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

[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.

Not so.

Before I have any children, the probability that my first two are both girls is 25% because I haven't had any yet. However, the fact that my first child is a girl does not change the probability that the next one is as well. That is, after I have one child, the probability that my next child is a girl is still 50%, and not 25%.

You can toss a coin to see this. Toss a coin twice and repeat a bunch of times. You'll see that about 25% of the results are heads-heads. However, if you look at only the trials that start with heads, about 50% of them should also end with heads. The result of the first trial (the first child, or the first toss) DOES NOT affect the result of the second trial (the second child, or the second toss).

In the question, we are given that at least one of the children is a girl.

So we either have one boy and one girl, or two girls. These two cases do not have equal probability.

With two children, the probability of one boy and one girl is 50% and the probability of two girls is 25% (again, try it with coin tosses...in two tosses, about half the trials will result in either heads-tails or tails-heads, while only about a quarter of them will result in tail-tail). It is twice as likely to get boy-girl (or girl-boy) as it is to get girl-girl. Since these are the only two possibilities, and one must be the case, the probability of boy-girl or girl-boy is 2/3 and the probability of girl-girl is 1/3.

We know that P(A|B), the probability of event A given event B has already occurred, is the probability of A and B occurring together divided by the probability of B occurring on its own:

P(A|B) = P(A and B occurring on their own)/P(B)

So we are given that we have at least 1 girl, and we want to know the probability of both children being girls, given that at least one of them is, so let A = both children are girls.

P(A|B) = P(both children are girls AND at least one is a girl) / P(at least one girl)

Since the event that both children are girls AND at least one is a girl reduces to the event that both children are girls independently...

P(both are girls given at least one is) = P(both children are girls independently) / P(at least one girl)

The probability that both children are girls independently = 1/2 * 1/2 = 1/4

The probability that we have at least one girl = 1-P(no girls) = 1 - P(two boys) = 1 - 1/2*1/2 = 1-1/4 = 3/4

P(both are girls given at least one is) = (1/4)/(3/4) = 1/3

[Guest]

It's 50/50. I like Chuck's reasoning kinda, and I see where it's all coming from, but every time a coin is tossed, its a whole new ball game. The previous toss does not determine the next. But apparently it depends on how and when the baby is conceived as to whether it is a girl or boy.

[Guest]

I have not read every single post, but I have read enough to get the gist of the 1/2 versus 1/3 arguments.

Here's my take:

It has been stated (by a member of the "1/3" coalition) that the two children are already born and that the puzzle is not about the probability of the gender of the next child to be born.

This is true. It is also true that the order of said children has also been established. We know that one of the children is a girl, but we don't know if it's the first or 2nd child. However, as I said, that order has already been established. So...

If the girl that we know about is the first child, then the possibilities are:

GB

GG

If the girl that we know about is the second child, then the possibilities are:

BG

GG

Thus, we see that the probability that the unknown child is a girl is 1/2.

[Guest]

If the girl that we know about is the first child, then the possibilities are:

GB

GG

If the girl that we know about is the second child, then the possibilities are:

BG

GG

Thus, we see that the probability that the unknown child is a girl is 1/2.

It doesn't work that way. You are correct that if we know the sex of the first child then the probability is 1/2. But we don't know the sex of the first child. You just separated three possible outcomes and repeated one (GG).

What do you think would happen if you tried the following experiment?:

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?

[Guest]

It doesn't work that way. You are correct that if we know the sex of the first child then the probability is 1/2. But we don't know the sex of the first child. You just separated three possible outcomes and repeated one (GG).

What do you think would happen if you tried the following experiment?:

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?

The girl we know about can be child #1 or child #2. Only one of these conditions can be true. The fact that we don't know which one doesn't mean that that condition is not established. She is either child #1 or child #2, but she can't be both.

Let's use uppercase (G) for the girl we know. We'll use lowercase (b or g) for the child we don't know. Here are the possibilities:

Gg

Gb

bG

gG

We have 4 distinct possibilities based on the information we have, with a 50% probability that the unknown child is a girl.

