Logic problem (1 Viewer)

[Shutterbug]

Ok, someone in the 2/3 group explain why this isn't accurate.

RNG is a random number generator function.

y = 0

x = {RNG}

if x = odd #

then girl

else boy

x is an even number, so we have a boy. This is the point we are at when we begin the problem.

#ofgirls = 0

#ofboys = 0

do until #ofboys OR #ofgirls = 1000

y = {RNG}

if y = odd #

then #ofgirls = #ofgirls + 1

else #ofboys = #ofboys + 1

loop

When one hits 1000 and the loop exits, the other will be much closer to 1000 than 500, thus proving there is close to a 50/50 chance for each after the initial result of boy.

I haven't read the thread, but common sense alone tells me it is over 50% of a chance. If the guy had two sons, then he would refer to them as "oldest son" and "youngest son", not "my son". The father saying "my son" implies he only has one.

Read the thread.

Can I get cliff notes?

 

[JBPH]

I'm willing to admit I was wrong about this, but I'd sure appreciate someone ripping my logic below apart a bit just so it makes mathematical sense to me.

I know the guy has two kids, and he's talking about one of them being a boy. Dunno if it's the older or younger.

That means the possibilities are equally:

BB, BG, GB

From our perspective, it's 50/50 whether he's talking about the older or the younger.

If he's talking about the older kid, which happens 50% of the time: We can eliminate GG, GB. That leaves BB, BG.

If he's talking about the younger kid, which happens the other 50%: We can eliminate GG, BG. That leaves BB, GB.

This is going to boil down to 50/50.

Again - I'm cautiously agreeing the answer should be 2/3 but it seems like the math above must be flawed if that is true, and I'm not seeing where.

 

[Sweet J]

By the way, does anybody else want a poll:

I'm a socially concervative Christian and I believe the answer is 50%.

I'm a socially Concervative Christian and I believe the answer is 66%.

I'm not a socially concervative Christian and I believe the answer is 50%

I'm not a socially concervative Christian and I believe the asnwer is 66%

For the life of me, I'm convinced the Christians are skewing the results to the 50% number.

Easy there chief.

LOL. I'm trying to find the agnostic who believes the answer is 50%. But I take your point: I should have said "feels stronly about his religion" instead of "concervative Christian." I'd love to see if there was correlation between feeling very strongly about your religion and the answer to this question.Edit because I continue to use the wrong words.

I wonder if there's a correlation between a lack of religious belief as well as spelling ability even after an edit?

LOL. finger

 

[Sweet J]

By the way, does anybody else want a poll:

I'm a socially concervative Christian and I believe the answer is 50%.

I'm a socially Concervative Christian and I believe the answer is 66%.

I'm not a socially concervative Christian and I believe the answer is 50%

I'm not a socially concervative Christian and I believe the asnwer is 66%

For the life of me, I'm convinced the Christians are skewing the results to the 50% number.

Easy there chief.

LOL. I'm trying to find the agnostic who believes the answer is 50%. But I take your point: I should have said "feels stronly about his religion" instead of "concervative Christian." I'd love to see if there was correlation between feeling very strongly about your religion and the answer to this question.Edit because I continue to use the wrong words.

I'm agnostic and at least vacillated. HTH.

Absolutely. As did lots of people. You can think one thing at first, then think about it some, and then think another. The key is how hard it is to change your mind on something that you think is true because it "feels" right.

 

[videoguy505]

Ok, someone in the 2/3 group explain why this isn't accurate.

RNG is a random number generator function.

y = 0

x = {RNG}

if x = odd #

then girl

else boy

x is an even number, so we have a boy. This is the point we are at when we begin the problem.

#ofgirls = 0

#ofboys = 0

do until #ofboys OR #ofgirls = 1000

y = {RNG}

if y = odd #

then #ofgirls = #ofgirls + 1

else #ofboys = #ofboys + 1

loop

When one hits 1000 and the loop exits, the other will be much closer to 1000 than 500, thus proving there is close to a 50/50 chance for each after the initial result of boy.

I haven't read the thread, but common sense alone tells me it is over 50% of a chance. If the guy had two sons, then he would refer to them as "oldest son" and "youngest son", not "my son". The father saying "my son" implies he only has one.

Read the thread.

Can I get cliff notes?

8 pages of arguing whether the OP meant scenario 1 or scenario 2 here. Interspersed with math-challenged arguments that argued entirely different scenarios ignoring the facts in the OP.

 

[mon]

Ok, someone in the 2/3 group explain why this isn't accurate.

RNG is a random number generator function.

y = 0

x = {RNG}

if x = odd #

then girl

else boy

x is an even number, so we have a boy. This is the point we are at when we begin the problem.

#ofgirls = 0

#ofboys = 0

do until #ofboys OR #ofgirls = 1000

y = {RNG}

if y = odd #

then #ofgirls = #ofgirls + 1

else #ofboys = #ofboys + 1

loop

When one hits 1000 and the loop exits, the other will be much closer to 1000 than 500, thus proving there is close to a 50/50 chance for each after the initial result of boy.

I haven't read the thread, but common sense alone tells me it is over 50% of a chance. If the guy had two sons, then he would refer to them as "oldest son" and "youngest son", not "my son". The father saying "my son" implies he only has one.

Read the thread.

Can I get cliff notes?

This is akin to asking for cliff notes of War and Peace. I could use some cliff notes, myself. Somewhere back there, it was said that the wording or how he referred to them didn't matter.

 

[Sweet J]

I'm willing to admit I was wrong about this, but I'd sure appreciate someone ripping my logic below apart a bit just so it makes mathematical sense to me.

I know the guy has two kids, and he's talking about one of them being a boy. Dunno if it's the older or younger.

