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

[Yoruichi-san]

Yeah...probability is confusing like that...but the main thing people don't realize (or realize they know) is that probability is dependent on observation...i.e. the probability of an event changes after it has been observed (or becomes statistics instead of probability)...

Ever heard of Schrodinger's Cat or the collapse of the wavefunction? Well, to put it in a simple example: before you toss a coin, there is a 1/2 chance of heads and same for tails. Once you toss the coin, if you observe it turns up heads, that eliminates the possibility of tails, and then the probability becomes 1 that it was heads and 0 that it was tails. If you don't observe it (i.e. you don't know the result), the probability is still 1/2 1/2.

If you toss two coins, the probability of heads for each is 1/2. So if you ask "what is the probability of coin i being heads", the probability is 1/2. If you toss one coin first and observe the result, then ask what the probability of heads for the coin 2 is, the answer is 1/2.

However, if you toss 2 coins, the probability is 1/4 for each of the combos: HH HT TH TT. Now you observe one of the two is a head, so that eliminates the possibility of TT, but you don't know whether it is coin 1 or coin 2 and you don't have an observation for the other coin, so you have three equally likely possibilities, so the probability of any of the three is 1/3.

[Guest]

However, if you toss 2 coins, the probability is 1/4 for each of the combos: HH HT TH TT. Now you observe one of the two is a head, so that eliminates the possibility of TT, but you don't know whether it is coin 1 or coin 2 and you don't have an observation for the other coin, so you have three equally likely possibilities, so the probability of any of the three is 1/3.

This is wrong. If you randomly observed one of the two and see that it landed on heads, then the probability the other landed on heads is 1/2. In this example, you can just replace "first coin' with "observed coin". Your probabilities are now:

observed coin:H

unobserved coin:H

observed coin:H

unobserved coin:T

This is different than the situation in the riddle. However, if the riddle stated that the couple has two children and you went to visit their house and one of the children answered the door and the child was a girl, the probability that the other child is also a girl is 1/2.

Now, you may be thinking, "Wait a minute here. What's the difference if you see one of the children and it's a girl vs. you don't see one and are told at least one is a girl? Why can't we do the same thing in the riddle and claim one child is a girl and the probability that the other child is also a girl is 1/2?"

Well, that's pretty much been explained above. Once you observe a child randomly (we'll assume there's just as much of a chance of either boy or girls answering the door), the probability that her sibling is a girl is 1/2. We are zeroing in on a specific individual, while in the riddle we are picking the physical possibilities from the expanded set of all possible permutations. These are your probabilities:

- Girl at the door, girl in the backyard

- Girl at the door, boy in the backyard

Now, lets suppose you didn't visit the family yourself. A friend of yours did, and reported: "The couple has two children. One answered the door. The other was out in the backyard playing at the time. At least one of them is a girl." These are your probabilities:

- Boy at the door, girl in the backyard

- Girl at the door, boy in the backyard

- Girl at the door, girl in the backyard

If you know that one specific child is a girl, it's different from knowing that 'the kids are not both boys', (which is the same thing as saying that at least one of the children is a girl but not specifying which in any way.)

[Yoruichi-san]

This is wrong. If you randomly observed one of the two and see that it landed on heads, then the probability the other landed on heads is 1/2. In this example, you can just replace "first coin' with "observed coin". Your probabilities are now:

observed coin:H

unobserved coin:H

observed coin:H

unobserved coin:T

This is different than the situation in the riddle. However, if the riddle stated that the couple has two children and you went to visit their house and one of the children answered the door and the child was a girl, the probability that the other child is also a girl is 1/2.

Now, you may be thinking, "Wait a minute here. What's the difference if you see one of the children and it's a girl vs. you don't see one and are told at least one is a girl? Why can't we do the same thing in the riddle and claim one child is a girl and the probability that the other child is also a girl is 1/2?"

Well, that's pretty much been explained above. Once you observe a child randomly (we'll assume there's just as much of a chance of either boy or girls answering the door), the probability that her sibling is a girl is 1/2. We are zeroing in on a specific individual, while in the riddle we are picking the physical possibilities from the expanded set of all possible permutations. These are your probabilities:

- Girl at the door, girl in the backyard

- Girl at the door, boy in the backyard

Now, lets suppose you didn't visit the family yourself. A friend of yours did, and reported: "The couple has two children. One answered the door. The other was out in the backyard playing at the time. At least one of them is a girl." These are your probabilities:

- Boy at the door, girl in the backyard

- Girl at the door, boy in the backyard

- Girl at the door, girl in the backyard

If you know that one specific child is a girl, it's different from knowing that 'the kids are not both boys', (which is the same thing as saying that at least one of the children is a girl but not specifying which in any way.)

No, but you didn't observe a particular coin, what you observed (or in this case, was told) was that *one* of the two coins was heads, sorry, I guess the phrasing makes it confusing (I'm used to talking about this in terms of electrons and wavefunctions...), I guess think of it in terms of a computer simulation of a coin toss where the results screen tells you that one of the coins (not a particular coin) was heads...

