Daughter(s)

[ydoaPs]

A woman has two children. After talking with her for a while, you figure out that at least one of them is a daughter. Assuming any given child has a 50-50 chance of being a girl, what is the probability of both children being girls?

 

Now, you keep talking to her and you're able to figure out that this girl is the youngest child. Does the probability that they're both girls change? If so, what's the new probability?

 

 

 

 

 

 

 

 

 

 

[Litenoumjuq]

 

  Reveal hidden contents

 

The probability of her having two daughters is 1/3.

 

There are 3 possible combinations; only the oldest child is a girl, only the youngest child is a girl, both children are girls. Given that each child is a girl by a probability of 50%, all these combinations have equal probability.

 

In the second case, the probability of her having two daughters is 1/2

 

If we do know that the her youngest child is a girl, the possible combinations are reduced to two; only the youngest child is a girl and both children are girls. This means that the problem can be reduced to whether the oldest child is a girl or not, the probability of which is given to 50%

 

 

[MonDie]

  On 1/9/2013 at 3:15 PM, Litenoumjuq said:

 

  Reveal hidden contents

There are 3 possible combinations; only the oldest child is a girl, only the youngest child is a girl, both children are girls. Given that each child is a girl by a probability of 50%, all these combinations have equal probability.

 

 

 

  Reveal hidden contents

I was about to criticize your solution, but then I found it was correct.

 

There are 4 equally likely possibilities.

1. boy girl

2. boy boy

3. girl girl

4. girl boy

 

With the information of the first paragraph, you eliminate possibility number 2, leaving one possibility in which both are girls and two possibilities in which the other is a boy.

 

 

Edited January 9, 2013 by Mondays Assignment: Die

[michel123456]

  On 1/9/2013 at 1:31 PM, ydoaPs said:

A woman has two children. After talking with her for a while, you figure out that at least one of them is a daughter. Assuming any given child has a 50-50 chance of being a girl, what is the probability of both children being girls?

 

 

 

 

 

 

 

 

 

 

 

  Reveal hidden contents

the answer added to the probability for the 2 children not being girls must give 100% result. We know that one is a girl already. What are the odds for the second one to be a girl? that is 50%

 

Edited January 9, 2013 by michel123456

[ydoaPs]

Litenoumjuq is correct.

 

It's one of the many counterintuitive problems with probability. Many people (myself, the first time I saw the problem, included) want to say that you just multiply the probability that the one is a girl times the probability that the other is a girl. So, 1 times .5 to get an answer of 1/2 and that knowing the order has no bearing on the answer. But this intuition is wrong.

[J.C.MacSwell]

  On 1/9/2013 at 5:54 PM, ydoaPs said:

Litenoumjuq is correct.

 

It's one of the many counterintuitive problems with probability. Many people (myself, the first time I saw the problem, included) want to say that you just multiply the probability that the one is a girl times the probability that the other is a girl. So, 1 times .5 to get an answer of 1/2 and that knowing the order has no bearing on the answer. But this intuition is wrong.

The problem with this is that we only have information that there are 2 children, and that the youngest is a girl. You have no information on the sex of the elder. It's still 50/50

 

If you knew at least one was a girl and you asked if the youngest was a girl, that would be different.

 

Note your first post:

 

  On 1/9/2013 at 1:31 PM, ydoaPs said:

A woman has two children. After talking with her for a while, you figure out that at least one of them is a daughter. Assuming any given child has a 50-50 chance of being a girl, what is the probability of both children being girls?

 

Now, you keep talking to her and you're able to figure out that this girl is the youngest child. Does the probability that they're both girls change? If so, what's the new probability?

 

It depends on how the information came out, but based on the fact that it was information on this girl, you still have nothing to indicate any change in possibility for the other.

 

edit: I reread Litenoumjuq's post looks correct

Edited January 10, 2013 by J.C.MacSwell

[michel123456]

Sorry, I read the question as "what is the probability for the child to be a girl?"