That means the possibilities are equally:

BB, BG, GB

From our perspective, it's 50/50 whether he's talking about the older or the younger.

If he's talking about the older kid, which happens 50% of the time: We can eliminate GG, GB. That leaves BB, BG.

If he's talking about the younger kid, which happens the other 50%: We can eliminate GG, BG. That leaves BB, GB.

This is going to boil down to 50/50.

Again - I'm cautiously agreeing the answer should be 2/3 but it seems like the math above must be flawed if that is true, and I'm not seeing where.

See bolded. See how 3 out of the four bolded are Bs?

 

[videoguy505]

I'm willing to admit I was wrong about this, but I'd sure appreciate someone ripping my logic below apart a bit just so it makes mathematical sense to me.

I know the guy has two kids, and he's talking about one of them being a boy. Dunno if it's the older or younger.

That means the possibilities are equally:

BB, BG, GB

From our perspective, it's 50/50 whether he's talking about the older or the younger.

If he's talking about the older kid, which happens 50% of the time: We can eliminate GG, GB. That leaves BB, BG.

If he's talking about the younger kid, which happens the other 50%: We can eliminate GG, BG. That leaves BB, GB.

This is going to boil down to 50/50.

Again - I'm cautiously agreeing the answer should be 2/3 but it seems like the math above must be flawed if that is true, and I'm not seeing where.

http://mathforum.org/dr.math/faq/faq.boygirl.choose.htmlBy making the second distinction, regarding older/younger, you're changing the problem to scenario 2 on the Dr. Math article.

You don't have to make that distinction. Once you get to "the possibilities are equally BB, BG, GB" you can stop. By further asking which is older, you're now asking "From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?" which is 50%.

 

[Dave Baker]

Had the question been stated "You ask him if he has kids, and he says '2.' You ask huim if he has a son and he says 'yes.'"The answer would be 2/3.

But that's what happened, but instead of you asking him, he offered the info...in the same exact order where you agree the answer is 2/3.

exactly - he offered the info and created a sequence or ordering for his children, the same as if they were sequenced by age.

No, he did not create a sequence or order.He provided more information after you had an initial data set.

 

[Dave Baker]

That means the possibilities are equally:BB, BG, GB

Wrong. The possibilities for having one boy and one girl are not equal to that of having two boys or two girls. This is a long thread, but it's been explained many times. If it is agreed that the odds of having a boy vs a girl are 50/50 (50%), and if you know already someone has two kids, then you have the following:50% chance they are a boy and a girl25% they are two boys25% chance they are two girlsIf that doesn't sound right to you, consider your entire options:BBBGGBGG50% of those options result in having one boy and one girl. The rest of it plays out accordingly.

 

[JBPH]

I'm willing to admit I was wrong about this, but I'd sure appreciate someone ripping my logic below apart a bit just so it makes mathematical sense to me.

I know the guy has two kids, and he's talking about one of them being a boy. Dunno if it's the older or younger.

That means the possibilities are equally:

BB, BG, GB

From our perspective, it's 50/50 whether he's talking about the older or the younger.

If he's talking about the older kid, which happens 50% of the time: We can eliminate GG, GB. That leaves BB, BG.

If he's talking about the younger kid, which happens the other 50%: We can eliminate GG, BG. That leaves BB, GB.

This is going to boil down to 50/50.

Again - I'm cautiously agreeing the answer should be 2/3 but it seems like the math above must be flawed if that is true, and I'm not seeing where.

http://mathforum.org/dr.math/faq/faq.boygirl.choose.htmlBy making the second distinction, regarding older/younger, you're changing the problem to scenario 2 on the Dr. Math article.

You don't have to make that distinction. Once you get to "the possibilities are equally BB, BG, GB" you can stop. By further asking which is older, you're now asking "From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?" which is 50%.

Thanks. So, the difference between the 50/50 version and the 2/3 version is the difference between "You have a 2 child family that has at least one boy..." and "You have a two child family. (Though GG was originally a possibility,) you discover next that one of the kids is a boy...". One is focused on the sex of the boy and the second kid, one is about a family in which you have one clue.

 

[GoSensGo]

Take 100 families with 2 children.25% should have 2 girls25% should have 2 boys50% should be 1 and 1If you dispute this, the argument is over.Now disregard the families with 2 girls.How man familes are left?How many have 2 boys? 1 boy and 1 girl?What are the respective percentages?QED

100 random families would not have a perfectly even distribution.

 

[FlapJacks]

he offered the info and created a sequence or ordering for his children

I'm trying to see how this creates order. Not coming up with anything.

you've established that his "1st" child is a boywhetehr it's first in birth order, first in his heart or the first one he talks about doesn't matter. The sex of his "2nd" child becomes an independent event.

 

[GoSensGo]

Okay I'm rethinking this;

The way I wee it, there are four possibilities; we're talking about a specific kid (TOM) and he has either

A - older sister

B - older brother

C - younger sister

D - younger brother

In two cases he has a sister, and two a brother. So now I'm thinking 50/50. What am I missing?

These are the possiblities of the SEQUENCE:BB

BG

GB

GG

The middle two are the same when talking about PROBABILTY.

Correct. There is an equal chance of each of those 2-child families resulting.If the question was, "What are the odds that any child in a 2-child family is a girl", then any of the children could be the one referred to and there are 4 chances at being a boy and 4 at being a girl, and the answer would be 50/50. (This wasn't the question asked).

If the question was "What are the odds that the younger child is a girl", then there are 2 younger boys and 2 younger girls it could refer to, and the odds would be 50/50. (This wasn't the question asked either).