[Guest]

She is either child #1 or child #2, but she can't be both.

Let's use uppercase (G) for the girl we know. We'll use lowercase (b or g) for the child we don't know. Here are the possibilities:

Gg

Gb

bG

gG

We have 4 distinct possibilities based on the information we have, with a 50% probability that the unknown child is a girl.

You're using the GG combination twice. You're claiming it's okay do do this because we know one of the girls? We don't. According to your logic, in my penny experiment, half of the penny pairs will be GG (tail- tail).

In the sticky in the red box titled "Important: READ BEFORE POSTING" I linked to a wikipedia page explaining what a brain teaser is. It uses this very riddle as an example:

If we encounter someone with exactly two children, given that at least one of them is a boy, what is the probability that both of her children are boys?

(Of course, for the purpose of simplicity, we will disregard hermaphrodites and assume that boys and girls are born with equal probability.) The common intuitive way of thinking is that the births of the two children are independent of each other, and so the answer must be the absolute probability of one child being a boy, 1/2. However, the correct answer is 1/3 as shown by the following argument:

* For a single birth, there are two possibilities (a boy or a girl) with equal probability.

* Therefore, for two births, there are four possibilities: 1) two boys, 2) two girls, 3) first a boy, then a girl, and 4) first a girl, then a boy; all of them have equal probability.

* We are given that one of the children is a boy. Thus, only one of the four possibilities -- two daughters -- is eliminated. Three possibilities with equal probabilities (1/3) remain.

* Out of those three, only one -- two sons -- is what we are looking for. Hence, the answer is 1/3.

If you're still not convinced, answer my question about what you think would happen if you laid out pennies in the manner I described in my last post. If your answer is that about half of the pairs will be tails, perform the experiment yourself.

After you lay out the pennies, 1/4 will be BB, 1/4 will be GG, 1/4 will be BG and 1/4 will be GB (approximately). After removing the BB pairs, you will have about 1/3 of each. According to your logic, this won't happen. You are under the impression that 1/4 will be Gg and another 1/4 will be gG- or that 1/2 of the pairs left will be all tails. Try it and see.

[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

How about we arrange the possibilities like this (G1 means Girl was born first, G2 means Girl was born second, etc)

G1 - G2

G1 - B2

B1 - G2

B1 - B2

Since we do not know if the girl they have was the first or second (G1 or G2) we compute the probability for each case

Girl 1

The only combinations with G1 are G1-G2 and G1 - B2, so the odds of having a second girl are 50/50

Girl 2

The only combinations with G2 are G1-G2 and B1 - G2, so the odds of having a second girl are 50/50

The other way to look at this is that there are only 3 possibilities (when order does not matter)

G - G

B - G

B - B

Since there cannot be a B - B, the probability of a second girl is again 50/50

[Guest]

Okay, so he does not say that the first child is a girl, but one is. I still think it is still a 50% chance. You see that there are still four choices and at the beginning there are eight.

SORRY IF THIS IS WRONG BUT THIS IS WHAT I GET!!!

There are 4 choices after the first girl, in capitals is the first girl:

Child 1 - Child 2

GIRL ------- Girl

Girl ------- GIRL

GIRL ------- Boy

Boy ------- GIRL

Since you are saying that the GIRL can either be child 1 or child 2, so you must have girl - boy AND boy - girl then you must also have GIRL - girl and

girl - GIRL, otherwise you are also think of the GIRL as known, so you know if it is 1 or 2.

Another way to think about it is like this. say the question told you that GIRL is the first chil, but they have 2 children, then the possible choices for the children are:

GIRL - girl OR.......

