A simple logic problem

[Geoff101]

I'm not actually good at math - so I'd take what I've typed with a grain of salt, it just makes the most sense to me that the gender of the first child is irrelevent to the question - and thus you have a simple set, either {B} or {G}. I can't see how there is a difference between the question the father asks and the question "I have one child, what is the probability it's a girl?"

But like I said, grain of salt.

celtic, you got it right, it would be 2/3 IF the question asked what are the odds the boy is the oldest....
But ORDER was never requested in the question: therefore the equation he is using is wrong. it is 50%

 

[billsgirl]

50%

 

[shdw01]

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?

I can actually see the reasoning behind both 2/3 and 1/2....the answer lies in the heart of the question....


as much as probability says 2/3 (if we know one of the two are a boy)...i think common sense has to outweigh here...

I will go on record as saying 50% (we know child A is a boy...regardless of older or younger)......if I'm wrong, it is because of how the question was interpreted by me.

 

[Geoff101]

exactly...if the wording PLACES the boy into the equation by asking What are the odds on the OLDEST/YOUNGEST being a boy...THEN it is 2/3....but the wording ELIMINATES the boy from the equation....

 

[MattRain]

I get it now. There's twice as many families with children of different gender than there are families with two boys (or two girls). 1/4 of 2-children families have two boys, 1/4 have two girls, and 2/4 have children of different gender. Remove the possibility of two girls and you're left with 2 out of 3 families having children of different gender.

I need a beer.

 

[Geoff101]

matt, you wife is having twins....the doctor tells he did a sonogram....one of the fetuses had a pee pee...what are the odds the othere one has a pee pee too?

 

[billsgirl]

OK I have a question, did the father buy a chicken with the intention of reselling it at any point in this thread?

 

[celticGreen+]

Using probability you'd get 50% too.

The only way you can use the set {BB GG BG GB} is if the ordering matters (for instance, if we were unlocking a lock then the order of the digits would most certainly matter). But because the order of the children (be it ordered by age, height, weight, etc) doesn't matter, you have only three options in the set.

Funny problem - I mean, how could oldest or youngest make a difference? It's unnecessary information put out there to make the reader think about age, and order. It'd be like me saying:

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 fattest child or the thinnest child. What is the probability that his other child is a girl?

 

[Geoff101]

OK I have a question, did the father buy a chicken with the intention of reselling it at any point in this thread?

bwahahaha....he sold the chicken to the boy, who sold it to the other sibling for $20...now, the question becomes, how do you divide the chicken equally amoung the father, son and the hemaphrodite, knowing the full chicken was 20 dollars and the chicken can only be cut into quarters.

 

[Geoff101]

Using probability you'd get 50% too.

The only way you can use the set {BB GG BG GB} is if the ordering matters (for instance, if we were unlocking a lock then the order of the digits would most certainly matter). But because the order of the children (be it ordered by age, height, weight, etc) doesn't matter, you have only three options in the set.

Funny problem - I mean, how could oldest or youngest make a difference? It's unnecessary information put out there to make the reader think about age, and order. It'd be like me saying:

It's 50%....

 

[MattRain]

matt, you wife is having twins....the doctor tells he did a sonogram....one of the fetuses had a pee pee...what are the odds the othere one has a pee pee too?

Following the same logic, my pair of twins is either part of the group that has two pee pees (1/4) or the group that has one pee pee and one non-pee pee, which is twice as populous (2/4, or 1/2) as either of the other possibilities. You remove the possibility of two non-pee pees, so there's 1 in 3 chances that my other kid has a pee pee, 2 in 3 that he does not.

 

[Geoff101]

ok, here's where the math goes wrong....it is not twice as populous...here's why:
you have a known B,
you have either B+G OR B+B....
you CANNOT have G+B or a G+G.....
so you only have 2 possibilities...you do not have 3 choices...
your equaltion has you choosing from B+G (valid) B+B(valid) AND a G+B (invalid, because B is a control and is already known)
so, given 2 choices, B+G and B+B...you have a 50/50 shot at being correct. BECAUSE YOU ALREADY KNOW ONE....

 

[celticGreen+]

Okay, so I'm going to give up and say I was wrong. Some more reading I found on a reputable site:

Suppose there are 100 fathers in an auditorium, and each is the father
of two children. Each father is instructed to tell you (truthfully)
if at least one of his children is a boy. This will apply to about
75 of the fathers. Now, of those 75 Dads, 2/3 (i.e., 50) have a
daughter, and 1/3 (i.e., 25) have two sons. Thus, if you want to
guess the gender of their "other" child, the chances are 2/3 that
it is a girl. (Of course, for the remaining 25 fathers - those
who did not report at least one son - you know immediately they
have two daughters.)

Sorry for adding my incorrect thoughts to the conversation. Zit you are correct and I was, fairly sure at this point, wrong.

 

[billsgirl]

My mother had 3 girls. After my older sister was born, it didn't matter that she was a girl. The odds start fresh after they try again. They got 3 girls. The guilt I carry for my father after all these years saddens me, but that's the way it works, %^_

If you draw out of an unlimited deck of cards and draw a red card, the odds that you draw a red again is 50%, the deck is unlimited.

 

[Geoff101]

celtic, that is a completely different scenario...

 

[billsgirl]

This reminds me of Roulette, if RED comes up 5 times in a row, do you bet black because you think the odds lean that way?

I don't even gamble.

 

[MattRain]

ok, here's where the math goes wrong....it is not twice as populous...here's why:
you have a known B,
you have either B+G OR B+B....
you CANNOT have G+B or a G+G.....
so you only have 2 possibilities...you do not have 3 choices...
your equaltion has you choosing from B+G (valid) B+B(valid) AND a G+B (invalid, because B is a control and is already known)
so, given 2 choices, B+G and B+B...you have a 50/50 shot at being correct. BECAUSE YOU ALREADY KNOW ONE....

It is twice as populous. Your chances of having two boys are 1/4. Same for your chances of having two girls. You're twice as likely to have children of different genders. That's where Zit's coin flip example comes in handy.

You had a boy, then you had a girl.
You had a girl, then you had a boy.
You had a boy, then you had another boy.
You had a girl, then you had another girl.

The first two statements unite to form 50% of all probabilities. Remove the last one and you're left with:

You had a boy, then you had a girl.
You had a girl, then you had a boy.
You had a boy, then you had another boy.

 

[LogansRun]

2/3

 

[billsgirl]

The odds start fresh after they try again.

If you draw out of an unlimited deck of cards and draw a red card, the odds that you draw a red again is 50%, the deck is unlimited.

Someone please tell me how my explanation isn't relevant here, as there is no trick question.

 

[LogansRun]

Forgot to say that this reminds me of the Monty Hall Paradox.