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# One Girl - One Boy

Best Answer Riddari , 13 July 2007 - 07:38 PM

Spoiler for solution
Go to the full post

348 replies to this topic

### #311 itsclueless

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Posted 10 August 2008 - 09:50 PM

Why it can't possibly be 1/3:

There are two ways of looking at this problem:

Method 1: You are looking at the first child, and only the first child.

Following your second spoiler:

Girl - Girl
Girl - Boy
Boy - Girl
Boy - Boy

You are looking at the first child, and it is a girl, so the third case is invalid as well as the fourth, leaving the first two choices, 50% each boy and girl. This has been mentioned several times, and is at the root of the "Gambler's fallacy" and other people's explanations in which the first event is already decided and therefore the second is independent, and therefore a probability of 1/2.

Girl - Girl
Girl - Boy
Boy - Girl
Boy - Boy

Method 2: You are looking at any one of the children in the original spoiler diagram.

Girl - Girl
Girl - Boy
Boy - Girl
Boy - Boy

This is how the author of the problem explained his 1/3 solution, which is incorrect. Take a look at the diagram: by the author's logic, you can pick a girl from the left or the right. This invalidates picking any of the boys, not just the last possibility, although of course that one is completely invalidated, while the middle 2 are only halfway invalidated.

The possibilities now look like this:

Girl - Girl
Girl - Boy
Boy - Girl
Boy - Boy

Knowing that you can pick the left or the right child, there are 4 girls to pick from! If you pick either of the girls from the first choice, you have another girl child. If you pick either of the girls from the middle two rows, you have a second boy child. 50-50%, 1/2 probability, not 1/3.

In other words, while by specifying that the first choice is not a boy you have completely eliminated one outcome, you have also made it half as likely to pick either of the two middle solutions.
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### #312 bonanova

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Posted 11 August 2008 - 07:56 AM

There are two ways of looking at this problem:

Method 1: You are looking at the first child, and only the first child.
You are looking at the first child, and it is a girl

Method 2: You are looking at any one of the children in the original spoiler diagram.
you can pick a girl from the left or the right.

Construct a truth table of the premise "One child is a girl" for the 4 cases BB/BG/GB/GG.
What you find is that "One is a girl" is false for BB and true otherwise.
"One is a girl" is logically equivalent to "they are not both boys".

Included among the things that premise does not do is to assign the gender of a selected child.
Both of your methods scrutinize individual children, which ends up trivializing the OP to:
"The probability that a child is a girl is 1/2. What is the probability that child X is a girl?"

Because "one is a girl" means only "not both are boys" we're told [only] about gender combinations.
And we're asked about combinations. Specifically, the case "both are girls." [one + the other = both]

The expectation for 100 two-child families is: 25 have 2 girls; 25 have 2 boys; 50 are mixed gender.
Removing "both are boys" shrinks the sample space to 75 -- 25 with two girls, and 50 with mixed gender.

These probabilities are 1/3 and 2/3, respectively.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #313 itsclueless

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Posted 11 August 2008 - 08:13 AM

Construct a truth table of the premise "One child is a girl" for the 4 cases BB/BG/GB/GG.
What you find is that "One is a girl" is false for BB and true otherwise.
"One is a girl" is logically equivalent to "they are not both boys".

Included among the things that premise does not do is to assign the gender of a selected child.
Both of your methods scrutinize individual children, which ends up trivializing the OP to:
"The probability that a child is a girl is 1/2. What is the probability that child X is a girl?"

Because "one is a girl" means only "not both are boys" we're told [only] about gender combinations.
And we're asked about combinations. Specifically, the case "both are girls." [one + the other = both]

The expectation for 100 two-child families is: 25 have 2 girls; 25 have 2 boys; 50 are mixed gender.
Removing "both are boys" shrinks the sample space to 75 -- 25 with two girls, and 50 with mixed gender.

These probabilities are 1/3 and 2/3, respectively.

I see what you are saying. The problem states that "one of them is a girl". You interpret that to mean that if either child is a girl, you are told about it, and left to guess about the other one. With three cases of equal probability, and if you are guaranteed to pick the girl of the family if there is one, then I agree. If, however, you choose a random child and it is a girl, I still maintain that the probability that the second child will also be a girl is 1/2 as shown above.

This reminds me of a two-sided card puzzle which I will post right away.
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### #314 Scraff

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Posted 11 August 2008 - 03:59 PM

Yes. By changing the statement to "Given that one of them is a girl" adds structure to the information. The information is structured in such a way that we will always be given that their is a girl.

You're making no sense. "Given that" does nothing to the statement "One of them is a girl". "We will always be given that their is a girl" either way.

However if the information being supplied is as random as the couple then half the time when the have a GB/BG mix we will be told that one of them is a boy. Why because their is no structure to it.

