A simple logic problem

[JaBrady]

Out of those 99 dads, with 2 kids, ALL 99 dads having at least 1 boy....
You are honestly sitting here tellimg me that you thoroughly believe that the out of 99 second children, 66 will be girls and 33 will be boys?????

Given your mathematics....this world would be overrun with women...why isnt that the case?

OK...
Example A
99 dads have 2 kids = 198 kids total
99 you are told are boys (not oldest or youngest) = 99 boys
33 based on probability are boys = 33 boys
66 based on probability are girls = 66 girls

right?...

so according to what I have the world would be overrun with boys (132 out of 198) not girls (66 out of 198) as you stated right?...NO

why????...

because you have another set of 2 children homes that have 2 girls exclusively and the numbers based on my example would be exactly the same but swap girl for boy in every line...
EXAMPLE B
99 dads have 2 kids = 198 kids total
99 you are told are girls (not oldest or youngest) = 99 girls
33 based on probability are girls = 33 girls
66 based on probability are boys = 66 boys

so according to what I have the world would be overrun with girls (132 out of 198) not boys (66 out of 198)?...NO


EXAMPLE A amt of boys/girls is 132/66
EXAMPLE B amt of boys/girls is 66/132

TOTAL of all sets of families with 2 children home are:
boys = 198 (132+66)
girls = 198 (66+132)

So look at each example set (A and B) and you see that the probability is 2/3 but you look at the total and it is 50/50 when all things taken into account.

Those that continue to see 50/50 are thinking only as a TOTAL while the original questions is counter intuitive because it is asking for EXAMPLE A infromation.

does this make sense?

 

[festeringZit]

No, because of all dads that have two and only two kids, 25% of them are going to have two girls. This group doesn't include any of those dads, because all of these dads have at least one son. So since every other kid in the world is a male and every other one is a female, two thirds of these dads are going to have girls as their other child.

However, in my scenario that I just laid out.......it's still 50%. And it doesn't read any differently than the original question.

2/3 or 50% both are correct, I believe. And my mind is fucked and I'm done thinking about it.

No. The two problems are different.

The answer to "given that at least one child is male, what is the
probability the other child is a girl?" is 2/3

The answer to "given that this child I'm looking at right here
(e.g. Kobe Bryant) is male, what is the probability that the
other child is a girl?" is 1/2

 

[quantumleap]

because this series has a known outcome....

in other words:

if the boy was the oldest:

BG
BB

if the boy was the youngest:

BB
GB

there are no other possibilities....so, based on the information, the result would fall under the first OR second series...you cannot combine the 2

if it falls under the first, you have a 50/50 chance the other child is a girl
if it falls under the second, you have a 50/50 chance the child is a girl

everyone in this thread is wanting to COMBINE the two into one equation, you cant because you have a KNOWN RESULT ....

it's 50/50....

You're taking 2 outcomes and combining them into one. You have combined BG and BB if the boy is the oldest and you have combined BB and GB if the boy is the youngest. In probability, each outcome must be taken separately. There are 3 possibilities here: BG, GB and BB, not 2: (BG & BB) and (BB & GB).

Here are the 3 possibilities copied from my earlier post:

What are the only possibilities?

older girl, younger boy: fits
older boy, younger girl: fits
older boy, younger boy: doesn't fit

How many possibilities? 3
How many fit? 2
What is the probability? 2 out of 3

 

[Geoff101]

what if he tells you he has 4 children....3 are boys...what are the odds that the 'other' child is a girl? Show me the math...50/50 still....the known boys have no bearing or relevance to the outcome of the unknown child's sex....what if he said he had NO OTHER CHILDREN...only ONE...WHAT IS ARE THE ODDS? 50/50

what if he said he has 20 children...19 are boys, WHAT IS ARE THE ODDS ON THE 'OTHER' child being a girl...50/50

the fact that he stated he has one boy, has NO BEARING ON THIS QUESTION

 

[MinnesotaEconMajor]

what if he tells you he has 4 children....3 are boys...what are the odds that the 'other' child is a girl? Show me the math...50/50 still....the known boys have no bearing or relevance to the outcome of the unknown child's sex....what if he said he had NO OTHER CHILDREN...only ONE...WHAT IS ARE THE ODDS? 50/50

what if he said he has 20 children...19 are boys, WHAT IS ARE THE ODDS ON THE 'OTHER' child being a girl...50/50

the fact that he stated he has one boy, has NO BEARING ON THIS QUESTION

:money8:

 

[timotinbowen]

what if he said he has 20 children...19 are boys, WHAT IS ARE THE ODDS ON THE 'OTHER' child being a girl...50/50

Again, you're not being lied to. Knowing the gender of the other child IS VERY significant. And in your example, if you were told that a man has 20 children, and told that at least 19 of them are male, there is a 20 to 1 chance that the other child is a female. I know it is counterintuitive, but look at the math. There have been a half dozen proofs outlined in this thread. In your example, there are 20 ways to create a permutation with 19 B and 1 G, and only 1 way to have 20 B.

