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

[roolstar]

The gender of children is not affected by the gender of their sibilings*, but is surely affected by biological and physical conditions for both parents.

It was determined that the pH level of the mother's body can have an effect on the gender of her children. pH as you all know moves along the 0 – 14 spectrum.

Although our body pH is rigorously kept stable by the water and minerals in our body, a more acid pH in the woman's tract (pH toward the 0 end of the spectrum) favors female fetus formation and vice-versa…

The only thing I’m not sure of is if acid favors male or female. But it was proven to favor one….

If I had to guess, I would say acid is for female at least the one I know… Just kidding ladies…

That’s why consuming some kinds of vegetables may favor one gender or the other...

Of course there are lots of other biological and physical factors as well (you can google them if you're interested).

What's interesting is that "affecting" the gender of a child is calculated by probability: if they find that in a certain “condition” the number of girls v/s boys is tilted either way (probability different than 1/2) on a big population, they consider this “condition” to have an impact.

*Even more interesting: Having 10 daughters may be an indicator of the presence of one or more of these "affecting" factors and thus increase the probability of having another girl. So the gender of a child may be in a way indirectly affected by the gender of his/her sibilings...

Now that being said, I don’t think this is of any relevance to the puzzle at hand… (Now that I think of it, I probably should have said that right at the beginning and saved you this boring reading).

Simply put, the puzzle can be read: “Knowing that one of the kids is a girl, does this affect the probability of the other one being a girl? (It doesn't matter whether the probabiblity before this information was 50% or 92%)”

[Guest]

That’s why consuming some kinds of vegetables may favor one gender or the other...

Nonsense! If you'd like to debate this further, start a thread in "Others".

What's interesting is that "affecting" the gender of a child is calculated by probability: if they find that in a certain “condition” the number of girls v/s boys is tilted either way (probability different than 1/2) on a big population, they consider this “condition” to have an impact.

*Even more interesting: Having 10 daughters may be an indicator of the presence of one or more of these "affecting" factors and thus increase the probability of having another girl. So the gender of a child may be in a way indirectly affected by the gender of his/her sibilings...

There's nothing interesting about you supposing something without any evidence and you don't mention who "they" are. But again, start a thread somewhere else if you'd like to debate this.

[bonanova]

The OP says ... Assume safely that the porbability [sic] of each gender is 1/2.

All the "affecting factor" discussion does not pertain.

This is a calculation of probability, not a study of the human reproductive process.

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

It has the same answer as:

I flipped a fair coin twice. One of the outcomes was heads. What's the probability the other outcome was heads?

[Guest]

I flipped a fair coin twice.

Had you been eating your vegetables before flipping the coin?

[Guest]

uhhh.

okay

[roolstar]

Now that being said, I don’t think this is of any relevance to the puzzle at hand… (Now that I think of it, I probably should have said that right at the beginning and saved you this boring reading).

Simply put, the puzzle can be read: “Knowing that one of the kids is a girl, does this affect the probability of the other one being a girl? (It doesn't matter whether the probabiblity before this information was 50% or 92%)”

All the "affecting factor" discussion does not pertain.

This is a calculation of probability, not a study of the human reproductive process.

I couldn't agree more bonanova...

[Guest]

NO matter what maths you apply, it will always be 50% noted from basic genetics

[Guest]

NO matter what maths you apply, it will always be 50% noted from basic genetics

The math does matter. Look at the probability and you'll see it's not 50% but 33%

We are NOT saying that the chance of having a girl changes. Re read carefully.

Edited March 24, 2008 by Noct

[Guest]

It's a 50-50 chance

You said there was three possibilities

Girl-Girl

Girl-Boy and

Boy-Girl

But Boy-Girl and Girl-Boy is the same but in a reverse order

So there is only 2 possible possibilities.

Understood

Edited March 25, 2008 by Idienasty

[Guest]

It's a 50-50 chance

You said there was three possibilities

Girl-Girl

Girl-Boy and

Boy-Girl

But Boy-Girl and Girl-Boy is the same but in a reverse order

So there is only 2 possible possibilities.

Understood

they're not the same. they are in reverse order, like you just said. so it's 1 out of three possibilities.

[bonanova]

It's a 50-50 chance

You said there was three possibilities

Girl-Girl

Girl-Boy and

Boy-Girl

But Boy-Girl and Girl-Boy is the same but in a reverse order

So there is only 2 possible possibilities. Yes, but be careful where you go from here. See below

Understood

Suppose the OP did not say "at least one of my children is a girl."

Then I would say the possible cases are

GG

GB

BG

BB

But you will say that's wrong - there really are only three cases: two girls, a boy and a girl, and two boys.

OK, let's go your way. Three cases. Then answer the question:

What is the probability, for families with two children, that there is one boy and one girl?

Your answer has to be one-third; one case out of three.

Are you really comfortable with that answer?

Boy [then] Girl is not the same case as Girl [then] Boy.

Before the second child is born, these are clearly different, equally probable cases.