What? We don't know what we would be told half the time. You've even admitted that we don't know why we were supplied the information. All we know is that there is a particular couple that has two children and that one of them is a girl.

Yes and no. If you walk up to a guy on the street and he says.
"Hi my name is Teanchi and this is my wife Beanchi. We have two kids, one of them is a girl" (aka the riddle)

No, that's not the "aka the riddle". Stop making things up already! No representative from the couple has made any statements.

Then your odds of the other one being a girl is best guess 50%. Why? Because in a GB/BG mix it is just as likely we will be informed that one of them is a boy. Their is no structure to the information.

See above. We know nothing about the likelihood of receiving any information in the riddle. All we know is that the couple has two children and one of them is a girl. We know NOTHING about how the information was obtained or why it was obtained.

But if you walk up to a guy and he says.
"Hi my name is Teanchi and this is my wife Beanchi. We have two kids."
"Is at least one of them a girl"
"Yes"

Then the odds are 1/3 because their is structure to the information.

Good. Then you should have to agree that 1/3 is the correct answer in this riddle also as we no nothing about how the information is obtained.

"In a two-child family, one child is a boy. What is the probability that the other child is a girl?"
Why is 2/3 the correct answer in that case and 1/3 not in the case of the OP's riddle?

Because "In a two-child family, one child is a boy" is a statement (however false) that sets the guidelines for the question. It says that in any given two-child family, one child is a boy.

No, it doesn't. "In a two-child family, one child is a boy" is talking about one family. It's not a statement "however false" about all two child families.

Even in your made up example where Teanchi makes a statement, we have no reason to assume all two-child families have one boy.

Kind of like saying "In a bathroom, their is toilet paper". Their a plenty of bathrooms out their without toilet paper (I hate it when that happens) but the initial statement sets the condition that all bathrooms have toilet paper. Now I know what you are going to say. "how is the statement in the riddle any different". Well because the statement riddle pertains only to Teanchi and Beanchi ("they have two kids"). So it more like saying "In your bathroom, their is toilet paper. What is the probability that I have toilet paper in mine".

Teanchi and Beanchi have toilet paper in their bathroom. What is the probability that they have toilet paper in their bathroom?

A two child family has a boy. What is the probability that they have a boy?

It's the same. It doesn't make any difference whatsoever that the couple has been named.

Your example above is equivalent to saying the following.:

Kind of like saying "In a family with two children, one is a boy". Their a plenty of two-child families out their without a boy (I hate it when that happens) but the initial statement sets the condition that all two-children families have one boy. Now I know what you are going to say. "how is the statement in the riddle any different". Well because the statement riddle pertains only to Teanchi and Beanchi ("they have two kids"). So it more like saying "in your two-child family there is a boy. What is the probability that my two-child family has a boy".

If that were how the riddle were framed, the answer would be 1/1 because your version already stated "the condition that all two-children families have one boy". But even if the riddle did state that all two-child families have one boy, the answer would still be 1/3 because the riddle asks what is the probability the other child is a boy (actually, we're dealing with girls in the riddle). It doesn't ask what the chance of having a boy is. Again, if it did and it were framed for us to believe that all two-child couples have one boy (and the riddle is not framed that way) and asked what the probability of having one boy were, the answer would be 1/1.

I see what you are saying. The problem states that "one of them is a girl". You interpret that to mean that if either child is a girl, you are told about it, and left to guess about the other one.

There's no interpreting how anything was told to us; we just deal with what we were told. With the information that was give, the answer is 1/3.

With three cases of equal probability, and if you are guaranteed to pick the girl of the family if there is one, then I agree. If, however, you choose a random child and it is a girl, I still maintain that the probability that the second child will also be a girl is 1/2 as shown above.

We've all agreed with this throughout the thread. But the riddle is not about "the second child".
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### #315 bonanova

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Posted 11 August 2008 - 08:54 PM

I see what you are saying. The problem states that "one of them is a girl". You interpret that to mean that if either child is a girl, you are told about it, and left to guess about the other one. With three cases of equal probability, and if you are guaranteed to pick the girl of the family if there is one, then I agree. If, however, you choose a random child and it is a girl, I still maintain that the probability that the second child will also be a girl is 1/2 as shown above.

This reminds me of a two-sided card puzzle which I will post right away.

Well put.

Is this your card puzzle?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #316 ShawnInToronto

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Posted 15 August 2008 - 03:27 AM

Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

Assume safely that the porbability of each gender is 1/2.