[Guest]

No, but you didn't observe a particular coin, what you observed (or in this case, was told) was that *one* of the two coins was heads

If that had been what you actually wrote, then I would have never claimed you were wrong as that would have been a rewording of the riddle.

[Guest]

It is really amusing that some people call this a lousy puzzle and yet there are 23 pages of replies and debates for this "Lousy Puzzle" , So we can assume it is not so Lousy after all

[Yoruichi-san]

If that had been what you actually wrote, then I would have never claimed you were wrong as that would have been a rewording of the riddle.

Umm, sorry for being confusing. I understand where you are coming from, but the way I see it, it's not the wording, it's how you read the wording:

"You observe one of the two is heads" can be read either:

"You observe one of the two is heads" -i.e. you observe one coin, or

"You observe one of the two is heads" -i.e, you observe the phenomenon that one out of two coins is heads...I know the perspective of a person who has not suffered through as much quantum physics as I have ;P this does not make much sense, since when usually talking about coins, "observe" means to "see", but this is not necessarily the case in probability theory, especially when applied to physics, for example, you can't "see" an electron...but scientists "observe" phenomenon by measuring certain properties which are indicative of that phenomenon...

So in summary, I apologize for being confusing and inconsiderate of the perspective of normal people, but I don't think I was wrong in what I said, based on my own experiences (which have been difficult...which is probably why I feel the need to say this :/)

[Yoruichi-san]

Umm, sorry for being confusing. I understand where you are coming from, but the way I see it, it's not the wording, it's how you read the wording:

"You observe one of the two is heads" can be read either:

"You observe one of the two is heads" -i.e. you observe one coin, or

"You observe one of the two is heads" -i.e, you observe the phenomenon that one out of two coins is heads...I know the perspective of a person who has not suffered through as much quantum physics as I have ;P this does not make much sense, since when usually talking about coins, "observe" means to "see", but this is not necessarily the case in probability theory, especially when applied to physics, for example, you can't "see" an electron...but scientists "observe" phenomenon by measuring certain properties which are indicative of that phenomenon...

So in summary, I apologize for being confusing and inconsiderate of the perspective of normal people, but I don't think I was wrong in what I said, based on my own experiences (which have been difficult...which is probably why I feel the need to say this :/)

Anyways, my original point was that there is a difference between the probability of the second coin being heads based on whether you "observe" a particular coin or whether you "observe" the phenomenon of one out of the two coins being heads, because the fact that you have "observed" something changes the probability (and differently depending on what you observed).

In the first case, you only "observed" one coin, which now has a probability of 1 of being heads and 0 of being tails. So the probability of the second coin is still 1/2 1/2. In the second case, you "observed" a fact about both coins, and that fact (that one coin is heads) changes the probability to 1/3 2/3 since it removes the possibility of TT.

The arguments for 1/2 were the first case, and the arguments for 1/3 were the second case. This problem, as far as I can tell, is the second case.

Edited July 25, 2008 by Yoruichi-san

[bonanova]

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

GG - that would be 1.

GB - that would be 2.

And BG - that would be - (drum roll, please) - 3.

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.

The problem you're describing is this:

I have two children with androgenous names of Taylor and Alison.

I tell you that Taylor is a girl.

I ask you the probability that Alison is a girl.

You reply 1/2.

You are correct.

That's not really much of a puzzle now, is it? And it's not what the OP asks.

Here's a hint:

The statements "one of the children is a girl" and "child number one is a girl" are different.

[1] Try to figure out why they are different.

[2] Then try to figure out which statement is in the OP.

Here's another hint:

If you believe the answer is 1/2, then you must believe that two-girl families and mixed-gender families occur with equal probability.

Why?

Because if one is a girl and the other is a girl has a 1/2 probability,

then the only other case one is a girl and the other is a boy must have the remaining 1/2 probability.

That's a tough position to defend.

But something tells me you might try.

[Guest]

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.

We are dealing with probability. Probability is mathematical formulation of probable outcomes and in such a formula is required. In the math world is is called a "proof" and for a good reason. What is the mathematical process behind this elimination? Because if this is true then surly to god their is a guy out their that has written it down. For example my formula

P=n/N ( well it's not mine... it's actually the formula for probability )

Kidx/Pos * Kidy/Pos = Prob

because we one the value of Kidx then Kidx has a 100% chance of being that value.

1/1 * 1/2 = 50%

This formula works with cards, dice, kids, anything.

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."

Let me ask you a question. What if the riddle where asking the possibility of the other child being a boy? Would the answer still be 1/3? or 2/3? If it is 2/3 then you have to agree that GB/BG are equal. That being said if GB/BG are equal then one has to be removed along with BB because it can only be one or the other and it dose not matter what one because they are equal! But if it is 1/3 then their is something wrong with your math. Because 1/3 possibility of a girl + 1/3 possibility of a boy dose not equal 100%.

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.

If you read my post I didn't say that this was wrong. I said that

GG = 1/3

GB = 1/3

BG = 1/3

is where it go's wrong.

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?