The question asked had the father telling us that one of his children (we don't know which) is a boy. This means it can't be a GG family and the potential set of children has now become:

BB

BG

GB

GG

Further, we are asked the sex of the child that is not the boy the father referred to. So we can eliminate one boy from each family as being the boy that the father referred to:

BB

BG

GB

GG

There are only 3 children remaining with the same probability of being the one the father is referring to. 1 is a boy, and 2 are girls. Thus the 2/3 chance she is a girl, when we know the information that the father gave us. If we didn't have that information that he has a different child that is a son, then we couldn't eliminate those possibilities and it would just be a 50/50 chance.

IncorrectBG = GB you can't count them twice.

There is one boy therefore your options are

BB

BG

 

[Short Corner]

he offered the info and created a sequence or ordering for his children

I'm trying to see how this creates order. Not coming up with anything.

The term order is being used a little loosely here and in terms of our sample space doesn't matter. GB=BG. But the fact that BG=GB tells us that he P(BG) = 2*P(BB)=2*P(GG), ie effecting the distribution of our sample space, which is THE key point that people are leaving out.

 

[FlapJacks]

8 pages of arguing whether the OP meant scenario 1 or scenario 2 here. Interspersed with math-challenged arguments that argued entirely different scenarios ignoring the facts in the OP.

Nailed it.The wording of the OP is really scenario 2 IMO - "Choosing the Child First"

Choosing the Child First

Supposing, on the other hand, that we randomly pick a child from a two-child family. We see that he is a boy, and want to find out whether his sibling is a brother or a sister. (For example, from all the children of two-child families, we select one at random who happens to be a boy, and ask how many children are in his family and he responds \"two.\") In this case, an unambiguous statement of the question could be:

From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?

From the possible combinations of all two-child families we can again eliminate the GG combination, since we know that one child is a boy. As before, the three remaining possible combinations are:

In these combinations there are four boys, of whom we chose one. Let\'s identify them from left to right as B1, B2, B3 and B4. Of these four boys, only B3 and B4 have a sister, so our chance of randomly picking one of them is 2 in 4, and the probability is 1/2.

A man tells you he has two children. He then starts talking about his son.

 

[Short Corner]

Ok, someone in the 2/3 group explain why this isn't accurate.

RNG is a random number generator function.

y = 0

x = {RNG}

if x = odd #

then girl

else boy

x is an even number, so we have a boy. This is the point we are at when we begin the problem.

#ofgirls = 0

#ofboys = 0

do until #ofboys OR #ofgirls = 1000

y = {RNG}

if y = odd #

then #ofgirls = #ofgirls + 1

else #ofboys = #ofboys + 1

loop

When one hits 1000 and the loop exits, the other will be much closer to 1000 than 500, thus proving there is close to a 50/50 chance for each after the initial result of boy.

I haven't read the thread, but common sense alone tells me it is over 50% of a chance. If the guy had two sons, then he would refer to them as "oldest son" and "youngest son", not "my son". The father saying "my son" implies he only has one.

Read the thread.

Can I get cliff notes?

8 pages of arguing whether the OP meant scenario 1 or scenario 2 here. Interspersed with math-challenged arguments that argued entirely different scenarios ignoring the facts in the OP.

I don't recall anyone using the logic in scenario 2 to come up with their 50% answer. Everyone is pretty much clueless about conditional probability.

 

[Short Corner]

Take 100 families with 2 children.25% should have 2 girls25% should have 2 boys50% should be 1 and 1If you dispute this, the argument is over.Now disregard the families with 2 girls.How man familes are left?How many have 2 boys? 1 boy and 1 girl?What are the respective percentages?QED

100 random families would not have a perfectly even distribution.

Did I say they would?

 

[mon]

Ok, someone in the 2/3 group explain why this isn't accurate.

RNG is a random number generator function.

2ndkid = 0

1stkid = {RNG}

if 1stkid = odd #

then girl

else boy

1stkid is an even number, so we have a boy. This is the point we are at when we begin the problem.

#ofgirls = 0

#ofboys = 0

do until #ofboys OR #ofgirls = 1000

2ndkid = {RNG}

if 2ndkid = odd #

then #ofgirls = #ofgirls + 1

else #ofboys = #ofboys + 1

loop

When one hits 1000 and the loop exits, the other will be much closer to 1000 than 500, thus proving there is close to a 50/50 chance for each after the initial result of boy.

:tapsfoot:

 

[Short Corner]

Okay I'm rethinking this;

The way I wee it, there are four possibilities; we're talking about a specific kid (TOM) and he has either

A - older sister

B - older brother

C - younger sister

D - younger brother

In two cases he has a sister, and two a brother. So now I'm thinking 50/50. What am I missing?

These are the possiblities of the SEQUENCE:BB

BG

GB

GG

The middle two are the same when talking about PROBABILTY.

Correct. There is an equal chance of each of those 2-child families resulting.If the question was, "What are the odds that any child in a 2-child family is a girl", then any of the children could be the one referred to and there are 4 chances at being a boy and 4 at being a girl, and the answer would be 50/50. (This wasn't the question asked).

If the question was "What are the odds that the younger child is a girl", then there are 2 younger boys and 2 younger girls it could refer to, and the odds would be 50/50. (This wasn't the question asked either).

The question asked had the father telling us that one of his children (we don't know which) is a boy. This means it can't be a GG family and the potential set of children has now become:

BB

BG

GB

GG

Further, we are asked the sex of the child that is not the boy the father referred to. So we can eliminate one boy from each family as being the boy that the father referred to:

BB

BG

GB

GG

There are only 3 children remaining with the same probability of being the one the father is referring to. 1 is a boy, and 2 are girls. Thus the 2/3 chance she is a girl, when we know the information that the father gave us. If we didn't have that information that he has a different child that is a son, then we couldn't eliminate those possibilities and it would just be a 50/50 chance.

IncorrectBG = GB you can't count them twice.

