we find the solution. It's found by looking at the probability in the one situation the riddle deals with. These are the equal possibilities:

GG

GB

BG

See those links I provided if you're still not convinced.

Why do you guys have to over complicate this question.

We're not, you are.

This question is lousy and it is asking what's the chance of a baby being a girl. The fact that 1 has to be the girl make one 100% and the other one has 50% to be a girl so 1/2.

Sorry, probability doesn't work that way.

I've read the question at least 4 times before posting. I don't like making mistakes. Stoping accusing me of not reading the problem, you just have to understand the problem.

Now I'm going to accuse you of not paying attention to my post because I didn't accuse you of not reading the problem. I wrote, "You didn't read all the posts in this thread, did you?".

You ignored my rebuttal to your following statement:

But it does not matter because if it were premutation there would be 4 possilbilites. boy-girl, girl-boy and girl-girl and girl-girl, you flip the girl-girl so you get it 2 times, just like what you did for boy-girl. The answer would be 2/4 = 1/2 = 50%

Do you understand now that you don't count GG twice?

[Guest]

To all 1/2 supporters.

I have read through all the posts on this topic (yes all 27 pages), and prior to posting this I looked at the links Scraff last posted, they are worth taking a look at.

In my previous posts I used th following combinations in the sample space.

BB

GG

GB

BG

I eliminated the BB option because I know that at least one of the children is a girl, which leaves me 3 combinations that the family could have.

From these 3 combinations 1 is GG. 1/3 probability

Lets take another look.

You know that in this family there are 2 children.

yes? - yes!

You know that one child is a girl.

yes? - yes!

Do you know if this girl is the oldest child?

yes!

Okay, so your possible combinations for the two child family are:

GG - oldest child Girl, youngest child Girl

GB - oldest child Girl, youngest child Boy

Easy, the probability that the second child is a girl is 1/2.

You know in this family there are two children.

yes? yes!

You know that one child is a girl.

yes? - yes!

Do you know if this girl is the youngest child?

yes!

Okay, so your possible combinations for the two child family are:

GG - oldest child Girl, youngest child Girl

BG - oldest child Boy, youngest child Girl

Easy, the probability that the second child (in this case the first child born) is a girl is 1/2

You know in this family there are two children.

yes? yes!

You know that one child is a girl.

yes? - yes!

Do you know if the girl is the oldest child?

No!

Do you know if the girl is the youngest child?

No!

Okay, so lets look at all the possible combinations for the two child family:

BB - oldest child Boy, youngest child Boy

GG - oldest child Girl, youngest child Girl

GB - oldest child Girl, youngest child Boy

BG - oldest child Boy, youngest child Girl

Okay we know that at least one of the children have to be a girl.

yes? - yes!

So lets eliminate the BB combination. That now leaves us with 3 combinations:

GG - oldest child Girl, youngest child Girl

GB - oldest child Girl, youngest child Boy

BG - oldest child Boy, youngest child Girl

You could argue that the last 2 combinations are the same because they are both a combination of 1 boy and 1 girl and therefore there are only two combinations, right? - wrong!.

They are not the same because you have to consider the birth order, you don't know if the first girl is the oldest or youngest child and because of this very fact the probability is 1/3.

Again please read the links Scraff has posted before posting another argument for 1/2. And if you still believe the answer is 1/2 then maybe you need to contact wikipedia and let them know they are wrong. (I believe they are right).

And if any 1/2 supporters have jumped ship and are now sailing with us (1/3), welcome aboard!

[Guest]

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

Please remind me how the girl is older?

There are 3 probabilities but the chance of the girl girl probability is actually 1/2. the boy-girl is 1/4 and girl boy is 1/4

Edited August 6, 2008 by noobsauce

[Guest]

Please remind me how the girl is older?

There are 3 probabilities but the chance of the girl girl probability is actually 1/2. the boy-girl is 1/4 and girl boy is 1/4

No one said the girl is older. By not responding to posts directed at you, you're dragging this out more than is necessary.

How are you getting 1/2 for GG?

Do you agree with the following possibilities for a couple with two children, yes or no?

GG- 1/4

BB- 1/4

BG- 1/4

GB- 1/4

Do you agree that if we remove all possibilities where at least one of the children is not a girl, the following is the remaining possibilities, yes or no?