You missed the point of the experiment. Give it a try and see what happens. Get another person (wife/friend/child...) to toss a coin in the air twice and record the outcome on a piece of paper making sure that you never know what they are. Then ask them to pick one of the two at random (in fact they could use the coin for this as well) and tell you what the value is. Now this has the possibility of being toss one or toss two / heads or tails... we don't know. But what are the odds that the other toss is going to be of the same value? example if they get heads for toss one and tails for toss two and picks tails at random what are the odds the the other coin was tails. If she picks heads what is the odds that the other one is going to be heads. Try it out. That is what my experiment was about. What you will find is that 50% of the time what ever toss they pick will be the same as the other toss. This is what is asked in the riddle. We know that one of them is a girl what is the odds of the other one being a girl aka in the coin toss the same value. I would be vary interested in knowing your results.

The statements "one of the children is a girl" and "child number one is a girl" are different.

Not for the purpose of this equation.

If you believe the answer is 1/2, then you must believe that two-girl families and mixed-gender families occur with equal probability.

Why?

No I believe that same gender families are just as likely as mixed-gender families.

but you must believe that it is imposable for a woman to have two boys?

[Guest]

We are dealing with probability. Probability is mathematical formulation of probable outcomes and in such a formula is required. In the math world is is called a "proof" and for a good reason. What is the mathematical process behind this elimination? Because if this is true then surly to god their is a guy out their that has written it down. For example my formula

P=n/N ( well it's not mine... it's actually the formula for probability )

And bonanova told you that he used the same formula. You asked what the formula for knowing the value of N is, not P. I told you that N is derived by counting the number of probable outcomes. What formula did you use for getting the value of N?

Let me ask you a question. What if the riddle where asking the possibility of the other child being a boy? Would the answer still be 1/3? or 2/3? If it is 2/3 then you have to agree that GB/BG are equal.

Equal to what? GB is equal to BG. If a couple has two children, they each have an equal probability of occurring (1/4). Which means the probability of a couple having one girl and one boy in any order is derived by adding the two (1/2).

If we know a couple has two children but none are boys (rephrased as at least one is a girl) then GB and BG still have an equal probability as the other along with GG. Each has a 1/3 probability. The probability that the couple has one girl and one boy in any order is derived by adding GB and BG (2/3).

You already admitted that the following is correct:

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

If you combine GB and BG, the probability that a couple with two kids having one of each is 1/2.

If we know that one of the children is a girl, only BB gets removed and the remaining possibilities have an equal chance of existing:

GG = 1/3

GB = 1/3

BG = 1/3

Since the other child being a girl means both must be girls, that only happens 1/3 of the time.

That being said if GB/BG are equal then one has to be removed along with BB because it can only be one or the other and it dose not matter what one because they are equal! But if it is 1/3 then their is something wrong with your math. Because 1/3 possibility of a girl + 1/3 possibility of a boy dose not equal 100%.

Of course it's only one or the other. But BOTH are still possibilities and neither can get removed.

They are both equal in what way? Do you mean they are the same thing because it's a combination of boy/girl? That doesn't matter. If it did matter, then you wouldn't have agreed that the following are the probabilities for a couple with two children:

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

According to your rationale, since GB and BG are the same, it should look like this (since you say we can remove either GB or BG, I removed BG):

GG = 1/3

BB = 1/3

GB = 1/3

If you read my post I didn't say that this was wrong. I said that

GG = 1/3

GB = 1/3

BG = 1/3

is where it go's wrong.

Then why don't you think this is wrong:

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

Why aren't you making the same claim as above and claiming that GB/BG are equal to the rest?

You missed the point of the experiment. Give it a try and see what happens. Get another person (wife/friend/child...) to toss a coin in the air twice and record the outcome on a piece of paper making sure that you never know what they are. Then ask them to pick one of the two at random (in fact they could use the coin for this as well) and tell you what the value is. Now this has the possibility of being toss one or toss two / heads or tails... we don't know. But what are the odds that the other toss is going to be of the same value? example if they get heads for toss one and tails for toss two and picks tails at random what are the odds the the other coin was tails. If she picks heads what is the odds that the other one is going to be heads. Try it out. That is what my experiment was about. What you will find is that 50% of the time what ever toss they pick will be the same as the other toss. This is what is asked in the riddle. We know that one of them is a girl what is the odds of the other one being a girl aka in the coin toss the same value. I would be vary interested in knowing your results.

This is plain silly. Of course the probability of one coin toss being compare to random coin toss matching up is 1/2. That's not at all reflective of the riddle. What Martini did with your coin tosses is, and the result was that the other coin was tails 37.5% of the time.

What you're asking us to do has zero to do with the riddle.

If we do your experiment, the outcome is going to include HH (BB) outcomes that you didn't bother removing and they must be removed to adhere to the conditions of the riddle (BB is an impossibility since we know one child is a girl).

Not for the purpose of this equation.

Are you serious? They are totally different statements and will of course affect the equation!

One of the children is a girl:

P = n/N

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

Child number one is a girl

P = n/N

N = 2 [GG GB] - if you know the first child is a girl, these are the possible outcomes.

