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Logic problem (1 Viewer)

Choose the answer

  • 2/3 (66.66%)

    Votes: 37 27.8%
  • 50 %

    Votes: 90 67.7%
  • No idea

    Votes: 6 4.5%

  • Total voters
    133

biggamer3

Footballguy
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 would like to add the term his "son" does not mean the other is not a son. Its a tough question to word but its not a play on words, just a logic question)

 
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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?
Population statics show there are slightly more likely of a chance that he has another boy.
 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:

boy-boy

boy-girl

girl-boy

girl-girl

We know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.

 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.
What if the question was phrased as:A man and his son walk up to you and say hello. The man says he has another child at home. What are the odds the other child is a girl?Wouldn't it be 50/50 in that case?
 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.
What if the question was phrased as:A man and his son walk up to you and say hello. The man says he has another child at home. What are the odds the other child is a girl?Wouldn't it be 50/50 in that case?
No, it's still 2/3. If the guy told you that his younger child was at home, then it would be 50/50.
 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.
I know this has been done before and this is the "right" answer, but I'm still not sure I agree with it. 50% still seems more logical to me.
 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.
What if the question was phrased as:A man and his son walk up to you and say hello. The man says he has another child at home. What are the odds the other child is a girl?Wouldn't it be 50/50 in that case?
I concur. The sex of a child should be independent of all other children.
 
99%. He talked about his son. If the other was also a son, he would of said my eldest son or my youngest son. I left the 1% open because we might be dealing with an idiot.

 
This example is a old chestnut that math teachers like to use when introducing conditional probabilities. I think I've probably seen this exact hypothetical at least three times in various sources.

 
This question is a very tough one and there seems to be a large split on the 50% and 2/3 whenever its asked.

FWIW i believe Wikipedia says 2/3 is the correct answer:

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

* A random two-child family with at least one boy is chosen. What is the probability that it has a girl?

An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"

Neither order nor age is important. There are four possible child combinations for a two-child family as seen in the sample space above. Three of these families meet the criteria of having at least one boy. The set of possibilities (possible combinations of children that meet the given criteria) is:

Older child Younger child

Girl Girl

Girl Boy

Boy Girl

Boy Boy

[edit] Bayesian approach

Consider the sample space of 2-child families.

* Let X be the event that the family has one boy and one girl.

* Let Y be the event that the family has at least one boy.

* Then:

o P( X | Y ) = \frac{ P( X \cap Y ) } { P( Y ) } = \frac{ \frac{2} {4} } { \frac{3} {4} } = \frac{2}{3}

Or, the set {GB, BG, BB}, in which two out of the three possibilities includes a girl.

Therefore the probability is 2/3.
 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.
The answer is 1/2. If a guy has two children, there are six equally-likely possibilities for their sequence of birth and genders:boy A-boy Bboy B-boy Aboy-girlgirl-boygirl A-girl Bgirl B-girl AWe know the last two didn't happen, but the remaining four are all still equally likely, so 1/2 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men. :mellow:
 
This question is a very tough one and there seems to be a large split on the 50% and 2/3 whenever its asked.

FWIW i believe Wikipedia says 2/3 is the correct answer:

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

* A random two-child family with at least one boy is chosen. What is the probability that it has a girl?

An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"

Neither order nor age is important. There are four possible child combinations for a two-child family as seen in the sample space above. Three of these families meet the criteria of having at least one boy. The set of possibilities (possible combinations of children that meet the given criteria) is:

Older child Younger child

Girl Girl

Girl Boy

Boy Girl

Boy Boy

[edit] Bayesian approach

Consider the sample space of 2-child families.

* Let X be the event that the family has one boy and one girl.

* Let Y be the event that the family has at least one boy.

* Then:

o P( X | Y ) = \frac{ P( X \cap Y ) } { P( Y ) } = \frac{ \frac{2} {4} } { \frac{3} {4} } = \frac{2}{3}

Or, the set {GB, BG, BB}, in which two out of the three possibilities includes a girl.