 

[quantumleap]

what if he tells you he has 4 children....3 are boys...what are the odds that the 'other' child is a girl? Show me the math...50/50 still....the known boys have no bearing or relevance to the outcome of the unknown child's sex....what if he said he had NO OTHER CHILDREN...only ONE...WHAT IS ARE THE ODDS? 50/50

what if he said he has 20 children...19 are boys, WHAT IS ARE THE ODDS ON THE 'OTHER' child being a girl...50/50

the fact that he stated he has one boy, has NO BEARING ON THIS QUESTION

This is where I believe you are not following all the possibilities. You are taking it as "the odds on the other child being a girl". We all agree that is 50%. But it's not just the "other child". It is also whether the boy is the oldest or the youngest that has to be taken into the total possibilities as well as stated in this situation.

 

[Geoff101]

Again, you're not being lied to. Knowing the gender of the other child IS VERY significant. And in your example, if you were told that a man has 20 children, and then given the names of 19 of them (all male names), there is a 20 to 1 chance that the other child is a female. I know it is counterintuitive, but look at the math. There have been a half dozen proofs outlined in this thread. In your example, there are 20 ways to create a permutation with 19 B and 1 G, and only 1 way to have 20 B.

So you're telling me that the odds on one unknown child's sex is predicated on HOW MANY CHILDREN ARE IN THE FAMILY????? You have lost your mind....

Show me the math as you have deltailed it for 2 children....so if someone has 10 kids...and 9 of them are boys...you would be willing to lay me 10-1 odds that the 10th child was a girl/boy????
You would go broke using this logic....

 

[Geoff101]

This is where I believe you are not following all the possibilities. You are taking it as "the odds on the other child being a girl". We all agree that is 50%. But it's not just the "other child". It is also whether the boy is the oldest or the youngest that has to be taken into the total possibilities as well as stated in this situation.

NO...the question says NOTHING about whether the boy was the oldest or not....THAT'S WHY IT'S 50/50

 

[timotinbowen]

there are NOT 3 possible out comes...there are 4 if you include the known in each set....
the first set puts the known in the first of the series....the second set puts the known in the second place in the series...it is still 50/50

BB
BG

is a possible outcome OR

GB
BB

is a possible outcome....all 4 have an even chance of happening

since BB is the same as BB in each, and GB is the same as BG in each...you have a 50/50 chance that the unknown child is a boy or a girl....

in closing, I leave my challenge open to anyone who wants to put this to practical use....we go out to an event, we survey people who have 2 children, one being a boy.....you pay me 2:1 every time the unknown child is a boy, and I will pay you 1 to 2 everytime it is a girl....

Your challenge has 0 EV. Do something where you think you are getting the best of it, and I think I am getting the best of it. I will give you 1.7:1. But I promise, you will lose. Ask any high school math teacher. This is simple set theory.

Remember, this is how we would ask the parents:
Do you have exactly 2 children?
If yes: Is at least 1 of them a boy?
If yes (We would bet on this next question): Is the gender of your other child a female?

If we asked this 3rd question to 1,000,000 couples, 2/3 of them would say yes.

 

[GoForBroke]

I can't believe there are still people that think it's 50% after everything has been explained. Also to think that it's irrelevant that we know one is a boy out of a set of two children is just silly :monsters-

 

[Geoff101]

Your challenge has 0 EV. Do something where you think you are getting the best of it, and I think I am getting the best of it. I will give you 1.7:1. But I promise, you will lose. Ask any high school math teacher. This is simple set theory.

Remember, this is how we would ask the parents:
Do you have exactly 2 children?
If yes: Is at least 1 of them a boy?
If yes (We would bet on this next question): Is the gender of your other child a female?

If we asked this 3rd question to 1,000,000 couples, 2/3 of them would say yes.

I would TAKE 1.7 to one....you will go broke using that scenario....why dont you give me the 2:1, true odds?