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

p = 1/2

No I believe that same gender families are just as likely as mixed-gender families.

Well, then tell us what probabilities exist after you remove the BB possibility from this:

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

You tell us it isn't:

GG = 1/3

GB = 1/3

BG = 1/3

Then what is it?

but you must believe that it is imposable for a woman to have two boys?

Huh? No one ever said that. If you mean we believe it's impossible for a woman to have two boys when we know she only has two children and one of them is a girl, then yes, we do believe it's impossible.

Did you look at the Ask Dr. Math article? That's wrong too?

Here's another math website:

http://www.mathpages.com/home/kmath036.htm

Read the third to last paragraph starting with "Suppose there are 100 fathers". Do you agree or disagree with the result? If you agree, then you should agree with our answer here also as it's the same problem, only reworded. If you don't agree, shouldn't you maybe reconsider since two math websites, an article on brain teasers, and two moderators on an internet forum that specializes in brain teasers all agree with one another and the conclusion is different than yours?

[Guest]

And bonanova told you that he used the same formula. You asked what the formula for knowing the value of N is, not P. I told you that N is derived by counting the number of probable outcomes. What formula did you use for getting the value of N?

That's my point I didn't. If you are going to remove BB as a possibility then you must remove one of the GB or BG. Why? Because it can only be one or the other not both. "but we don't know what one". That's fine because they are equal so it doesn't matter what one.

Equal to what? GB is equal to BG. If a couple has two children, they each have an equal probability of occurring (1/4). Which means the probability of a couple having one girl and one boy in any order is derived by adding the two (1/2).

Thank you that's my point. Now tell me why when we know that it can only be ether GB or GB we can not remove one of them if they are equal.

GG = Can be

BG = Can be only if it's not GB

GB = Can be only if it's not BG

BB = Cant be

But if GB and BG are equal then one can safely be removed.. They are the same thing. But It can only be one of them. So BAM! One of them is gone! If they are not the same thing then you can't combine them at the end. So your probability of the other kid being a boy would be 1/3 as well. And this is not correct.

If we know a couple has two children but none are boys (rephrased as at least one is a girl) then GB and BG still have an equal probability as the other along with GG. Each has a 1/3 probability. The probability that the couple has one girl and one boy in any order is derived by adding GB and BG (2/3).

I think you mean can't be two boys. But again thank you. The question I have is why are they only equal only after we remove BB? Why can we not remove one of them with BB?

Of course it's only one or the other. But BOTH are still possibilities and neither can get removed.

Why not if they are both the same?

They are both equal in what way? Do you mean they are the same thing because it's a combination of boy/girl? That doesn't matter. If it did matter, then you wouldn't have agreed that the following are the probabilities for a couple with two children:

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

According to your rationale, since GB and BG are the same, it should look like this (since you say we can remove either GB or BG, I removed BG):

GG = 1/3

BB = 1/3

GB = 1/3

What? What I am saying is that they are equal like

GB = 1/4

BG/GB = 1/2

BB = 1/4

Let's do it with single variables so we don't think order has something to do with it.

GG = X BB = Y GB = Z GB = Z

so it looks like this

X = 1/4

Y = 1/4

Z = 1/4

Z = 1/4

note their are two Z's because they are the same thing. Now we know that one of them is a girl. That means it can't be "Y" because Y is only boys. But we also know that it can only be one of the "Z"'s. So we remove one of them. So it now looks like

X = 1/2

Z = 1/2

Then why don't you think this is wrong:

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

Why aren't you making the same claim as above and claiming that GB/BG are equal to the rest?

Lol GB and BG are equal to each other. Where you go wrong is not removing one of them. Because according to you they are only equal after we remove BB. I agree that GG BB BG GB all have a 1/4 chance. It's your next step. The removal of BB without ether GB or BG.

(I ran out of quote tags)

"What Martini did with your coin tosses is, and the result was that the other coin was tails 37.5% of the time.

What you're asking us to do has zero to do with the riddle.

If we do your experiment, the outcome is going to include HH (BB) outcomes that you didn't bother removing and they must be removed to adhere to the conditions of the riddle (BB is an impossibility since we know one child is a girl)."

OOoo I see! So a pack of wild dogs swept through the world and killed every family with two boys? What Martini did was ignore the experiment and tell me I'm Wong. It has everything to do with the riddel! It is just as likely that we could have been told that one of them was a boy is it not? What you are assuming is the it is imposable for anyone to have a boy. The only reason why we can remove two boys a** a possibility is because we know that one of them is a girl. That also stipulates that it can't be GB and GB but only one of them.

"Are you serious? They are totally different statements and will of course affect the equation!

One of the children is a girl:

P = n/N

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

Child number one is a girl

P = n/N

N = 2 [GG GB] - if you know the first child is a girl, these are the possible outcomes.

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

p = 1/2 "

So GB and GB are not equal?

Well, then tell us what probabilities exist after you remove the BB possibility from this:

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

You tell us it isn't:

GG = 1/3

GB = 1/3

BG = 1/3

Then what is it?

lol. This is good. Yes if you remove BB it is correct..... But what makes you think that you can only remove BB? Because GB and GB and not equal to each other?