Therefore the probability is 2/3.
I really don't think this is the same situation given in the original post.
 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.
What if the question was phrased as:A man and his son walk up to you and say hello. The man says he has another child at home. What are the odds the other child is a girl?Wouldn't it be 50/50 in that case?
No, it's still 2/3. If the guy told you that his younger child was at home, then it would be 50/50.
I still think once you know that you are talking about one child, the odds are 50/50. Nothing else matters.What if the guy had 100 children and 99 of them were boys (and there is no tricky stuff about he can only have boys - it just happened to work out that way). The odds of the last child being a girl is still 50/50. You don't care about the previous trials.
 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.
I know this has been done before and this is the "right" answer, but I'm still not sure I agree with it. 50% still seems more logical to me.
I agree with Ivan, it's 2/3. I've used random numbers in Excel to set up trials for this. It always works out to be around 2/3.Yet, 50% just "feels" right.
 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.
What if the question was phrased as:A man and his son walk up to you and say hello. The man says he has another child at home. What are the odds the other child is a girl?Wouldn't it be 50/50 in that case?
No, it's still 2/3. If the guy told you that his younger child was at home, then it would be 50/50.
I still think once you know that you are talking about one child, the odds are 50/50. Nothing else matters.What if the guy had 100 children and 99 of them were boys (and there is no tricky stuff about he can only have boys - it just happened to work out that way). The odds of the last child being a girl is still 50/50. You don't care about the previous trials.
I'm with kutta. Each child is an independent event.Isn't this like asking the following: "a man flips a coin twice, one of the results is heads, what are the odds that the other flip resulted in tails?"It doesn't matter what the other result was, any given toss of the coin yields a 50% chance of tails. Same with the sex of a child.
 
"I have two kids. My son just went off to Harvard..."

C'mon. His other kid is a daughter.
This is not a play on words, the question doesnt intend to say he talks about "son" so the other must be a "daughter" at all
I don't see a play on words.
A man tells you he has two children. He then starts talking about his son.
Perhaps this problem should be rephrased.
Tell me how i should rephrase it exactly?
 
The only way it can be 2/3's is if you include hermaphrodite as an acceptable answer.

His other child is either a boy or girl. You can spin the rest any other way but the odds are 50/50. Insert coin flip analogy.

 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.
This is for before they were born. You have to use current info, we KNOW he has a son.He has a son. So choices are:Son-SonSon-Daughter50/50
 
"I have two kids. My son just went off to Harvard..."

C'mon. His other kid is a daughter.
This is not a play on words, the question doesnt intend to say he talks about "son" so the other must be a "daughter" at all
I don't see a play on words.
A man tells you he has two children. He then starts talking about his son.
Perhaps this problem should be rephrased.
Tell me how i should rephrase it exactly?
This is kind of wordy, but maybe more ideal as far as laying out the problem clearly:
A man tells someone he has two children. He then starts talking about one of his two children without mentioning that the child he is talking about is his son. He does not say whether the child is the oldest or the youngest. What is the probability that his other child is a girl?
 
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This example is a old chestnut that math teachers like to use when introducing conditional probabilities. I think I've probably seen this exact hypothetical at least three times in various sources.
It is another variation of the Monte Hall problem. There was a long thread on old yeller about this. 2/3 is the correct answer.
 
Here are the coin flip analogies depending on how the question is asked/interpreted:

You flip a pair of coins 100 times. How many sets are head/head if you only use the sets with at least one head in them - 66.66%

You flip a pair of coins and look at one of them and its a "head". Whats the odds the other one is a "head" - 50.00%

 
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"I have two kids. My son just went off to Harvard..."

C'mon. His other kid is a daughter.
This is not a play on words, the question doesnt intend to say he talks about "son" so the other must be a "daughter" at all
I don't see a play on words.
A man tells you he has two children. He then starts talking about his son.
Perhaps this problem should be rephrased.
Tell me how i should rephrase it exactly?
This is kind of wordy, but maybe more ideal as far as laying out the problem clearly:
A man tells someone he has two children. He then starts talking about one of his two children without mentioning that the child he is talking about is his son. He does not say whether the child is the oldest or the youngest. What is the probability that his other child is a girl?
As you can see this is very wordy.Your rephrasing causes the same "son" problem

 
Here are the coin flip analogies depending on how the question is asked/interpreted:

You flip a pair of coins 100 times. How many sets have a "head" in them if you only use the sets with at least one head in them - 66.66%

You flip a pair of coins and look at one of them and its a "head". Whats the odds the other one is a "head" - 50.00%
:hophead:
 
This example is a old chestnut that math teachers like to use when introducing conditional probabilities. I think I've probably seen this exact hypothetical at least three times in various sources.
It is another variation of the Monte Hall problem. There was a long thread on old yeller about this. 2/3 is the correct answer.
I use this in Critical Thinking courses all the time.The kids lose their minds.

Monty Hall

 
The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.
What if the question was phrased as:A man and his son walk up to you and say hello. The man says he has another child at home. What are the odds the other child is a girl?Wouldn't it be 50/50 in that case?
No, it's still 2/3. If the guy told you that his younger child was at home, then it would be 50/50.
Why do we care about the sequence?
 
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?
You must read the same blogs I do.
 
I set up two columns of one thousand rows in Excel, with "BOY" and "GIRL" randomly assigned to each cell. Then I eliminated any rows that had "GIRL" in both cells. I was left with 751 rows. 265 rows (35.3%) had "BOY" and "BOY", 486 rows (64.7%) had "BOY" and "GIRL".

Try it yourself. I didn't believe it myself until I did it.

