A simple logic problem

[Kornholio]

Zit-



On the other hand, it's 2/3, if you are comparing it with all other 2 child families.

.

This part is irrelevant to the question.

 

[JakeT.]

I still say it's 50%. The 2nd child is either a boy or a girl and another unrelated child's sex has no bearing on that IMO. I'm not changing my answer.

 

[Illini]

This part is irrelevant to the question.

I do see what Zit's saying.

If you talk to 1 million dads that have two kids, and they all mention one son and nothing about the other child, you can bet your bottom dollar that 2/3 of the children not mentioned are going to be girls.

But on the other hand, if one guy out there tells you he has two kids, and then waves at the driver of a car that passes by and says "there's my son," the probability that his other child is a girl is 50%.

I think both answers are correct.

 

[festeringZit]

I do see what Zit's saying.

If you talk to 1 million dads that have two kids, and they all mention one son and nothing about the other child, you can bet your bottom dollar that 2/3 of the children not mentioned are going to be girls.

But on the other hand, if one guy out there tells you he has two kids, and then waves at the driver of a car that passes by and says "there's my son," the probability that his other child is a girl is 50%.

I think both answers are correct.

Illini, both those answers are correct, but they are not the same
problem.

 

[festeringZit]

I'm going to bed. Enough of this for one day. :toast:

 

[Illini]

Illini, both those answers are correct, but they are not the same
problem.

There is nothing in my scenario that is different than the original question.

 

[Geoff101]

Zit is right here.

I didn't see it till I broke it down in to coin tosses.


We know a coin was tossed twice. We know that heads came up at least once but we don't know on which toss. What are the odds that tails came up on the other toss.


HT
TH
HH
TT coulnd't have happened.

NO the coin was not tossed twice that's where everyone is getting it wrong....the coin is tossed ONCE....

If you toss the coin and it comes up heads, (girl) then you can not toss the second coin because it HAS to be tails (boy)...therefore you cannot flip two voins...only one...giving it a 50/50 probability...

 

[RobFunk]

Yo, prolly like 2 out of 3 times dat otha seed gunna be a gurl. I'm telln' you son.

:lolBIG:

 

[JakeT.]

Alright I'm going to go find two chicks to impregnate and I will let you guys know what the two children are in 9 months. It's the only way to know for sure.

 

[Kornholio]

Ok so if you take 1 million dads that all say, "hey this be my son, and my wife just gave birth...we found out it's a ????" It is truly 50/50, 50% of those will be girls. If it doesn't work out like this then life has turned upside down on itself because it is the only option.

 

[Illini]

Ok so if you take 1 million dads that all say, "hey this be my son, and my wife just gave birth...we found out it's a ????" It is truly 50/50, 50% of those will be girls. If it doesn't work out like this then life has turned upside down on itself because it is the only option.

No, it's like this.

All of the dads already have 2 and only 2 kids. And every dad mentions one son and nothing about the other child.

This means that none of these dads have two girls. They all have at least one son.

So GG is out for all of them.
The remaining combos are BG, GB, and BB. 2/3 of the unspoken children are going to be girls.

However, I think 50% is also correct, because the question posed could read exactly like this:
A guy tells you that he has two children and then waves at the driver of a car passing by saying, "there's my son."

In this case, the probability that his other child is a girl is 50%, and this doesn't read any different than the original post.

 

[timotinbowen]

Guys, I wouldn't have posted this if I didn't know the answer. I promise you, the answer is 2/3. It is not due to tricky wording. You do not need to have an understanding of genetics to get the answer. I just thought it would be fun to get some answers to a seemingly counterintuitive promlem.

Geoff, you are completely wrong and should not bet (and $200 to win $300 is not the proper amounts for a 2:1 bet anyways, but that's beside the point). This problem revolves around set theory. THE QUESTION IS NOT ASKING WHAT THE GENDER OF THE OLDEST CHILD IS. It is asking what the second child's gender is. It is VERY relevant to know that one of the children's gender's is male.

Here's how to disprove the 50% answer.

1) What is the probability of having at least one boy?
Answer: 75%

2) What is the probability of having exactly one boy and one girl?
Answer: 50%

Given one boy, what is the probability of the other child being a girl?

THIS IS THE ORIGINAL QUESTION, and to answer it we know
.5 (Which is the prob. of having one of each) = .75 (which is the prob. of having at least one boy) * Probability of having a girl under these circumstances

when you solve this equation, you get the answer is 2/3.



This is not the same as seeing red on a roulette wheel, and then being asked what the next spin will be! This is the same as being told that out of two spins, at least one of them is red. What is the probability that the other is black? The answer here would also be 2/3.


Look, I know the problem is counterintuitive, but it is a commonly known problem and the answer is not in question. FesteringZit explained it well, and the people ridiculing him should know that he is correct.

Finally, if you don't believe any of the explanations, do this simple experiment (should take about 20 minutes).

Get two coins. Toss them, and only record the data if at least one is heads. When at least one lands heads, record the number of times that you get one Tails. Do this until you have 300 trials. You should see about 200 instances where there was a Tails, and 100 instances where there was two Heads. Please do this experiment if you still don't believe the explanations!