Did you look at the Ask Dr. Math article? That's wrong too?

Yup and Yup.

Here's another math website:

http://www.mathpages.com/home/kmath036.htm

Read the third to last paragraph starting with "Suppose there are 100 fathers". Do you agree or disagree with the result? If you agree, then you should agree with our answer here also as it's the same problem, only reworded. If you don't agree, shouldn't you maybe reconsider since two math websites, an article on brain teasers, and two moderators on an internet forum that specializes in brain teasers all agree with one another and the conclusion is different than yours?

Did you read this? He says that im right about the kids thing.

However, by the "unwritten rule of probability riddles" (the UROPR),

let's assume that all specified choices are to be made randomly from

the available options. Thus we assume that if the man has one child

of each gender he is equally likely to report "At least one boy" as

to report "At least one girl". It follows that each of these reports

has a probability of 1/2. For clarity, let's denote C1 as [boys>=1],

and C3 as [told:boys>=1]. On this basis we can use Bayes' formula

to compute the conditional probability as follows

Pr{ [told:boys>=1] AND total=3 }

Pr{ total=3 | [told:boys>=1] } = ---------------------------------

Pr{ told:boys>=1 }

The denominator is known to be 1/2, and the numerator is 1/4 because

the total equals 3 exactly half the time, and (by assumption) the

father reports "At least one boy" exactly half of those times. Thus

the conditional probability of "total=3" given that the father tells

us "At least one boy" is exactly 1/2.

Wow........ um..... you posted the link not me. Now let's look at the next part.

Suppose there are 100 fathers in an auditorium, and each is the father

of two children. Each father is instructed to tell you (truthfully)

if at least one of his children is a boy. This will apply to about

75 of the fathers. Now, of those 75 Dads, 2/3 (i.e., 50) have a

daughter, and 1/3 (i.e., 25) have two sons. Thus, if you want to

guess the gender of their "other" child, the chances are 2/3 that

it is a girl. (Of course, for the remaining 25 fathers - those

who did not report at least one son - you know immediately they

have two daughters.)

However, suppose instead that all 100 fathers were instructed to tell

you either (a) "At least one of my children is a boy" or (b) "At least

one of my children is a girl". Based on what each father tells you,

you try to guess the gender of his "other" child. Strictly speaking

this problem is indeterminate, but if it's also stipulated that fathers

with both a son and a daughter should flip a coin to decide what to

tell you, then the probability that the "other" child is of the

opposite gender is exactly 1/2. The breakdown is

25 have two sons, and they report at least one son

25 have a son and daughter, and report at least one son

25 have a son and daughter, and report at least one daughter

25 have two daughters, and they report at least one daughter

Thus, regardless of what a particular father reports, you have only a

50% chance of correctly guessing the gender of his "other" child.

So. What's the problem? What he is saying is that if instructed to tell you if he has a least one boy then it is 1/3.... Nothing wrong with that. But if instructed to tell you at least one girl or at least one boy...... it's 1/2. Well he says "stricly speeking this problem is indertminate" and i'll take that because we don't know why he is suplying the information. But im going on the "unwritten rule of probability riddles".

Edited July 26, 2008 by toxicoxyde

[Guest]

Thank you that's my point. Now tell me why when we know that it can only be ether GB or GB we can not remove one of them if they are equal.

They have equally possibility of being the situation, but they are still different situations. They are each equal to GG also. Can we remove GG? No. It's still a possibility. They are equal in that they all have an equal probability of being the situation- that's why we divide them equally and assign a 1/3 probability to each.

GG = Can be

BG = Can be only if it's not GB

GB = Can be only if it's not BG

BB = Cant be

But if GB and BG are equal then one can safely be removed.

All three are equal to one another (BB is removed), but none of the remaining three can be removed. What you wrote should have looked like this:

GG = will be if it's not BG or GB

BG = will be if it's not GG or GB

GB = will be if it's not GG or BG

BB = Cant be

They are the same thing. But It can only be one of them. So BAM! One of them is gone!

No, magic does not apply here. They are not the same thing. They have equal probability...along with GG, but we can't just throw out GG because of that, can we?

If they are not the same thing then you can't combine them at the end. So your probability of the other kid being a boy would be 1/3 as well. And this is not correct.

You can combine them because we are interested in situations where both children are girls. Any other situation can be added together and will look like this:

GB or BG: 2/3

GG: 1/3

I think you mean can't be two boys. But again thank you. The question I have is why are they only equal only after we remove BB? Why can we not remove one of them with BB?

They have equal probability. THIS DOES NOT MEAN ONE CAN BE REMOVED! GG also has an equal probability, but as I said above, that doesn't mean we can throw GG out as it is still a possibility. You can't throw GB or BG out either, but if we're not interested in any situations that aren't GG, we can add them together and say they represent situations in which both children are not girls (2/3).

Why not if they are both the same?

They have equal probability. That's not the same as being the same.

Let's do it with single variables so we don't think order has something to do with it.