GIRL - boy

OR .... say they told you that the 2nd child is in fact the girl, then the possible outcomes for both children are:

girl - GIRL OR......

boy - GIRL

So if you don't know which one it is, the first or the second then the outcomes can be any of these, so the possible outcomes are:

GIRL - girl

girl - GIRL

GIRL - boy

boy - GIRL

BUT... If you only include girl - girl once, then that means that you know which way round it is, that means that you are sayin that GIRL is known. BUT... HAVING IT TWICE SHOWS THAT THERE IS A 50% CHANCE OF BOTH CHILDREN BEING GIRLS AS YOU CAN SAY THAT THERE IS A CHANCE THAT THE FIRST CAN ALSO BE A GIRL IF THE KNOWN IS THE YOUNGER, AND THERE IS CHANCE THAT THE SECOND CHILD IS ALSO A GIRL IF THE KNOW IS THE OLDER OF THE TWO.

HOPE THIS MAKES SENSE!!!

ALSO SORRY!!!!! I FORGOT THAT YOU DIDN'T SPECIFY WHICH OF THE TWO THE GIRL IS!!!!

[Guest]

The other way to look at this is that there are only 3 possibilities (when order does not matter)

G - G

B - G

B - B

Since there cannot be a B - B, the probability of a second girl is again 50/50

I'll defer to my example with pennies again. Do you think if you laid out 100 pairs of pennies, only about one-third will be one heads and one tails? Try it and see. About 1/4 will be H-T and 1/4 will be T-H.

Okay, so he does not say that the first child is a girl, but one is. I still think it is still a 50% chance. You see that there are still four choices and at the beginning there are eight.

SORRY IF THIS IS WRONG BUT THIS IS WHAT I GET!!!

There are 4 choices after the first girl, in capitals is the first girl:

Child 1 - Child 2

GIRL ------- Girl

Girl ------- GIRL

GIRL ------- Boy

Boy ------- GIRL

This means you think the possibilities including having two boys look like this:

GIRL ------- Girl

Girl ------- GIRL

GIRL ------- Boy

Boy ------- GIRL

BOY ------- boy

Boy ------- BOY

Do you think a couple planning on having two children has a 1/3 chance of having two boys?

Are you using G-g and g-G because the girls could have been born in a different order? Your chance of coming up heads twice in a row on two successive coin flips is 25%. 50% on the first flip and 50% on the second flip (.50 x .50 = .25). The possibilities are H-H, T-T, H-T, and T,H (1 out of 4 [25%] you will have both heads). It doesn't matter that one coin could have been used before the other.

If we use your logic, the possibilities are H-h, h-H, H-t, T-t, t-T and T-h (2 out of 6 [33.3...%] you will have both heads).

Why don't you try my experiment with pennies? See post #182.

[Guest]

What a thread in which to record my first post! And yes, I've read all 19 pages - ah, the benefits of self-employment <_<

Whilst I intuitively agree with the 1/3 Camp, I can't get the following analysis out of my mind, so I'd like someone to put it to rest please!

We know the gender of one child is female.

We know there are 2 children in total.

That gives the following possible combinations of kids:

a) B1-G2

b) G1-B2

c) G1-G2

In other words (and others' words many times!) B1-B2 is ruled out.

That leaves 4 possible 'roles' in the above list for our known girl, respectively:

Younger sister of older brother, as per (a)

Older sister of younger brother, as per (b)

Older of 2 sisters, as per ( c)

Younger of 2 sisters, as per ( c)

Of those 4 'roles', and she can only occupy one of them:

2 result in the other child being a brother [(a) and (b)]

2 result in the other child being a sister [( c) and, erm ( c)]

Therefore, given all that we know (figuratively, "that the girl must fill one of those 4 roles"), then the likelihood of the other child being a brother or sister is equal. 50:50, 1/2, 50% aaaargh!

Someone please explain what I'm missing!

Thanks

[Guest]

What a thread in which to record my first post! And yes, I've read all 19 pages - ah, the benefits of self-employment <_<

Whilst I intuitively agree with the 1/3 Camp, I can't get the following analysis out of my mind, so I'd like someone to put it to rest please!

We know the gender of one child is female.

We know there are 2 children in total.