 

The other solution is that the child is a boy. There is no third choice.

And there is no dependence from the fact that the one child is a girl.

 

It is not asked "what is the probability for both children to be girls knowing nothing".

Edited January 10, 2013 by michel123456

[MonDie]

Maybe not. Try looking at it this way.

 

Now the answers are:

1) 50 50

2) 1/3 2/3 split between GB and GG.

 

 

 

Why does #2 come out to a 1/3 2/3 split after #1 was a 50/50 split?

 

1) If we only know that she has two children, we know that she is more likely to have a boy/girl combo than she is to have two girls. A boy/girl combo is more likely for the same reason you're more likely to flip a 50/50 split of heads and tails. However, when she tells us she has a girl, things are evened back out because she would be more likely to say she has a girl if she has two girls.

 

2) When we find out that the girl in question is the younger, we can overturn the original 50/50 split logic. Before, we had this:

BG + GB > GG

Now, we have this:

G? - where ? can be B or G.

 

The purple font reasoning can be applied to the Monty Hall problem. The mother mentioning a daughter is equivalent to the host opening a door. See following quote.

 

  Mondays Assignment: Die said:

Assume you chose door one, and door two opened to reveal a goat. This is displayed in the top diagram of my previous image. One way we could explain the 1/3 2/3 split is by saying this, "If door one was right, the host could have chosen to open either door two or door three. However, if door three was right, the host could only open door two. Therefore, door three is more likely the right door."

Consider another example that applies this logic in an exaggerated way. People are sending two-digit numbers to a program. The program randomly chooses a person and displays their number. You type in "62," and the program displays "62."
We don't know the chances of it choosing you randomly. However, if it chose you, it would have to display "62." If it chose someone else, the chances of it displaying "62" would be 1/100. This makes it more probable that the program chose you and not someone else. This is the same logic, but it's exaggerated. The number displayed by the program is analogous to the door opened by the host.

http://www.scienceforums.net/topic/65398-why-is-the-monty-hall-problem-so-controversial/?p=687036

Edited January 10, 2013 by Mondays Assignment: Die

[AdamYax]

I want to say 50% chance....

[md65536]

  On 1/9/2013 at 10:58 PM, J.C.MacSwell said:

It depends on how the information came out, but based on the fact that it was information on this girl, you still have nothing to indicate any change in possibility for the other.

This is a very important thing to consider in these types of puzzles. For example:

 

A woman has two children. You figure out that at least one of them is a girl. You ask her what it's like having a daughter, and she tells you a funny story about her daughter. Now, you keep talking to her and you're able to figure out that this girl is the youngest child.

 

What is the probability of both children being girls?

 

Why is the answer...

  Reveal hidden contents

different from the answer to the second half of the original post?

 

Edited March 4, 2013 by md65536

[MonDie]

I guess we're stilll using the spoiler thing then.

 

  Reveal hidden contents

 

Shortly after taking my leave, I realized that the probabilities depend on how you came to the awareness that one child is a daughter. The above image gives different probability charts for two situations in which you arrive at the awareness via a different question.

 

 

As for the probabilities changing after you realize the daughter is younger, the information is only relevant if first children are relatively more likely to be boys or girls. We have no reason to think so, so it's irrelevant.

If you repetitively ask, "Is the younger child a girl?" and you only guess when her answer is yes—a duncical move if you're guessing boy—only 50% of the BG/GB situations (and 100% of the GG situations) will meet your criteria. Thus you would see the effect. However, that isn't the situation posed above. There is a difference between asking "Is the younger child a girl?" and "Is the girl in question the younger child?" The latter would eliminate 50% of the BG/GB situations as well as 50% of the GG situations, resulting in no net change.

But, really, the finding is no more relevant than the finding that the duaghter is the lonelier child or the angrier child.

 

Edited April 15, 2013 by Mondays Assignment: Die