GG = X BB = Y GB = Z GB = Z

so it looks like this

X = 1/4

Y = 1/4

Z = 1/4

Z = 1/4

note their are two Z's because they are the same thing. Now we know that one of them is a girl. That means it can't be "Y" because Y is only boys. But we also know that it can only be one of the "Z"'s. So we remove one of them. So it now looks like

X = 1/2

Z = 1/2

You can not just remove a Z just because there is another Z. They are both represented by a Z, but each represents a separate possibility (so you should not have labeled them both as Z). In one the boy is born first, in the other the girl is born first . But we can add them together, because we're not interested in order. But we can't just throw one out and pretend that possibility doesn't exist.

You didn't throw one out before we knew that two boys weren't possible, why throw one out after? Read my last question again if you have to.

Why did you keep both Zs when you divided the four possibilities evenly? Because you should have! And you do the same thing after throwing out the Y, which is done after receiving the information that Y is an impossibility.

So, after we throw out the Y, our possibilities should be divided evenly and look like this:

X = 1/3

Z = 1/3

Z = 1/3

Since we're only interested in X situations, we can add all other situations together and define them as "situations in which both children are not girls" (or non X situations). It will then look like this:

X = 1/3

all others situations: 2/3

[Guest]

Lol GB and BG are equal to each other. Where you go wrong is not removing one of them. Because according to you they are only equal after we remove BB. I agree that GG BB BG GB all have a 1/4 chance. It's your next step. The removal of BB without ether GB or BG.

You didn't remove either GB or BG when putting your possibilities together before you knew at least one child was a girl, why do it after?

Please, pay attention to this part because I believe it's where you're getting tripped up.

After we find out BB is not possible, we remove it and it alone. That is all the information allows us to remove. Why is it you want to remove a GB or BG only after this information? If you're claiming they are the same, why not remove it when looking at the four possibilities a couple can have? You didn't do that; you said the following was correct:

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

You also correctly stated that we can combine BG and GB, but we have to add there probabilities together to look like this:

GG = 1/4

BG/GB = 1/2

BB = 1/4

But we can't just throw BG or GB away. Notice that the BG/GB combo have twice the probability of each of the others. Once we are told that the BB pair is impossible, it seems you want to do this:

GG = 1/2

BG/GB = 1/2

Now, what's wrong with that? What's wrong is that before we removed BB, the BG/GB pair had twice the chances of each of the others, now it magically has the same probability of the remaining possibility? Not only that, but it remains at 1/2 while only the GG pair increases? It shouldn't. It should still have twice the probability of each remaining set:

GG = 1/3

BG/GB = 2/3

It's more intuitive if you never bother combining BG with GB. First it's this:

GG = 1/4

BB = 1/4

GB = 1/4

BG = 1/4

All of the above have an equal probability. We know that we can remove BB, so we divide the three remaining possibilities evenly:

GG = 1/3

GB = 1/3

BG = 1/3

Both children are girls 1/3 of the time.

OOoo I see! So a pack of wild dogs swept through the world and killed every family with two boys?

You're getting sillier. A condition of the riddle is that the couple can't have two boys. Because of that we remove coins that represent BB pairs.

What Martini did was ignore the experiment and tell me I'm Wong.

He didn't ignore it. He used your own 20 coin tosses and removed HH pairs because he chose Tails (in an earlier part of this thread) to represent boys.

It has everything to do with the riddel! It is just as likely that we could have been told that one of them was a boy is it not?

It's irrelevant what else we could have been told or how likely that is. We work with what we were told. We first look at all the possibilities a couple can have regarding the sex of children when having two of them. You get this part just fine. If we're told that at least one is a girl, we remove the BB pair. You seem to get this too, so I don't understand why you're now claiming since we could have been told something different, it shouldn't get removed and you make the comment "So a pack of wild dogs swept through the world and killed every family with two boys?".

The big mistake you make is thinking that getting the information that a BB pair can't exist means you can remove a BG or GB pair. It makes no sense to randomly remove one as both pairs are still possibilities.

What you are assuming is the it is imposable for anyone to have a boy.

What? Where on Earth have I done that? Had I done that, I would have removed BB, BG, and GB and stated that the possibility that the other child is a girl is 1/1.

The only reason why we can remove two boys a** a possibility is because we know that one of them is a girl.

Correct.

That also stipulates that it can't be GB and GB but only one of them.

No, common sense stipulates that only one situation is possible. Of course the children can't be both GB and BG. They also can't be GB and GG at the same time either, can they? But you do not remove any combination that is still possible.

Did you read this? He says that im right about the kids thing.

Sigh. I can't believe I have to go through this.

No, he clearly says you're wrong. What you've done is gone ahead and skipped to the next paragraph which describes a different situation. Let's go through it.

Suppose there are 100 fathers in an auditorium, and each is the father

of two children. Each father is instructed to tell you (truthfully)

if at least one of his children is a boy. This will apply to about

75 of the fathers. Now, of those 75 Dads, 2/3 (i.e., 50) have a

daughter, and 1/3 (i.e., 25) have two sons. Thus, if you want to

guess the gender of their "other" child, the chances are 2/3 that

it is a girl. (Of course, for the remaining 25 fathers - those

who did not report at least one son - you know immediately they

have two daughters.)