That gives the following possible combinations of kids:

a) B1-G2

b) G1-B2

c) G1-G2

In other words (and others' words many times!) B1-B2 is ruled out.

That leaves 4 possible 'roles' in the above list for our known girl, respectively:

Younger sister of older brother, as per (a)

Older sister of younger brother, as per (b)

Older of 2 sisters, as per ( c)

Younger of 2 sisters, as per ( c)

Of those 4 'roles', and she can only occupy one of them:

Welcome to the boards, DrRhythm. Who is this "she" and "known girl" you're talking about? We know that a couple has two kids and that at least one is a girl. That is why we look at the possibilities exactly like you laid it out the first time:

a) B1-G2

b) G1-B2

c) G1-G2

Let's use names instead of G1 and G2. Lets say a couple plans on having two kids. If they have one girl, they will name her Sara. If they have a second girl, they will name her Barbara. If they have one boy, they will name him Bob. If they have a second boy, they will name him Mark. There are four equally possible outcomes:

Sara-Barbara

Bob-Mark

Bob-Sara

Sara-Bob

If we are told that the couple had at least one girl, we delete the possibility of Bob-Mark.

Their kids must be:

Sara-Barbara or

Bob-Sara or

Sara-Bob

One out of three chance that both kids are girls.

[Guest]

We know the gender of one child is female.

We know there are 2 children in total.

That gives the following possible combinations of kids:

a) B1-G2

b) G1-B2

c) G1-G2

In other words (and others' words many times!) B1-B2 is ruled out.

That leaves 4 possible 'roles' in the above list for our known girl, respectively:

Younger sister of older brother, as per (a)

Older sister of younger brother, as per (b)

Older of 2 sisters, as per ( c)

Younger of 2 sisters, as per ( c)

Of those 4 'roles', and she can only occupy one of them:

2 result in the other child being a brother [(a) and (b)]

2 result in the other child being a sister [( c) and, erm ( c)]

Therefore, given all that we know (figuratively, "that the girl must fill one of those 4 roles"), then the likelihood of the other child being a brother or sister is equal. 50:50, 1/2, 50% aaaargh!

Someone please explain what I'm missing!

Thanks

I think that what is confusing people about this question is matter of comparing apples to oranges. If we were interested in which of the children our known girl was, then 50% might make sense (I haven't done the math,) but we're only interested in which situation has occurred. As DrRhythm stated in the above post, there are only three possible situations for the information provided. The probability that we are seeking is the probability of which situation we are in. It does not matter which girl she happens to be, she can only exist in one of (a), (b), or ( c). And since all three situations should occur with equal probability, the probability of the situation with two girls occurring is 1/3.

I'm hoping that this explanation might make the answer clearer for the people who are confused and don't have the time (or pennies) to try Martini's well-advertised experiment.

[Guest]

I'll defer to my example with pennies again. Do you think if you laid out 100 pairs of pennies, only about one-third will be one heads and one tails? Try it and see. About 1/4 will be H-T and 1/4 will be T-H.

I would agree that if you laid out 100 pairs of pennies you would see 1/4 H-T and 1/4 T-H. Now if order does not matter we see 1/2 of all pairs contain an H and a T, 1/4 2 T's and 1/4 2 H's. So given a T what is the probability of another T? I would say 25%. Given an H the probability of another H is 25%. So given an H or a T the probability of another is 50/50. Can we not then conclude that given a boy or a girl the probability of having the other being the same is also 50/50? Ah...so confusing! HELP.

I was also thinking that if the answer is actually 1/3 could we extend this thought to roulette? There were 2 spins of the wheel and one came up red, the probability of the other spin being red as well is 1/3. Can we make money knowing this if we get about even odds on betting red or black? NEED MORE HELP!

[Guest]

Can we not then conclude that given a boy or a girl the probability of having the other being the same is also 50/50? Ah...so confusing! HELP.