 
"I have two kids. My son just went off to Harvard..."

C'mon. His other kid is a daughter.
This is not a play on words, the question doesnt intend to say he talks about "son" so the other must be a "daughter" at all
I don't see a play on words.
A man tells you he has two children. He then starts talking about his son.
Perhaps this problem should be rephrased.
Tell me how i should rephrase it exactly?
This is kind of wordy, but maybe more ideal as far as laying out the problem clearly:
A man tells someone he has two children. He then starts talking about one of his two children without mentioning that the child he is talking about is his son. He does not say whether the child is the oldest or the youngest. What is the probability that his other child is a girl?
As you can see this is very wordy.Your rephrasing causes the same "son" problem
But in this case, he's telling someone else, not "you." That someone else does not know that the man is talking about his son, but we, the problem-solvers, do.
 
Isn't this like asking the following: "a man flips a coin twice, one of the results is heads, what are the odds that the other flip resulted in tails?"
Yes, although your example is an even better one at coming up with 66%. If you know a man flipped a coin twice (i.e. it's already happened), and he tells you that one result was heads, the chances of the other being tails is 66% (per Ivan's example of four choices).When dealing with boys and girls, some statistical accountability should likely reside in the odds between having boys and girls. For example, if you have two kids, and the question involves children with a rare disease (say 1 in 10,000), if you say the first child does not have that rare disease, it's not 66% chance that the other does. But the idea in the OP example is taking for granted that there is an even likelihood in having boys or girls. There isn't, although it's close enough where I would say the right answer is a lot closer to 66% than 50%.
 
Here are the coin flip analogies depending on how the question is asked/interpreted:

You flip a pair of coins 100 times. How many sets have a "head" in them if you only use the sets with at least one head in them - 66.66%

You flip a pair of coins and look at one of them and its a "head". Whats the odds the other one is a "head" - 50.00%
Incorrect. As I indicated, this is an even better example than the boys/girls one of 66%.Again, the results are complete before you looked at one of them. The possible results before looking is:

1. heads/heads

2. heads/tails

3. tails/tails

4. tails/heads

After looking at one of them, you know that option three could not have happened. With the other available three options, tails as the opposite probability to heads happens 66% of the time.

 
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?
You must read the same blogs I do.
Nice post, never read that blog, got the question from another poster from a different forum.2/3 is the answer and we have a majority 50% here in the poll so far!

 
Isn't this like asking the following: "a man flips a coin twice, one of the results is heads, what are the odds that the other flip resulted in tails?"
Yes, although your example is an even better one at coming up with 66%. If you know a man flipped a coin twice (i.e. it's already happened), and he tells you that one result was heads, the chances of the other being tails is 66% (per Ivan's example of four choices).When dealing with boys and girls, some statistical accountability should likely reside in the odds between having boys and girls. For example, if you have two kids, and the question involves children with a rare disease (say 1 in 10,000), if you say the first child does not have that rare disease, it's not 66% chance that the other does. But the idea in the OP example is taking for granted that there is an even likelihood in having boys or girls. There isn't, although it's close enough where I would say the right answer is a lot closer to 66% than 50%.
How about if he has 100 kids and 99 of them are boys?
 
December 31, 2008Finishing The GameIn yesterday's post, I asked this question: Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl? Most people answer 50%.Unfortunately, this isn't correct.This problem, although seemingly simple, is hard to understand. For cognitive reasons that are not fully understood, while our intuitions regarding a priori possibilities are fairly good, we are easily misled when we try to use probability to quantify our knowledge. This is a fancypants way of saying there were almost a thousand comments on that post, with not a lot of agreement to be found.The key thing to bear in mind here is that we have been given additional information. If we don't use that information, we arrive at 50% -- the odds of a girl or boy being born to any given pregnant woman. That's true insofar as it goes, but it's the answer to a different, much simpler question, and certainly not the answer to the question we asked.Our question contains additional information: 1. The person has two children. 2. One of those children is a girl. We can use that information to come up with a better, more correct answer. We know this person has two children. What are all possible combinations of two children?BB, GB, BG, GGtwo children possibility matrix: BB, GB, BG, GGWe know that one of the children is a girl. This rules out one of those possible combinations of two children (BB), so we're left with:GB, BG, GGtwo children possibility matrix, has a girl: GB, BG, GGOf the remaining three possibilities, two include boys.GB, BGThus, the odds of this person having a boy and a girl is 2/3 or 66%.I noticed a few comments where people complained that the GB and BG possibilities are the same thing, and should have been reduced toBG/GB, GGWhich equates to 1/2 or 50%.
 
Order does not matter. I know he had two kids and one is a male. The answer is..........

M x

M x

Possiblities

are only

M F

M M

When he had who does not matter. The logic you are using for 2/3 is flawed.

 

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