Spoiler for solution

Spoiler for solution

Your spolier logic regarding sample space is faulty. The sample space you present are the combinations possible before any children are born. Once one child is known to be a girl, the sample space for a family of two becomes:

Girl - Girl
Girl - Boy

And the chance that the second child is a boy is 50%. Probability only applies to unknowns, not things already known. To argue otherwise, you would have to explain how the birth of one childs affects the gender of the next.
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### #317 Martini

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Posted 15 August 2008 - 03:37 AM

ShawnInToronto, please read more of the posts in this thread. What you've described has been explained ad nauseum. The riddle does not ask about the probability of "the second child". We don't know if a boy was born first or not, so there are three equal possibilities:

GG
GB
BG

See the following links for further information.

http://mathforum.org...q.boy.girl.html
http://www.mathpages...me/kmath036.htm

The rationale behind the OP's answer has been explained more than enough.
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### #318 syonidv

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Posted 10 July 2012 - 01:05 AM

Let mine be one more vote for itsclueless's correct (also, rational ) answer. The probability of the "other child" being a girl is 1/2.

The answer offered by brainden.com is incorrect, and even the cusory consideration that observing one sibling changes the probability of the other's gender should set off red flags in people's minds.

The proof that I give to people is as follows:

The two facts we know are:

two children have been born (fact A)
at least child is female (fact B)

Fact A gives us the four possible populations (male/female configurations):

M-M
M-F
F-M
F-F

It then also seems logical to eliminate any combination where at least one female is not present (i.e. M-M), leaving,

M-F
F-M
F-F

and then concluding a 67% probability of 2M and 1F. The problem is that these three remaining combinations are not equally likely. That is, we've botched the application of Bayes' rule.

The correct approach is to start with as our four equally-likely populations:

M-M
M-F
F-M
F-F

We now sample randomly from one of these four possible populations. We sample a 'F'. (This sampling process is completely equivalent to the statement "there is at least one female in the population". No more and no less information is gained.)

Since we know the a priori probabilities of each of the four configurations (populations) and the probability of sampling an F given each configuration, Bayes' rule tells us what the (conditional) probability of each configuration is after we've observed an 'F'.

I turns out that prob( M-M ) = 0, as expected (i.e. you can't sample a male from female-only population), but for the remaining three,

prob( M-F ) = 1/4
prob( F-M ) = 1/4
prob( F-F ) = 1/2

Or stated more intuitively: the fact that we sampled a female member gives us a higher probability that we sampled it from a female-rich population. In this case, the probability of the F-F config is exactly the same as the combined probabilities of the F-M and M-F configs, giving 50%/50% as expected.

Hence, to summarize, we started out with four populations of equal a priori probability:

M-M : probability 1/4
M-F : probability 1/4
F-M : probability 1/4
F-F : probability 1/4

We observed that at least one of the two children is female, which, through a straightforward application of Bayes' rule, yields our post-observation probabilities:

M-M : probability 0
M-F : probability 1/4
F-M : probability 1/4
F-F : probability 1/2

In configurations 2 and 3, the "other child" is always male. In configuration 4 (which is twice as likely given our observation), the "other child" is always female. Hence:

probability( second child is male ) = 1/4 + 1/4 = 1/2
probability( second child is female ) = 1/2 = 1/2

Q.E.D.

Edited by syonidv, 10 July 2012 - 01:06 AM.

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### #319 benjer3

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Posted 10 July 2012 - 02:09 AM

Wow, I was a doubter, but I have to say I'm impressed. The question I will ask all of you, before all of your bickering and mathematics, is have you tried it out? I wrote a simple program to chuck out the numbers, and the results are pretty astounding. (If someone has already done this, forgive me, but I didn't feel like going through 32 pages checking for it.) I also have to say that in writing the program, I see why it gets the results it gets.

Here's the code (in C#):
```
static void Main(string[] args)
{
int secondGirls = 0, secondBoys = 0;

for (int count = 0; count < 1000; ++count)
{
Family family = new Family();

if (family.Child1 || family.Child2)
{
Console.Write("Family1 has a girl. ");
if (family.Child1 && family.Child2)
{
Console.Write("The other child is a girl.\n");
++secondGirls;
}
else
{
Console.Write("The other child is a boy.\n");
++secondBoys;
}
}
}

Console.WriteLine("Of the second children, {0} are girls and {1} are boys", secondGirls, secondBoys);

}
}

public class Family
{
private static readonly Random random = new Random();

public bool Child1 = random.Next(2) == 1;
public bool Child2 = random.Next(2) == 1;
}
```

Spoiler for The Results for 1000 iterations

Think of it as you will, but this program is short, sweet, and to the point, and I think full-proof.

Spoiler for Hmm, looking back at the code...

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### #320 fabpig

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Posted 10 July 2012 - 10:27 AM

This teaser is about 4 years old now. It was resurrected about a year ago as: http://brainden.com/...8--/&#39; . On the last couple of pages, you'll see there are programs which come to the same conclusion as benjer3's. (including a rare excursion into programming by one Mr F.Pig )
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