No. The reasons why have been explained multiple times. I don't know how else to help you. You just stated above:

"Now if order does not matter we see 1/2 of all pairs contain an H and a T, 1/4 2 T's and 1/4 2 H's."

Isn't it obvious if we call tails girls and heads boys, and we are told there is at least one tail, the three remaining equally probable results are H-T, T-H and T-T?

I was also thinking that if the answer is actually 1/3 could we extend this thought to roulette? There were 2 spins of the wheel and one came up red, the probability of the other spin being red as well is 1/3. Can we make money knowing this if we get about even odds on betting red or black? NEED MORE HELP!

I wish! All spins are independent of another. Had the original riddle been: "They have two kids, the first child is a girl, what is the probability that the other kid is also a girl?", the answer would be 1/2.

The possibilities would be:

GB

GG

If we followed NintendoMad's logic (also DrRhythm's and your logic), the possibilities would be:

GIRL - girl

girl - GIRL

GIRL - boy

Likewise, if the question was, "A roulette wheel (with no green spots, just an equal amount of black and red) was spun twice. One of the spins came up red. What is probability the other spin came up red?", the answer would be 1/3.

R-B

B-R

R-R

But there's no way to make money on roulette knowing this probability (even if you could find a wheel without green) because it's only relevant after the fact. We couldn't know in advance that one of the spins will be red.

"A roulette wheel just landed on red four times in a row. Will the next spin most likely be black?" Nope. Assuming it's a fair wheel, prior spins are irrelevant.

[Guest]

sample dpace for the 2 children

BB/BG/GB/GG ( we dont know which one is girl). As one is girl we drop BB combination

leaving GB/BG/GG.

therefore p(other is also girl) is 1/3

[Guest]

No. The reasons why have been explained multiple times. I don't know how else to help you. You just stated above:

"Now if order does not matter we see 1/2 of all pairs contain an H and a T, 1/4 2 T's and 1/4 2 H's."

Isn't it obvious if we call tails girls and heads boys, and we are told there is at least one tail, the three remaining equally probable results are H-T, T-H and T-T?

I wish! All spins are independent of another. Had the original riddle been: "They have two kids, the first child is a girl, what is the probability that the other kid is also a girl?", the answer would be 1/2.

The possibilities would be:

GB

GG

If we followed NintendoMad's logic (also DrRhythm's and your logic), the possibilities would be:

GIRL - girl

girl - GIRL

GIRL - boy

Likewise, if the question was, "A roulette wheel (with no green spots, just an equal amount of black and red) was spun twice. One of the spins came up red. What is probability the other spin came up red?", the answer would be 1/3.

R-B

B-R

R-R

But there's no way to make money on roulette knowing this probability (even if you could find a wheel without green) because it's only relevant after the fact. We couldn't know in advance that one of the spins will be red.

"A roulette wheel just landed on red four times in a row. Will the next spin most likely be black?" Nope. Assuming it's a fair wheel, prior spins are irrelevant.

Thanks for your help! It might slowly be sinking in now, but I have a problem getting over the order thing:

If the first child is a girl then the odds of the second being a girl is 50/50...we agree on this

If the second child is a girl then the odds of the first child being a girl are also 50/50...we should agree on this.

So if either the first or second child is a girl the odds of the other being a girl is 50/50....is there a fallacy here?

I was also thinking the problem could be restated as a couple has 2 children and both are not boys, what is the probability of 2 girls?

Now it makes sense as I see 1/4 * 4/3 = 1/3 EUREKA

[Guest]

Thanks for your help! It might slowly be sinking in now, but I have a problem getting over the order thing:

If the first child is a girl then the odds of the second being a girl is 50/50...we agree on this

If the second child is a girl then the odds of the first child being a girl are also 50/50...we should agree on this.

So if either the first or second child is a girl the odds of the other being a girl is 50/50....is there a fallacy here?

Yes, the possibilities say it all:

First child a girl, second unknown:

GB

GG

At least one of two children a girl:

GB

GG

BG

I was also thinking the problem could be restated as a couple has 2 children and both are not boys, what is the probability of 2 girls?

