probability problem

[Flammable]

You see a woman in the park with her son. She turns to you and says, "The other day, I came to this park with my two children, as they are all that I have."

What is the probability that her other child is a boy?

not homework

 

[Leros]

It would seem 50/50 if the gender of a child is independent of other children the parents have had.

 

[Hacp]

I would guess it's 50/50?

 

[DisgruntledVirus]

Originally posted by: Leros
It would seem 50/50 if the gender of a child is independent of other children the parents have had.

Well, that's if you only have male/females. What about transgendered? That just fucked your ratio all up...

 

[SonnyDaze]

1 in 2 chances that the other is a boy.

 

[Leros]

Originally posted by: DisgruntledVirus

Originally posted by: Leros
It would seem 50/50 if the gender of a child is independent of other children the parents have had.

Well, that's if you only have male/females. What about transgendered? That just fucked your ratio all up...

Lets assume she is crazy and would have drowned any transgendered children in her bathtub. There my 50/50 ratio is back. Take that!

 

[DrawninwarD]

Originally posted by: SonnyDaze
1 in 2 chances that the other is a boy.

This...

But there's gotta be a twist or something.

 

[Special K]

I'm going to go with 1/3. The starting sample space would be: {BB, BG, GB, GG}. You already know it can't be GG, so eliminate that possibility, leaving you with: {BB, BG, GB}.

The answer is therefore 1/3.

 

[DrawninwarD]

Originally posted by: Special K
I'm going to go with 1/3. The starting sample space would be: {BB, BG, GB, GG}. You already know it can't be GG, so eliminate that possibility, leaving you with: {BB, BG, GB}.

The answer is therefore 1/3.

I think that would be the answer is you were trying to find out the probability that BOTH her kids are boys given that one of them is a boy.

You would be correct if the question was: What is the probability that her other child is also a boy?

I dunno, whatever.

 

[sactoking]

Originally posted by: DrawninwarD
[I think that would be the answer is you were trying to find out the probability that BOTH her kids are boys given that one of them is a boy.

You would be correct if the question was: What is the probability that her other child is also a boy?

I dunno, whatever.

But that is the question.

You see a woman in the park with her son.

"The other day, I came to this park with my two children

What is the probability that her other child is a boy?

One boy is a given. Two children is given. Solve for P(child2)=boy s.t. child1=boy. You don't know if child1 or child2 is older. I believe Special K's answer is correct.

If you knew which child was eldest the probability would be 50%, but since you don't you have to include both GB AND BG as viable combinations, which changes the probability.

 

[sponge008]

Originally posted by: Special K
I'm going to go with 1/3. The starting sample space would be: {BB, BG, GB, GG}. You already know it can't be GG, so eliminate that possibility, leaving you with: {BB, BG, GB}.

The answer is therefore 1/3.

That's only true though if people with any sons at all always go out to walk with their sons and not their daughters. Here, we can make the following reasonable assumptions:

-Of the people who came to the park that have two kids, half of these people have a boy and a girl, one quarter have two girls, and one quarter have two boys

-All of the people with two boys (1/4 of total) and half of the people with a boy and a girl (1/4 of total) that take one child with them will have a boy with them.

Thus, of those in the park with two children and one of them being a boy, half of them will have the other child being a boy, and half will have the other being a girl.

1/2 chance.


This seems like a degenerate case of "unfinished game" probability; try Googling it for more in-depth cases.

 

[Epic Fail]

0, a woman with children will never talk to the OP.

 

[DrawninwarD]

Originally posted by: sactoking

Originally posted by: DrawninwarD
[I think that would be the answer is you were trying to find out the probability that BOTH her kids are boys given that one of them is a boy.

You would be correct if the question was: What is the probability that her other child is also a boy?

I dunno, whatever.

But that is the question.

If that was the question, it would've said: What is the probability that her other child is also a boy? The way I understand it, it's just asking what the missing kid's gender is regardless of the kid at the park, even though I'm guessing that wasn't the intended question.

 

[sactoking]

Rather than edit my prior post, I'll just say that upon further review I believe the answer is 1/2. Special K is solving in the proper fashion but he, and I initially, didn't consider all possible options.

We know that the woman has 2 children. We know that one of them is a boy. That is all we know. There are no 'reasonable assumptions' can be made.

Special K posited that the possible combinations of children is {BB, BG, GB}. I agreed, under the thinking that since we don't know their relative ages BG and GB are distinct options. What I failed to consider is that under that same scenario, BB and BB are both distinct options. More specifically, B1B2 is different than B2B1. Thus, the full set would be {BB, BB, BG, GB}. The probability of two boys is 2/4 or 1/2.

This is using Special K's original set, which makes a bit of a logical fallacy by including both GB and BG. The logical fallacy is that he's introduced a new variable, age. Age wasn't part of the original scenario. Thus, the problem can be simplified down to a set of {B, G} which yields the expected result.

 

[Special K]

Originally posted by: sponge008

Originally posted by: Special K
I'm going to go with 1/3. The starting sample space would be: {BB, BG, GB, GG}. You already know it can't be GG, so eliminate that possibility, leaving you with: {BB, BG, GB}.

The answer is therefore 1/3.

That's only true though if people with any sons at all always go out to walk with their sons and not their daughters. Here, we can make the following reasonable assumptions:

-Of the people who came to the park that have two kids, half of these people have a boy and a girl, one quarter have two girls, and one quarter have two boys

-All of the people with two boys (1/4 of total) and half of the people with a boy and a girl (1/4 of total) that take one child with them will have a boy with them.

Thus, of those in the park with two children and one of them being a boy, half of them will have the other child being a boy, and half will have the other being a girl.

1/2 chance.


This seems like a degenerate case of "unfinished game" probability; try Googling it for more in-depth cases.

It seems the answer depends to some extent on how you interpret the question:

Ask Dr. Math Example
Boy or Girl Paradox

 

[sponge008]

Originally posted by: Special K


It seems the answer depends to some extent on how you interpret the question:

Ask Dr. Math Example
Boy or Girl Paradox

Essentially, if anyone that has a boy shows the boy, the chances of the other kid being a boy are 1/3. If the kid anyone shows is chosen at random, and you are looking at someone who randomly chose to show a boy, the chances of the other kid being a boy are 1/2. The OP's problem was in the latter category.

Here's an example from the former category:
http://www.codinghorror.com/blog/archives/001203.html

 

[darkxshade]

I'm 100% certain it's a boy.

 

[Paperdoc]

At first glance, the probability Is 0.50000.... because the gender of the second child is completely independent of that of the first; hence, the known gender of one child has absolutely no impact on the gender of the other.

However, it is known that, although males outnumber females at birth about 53:47, they also die off slightly more so that the total population has a male:female ratio of something like 46:54. At what age (during early childhood?) does that ratio pass through 50:50? I don't know. And I suppose that, since this is a Statistics problem, reality is not part of the discussion.

 

[blinky8225]

1/3. Sample space is {BB, BG, GG}
P[BB] = 1/4
P[BG] = 1/2
P[GG] = 1/4
The prior probability is 1/4. However, we know that she has at least one boy.
The probability of BB given BX, where X can be any gender.
By Baye's Rule...
P[ BB | BX] = P[BB] / (P[BB] + P[BG]) = (1/4)/(3/4) = 1/3.

 

[lizardth]

Darn I thought this thread was going to be about a pot of petunias and a whale...

 

[Fullmetal Chocobo]

Originally posted by: lizardth
Darn I thought this thread was going to be about a pot of petunias and a whale...

This.

Wait, no. lizardth, that was the improbability drive. This is a problem operating in normality.