GG- 1/3

BG- 1/3

GB- 1/3

This can be proven with the penny experiment. For every pair of pennies that has at least one penny heads up, 1/3 of the time the other penny will be heads up. This is exactly analogous to the riddle. Tell me how it isn't.

[Guest]

We don't know if the girl is older. If you are referring to my post (the one prior to your last post), then you need to read the whole post. I listed three senarios, one with the possiblity of the oldest being a girl, one with the possiblity of the youngest being a girl, and one answering the actual question where we don't know if the girl is the oldest or youngest child.

I can see you are almost in agreement with us noobsauce. Your last post has agreed that there are three possible combinations:

"There are 3 probabilities but the chance of the girl girl probability is actually 1/2. the boy-girl is 1/4 and girl boy is 1/4"

However I can't see how you have come up with the 1/2, 1/4, 1/4. Surely if there are 3 possiblities then they are all equally 1/3.

[Guest]

For girl-girl you have to consider if the baby is the oldest sibling or the younger, just like what you did for B-G and G-B

[Guest]

For girl-girl you have to consider if the baby is the oldest sibling or the younger, just like what you did for B-G and G-B

noobsauce, it's time to stop ignoring posts directed at you or just finally admit you're wrong. What you wrote above is irrelevant regarding your claim that "the girl girl probability is actually 1/2. the boy-girl is 1/4 and girl boy is 1/4". You're plain wrong here and you've been told why.

Do you agree with the following possibilities for a couple with two children, yes or no?

GG- 1/4

BB- 1/4

BG- 1/4

GB- 1/4

The correct answer is that the above is correct but according to your rationale it shouldn't be because you wrote this:

But it does not matter because if it were premutation there would be 4 possilbilites. boy-girl, girl-boy and girl-girl and girl-girl, you flip the girl-girl so you get it 2 times, just like what you did for boy-girl. The answer would be 2/4 = 1/2 = 50%

Using that rationale, you think the following are the probabilities since GG has twice the probability of BG and GB:

GG- 1/3

BB- 1/3

BG- 1/6

GB- 1/6

Do you agree that if we remove all possibilities where at least one of the children is not a girl, the following is the remaining possibilities, yes or no?

GG- 1/3

BG- 1/3

GB- 1/3

This can be proven with the penny experiment. For every pair of pennies that has at least one penny heads up, 1/3 of the time the other penny will be heads up. This is exactly analogous to the riddle. Tell me how it isn't.

In the penny experiment, we lay out 100 pair of pennies to simulate all couples with two children.

When doing this, you'll find the following are the probabilities of each possibility:

HH - 1/4

TT - 1/4

HT - 1/4

TH - 1/4

If you think that's wrong, tell me what you think will happen and try the experiment for yourself.

When wanting to determine how many couples that have two children and have at least one girl have two girls, we look at what fraction of the pennies pairs that have at least one heads-up also have another heads-up. It will be 1/3 of the time. If you don't agree, tell me what you think the result be and then try the experiment for yourself. If you don't think this experiment is analogous to the riddle, tell me why, but first answer what you think the results would be if you bothered to try it.

[Guest]

For girl-girl you have to consider if the baby is the oldest sibling or the younger, just like what you did for B-G and G-B

Please read the links that Scraff posted earlier, even if it is only one. Maybe by reading what the "experts" say, may convince you to look at it how we do and hence the correct answer.

[Guest]

You are ignoring 1 fact that is there has to be a girl and the girl isn't specified where it is. By your calculation it would be:

Girl-Girl : 1/2 * 1/2 = 1/4 note that 1/2 is for the girl equal likely to being the 1st or second place.

boy-girl : 1/2 * 1/2 = 1/4

girl-boy : 1/2 * 1/2 = 1/4

3* 1/4 = 3/4 does 3/4 = 1?

so that adds up to 3/4. Where is the missing 1/4?

Simple the missin g 1/4 is at girl girl. You are ignoring the question and you are supposing that the girl could be the older sibling and the younger sibling at the sametime. No, only one baby has to be a girl not the other, so 1 is 100% and the other is 50% what you are doing now is supposeing that both baby are 50% for the girl-girl situations. Not to metion Your answer doesn't add up mathematically and is therefore incorrect.

Like i posted before, if you were actually reading. This question is not the same as a normal coin flipping question.

By eye, it is correct that there are 3 possible outcome, but the fact that girl-girl can appear 2 times was not metioned. Yet you continue to ignore my comment and posting the same thing for 10 posts. You should get yoiur facts straight.