This situation is exactly the same as the riddle. Out of 100 fathers that have two children, about 75 will have at least one boy. Notice that we throw out the fathers with two girls, because we're not interested in those situations, just as Martini threw out the coins that are HH pairs in your coin tosses.

Now, how many of these 75 fathers does the author claim has two sons? Is it half, which is what you're saying should happen in the riddle, or is it 1/3, which is what most of the rest of us are claiming, including another math website and an article on Brain Teasers?

Now, let's look at the next paragraph which you claim supports your conclusion to the riddle in the OP:

However, suppose instead that all 100 fathers were instructed to tell

you either (a) "At least one of my children is a boy" or (b) "At least

one of my children is a girl". Based on what each father tells you,

you try to guess the gender of his "other" child. Strictly speaking

this problem is indeterminate, but if it's also stipulated that fathers

with both a son and a daughter should flip a coin to decide what to

tell you, then the probability that the "other" child is of the

opposite gender is exactly 1/2. The breakdown is

25 have two sons, and they report at least one son

25 have a son and daughter, and report at least one son

25 have a son and daughter, and report at least one daughter

25 have two daughters, and they report at least one daughter

Thus, regardless of what a particular father reports, you have only a

50% chance of correctly guessing the gender of his "other" child.

This is a totally different situation. If it weren't, the author wouldn't give a different answer as he did in the first situation, would he?

In this situation, fathers with two sons (about 25% of the fathers) will tell you "At least one of my children is a boy". Fathers with two daughters (about 25 % of the fathers) will tell you "At least one of my children is a girl". Fathers that have one boy and one girl (about 50% of fathers) will flip a coin to decide what to tell you (but of course you're not aware of which fathers had to flip a coin). You have to guess what the gender of the child is after each statement. The author correctly states that you should guess correctly about 50% of the time.

Is this second situation the same as the riddle? No. Is the first situation; the one where the author comes up with the same answer as the OP and everyone arguing with you? YES.

[Guest]

Wow, I just found this site today and have read a few pages on this puzzle. I now feel I have to say something. I am not a mathmatician but do remember some stuff on probablity. I am going to have a go at explaining what I think the answer should be and why, gosh why not it looks like everyone else has

I do believe it is how you read the question. I have two examples to support my answer (see below).

If a family has two children, and the older child is a girl, there is a 50 percent chance the family will have two girls. This one is easy because we know the first child is a girl so the sex of the second child is 50/50 boy or girl.

This is an example of conditional (Baysian) probability. You are told additional information (a condition e.g oldest child is a girl), and this restricts the sample space, changing the probabilities.

Any two-child family has four possible outcomes: b:b, g:g, b:g, g:b.

In the first case where the known girl is also known to be the older child, we thereby eliminate the b:b and b:g combinations, leaving two left. And obviously one of those two is g:g so our odds of having two female children is 1/2.

By this same logic, in the 2nd instance all that is known is that one child is a girl, therefore of course the b:b combination cannot exist,

leaving three possible outcomes g:g, b:g, g:b. Hence, there is a 1/3 chance of two girls.

The more information you provide, the more you change the sample space of the probability calculation. In the case of sons or daughters in a family of two children, if you exclude bb or bg, then as you said you can only have gg or gb, so 1/2 probability that second child is a girl, whereas if one child (unspecified) is a girl, the probability space is now gg, bg, gb and the chance of two girls is now only 1/3.

The thing to remember is that CONDITIONAL probability can dramatically change the commonsense idea of what a particular probability should be.

Examples of both these possibilities can be found on wikipedia.com, just search boy or girl paradox and it will explain both probability theories.

Well I am glad I read on, the first two postings said this was a lousy puzzle. If all puzzles on this site are louzy then I will have fun! I am sure I have said what many others have said, I didn't read all 24 pages. I am in support for everyone who has said 1/3 - just for the record I have 3 daughters!

[Guest]

I do believe it is how you read the question. I have two examples to support my answer (see below).

If a family has two children, and the older child is a girl, there is a 50 percent chance the family will have two girls.

Welcome aboard, Tearz. It's not how you read the question; it's what the question actually states. Had the question stated that the older child was a girl, 50% would be the correct answer. It's stated in such a way where the only correct answer is 1/3.

[Guest]

Welcome aboard, Tearz. It's not how you read the question; it's what the question actually states. Had the question stated that the older child was a girl, 50% would be the correct answer. It's stated in such a way where the only correct answer is 1/3.

Yep thats what I said, I used two senarios the one that showed how you would get the 50% (1/2) and the one that stated the 1/3. I did both to show the difference between the two because I believed the 1/2 supporters needed to see the two side by side, so they could see where they went wrong, well how they could have misread the question

[Guest]

Yep thats what I said, I used two senarios the one that showed how you would get the 50% (1/2) and the one that stated the 1/3. I did both to show the difference between the two because I believed the 1/2 supporters needed to see the two side by side, so they could see where they went wrong, well how they could have misread the question

Like you said, you only read a few pages of the puzzle. The folks that have claimed the answer is 1/2 did not do so because they misread the question and thought the riddle stated that the older child is a girl. Read the posts and you'll see their answer is based on a misunderstanding of how probability works.

