One Girl - One Boy

[Guest]

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.

Ofcourse its not 1/2 else would make it a lousy puzzle...

Ans: 1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.

This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

[bonanova]

If the order of boy-girl or girl-boy doesn't matter in your counting,

then you're assuming that the distribution of all two-children families is:

2 boys - 33.33%

2 girls - 33.33%

boy-girl - 33.33%

at some point you have to realize that p[boy-girl] = 2 x p[2 girls].

Take a deep breath.

Relax.

Think.

Got it?

[edit - Martini already points this out ... never mind]

[Guest]

Are you saying:

2 boys 33.3...%

1 boy one girl (in any order) 33.3...%

2 girls 33.3...%

Aha!

Ok, you got me. The odds are 25/50/25.

I was caught up in my skewed interpretation of the OP and kept trying to validate it. I have always understood the "1/3" logic and agreed with it in principle, but thought that the OP could be interpreted in a way to allow for 50%. You've backed me into a corner though, and I must concede. Good discussion!

(BTW, although I understood the "1/3" logic, I have not been baiting you. My posts were sincere. I was just wrong.)

[Guest]

I don't want to beat an already mutilated dead horse b-u-u-u-t-t-t-t I do want to hop up and down on the saddle a bit.

Its been agreed that the correct answer is 1/3. I've read all the gazillion posts and I truly do understand why thats correct but I'm still confused.

Let me explain before everybodies heads explode.

One point made that everyone seems to agree with yet glosses over is this;

If child A is female the probability that child B is female is 50%.

If child B is female the probability that child A is female is 50%.

So obviously the probability of the "other" child being female is 50%.

However if you don't know which of the two children is female the probability of the "other" child being female changes to 33%.

Why? I'm not much of a mathematichian but it doesn't seem to me that the probability should change if the equation hasn't changed. If it has change please school me in simple words so that I might understand.

Let the beatings begin.

[Guest]

One point made that everyone seems to agree with yet glosses over is this;

If child A is female the probability that child B is female is 50%.

If child B is female the probability that child A is female is 50%.

So obviously the probability of the "other" child being female is 50%.

However if you don't know which of the two children is female the probability of the "other" child being female changes to 33%.

Why?

It's because of the "if" word. Once we add "if" child A is female, we have new information that removes a possibility.

Here are the equal possibilities of a couple with two children:

BB

GG

BG

GB

'If" we know the couple has at least one girl, the possibilities drop down to three.

GG

BG

GB

The other child has a 1/3 probability of being a girl.

"If" we know the couple's first child is a girl, the possibilities drop down to two.

GG

GB

The other child has a 1/2 probability of being a girl. "Other" child in this case, just means "second" child. No matter how many children a couple has, the probability of a specific one being a girl is always 1/2. In the riddle, "other" child can refer to either of the two.

[Guest]

alright you nincompoops, let me explain, unless sum1 has already said this (im too lazy to read 4 pages of this stuff

ok, now here are the choices:

Boy, girl

Boy, boy

Girl, girl

Girl, boy

now, scale, your answer says that since there is already a girl, you can drop the boy, boy combo, and that leaves 3 choices. BUT, if it leaves these 3 choices:

Girl, Boy

Boy, Girl

Girl, Girl

then you are saying that (Girl, Boy) (Boy, Girl) is different. It is, but since there is already a baby girl that came first, you are switching around the order of (Boy, Girl) which makes it:

Girl, Boy

Girl, Boy

Girl, Girl

Now, you have made it look like there are 2 possibilities of (Girl, Boy) when there is only one. Therefore it should look like:

Girl, Boy

Boy, Girl

Which leaves only 2 possibilities.

100(%) / 2 = 50(%)

So there is a 50% that there will be 2 girls since there is already one.

The whole reason for the order of the choices is because of what comes first, so

Boy, -----

means the boy came first, and that is wrong.

[bonanova]

alright you nincompoops, let me explain,

...

since there is already a baby girl that came first

...

Hi Asdfgh Zxcvbn, and welcome to the group.

Here are a couple of tips that might be helpful.

1. The OP does not say a girl came first.
   
2. Before you call Instead of calling people names, suggest to yourself that you might not have understood the question.
   

And if you want to keep your membership at BrainDen,

you might want to consider showing respect - even in those cases where you might turn out to be right. <_<

[Guest]

I'm going with 50/50. The second kids gender has nothing to do with the first being a girl. Unless it's a goat! Then it's still a 50/50 shot of being male or female

[Guest]

I'd have to say 1/2.

There is absolutely no correlation between the two outcomes, and they don't affect one another.