 

[Illini]

A guy is walking down the street with his dog. The dog reminds you of your childhood dog, so you strike up a conversation with the guy. You ask him is he has any children and he says "Actually, I have two. And there's my son right there in that Ford F-150! I don't think he saw me, though."

What is the probability that his other child is a girl?

Now, even Festering Zit has said that my scenario is 50%. Now this scenario is no different than the original scenario.

 

[Geoff101]

Guys, I wouldn't have posted this if I didn't know the answer. I promise you, the answer is 2/3. It is not due to tricky wording. You do not need to have an understanding of genetics to get the answer. I just thought it would be fun to get some answers to a seemingly counterintuitive promlem.

Geoff, you are completely wrong and should not bet (and $200 to win $300 is not the proper amounts for a 2:1 bet anyways, but that's beside the point). This problem revolves around set theory. THE QUESTION IS NOT ASKING WHAT THE GENDER OF THE OLDEST CHILD IS. It is asking what the second child's gender is. It is VERY relevant to know that one of the children's gender's is male.

Here's how to disprove the 50% answer.

1) What is the probability of having at least one boy?
Answer: 75%

wrong the answer is 100%

2) What is the probability of having exactly one boy and one girl?
Answer: 50%

Given one boy, what is the probability of the other child being a girl?

THIS IS THE ORIGINAL QUESTION, and to answer it we know
.5 (Which is the prob. of having one of each) = .75 (which is the prob. of having at least one boy) * Probability of having a girl under these circumstances

when you solve this equation, you get the answer is 2/3.



This is not the same as seeing red on a roulette wheel, and then being asked what the next spin will be! This is the same as being told that out of two spins, at least one of them is red. What is the probability that the other is black? The answer here would also be 2/3.


Look, I know the problem is counterintuitive, but it is a commonly known problem and the answer is not in question. FesteringZit explained it well, and the people ridiculing him should know that he is correct.

Finally, if you don't believe any of the explanations, do this simple experiment (should take about 20 minutes).

Get two coins. Toss them, and only record the data if at least one is heads. When at least one lands heads, record the number of times that you get one Tails. Do this until you have 300 trials. You should see about 200 instances where there was a Tails, and 100 instances where there was two Heads. Please do this experiment if you still don't believe the explanations!

ugh..here's where we disprove your theory....you dont flip coins til it turns up heads...you are only flipping ONE coin...it is given to you that it IS already heads...so the variable is in the second coin, the first is a constant (boy)

 

[Kornholio]

No, it's like this.

All of the dads already have 2 and only 2 kids. And every dad mentions one son and nothing about the other child.

This means that none of these dads have two girls. They all have at least one son.

So GG is out for all of them.
The remaining combos are BG, GB, and BB. 2/3 of the unspoken children are going to be girls.

However, I think 50% is also correct, because the question posed could read exactly like this:
A guy tells you that he has two children and then waves at the driver of a car passing by saying, "there's my son."

In this case, the probability that his other child is a girl is 50%, and this doesn't read any different than the original post.

I disagree. It is either BB or BG (GB means nothing unless you add another BB). Two options 50%.

 

[Kornholio]

ugh..here's where we disprove your theory....you dont flip coins til it turns up heads...you are only flipping ONE coin...it is given to you that it IS already heads...so the variable is in the second coin, the first is a constant (boy)

Correct

Take 2 coins.

One is heads (boy)...leave it there.

Flip the second coin 1000 times. avg. 500 heads(boy)/ 500 tails(girl).

Odds of girl 50%.

 

[Illini]

I disagree. It is either BB or BG (GB means nothing unless you add another BB). Two options 50%.

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.

 

[Geoff101]

ya know i offered to take 2/3 odds because that is what zit tells us his theory equates to and he wont take me up on it...and now you're telling me the odds are 2/3 but pays 2:1, FINE i'll take 2-1...lets go get a sample of people with 2 kids, one being a boy....for every non-boy child that is a girl, i will gladly take 2 dollars and for every 2nd boy in the family i will pay you one dollar....you will be broke by the time i finish....this is because we are only sampling families WITH A KNOWN MALE AND ONE UNKNOWN GENDER CHILD

 

[timotinbowen]

ya know i offered to take 2/3 odds because that is what zit tells us his theory equates to and he wont take me up on it...and now you're telling me the odds are 2/3 but pays 2:1, FINE i'll take 2-1...lets go get a sample of people with 2 kids, one being a boy....for every non-boy child that is a girl, i will gladly take 2 dollars and for every 2nd boy in the family i will pay you one dollar....you will be broke by the time i finish....this is because we are only sampling families WITH A KNOWN MALE AND ONE UNKNOWN GENDER CHILD

2/3 is not an expression of odds. If you did it your way, (300 win, 200 loss) and the and there was really a 2/3 chance of him winning, you would win 300 for every 400 you lose.

The odds of an event with probability 2/3 occurring is 2:1, and must be paid accordingly

 

[Illini]

2/3 probability is the same as 2 to 1.