In my post, I enumerated three distinct, equally likely cases, listing the eldest child first:

GG

GB

BG

The fourth, otherwise-possible case, BB, was eliminated by the conditions of the problem "at least one is a girl".

Now, look at the first two cases.

If the puzzle had said, "the elder child is a girl, what are the chances the younger is a girl?" the answer to that question is clearly 50%

Finally, you say GB and BG are the "same but in reverse order."

You are correct. So let's proceed using that reasoning.

Now you have two cases, shown as follows, where the first letter is not necessarily the older of the two children:

GG

GB

You reason that the likelihood of GB is 50% and the liklihood of GG is 50%.

But that reasoning is flawed.

You are looking at two possible outcomes and saying [incorrectly] that they have the same liklihood.

They don't.

GB is twice as likely as GG.

Try flipping a coin twice, repeatedly.

You will find that HT [which could be in either order remember you said] is TWICE as likely as HH.

If it's not, let me know, and we'll go to Las Vegas and make a lot of money.

So when you ask: "Given the options are GG and GB [in either order] what is the likelihood of GG?"

then since GB is twice as likely as GB, you see that the answer can't be 50%

My favorite way to dissuade people from calculating probability by just counting the number of possible

outcomes -- without regard to their likelihood -- and seeing how many are favorable -- is this.

Buy a lottery ticket. There are two possible outcomes.

- You win

- You lose

Surely you don't anticipate winning 50% of the time.

And the reason is the two outcomes have [vastly] different liklihood.

[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

At first I thought "it has to be 1/2", but two things helped snap me from it: too easy and (as I re-read it), it just says one kid is a girl, it doesn't say "the first kid is a girl". So, we aren't looking at independent events here. Since we know one kid is a girl, they could not have had two boys. Therefore, the only options are that the older kid is the girl and the younger a boy, the opposite, or that they have two girls. The question asks about the other kid being a girl as well, so that would be one option of the three remaining options that are viable.

Thus: 1/3

good one.

[Guest]

Ok I've just read every single post. All 14 pages. I flipped back and forth following 1/3 and 1/2. Then on my cigarette break I realized the nature of this question, because I unfortunately got a similar one wrong on a science exam. I'll sum up my arguments for both sides by simply referring to Post #101 by bonanova. For this particular question (as a native American English speaker) the answer is 1/3.

Now onto the smashing of post #127 by roolstar. He is actually correct. Couples that have multiple girls in a row, tend to continue having girls. I've been in the Army for 4 years now, and in my job field we work around a lot of communication equipment. I haven't met a single man who has been in the communication field longer than 5 years, whose wife gave birth to a male child (that he fathered). Something is making their children female. We joke that maybe the RF energy is frying our sperm. Enough defending of this ridiculous tangent. It's entirely irrelevant because the OP stated that the chance of each gender is 50%

This particular riddle's answer is 1/3 that both children are girls (ie: the other child is a girl). If asked differently, then maybe the answer would be different.

If you've made it this far in the forum, or just skipped to the end, and still think it's 50/50, please read post #101. And if your problem with the answer of 1/3 boils down to "semantics", "nuances of the English language", simultaneous, continuous, or whatever other words you like to use that change the question into one that fits your answer, pick up a textbook.

Quoting wikipedia for this (it was done back there somewhere) isn't the best thing to do (in my opinion). I can change any entry of Wikipedia into whatever I want. Anyone can. Who's to say that just because the majority (or wikipedia) agrees on 1/3 that it is in fact the true answer? No answer is assumed correct because of how many people think it. Collective beliefs don't change reality.

[Guest]

Ok, Im not reading all 14 pages, I dont think I really need to do that to know the answer. Each child has a 50% chance of being a girl. theres boy and theres girl, thats only 2 choices, 100% divided by 2 is 50%. that being said there is a different mathamatical percentage for having multiple same sex children in a row, but i dont know what it is for sure. Also, I have 3 children, all of them are girls. What is the mathematical percentage that I will be crazy long before I die.

[Guest]

1 out of 4 chances... 25% chance it'll be a girl.

[Guest]

Is it just me or is the 1/3 chance not the obvious answer. The wording of a riddle is paramount to the understanding of the solution.

It is known that there are four possible outcomes of having two children (MM, FF, MF, FM) in a particular order.

It is known that one of the two children is a girl. Therefore MM is not part of the equation, leaving three possibilites.

Two of these possibilities involve a second male and one involves a second female.

Therefore there is a 1/3 chance of the other child being a female.

The coin toss logic does not apply here because the question does not ask what the odds are of a single instance of a coin toss (always 50/50) but rather what the odds are, knowing that 1 of 2 outcomes is a girl, will the other also be a girl.

Is this not obvious or am I missing something?

[Guest]

I think the point IS being misses. The OP states that they already have the two children, they tell us that one is a girl, and ask the chance that the other is a girl. If the had no kids then we can say its a 1 in 4 chance. Example, say I have 2 pennies (whoo-hoo, I'm rich). I show you one penny on the table, and lets say its on tails. I have my hand covering the other penny so you cant see it. I ask you what is the chance it is on tails. Well, simple math says 50-50, heads or tails. This is not rocket science.