As an end result they are the same, but they have a different probability than the other two outcomes.

 

[Short Corner]

I was curious where this question came from, so I googled it. I giggled at a post on another board, but thought you guys may find it helpful.

--------------------------------------------------------------------------------Well, if you want to get technical, the odds of probabilities are incorrect before you bring out the basic statistics formulas.The odds of having 2 boys is 25.8%, the chances of 2 girls is 22.0%, and the chances of 1 of each is 52.2%. Besides, when one estimates the true trust value with some uncertainty,it is necessary to have a method of determining the standard error of expirimental outcomes. For a yes or no outcome or Bernoulli Distribution, the standard error of the estimated proportion p, based on random sample observations is given by: SE = [p(1-p)/N] 1/2 . The SE is at it's maximum when p=0.5. Under this condition,the sample size N,can than be expressed as the largest integer less than or equal to 0.25/SE2. So for SE to be 0.04, a sample size of 156 will be needed.

Someone want to translate that?

He is basically saying that as evidenced by an insane sample size, the probability of having a boy not equal to the probability of having a girl. He then states that for the actual statistics to have deviated that much from the true probability, the sample size would have to be no more than 156.

 

[bryhamm]

he offered the info and created a sequence or ordering for his children

I'm trying to see how this creates order. Not coming up with anything.

you've established that his "1st" child is a boywhetehr it's first in birth order, first in his heart or the first one he talks about doesn't matter. The sex of his "2nd" child becomes an independent event.

Did you see my example of the colored coins?

 

[Shutterbug]

I was curious where this question came from, so I googled it. I giggled at a post on another board, but thought you guys may find it helpful.

--------------------------------------------------------------------------------Well, if you want to get technical, the odds of probabilities are incorrect before you bring out the basic statistics formulas.The odds of having 2 boys is 25.8%, the chances of 2 girls is 22.0%, and the chances of 1 of each is 52.2%. Besides, when one estimates the true trust value with some uncertainty,it is necessary to have a method of determining the standard error of expirimental outcomes. For a yes or no outcome or Bernoulli Distribution, the standard error of the estimated proportion p, based on random sample observations is given by: SE = [p(1-p)/N] 1/2 . The SE is at it's maximum when p=0.5. Under this condition,the sample size N,can than be expressed as the largest integer less than or equal to 0.25/SE2. So for SE to be 0.04, a sample size of 156 will be needed.

Someone want to translate that?

He is basically saying that as evidenced by an insane sample size, the probability of having a boy not equal to the probability of having a girl. He then states that for the actual statistics to have deviated that much from the true probability, the sample size would have to be no more than 156.

Oh. Much clearer now. Thanks.

 

[BoltThrower]

8 pages of arguing whether the OP meant scenario 1 or scenario 2 here. Interspersed with math-challenged arguments that argued entirely different scenarios ignoring the facts in the OP.

Nailed it.The wording of the OP is really scenario 2 IMO - "Choosing the Child First"

Choosing the Child First

Supposing, on the other hand, that we randomly pick a child from a two-child family. We see that he is a boy, and want to find out whether his sibling is a brother or a sister. (For example, from all the children of two-child families, we select one at random who happens to be a boy, and ask how many children are in his family and he responds \"two.\") In this case, an unambiguous statement of the question could be:

From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?

From the possible combinations of all two-child families we can again eliminate the GG combination, since we know that one child is a boy. As before, the three remaining possible combinations are:

In these combinations there are four boys, of whom we chose one. Let\'s identify them from left to right as B1, B2, B3 and B4. Of these four boys, only B3 and B4 have a sister, so our chance of randomly picking one of them is 2 in 4, and the probability is 1/2.

A man tells you he has two children. He then starts talking about his son.

No, this is Scenario 1. An example of Scenario 2 would be "You approach a girl on the street (easy now, Homer J) and she tells you that she has 1 sibling..."

 

[moby fan]

interesting

 

[BoltThrower]

But I'm stuck on the fact that age is such an arbitrary identifying trait.

How is age mentioned at all? Age has zero to do with this. The only important thing is establishing your data set.

A. If he has two kids and his oldest is a boy, there is a 50% chance his youngest is a girlB. If he has two kids, one of which is a boy, there is a 2/3 chance the other is a girl

When we apply the trait, such as age, the data set and therefore the probability changes.

what if we stated it like this then:

If he has two kids and the kid he talked about is a boy, there is a ___ chance his youngest is a girl

should the resulting data sets look more like A above or B above?

ponderous....

If he has 2 kids and we know 1 is a boy:BB

BG

GB

are still the possibilities. Looks like a 1/3 chance that his youngest is a girl.

 

[Big Petey]

But I'm stuck on the fact that age is such an arbitrary identifying trait.

How is age mentioned at all? Age has zero to do with this. The only important thing is establishing your data set.

A. If he has two kids and his oldest is a boy, there is a 50% chance his youngest is a girlB. If he has two kids, one of which is a boy, there is a 2/3 chance the other is a girl

When we apply the trait, such as age, the data set and therefore the probability changes.

what if we stated it like this then:

If he has two kids and the kid he talked about is a boy, there is a ___ chance his youngest is a girl

should the resulting data sets look more like A above or B above?

ponderous....

If he has 2 kids and we know 1 is a boy:BB

BG

GB

are still the possibilities. Looks like a 1/3 chance that his youngest is a girl.

The reason that age (older/younger) keeps being brought up, is because people are trying to prove the 2/3 theory by using the BB, BG, GB, GG method. This is SEQUENCING. They are using "boy then girl" and "girl then boy" to describe the SAME combination.

 

[GregR_2]

Okay I'm rethinking this;

The way I wee it, there are four possibilities; we're talking about a specific kid (TOM) and he has either

A - older sister

B - older brother

C - younger sister

D - younger brother

In two cases he has a sister, and two a brother. So now I'm thinking 50/50. What am I missing?