It would be 1/3 if you were saying, what is the possibility they will get two girls should they receive twins ( Excluding of course, the mathematics behind fraternal and identical twins. ), and they know for sure that one of them is a girl.

But, as previously stated, these two have nothing to do with one another, so the possibility of the latter child being of the female gender is on a fifty percentile scale.

[Guest]

I'd have to say 1/2.

There is absolutely no correlation between the two outcomes, and they don't affect one another.

It would be 1/3 if you were saying, what is the possibility they will get two girls should they receive twins ( Excluding of course, the mathematics behind fraternal and identical twins. ), and they know for sure that one of them is a girl.

But, as previously stated, these two have nothing to do with one another, so the possibility of the latter child being of the female gender is on a fifty percentile scale.

There is a correlation. See, you are thinking that the puzzle is - what are the chances of out of one child it will be a girl. But that's not the puzzle.

The puzzle asks - given 2 children with the knowledge that one is a girl, what is the probability that they are both girls. And, in this case you have to have the complete chart of possible outcomes. The possibilities are: 1B+1B, 1B+1G, 1G+1B, 1G+1G.(25% chance of each occurrence) So, there are 4 distinct possibilities out of which only three pass the prerequisite. We already know that one of the children is a girl. So, out of the three remaining possibilities: 1B+1G, 1G+1B, 1G+1G only one affords 1G+1G and therefore your chances are 33%.

[Guest]

I'm going with 50/50. The second kids gender has nothing to do with the first being a girl. Unless it's a goat! Then it's still a 50/50 shot of being male or female

Justsnapd8, please read through this thread. It's been explained more than once that the riddle does not ask about the "second kid".

There is absolutely no correlation between the two outcomes, and they don't affect one another.

Please read through the thread. The riddle is not about two outcomes; it's about one.

It would be 1/3 if you were saying, what is the possibility they will get two girls should they receive twins ( Excluding of course, the mathematics behind fraternal and identical twins. ), and they know for sure that one of them is a girl.

You're right, it would be 1/3 if we were talking about twins. The fact that they may or may not be twins in the riddle is irrelevant. If you understand the probability is 1/3 for twins, you should see the probability remains the same for non-twins.

[Guest]

Hi! im new in the forum which i find very exciting!

ok, for this puzzle can i ask u something, does the order matter? the puzzle does not say if the first or the second child is the girl, so girl-boy, and boy-girl, does nt it count as the same?

Ans: 1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.

This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

[Guest]

It's a boy, because ONE of them is a girl.

[Guest]

Ok it's 1/2 and that's my final answer..... here's how.

Let's look what everyone is saying

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

Now because we can exclude BB from the equation it looks more like this

GG = 1/3

GB = 1/3

BG = 1/3

So the answer is

GG = 1/3 <--

GB = 2/3

It looks good at first glance.... and second glance..... and hell it just looks good. But this is why it's wrong

Let's start from the top.

GG = 1/4

BB = 1/4

BG = 1/4

GB = 1/4

Now we remove BB

GG 1/3

GB 1/3

BG 1/3

Now this is where it go's wrong. The assumption is that GB and BG are equal .... but they are not. In fact you have to remove one of the GB/BG combos. Why you ask because we are given that one of the two is a girl. The same thing that allows us to remove BB from the equation forces us to remove one of the GB/BG combos because it can only be one or the other. So we are left with only 2 possibilities. Not 3. For example let's say we have

G+x = KIDS

Because we have the value of G (girl) we only need figure out "x" and "x" has a value of 50%. Now let's look at the other combo....

x+G = KIDS

The same thing applies here. Because we have the value of G we only need to figure out "x". To prove this let's do the coin toss experiment and have a look at what happens...

HT , TH , HT , HT , TT , TT , HH , HH ,TH ,TT

TT = 30%

HH = 20%

HT/TH = 50%

This is the actual result of me tossing a coin 20 times. 30% of the time TT comes up. That's good. But what happens when we are given the value of one of the two.... What is the chance of the other being the same value. for this I am going to toss a coin to pick at random one of the two values of each set. Heads it's the first value tails it's the second....

HT , TH , HT , HT , TT , TT , HH , HH ,TH ,TT

H H H T H T T T T H

our first one is heads , so our equation looks like this

Hx , x = T = wrong

Now what is the probability that the other one is heads. 50% but this time we are wrong and it is tails. Let's try it again.... Heads again....

T+x , x = H wrong .

H+x , x = T wrong

x+T , x = H wrong

T+x , x = T right

x+T , x = T right

x+H , x = H right

x+H , x = H right

x+H , x = T wrong

T+x , x = T right

So let's see.... we where wrong 5 times.... and right ......... 5 times.... that's 50% ! Wow that worked better then I thought. I didn't think it would be exactly 50% but I'll take it.