 

[Geoff101]

I can't believe there are still people that think it's 50% after everything has been explained. Also to think that it's irrelevant that we know one is a boy out of a set of two children is just silly :monsters-

it only becomes relevant if we ask, "What are the odds on the OLDEST child being a boy/girl?"...At that point the odds are 2/3...but that was not the question...the boy's position in the series was not asked for....50/50

 

[Geoff101]

Show me the math if he says he has 5 children, 4 are boys...I dont wanna know the sex of the oldest or the youngest etc....just what are the odds that the 'other' child is a boy or girl?

 

[timotinbowen]

it only becomes relevant if we ask, "What are the odds on the OLDEST child being a boy/girl?"...At that point the odds are 2/3...but that was not the question...the boy's position in the series was not asked for....50/50

This is wrong. Asking about a specific child would lead to a 50/50 proposition.

 

[timotinbowen]

Show me the math if he says he has 5 children, 4 are boys...I dont wanna know the sex of the oldest or the youngest etc....just what are the odds that the 'other' child is a boy or girl?

We've explained how to do these problems about 12 different ways. This thread is riddled with links. If we know someone has 5 children, and either 4 or 5 boys, the permutations are:

BBBBB
BBBBG
BBBGB
BBGBB
BGBBB
GBBBB

5/6 chance that the "other" child is female. We know that because the "other" child could fill any of the 5 spots in the birth order as a female, or the set could be all males.

Again, there are over 100,000 google hits for this problem. We understand that you do not believe the math. I posted this problem to be counterintuitive and get people thinking, but if you don't accept the reasoning by now you probably will not accept it. This problem has fooled many bright people because people often think terms of individuals, and not sets. In fact, this problem is often asked at the beginning of introductory set theory classes, to show the thinking required for such seemingly abstract problems.

 

[quantumleap]

NO...the question says NOTHING about whether the boy was the oldest or not....THAT'S WHY IT'S 50/50

That's the point! That's why the condition of being older or younger HAVE to be factored in.

Read the original problem:

"A man tells you he has two children. He then starts talking about his son. He does not tell you whether the son is the oldest child or the youngest child. What is the probability that his other child is a girl?"

The fact that 2 conditions are entered into the scenario shows that these 2 conditions have to be taken separately.

 

[Geoff101]

That's the point! That's why the condition of being older or younger HAVE to be factored in.

Read the original problem:

"A man tells you he has two children. He then starts talking about his son. He does not tell you whether the son is the oldest child or the youngest child. What is the probability that his other child is a girl?"

The fact that 2 conditions are entered into the scenario shows that these 2 conditions have to be taken separately.

The question doesnt ASK you whether the boy is the oldest or youngest...it is irrelevant to the solution....it just says that he has 2 kids one is a boy....then asks...of the other child...what are the odds that it is a girl.... 50/50

 

[quantumleap]

We've explained how to do these problems about 12 different ways. This thread is riddled with links. If we know someone has 5 children, and either 4 or 5 boys, the permutations are:

BBBBB
BBBBG
BBBGB
BBGBB
BGBBB
GBBBB

5/6 chance that the "other" child is female. We know that because the "other" child could fill any of the 5 spots in the birth order as a female, or the set could be all males.

Again, there are over 100,000 google hits for this problem. We understand that you do not believe the math. I posted this problem to be counterintuitive and get people thinking, but if you don't accept the reasoning by now you probably will not accept it. This problem has fooled many bright people because people often think terms of individuals, and not sets. In fact, this problem is often asked at the beginning of introductory set theory classes, to show the thinking required for such seemingly abstract problems.

I think the difference between the 2 opinions is to what the problem is asking. The 50 percenters are reading the problem as what is the possibility of the child being a girl once 1 child has been determined to be a boy. But the problem is not stating that. It is stating that 1 possibility is that one child is a boy. It also includes 3 other possibilities also based upon whether the boy is the oldest or the youngest as explicity stated in the problem.

 

[GoForBroke]

it only becomes relevant if we ask, "What are the odds on the OLDEST child being a boy/girl?"...At that point the odds are 2/3...but that was not the question...the boy's position in the series was not asked for....50/50

This couldn't be more wrong. Please read the question again and just think about it. You have to remember, we know it's a SET of children. We KNOW one is a boy. We DO NOT KNOW whether he is the oldest or youngest.

Older boy-younger girl
Older girl-younger boy
Older boy-younger boy
Older girl-younger girl

We can eliminate older girl-younger girl because we know one is a boy. Leaves 2/3 chance other child is a girl. Yes I know this has been repeated. I just think this is the easiest way to explain it.