These are the possiblities of the SEQUENCE:BB

BG

GB

GG

The middle two are the same when talking about PROBABILTY.

Correct. There is an equal chance of each of those 2-child families resulting.If the question was, "What are the odds that any child in a 2-child family is a girl", then any of the children could be the one referred to and there are 4 chances at being a boy and 4 at being a girl, and the answer would be 50/50. (This wasn't the question asked).

If the question was "What are the odds that the younger child is a girl", then there are 2 younger boys and 2 younger girls it could refer to, and the odds would be 50/50. (This wasn't the question asked either).

The question asked had the father telling us that one of his children (we don't know which) is a boy. This means it can't be a GG family and the potential set of children has now become:

BB

BG

GB

GG

Further, we are asked the sex of the child that is not the boy the father referred to. So we can eliminate one boy from each family as being the boy that the father referred to:

BB

BG

GB

GG

There are only 3 children remaining with the same probability of being the one the father is referring to. 1 is a boy, and 2 are girls. Thus the 2/3 chance she is a girl, when we know the information that the father gave us. If we didn't have that information that he has a different child that is a son, then we couldn't eliminate those possibilities and it would just be a 50/50 chance.

IncorrectBG = GB you can't count them twice.

There is one boy therefore your options are

BB

BG

You can treat them the same if you wish but to do so correctly you have to acknowledge that the odds are twice of great of getting GB/BG in any order than they are of getting BB. Which results in the same exact answer of 2/3.

 

[FlapJacks]

8 pages of arguing whether the OP meant scenario 1 or scenario 2 here. Interspersed with math-challenged arguments that argued entirely different scenarios ignoring the facts in the OP.

Nailed it.The wording of the OP is really scenario 2 IMO - "Choosing the Child First"

Choosing the Child First

Supposing, on the other hand, that we randomly pick a child from a two-child family. We see that he is a boy, and want to find out whether his sibling is a brother or a sister. (For example, from all the children of two-child families, we select one at random who happens to be a boy, and ask how many children are in his family and he responds \"two.\") In this case, an unambiguous statement of the question could be:

From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?

From the possible combinations of all two-child families we can again eliminate the GG combination, since we know that one child is a boy. As before, the three remaining possible combinations are:

In these combinations there are four boys, of whom we chose one. Let\'s identify them from left to right as B1, B2, B3 and B4. Of these four boys, only B3 and B4 have a sister, so our chance of randomly picking one of them is 2 in 4, and the probability is 1/2.

A man tells you he has two children. He then starts talking about his son.

No, this is Scenario 1. An example of Scenario 2 would be "You approach a girl on the street (easy now, Homer J) and she tells you that she has 1 sibling..."

I would say this is scenario 2, since based the wording, we are starting with the child.why should the choice of sequencing field change the probability?

If I sequence based on the order in which they are mentioned, rather than the order in which they were born, the result should be the same, no?

BB

BG

GB

GG

 

[Dante Hicks]

How in the hell is this 8 pages?

 

[sirelfman]

[Lecture Mode]

There are three basic rules of probability that are in play here. Now these rules apply to events that are independent, ie the outcome of one tells you nothing about the outcome of the other. Every seems to agree that the sexes of siblings are independent, so these two rules would apply in this case.

1. The sum of the probabilities of all possible events must equal 1.00 (100%), no more, no less. (One of the events must happen, therefore there is a 100% that at least one of them will happen.)

2. To find the probability of any two events, A and B, happening at the same time, multiply the probabilities together. Using standard probability notation, this is P(A and B) = P(A) * P(B).

3. To find the probability of one of several events happening, add the probabilities together. Again, using standard probability notation, P(A or B) = P(A) + P(B).

If you don't understand these 3 rules, I'll wait. Don't go on until you understand and agree with these three rules. If you understand them, but don't agree with them, don't read any further. Instead, go take a basic probability course. I can't tell you anything that will make you see the reality of this situation.

Now that you understand and argree with the three rules above, let's start. It seems most everybody agrees that P(boy)=.50 and P(girl)=.50. True, experimental probability shows this not to be the case, but it does simplify matters. (Girls are slightly more than 50% of the world's population).

What is the probability of two boys?

#2 says P(boy AND boy) = P(boy) * P(boy) = .5 * .5 = .25

By a similar calculation, I can show that P(girl AND girl) = .25. There seems to be no dispute about these two facts. So let's move onto another question.

What is the probability of a boy and a girl?

There some that argue that this boy-girl and girl-boy should be consider the same outcome. So by #1, the two probabilities above, plus this one must have a sum of 1.00 (unless you think there is another possible outcome). What does that tell us about P(one of each)? That is must be .50 (1.00 - .25 - .25 = .50)

The answer here shows you that two boys, two girls and one of each are NOT equally likely. This is what has tripped all of the 50%'er up. If you see these as the only possibilities, that is fine. Just remember that these three are not equally likely and therefore cannot be assigned the same probabilities.

But wait you may say, I thought one of each would be .5 * .5 = .25! Well, that's right, but that's not what you have here. Here, you're saying the first event can be either a boy or a girl. By rule #3 P(boy or girl) = P(boy) + P(girl) = .5 * .5 = 1.00. Then, the second event must be different, ie, if A=boy, it must be that B=girl to have one of each. So P(second is different) = .5 (this works even if A=girl. Go ahead, try it, I'll wait). So now P(one of each) = P(either sex) * P(other sex) = 1.00 * .5 = .5

Let me put it another way. P(boy AND girl) = P(boy) * P(girl) = .5 * .5 = .25, and also P(girl AND boy) = P(girl) * P(boy) = .5 * .5 = .25. So this gives P(one of each) = P(boy AND girl) + P(girl AND boy) = .25 + .25 = .50 No matter how you address it, getting one of each sex is TWICE as likely as getting two boys or two girls. This is VERY crucial to understanding how the answer is arrived at in this problem. If you think there are three possibilities [bb, gg, 1 of each], you must agree that they are not equally likely. If you want all equally likely possibilities, you must use [bb, gg, bg, gb] as this is the only way to get the sum to be 1.00.