Now it makes sense as I see 1/4 * 4/3 = 1/3 EUREKA

Cool!

[itachi-san]

I was also thinking the problem could be restated as a couple has 2 children and both are not boys, what is the probability of 2 girls?

Now it makes sense as I see 1/4 * 4/3 = 1/3 EUREKA

Wouldn't that new variation just equal 1 since there's no possibility of there being a boy?

[Guest]

Wouldn't that new variation just equal 1 since there's no possibility of there being a boy?

That thought went through my head also. Then I realized what he meant by both not being boys is that BB is not possible, but BG, GB and GG are. Just a rewording of the riddle.

[Guest]

If you say that Gg and gG are the same, then Gb and bG would be the same too. Therefore, there are only 3 original possibilities:

GG

BB

GB/BG

We eliminate BB and then we have a 50% chance of the other child being a girl.

The puzzle doesn't specify order. Look at it this way: What if the two children are twins, born simultaneously (maybe by c-section)?

[Guest]

If you say that Gg and gG are the same, then Gb and bG would be the same too. Therefore, there are only 3 original possibilities:

GG

BB

GB/BG

We eliminate BB and then we have a 50% chance of the other child being a girl.

I went over this already with you and several others. I asked several times: If you think that's the case, try the penny experiment. If you lay out 100 pairs of pennies, what do you think the results will be? They will be the following (approximately):

BB 25%

GG 25%

GB 25%

BG 25%

You obviously disagree. Tell me what you think you'll get.

You wrote earlier:

Let's use uppercase (G) for the girl we know. We'll use lowercase (b or g) for the child we don't know. Here are the possibilities:

Gg

Gb

bG

gG

Using that logic, the possibilities including two boys would be:

Gg

Gb

bG

gG

Bb

bG

Do you think a couple planning on having two children has a 1/3 probability of having both be girls?

The puzzle doesn't specify order. Look at it this way: What if the two children are twins, born simultaneously (maybe by c-section)?

Exactly. Order doesn't matter. All that matters is that there's equal probability of having a boy or girl in each case and that at least one is a girl.

[Guest]

I went over this already with you and several others. I asked several times: If you think that's the case, try the penny experiment. If you lay out 100 pairs of pennies, what do you think the results will be? They will be the following (approximately):

This riddle is about two children, not 200.

So, let's place two pennies down on the table, covered with something. The orientation of each penny is set and cannot change, so we're not talking about the chance of a future coin toss.

We reveal one of the pennies. It is heads. What is the probability that the other is heads?

[Guest]

This riddle is about two children, not 200.

So, let's place two pennies down on the table, covered with something. The orientation of each penny is set and cannot change, so we're not talking about the chance of a future coin toss.

We reveal one of the pennies. It is heads. What is the probability that the other is heads?

You're not getting it. This riddle IS NOT about the chance of a future coin toss. It's about a couple that has two children. We know at least one of them is a girl. What are the chances the other is a girl? This is entirely different than what you're setting up with your two pennies.

My example using 200 pennies is to show you where you're wrong. If you you're not going to respond to my last post, I'm wasting my time.

[Guest]

I went over this already with you and several others. I asked several times: If you think that's the case, try the penny experiment. If you lay out 100 pairs of pennies, what do you think the results will be? They will be the following (approximately):

BB 25%

GG 25%

GB 25%

BG 25%

You obviously disagree. Tell me what you think you'll get.

I think I'll get:

2 boys

1 boy 1 girl (order doesn't matter, as you agree -- quoted below)

2 girls.

Exactly. Order doesn't matter. All that matters is that there's equal probability of having a boy or girl in each case and that at least one is a girl.

[Guest]

I think I'll get:

2 boys

1 boy 1 girl (order doesn't matter, as you agree -- quoted below)

2 girls.

Are you saying:

2 boys 33.3...%

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

2 girls 33.3...%