[Guest]

Now this is where it go's wrong. The assumption is that GB and BG are equal .... but they are not. In fact you have to remove one of the GB/BG combos. Why you ask because we are given that one of the two is a girl. The same thing that allows us to remove BB from the equation forces us to remove one of the GB/BG combos because it can only be one or the other. So we are left with only 2 possibilities. Not 3.

This is wrong. We remove BB because it is not a possibility at all (one of them is a girl). We don't remove GB or BG because both are still possibilities. Of course it can only be one or the other. That's not a reason to remove one as they both still have an equal chance (along with BB) of being the children in the scenario. How can you remove BG as a possibility when it still is one? Don't tell me, you think BG and GB are really the same, right? They're not.

You say:

Let's look what everyone is saying

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

Do you think that's wrong? It's not.

To prove this let's do the coin toss experiment and have a look at what happens...

HT , TH , HT , HT , TT , TT , HH , HH ,TH ,TT

TT = 30%

HH = 20%

HT/TH = 50%

If by the coin experiment, you mean my coin experiment, that's not it. Mine uses 200 coins, which will be more accurate than using 20 most of the time.

But as in mine, lets make Heads be boys and Tails be girls.

After removing the boy pairs, we're left with:

HT TH HT HT TT TT TH TT

The other child is a girl 37.5% of the time. Closer to 33.3% than 50%. Use more coins and it will be even closer.

[Guest]

You got it all wrong . The problem is that we know the sex of one of the two. So we only have to figure out what the other one is. The formula for probability is

P = n/N

Probability is equal to favorable outcomes divided by possible outcomes. To make sure this proof works let's use a coin toss. What are the odds that if I toss a coin it will be heads.

P = 1/2 = 50%

I think we can all agree that this is correct. Now what are the odds that the next coin is going to be heads.

P = 1/2 = 50%

So how do we figure out what the probability of two subsequent events are? We multiply the probabilities together. For this example I am going ask what are the chances of getting two heads in a row. Let's do the math.

P = 1/2 * 1/2 = 25%

What are the odds of getting three heads.....

P = 1/2 * 1/2 * 1/2 = 12.5%

What about 10 heads all in a row

P = (1/2)^10 = 0.09765625%

This also works with different things. Like what are the odds of me flipping a coin and getting heads, then pulling a random card from a deck and it being an ace.

P = (1/2) * (4/52) = 3.84%

In the end this formula works flawlessly.

So let's use this to solve our problem

P = Kid1/PosKid x Kid2/Poskid = 1/1 x 1/2 = 50%

How did I get 1/1 you ask. Because one of the kids is a girl. It's a fact. One of the kids only has the possibility of being one thing.... a girl. The other one has the possibility of being a girl or boy. This is the formula for probability. I am interested in knowing what formula you are using?

Edited July 24, 2008 by toxicoxyde

[bonanova]

You got it all wrong . The problem is that we know the sex of one of the two. So we only have to figure out what the other one is.

The formula for probability is P = n/N

Probability is equal to favorable outcomes divided by possible outcomes. This is the formula for probability. I am interested in knowing what formula you are using?

You have the right formula - just use it.

N = 3 [GG GB BG] - if you know one is a girl, [you don't know which one is certainly a girl] these are the possible outcomes

n = 1 [GG] - this is the favorable outcome.

p = 1/3

[Guest]

If this is a play on words...then there is 0% probablity that the other would be a girl becuase there was only "one" girl.

[unreality]

I can't believe this is still being disputed ;D the answer is 1/3, people (or 0% if it is a play on words like Pzlmenot said )

[bonanova]

Since the word only is not in the OP 0% does not work.

[Guest]

You have the right formula - just use it.

N = 3 [GG GB BG] - if you know that at least one is a girl, [read the OP: you don't know which it is] these are the possible outcomes

n = 1 [GG] - this is the favorable outcome.

p = 1/3

What is your formula for N = 3? What is the mathematical expression to describe this event?

Let's try your logic on a coin toss. The odds of a coin coming up heads on a coin toss is always 50% . So we flip a coin......

Heads. Now what are the odds that the next coin is going to come up heads again..... 50%. Why is it 50% because favorable outcomes divided by possible outcomes = probability. Favorable outcomes being 1 (heads) divided by possible outcomes 2 (heads or tails). Every coin that you toss into the air has a 50% chance of being heads or tails. Now lets look at what you are saying. We know that the first coin is Heads. So we can't possibly have two tails their for we have a 1/3 of it being heads again? This is wrong each coin toss is independent from one another. You can add order into random events.