If you know that one of two children is a boy, what is the probability that the other is a girl?

Let's look at the three outcome situation first. P(boy AND boy) = .25, P(girl AND girl) = .25, P(one of each) = .50 and we want P(girl if we know one is boy). Only the first has a boy, and P(boy AND boy) + P(one of each) = .75, and only the third has a girl, that is .50 out of .75, which is .50/.75 = 2/3 = .66 so there is a 66.6666.....% chance that if you know that one of two children is a boy, the other is girl.

If you prefer the four outcome situation, P(boy AND boy) = .25, P(girl AND girl) = .25, P(boy AND girl) = .25, P(girl AND boy) = .25 and the only the first, third and fourth outcomes come into play. Since these are equally likely, (ie, same probability), and 2 of the 3 have a girl, the probabilities is 2/3 = .6666... or 66.6.....%.

The lesson is you must know if your outcomes are equally likely or not. Once the probabilities of your chosen outcomes are known, be they equally or unequally likely, the answer will be the same.

[/lecture mode]

Oh, and by the way, I have two daughters, and rarely when talking to other do I identify them as "my older daughter" or "my younger daughter". I usually just say "my daughter", not really caring whether the listeners knows I have another one or not, so that semantic argument holds no water what so ever.

ETA: Try this one on for size. You and your two friends, Tom and Harry, are captured in Lithistania by Geristal extremists. The terrorist will execute two of you at dawn, just to prove they are a real threat to the Lithistanian government. What is the probability you will be executed at dawn? (This part is easy) But later, you overhear the guards arguing about who has to take Harry to the execution room to be beheaded at dawn. Now, what is the probability that you will be executed at dawn?

 

[Das Boot]

Ivan has convinced me.

I'm still fighting it.No one has addressed my question of a guy having 100 kids with 99 boys. Once I hear that explanation I might come around.

Well I can at least tell you what state the guy is from.

 

[videoguy505]

But I'm stuck on the fact that age is such an arbitrary identifying trait.

How is age mentioned at all? Age has zero to do with this. The only important thing is establishing your data set.

A. If he has two kids and his oldest is a boy, there is a 50% chance his youngest is a girlB. If he has two kids, one of which is a boy, there is a 2/3 chance the other is a girl

When we apply the trait, such as age, the data set and therefore the probability changes.

what if we stated it like this then:

If he has two kids and the kid he talked about is a boy, there is a ___ chance his youngest is a girl

should the resulting data sets look more like A above or B above?

ponderous....

If he has 2 kids and we know 1 is a boy:BB

BG

GB

are still the possibilities. Looks like a 1/3 chance that his youngest is a girl.

The reason that age (older/younger) keeps being brought up, is because people are trying to prove the 2/3 theory by using the BB, BG, GB, GG method. This is SEQUENCING. They are using "boy then girl" and "girl then boy" to describe the SAME combination.

It's just easier to look at BG & GB as separate, because individually, they're as likely as BB and GG. If you just wrote "the possibilites are boy-boy, girl-girl, or mixed", people would incorrectly assume that, of all 2-child families, the odds for the three possibilites are equal. They are not. In the total set of 2-child families, boy-boy and girl-girl would be in about 1/4th of cases, each, and the "mixed" possibility would appear in about half overall.

 

[Jack Burton]

As of right now 66 people chose 2/3, just thought it should be pointed out.

I would assume his other kid would be a girl because he says he has two kids and then talks about the boy in the family. Logically he would mention which boy he was referring to either by age or by name, he did neither. It seems more feasible that he would differentiate between the two boys, if he had two boys. Otherwise how would you know which child he is talking about.

Why is 100% not an option?

 

[FatMax]

ETA: Try this one on for size. You and your two friends, Tom and Harry, are captured in Lithistania by Geristal extremists. The terrorist will execute two of you at dawn, just to prove they are a real threat to the Lithistanian government. What is the probability you will be executed at dawn? (This part is easy) But later, you overhear the guards arguing about who has to take Harry to the execution room to be beheaded at dawn. Now, what is the probability that you will be executed at dawn?

Another variation of the Monte Hall problem. I like it.

 

[sirelfman]

ETA: Try this one on for size. You and your two friends, Tom and Harry, are captured in Lithistania by Geristal extremists. The terrorist will execute two of you at dawn, just to prove they are a real threat to the Lithistanian government. What is the probability you will be executed at dawn? (This part is easy) But later, you overhear the guards arguing about who has to take Harry to the execution room to be beheaded at dawn. Now, what is the probability that you will be executed at dawn?

Another variation of the Monte Hall problem. I like it.

Only if Monty Hall is done with the assumption that Monty knows where the real prize is. I hadn't considered this particular of the prisoner problems, but I see that it is the same.

 

[FatMax]

ETA: Try this one on for size. You and your two friends, Tom and Harry, are captured in Lithistania by Geristal extremists. The terrorist will execute two of you at dawn, just to prove they are a real threat to the Lithistanian government. What is the probability you will be executed at dawn? (This part is easy) But later, you overhear the guards arguing about who has to take Harry to the execution room to be beheaded at dawn. Now, what is the probability that you will be executed at dawn?

Another variation of the Monte Hall problem. I like it.

Only if Monty Hall is done with the assumption that Monty knows where the real prize is. I hadn't considered this particular of the prisoner problems, but I see that it is the same.