[akaslickster]

not 50/50 chance because that would not agree with the worlds last census or countries either. Stats have proven that there is more females in existence for a reason. I'll go with 60 girl and 40 a boy just from what I heard long ago. It's just more possible as tradition tends to repeat itself. I'm not exact but not far away.

[Guest]

not 50/50 chance because that would not agree with the worlds last census or countries either. Stats have proven that there is more females in existence for a reason. I'll go with 60 girl and 40 a boy just from what I heard long ago. It's just more possible as tradition tends to repeat itself. I'm not exact but not far away.

ok akaslickster, census aside, would you say 50-50 is the propper mathematical chance for 2 objects

[akaslickster]

ok akaslickster, census aside, would you say 50-50 is the propper mathematical chance for 2 objects

yes i would in flipping a coin but nature does not work with coins , it's far more complex thus why girls are bred more in human at least.

[Guest]

yes i would in flipping a coin but nature does not work with coins , it's far more complex thus why girls are bred more in human at least.

but the kids were already born, nature can no longer affect an out come. with the op, its just 50-50.

[Guest]

Ok, Im not reading all 14 pages, I dont think I really need to do that to know the answer.

That was your first mistake, not to mention a breach of message board etiquette. If you're going to argue against what the majority agrees is the correct answer, it would help if you read all of the arguments before arguing against it and having others explain something to you that was explained already.

The OP states that they already have the two children, they tell us that one is a girl, and ask the chance that the other is a girl. If the had no kids then we can say its a 1 in 4 chance.

That makes no sense.

Example, say I have 2 pennies (whoo-hoo, I'm rich). I show you one penny on the table, and lets say its on tails. I have my hand covering the other penny so you cant see it. I ask you what is the chance it is on tails. Well, simple math says 50-50, heads or tails. This is not rocket science.

What you just outlined is a different set-up than the one in the OP. In the OP, we know one of the kids is a girl, but not a specific one as we know a specific penny that is covered by your hand in your example.

If you want to see if you're right, try the following experiment with pennies that adheres to the conditions of the OP.

Lay out 100 pairs of pennies randomly. Consider pennies showing heads boys and pennies showing tails girls. Since we know the couple in the OP had at least one girl, remove all pairs that are both heads up (all boys). Count the pairs you now have. Now, how many of those pairs are both tails? Is it about half of the pairs showing or one-third?

not 50/50 chance because that would not agree with the worlds last census or countries either. Stats have proven that there is more females in existence for a reason. I'll go with 60 girl and 40 a boy just from what I heard long ago. It's just more possible as tradition tends to repeat itself. I'm not exact but not far away.

You're way off, akaslickster. The ratio of female births to male births is very close to 1/1. If you think 60% of the birth population is female, you're way off.

And it's a moot point. The OP states "Assume safely that the probability of each gender is 1/2."

[Guest]

ok, im gonna go put something cold on my cubes.

[Guest]

The answer is 25%. Before they had children the probability for a girl first is 50%. The probability for a girl second is 50%. BUT, to have 2 girls the probability moves to half of half, or 25%.

Comparing a coin toss is a common mislead. A single coin toss is of course 50/50. But 2 consecutive tosses is 50/50 for the first, and 25/75 for the second. Taken as a set (as you must in a family), you cannot discount the total probability by looking at the single toss.

Much like this:

I have 3 boxes (a - b - c).

2 have nothing in them.

1 has a million dollars.

If you guess the correct one, you get the money.

You pick one.

Out of the remaining 2, I show you that one is of course empty.

I give you the option of changing your answer.

Do you have better, worse, or the same odds if you DO change?

Better. The first guess you have a 1 in 3 chance of being right. Once I remove one, your odds move up to 2 in 3 by selecting the OTHER box, because it would have been like selecting the 2 boxes you didn't pick.

If I had offered for you to trade your ONE pick for the TWO you didn't pick, before showing one was empty would you?

Hi, could you explain this to me, because i dont get it.

i think you are saying that after you choose one and after the person takes an empty box out, then you should switch because then you will know what the box that was thrown had and you will know what the box that you chose will have. (if this is not what u meant when u said "you will know the two boxes you didn't pick", then please explain what you really meant). But, if that is what you meant, then wouldn't you know what was in 2/3 of the 3 boxes anyway because you would know what was in the box that was thrown and what was in the 1 box you chose.???

[Guest]

Hi, could you explain this to me, because i dont get it.

i think you are saying that after you choose one and after the person takes an empty box out, then you should switch because then you will know what the box that was thrown had and you will know what the box that you chose will have. (if this is not what u meant when u said "you will know the two boxes you didn't pick", then please explain what you really meant). But, if that is what you meant, then wouldn't you know what was in 2/3 of the 3 boxes anyway because you would know what was in the box that was thrown and what was in the 1 box you chose.???

isnt that also called the monty hall riddle, that i just read here a few posts ago - see whatcha ya gonna do (2 goats and a car)

Edited May 7, 2008 by johnclark