Edited July 24, 2008 by toxicoxyde

[Guest]

I think the important fact here is that the genders of the two childeren are Conditionally Independant. this means that the gender of one child does not effect the gender of the other. thus the probabilitys you are dealing with are:

child one: Boy/Girl 50% probability for either gender

child two: Boy/Girl 50% probability for either gender

if you were then told one of them is a girl, then you have:

child one: Girl 100% girl

child two: Boy/Girl 50% probability for either gender

or vice verse

in other words, the number of possible outcomes for one child is not effected by the other. knowing one child is a girl still only leaves 2 possible outcomes for the other childs gender. Thus the answer is 1/2 probability that the other child is a girl.

http://www.cs.princeton.edu/courses/archiv..._notes/0207.pdf (look at section 10.2)

[bonanova]

if you were then told one of them is a girl, then you have:

child one: Girl 100% girl

child two: Boy/Girl 50% probability for either gender

or vice verse

You are not told that child one is a girl, and then asked whether child two is a girl.

The OP does not single out either child.

The statement 'one of them is a girl' is true for three disjoint and equally likely cases: GG BG GB.

In one of those cases the other is a girl.

The probability is 1/3.

It's not all that hard. The OP rules out 1/2 as an answer. Read it.

[unreality]

Another explanation:

Hola 1 and Hola 2 are both holas. Holas can be only blue or red. Each hola has a 1/2 chance of being blue and a 1/2 chance of being red.

The EQUALLY LIKELY possibilities are:

* both red (1/4)

* Hola 1 is red, Hola 2 is blue (1/4)

* Hola 1 is blue, Hola 2 is red (1/4)

* both blue (1/4)

Now if we say: "One of the holas is red", it doesn't specify Hola 1 as red, or Hola 2 is red- just either of them. Thus the only thing that rules out is the last option (both blue)

we are left with:

* both red (1/4)

* Hola 1 is red, Hola 2 is blue (1/4)

* Hola 1 is blue, Hola 2 is red (1/4)

but they're 1/3 now, not 1/4, because they are all equally likely (1/4) but one of the chances has been axed out, leaving them all equally likely out of 3

[Guest]

What's with the Holas?

[Guest]

What is your formula for N = 3? What is the mathematical expression to describe this event?

No formula is needed; all you have to do is count to three. There are four combinations a couple can have two children and three if we know at least one is a girl.

You are doing what a lot of posters here have done: totally ignore posts that were directed at you. How about responding to Martini's post?

You said: "The same thing that allows us to remove BB from the equation forces us to remove one of the GB/BG combos because it can only be one or the other."

Did Martini's reply make any sense to you? Do you not see that you are totally wrong? You do not remove a GB/BG combo because they are BOTH still possibilities. BB is NOT a possibility.

You said:

Let's look what everyone is saying

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

Martini asked if you though this was wrong. You didn't answer.

He also showed that in your very own coin experiment, the TT combination happened 37.5% of the time and is much closer to 33.3% than 50%. Did that mean nothing to you?

Now lets look at what you are saying. We know that the first coin is Heads.

Bonanova never said the first coin is heads and the riddle never stated that the first child is a girl. If the riddle stated that, I'm sure he never would claim the probability of GG is 1/3 as if you've spent any time reading his posts, you'd know that he's quite good at math.

I think the important fact here is that the genders of the two childeren are Conditionally Independant. this means that the gender of one child does not effect the gender of the other. thus the probabilitys you are dealing with are:

child one: Boy/Girl 50% probability for either gender

child two: Boy/Girl 50% probability for either gender

if you were then told one of them is a girl, then you have:

child one: Girl 100% girl

child two: Boy/Girl 50% probability for either gender

or vice verse

in other words, the number of possible outcomes for one child is not effected by the other. knowing one child is a girl still only leaves 2 possible outcomes for the other childs gender. Thus the answer is 1/2 probability that the other child is a girl.

Tribbs, I'm going to repeat something that's been said quite a few times in this thread: Please read through all the posts.

NO ONE is claiming that the gender of one child affects the gender of another!

knowing one child is a girl still only leaves 2 possible outcomes for the other childs gender. Thus the answer is 1/2 probability that the other child is a girl.

That is incorrect. Two possible outcomes does not mean that each outcome is equally as likely as the other. If that were true, I'd play the lottery tonight as I'm either going to win or lose and I'd love for my odds to be 1:1.

Here is the wikipedia article on Brain Teaser.

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

It has this very riddle in it as an example. It also gives the answer.

Ever heard of "Ask Dr. Math"? It's from one of the most respected math sites on the internet. It also has this very riddle on it:

http://mathforum.org/dr.math/faq/faq.boy.girl.html

Do you think a website that specializes in math doesn't understand simple conditional probability?