Monty always knows were the prize is. He has to, or the show (choice) doesn't work.

 

[Short Corner]

How in the hell is this 8 pages?

The majority of FBGs seem to believe that the number of families with BB BG and GG are evenly distributed.

 

[Shutterbug]

[Lecture Mode]

There are three basic rules of probability that are in play here. Now these rules apply to events that are independent, ie the outcome of one tells you nothing about the outcome of the other. Every seems to agree that the sexes of siblings are independent, so these two rules would apply in this case.

1. The sum of the probabilities of all possible events must equal 1.00 (100%), no more, no less. (One of the events must happen, therefore there is a 100% that at least one of them will happen.)

2. To find the probability of any two events, A and B, happening at the same time, multiply the probabilities together. Using standard probability notation, this is P(A and B) = P(A) * P(B).

3. To find the probability of one of several events happening, add the probabilities together. Again, using standard probability notation, P(A or B) = P(A) + P(B).

If you don't understand these 3 rules, I'll wait. Don't go on until you understand and agree with these three rules. If you understand them, but don't agree with them, don't read any further. Instead, go take a basic probability course. I can't tell you anything that will make you see the reality of this situation.

Now that you understand and argree with the three rules above, let's start. It seems most everybody agrees that P(boy)=.50 and P(girl)=.50. True, experimental probability shows this not to be the case, but it does simplify matters. (Girls are slightly more than 50% of the world's population).

What is the probability of two boys?

#2 says P(boy AND boy) = P(boy) * P(boy) = .5 * .5 = .25

By a similar calculation, I can show that P(girl AND girl) = .25. There seems to be no dispute about these two facts. So let's move onto another question.

What is the probability of a boy and a girl?

There some that argue that this boy-girl and girl-boy should be consider the same outcome. So by #1, the two probabilities above, plus this one must have a sum of 1.00 (unless you think there is another possible outcome). What does that tell us about P(one of each)? That is must be .50 (1.00 - .25 - .25 = .50)

The answer here shows you that two boys, two girls and one of each are NOT equally likely. This is what has tripped all of the 50%'er up. If you see these as the only possibilities, that is fine. Just remember that these three are not equally likely and therefore cannot be assigned the same probabilities.

But wait you may say, I thought one of each would be .5 * .5 = .25! Well, that's right, but that's not what you have here. Here, you're saying the first event can be either a boy or a girl. By rule #3 P(boy or girl) = P(boy) + P(girl) = .5 * .5 = 1.00. Then, the second event must be different, ie, if A=boy, it must be that B=girl to have one of each. So P(second is different) = .5 (this works even if A=girl. Go ahead, try it, I'll wait). So now P(one of each) = P(either sex) * P(other sex) = 1.00 * .5 = .5

Let me put it another way. P(boy AND girl) = P(boy) * P(girl) = .5 * .5 = .25, and also P(girl AND boy) = P(girl) * P(boy) = .5 * .5 = .25. So this gives P(one of each) = P(boy AND girl) + P(girl AND boy) = .25 + .25 = .50 No matter how you address it, getting one of each sex is TWICE as likely as getting two boys or two girls. This is VERY crucial to understanding how the answer is arrived at in this problem. If you think there are three possibilities [bb, gg, 1 of each], you must agree that they are not equally likely. If you want all equally likely possibilities, you must use [bb, gg, bg, gb] as this is the only way to get the sum to be 1.00.

If you know that one of two children is a boy, what is the probability that the other is a girl?

Let's look at the three outcome situation first. P(boy AND boy) = .25, P(girl AND girl) = .25, P(one of each) = .50 and we want P(girl if we know one is boy). Only the first has a boy, and P(boy AND boy) + P(one of each) = .75, and only the third has a girl, that is .50 out of .75, which is .50/.75 = 2/3 = .66 so there is a 66.6666.....% chance that if you know that one of two children is a boy, the other is girl.

If you prefer the four outcome situation, P(boy AND boy) = .25, P(girl AND girl) = .25, P(boy AND girl) = .25, P(girl AND boy) = .25 and the only the first, third and fourth outcomes come into play. Since these are equally likely, (ie, same probability), and 2 of the 3 have a girl, the probabilities is 2/3 = .6666... or 66.6.....%.

The lesson is you must know if your outcomes are equally likely or not. Once the probabilities of your chosen outcomes are known, be they equally or unequally likely, the answer will be the same.

[/lecture mode]

Oh, and by the way, I have two daughters, and rarely when talking to other do I identify them as "my older daughter" or "my younger daughter". I usually just say "my daughter", not really caring whether the listeners knows I have another one or not, so that semantic argument holds no water what so ever.

ETA: Try this one on for size. You and your two friends, Tom and Harry, are captured in Lithistania by Geristal extremists. The terrorist will execute two of you at dawn, just to prove they are a real threat to the Lithistanian government. What is the probability you will be executed at dawn? (This part is easy) But later, you overhear the guards arguing about who has to take Harry to the execution room to be beheaded at dawn. Now, what is the probability that you will be executed at dawn?

 

[sirelfman]

[Lecture Mode]Big college style lecture I gave on probability

Did at least clear up the problem??

 

[DaVinci]

[Lecture Mode]

There are three basic rules of probability that are in play here. Now these rules apply to events that are independent, ie the outcome of one tells you nothing about the outcome of the other. Every seems to agree that the sexes of siblings are independent, so these two rules would apply in this case.

1. The sum of the probabilities of all possible events must equal 1.00 (100%), no more, no less. (One of the events must happen, therefore there is a 100% that at least one of them will happen.)

2. To find the probability of any two events, A and B, happening at the same time, multiply the probabilities together. Using standard probability notation, this is P(A and B) = P(A) * P(B).

3. To find the probability of one of several events happening, add the probabilities together. Again, using standard probability notation, P(A or B) = P(A) + P(B).

If you don't understand these 3 rules, I'll wait. Don't go on until you understand and agree with these three rules. If you understand them, but don't agree with them, don't read any further. Instead, go take a basic probability course. I can't tell you anything that will make you see the reality of this situation.

Now that you understand and argree with the three rules above, let's start. It seems most everybody agrees that P(boy)=.50 and P(girl)=.50. True, experimental probability shows this not to be the case, but it does simplify matters. (Girls are slightly more than 50% of the world's population).

What is the probability of two boys?

#2 says P(boy AND boy) = P(boy) * P(boy) = .5 * .5 = .25

By a similar calculation, I can show that P(girl AND girl) = .25. There seems to be no dispute about these two facts. So let's move onto another question.

What is the probability of a boy and a girl?

There some that argue that this boy-girl and girl-boy should be consider the same outcome. So by #1, the two probabilities above, plus this one must have a sum of 1.00 (unless you think there is another possible outcome). What does that tell us about P(one of each)? That is must be .50 (1.00 - .25 - .25 = .50)

The answer here shows you that two boys, two girls and one of each are NOT equally likely. This is what has tripped all of the 50%'er up. If you see these as the only possibilities, that is fine. Just remember that these three are not equally likely and therefore cannot be assigned the same probabilities.

But wait you may say, I thought one of each would be .5 * .5 = .25! Well, that's right, but that's not what you have here. Here, you're saying the first event can be either a boy or a girl. By rule #3 P(boy or girl) = P(boy) + P(girl) = .5 * .5 = 1.00. Then, the second event must be different, ie, if A=boy, it must be that B=girl to have one of each. So P(second is different) = .5 (this works even if A=girl. Go ahead, try it, I'll wait). So now P(one of each) = P(either sex) * P(other sex) = 1.00 * .5 = .5

Let me put it another way. P(boy AND girl) = P(boy) * P(girl) = .5 * .5 = .25, and also P(girl AND boy) = P(girl) * P(boy) = .5 * .5 = .25. So this gives P(one of each) = P(boy AND girl) + P(girl AND boy) = .25 + .25 = .50 No matter how you address it, getting one of each sex is TWICE as likely as getting two boys or two girls. This is VERY crucial to understanding how the answer is arrived at in this problem. If you think there are three possibilities [bb, gg, 1 of each], you must agree that they are not equally likely. If you want all equally likely possibilities, you must use [bb, gg, bg, gb] as this is the only way to get the sum to be 1.00.

If you know that one of two children is a boy, what is the probability that the other is a girl?

Let's look at the three outcome situation first. P(boy AND boy) = .25, P(girl AND girl) = .25, P(one of each) = .50 and we want P(girl if we know one is boy). Only the first has a boy, and P(boy AND boy) + P(one of each) = .75, and only the third has a girl, that is .50 out of .75, which is .50/.75 = 2/3 = .66 so there is a 66.6666.....% chance that if you know that one of two children is a boy, the other is girl.

If you prefer the four outcome situation, P(boy AND boy) = .25, P(girl AND girl) = .25, P(boy AND girl) = .25, P(girl AND boy) = .25 and the only the first, third and fourth outcomes come into play. Since these are equally likely, (ie, same probability), and 2 of the 3 have a girl, the probabilities is 2/3 = .6666... or 66.6.....%.

The lesson is you must know if your outcomes are equally likely or not. Once the probabilities of your chosen outcomes are known, be they equally or unequally likely, the answer will be the same.

[/lecture mode]

What's the probability that the younger child is actually Steve's?

 

[sirelfman]

[Lecture Mode]Scroll up a couple of posts and read it there

What's the probability that the younger child is actually Steve's?

Depends on what color hair the kid has.

 

[Shutterbug]

[Lecture Mode]Big college style lecture I gave on probability

Did at least clear up the problem??

I would say yes. That was impressive.ETA: I selected 2/3 anyway...

 

[biggamer3]

Wow the fact that 50% has almost 60% of the vote doesnt bode well for FBG

 

[TheAristocrat]

Wow the fact that 50% has almost 60% of the vote doesn't bode well for FBG

Is it really that surprising?

 

[FlapJacks]

Consider the population set of fathers with two kids A

Consider B the subset of A with at least one boy.

We can all agree that B is 75% of A

We can also agree that B consists of:

BB 25% of A

BG 25% of A

GB 25% of A

If we encounter a member of group B, and if that member starts talking about a child, of his we can see that the probablity that he talks about a boy is 2/3.

Now on to the OP:

We encounter a member of group B

2/3 of the time he mentions B, 1/3 of the time he mentions G

ASSUmPTION: If you encounter a member of group B and he mentions a son there is a 2/3 chance that his other kid is a daughter.

let's calculate C - the percentage of A that have two boys

If you are in C, then you are in the 2/3 that would mention the boy first.

Going by the ASSUmPTION above if you are in that 2/3, then there is 1/3 chance you are BB.

C = (2/3 * 1/3)B = 2/3 * 1/3 * 3/4 A = 1/6

1/6 <> 1/4, therefore the ASSUmPTION is proven false.

 

[Short Corner]

We encounter a member of group B2/3 of the time he mentions B, 1/3 of the time he mentions G

This has no effect on P(BB).eta---All you have done is change into problem 2 halfway through.

 

[FlapJacks]

I just modeled this on a spreadsheet and came up with 50%

 

[Shutterbug]

I just modeled this on a spreadsheet and came up with 50%

You are doing it wrong.

 

[TheAristocrat]

I just modeled this on a spreadsheet and came up with 50%

It funny how you and your spreadsheet are both wrong still.