[Guest]

Like you said, you only read a few pages of the puzzle. The folks that have claimed the answer is 1/2 did not do so because they misread the question and thought the riddle stated that the older child is a girl. Read the posts and you'll see their answer is based on a misunderstanding of how probability works.

Ok here are my two scenarios

First Girl is the Eldest,

Eliminates Boy-Boy and Boy-Girl

Leaves a 50% chance

Second Girl is the Youngest sibling

Eliminates Boy-Boy and Girl-Boy

Leaves a 50% chance

There are no other scenarios, so it must be a 50% chance

[Guest]

In any two child family there are four possible combinations:

First child_________Second child

Boy______________Boy

Boy______________Girl

Girl______________Girl

Girl______________Boy

With condition (Baysian) Probability you are given additional information so if you have been told the first child is a girl then it eliminates the two outcomes of where the familys first child could have been a boy leaving only two other options. And this also applies if you are told the second child is a girl.

However, because the original question asked, does not state if the first or second child is a girl, it only lets us know that one of the children is a girl. So we have three combinations that it could be. The Boy - Boy option is out because we know that one of the children has to be a girl. Because we haven't been given any additional (conditional) information there are three options of combinations that the family could have.

Girl____________Girl

Girl____________Boy

Boy___________Girl

All we know is one child is a girl, we don't know if it is the first or second child, so there are three possible combinations without knowing birth order.

[Guest]

Second Girl is the Youngest sibling

Eliminates Boy-Boy and Girl-Boy

Leaves a 50% chance

There are no other scenarios, so it must be a 50% chance

There are four scenarios in which a couple can have two children:

GB

BG

GG

BB

There are three possibilities if we know there is at least one girl:

GB

BG

GG

The other child is a girl 1/3 of the time.

What you're doing is claiming if the girl is born first, then her sibling has a 50% chance of being a girl and if the girl is born second, then her sibling has a 50% chance of being a girl, therefore, the other child is a girl 50% of the time.

It doesn't work that way. We have to look at all the possibilities at once and not look at if scenarios separately. The riddle is not talking about a specific girl to compare her sibling to. It's just saying that there is at least one girl in a pair of siblings. In this case, the three possibilities above are the choices

Read through this thread and the following links that were given in various posts:

http://www.learninghaven.com/articles/ridd...in-teasers.html

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

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

http://www.mathpages.com/home/kmath036.htm

[Guest]

There are four scenarios in which a couple can have two children:

GB

BG

GG

BB

There are three possibilities if we know there is at least one girl:

GB

BG

GG

The other child is a girl 1/3 of the time.

What you're doing is claiming if the girl is born first, then her sibling has a 50% chance of being a girl and if the girl is born second, then her sibling has a 50% chance of being a girl, therefore, the other child is a girl 50% of the time.

It doesn't work that way. We have to look at all the possibilities at once and not look at if scenarios separately. The riddle is not talking about a specific girl to compare her sibling to. It's just saying that there is at least one girl in a pair of siblings. In this case, the three possibilities above are the choices

Read through this thread and the following links that were given in various posts:

http://www.learninghaven.com/articles/ridd...in-teasers.html

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

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

http://www.mathpages.com/home/kmath036.htm

Also these are not if statements these are either or, which means one must be true

I disagree. since there are only two scenario, one were the girl is older and one is when the girl is younger. It has to be either or. It can't be something else.

Probability DEF:

Mathematics the extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases

It would matter if we cared if the girl was younger or older but we don't

Since both the scenarios have the same possibility that that it is a 50/50 shot of the other being a boy then the chance is 50/50.

Edited July 28, 2008 by Enigma

[Guest]

I disagree. since there are only two scenario, one were the girl is older and one is when the girl is younger. It has to be either or. It can't be something else.

How many times does this need to be explained?

There are three possibilities:

GG

GB

BG

Multiple reliable cites agree!

I disagree. since there are only two scenario, one were the girl is older and one is when the girl is younger.

There is no the girl! There is a couple with two children who have at least one girl.

If there were a specific girl (lets say her name is Jenny), there's more than two scenarios.

Jenny- older Benny- younger GB

Benny- older Jenny- younger BG

Jenny- older Cindy- younger GG (or vice-versa)

so the three equal possibilities are these:

GG

GB

BG

Probability DEF:

Mathematics the extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases

We went through this already. There are three possible events; one is favorable.

Why don't you try Martini's penny experiment and get back to us?

[Guest]

I disagree. since there are only two scenario, one were the girl is older and one is when the girl is younger. It has to be either or. It can't be something else.

There are not two senarios, there are three. There are two options where the girl could be older (Girl-Girl and Girl-Boy) and one option where the girl could be younger (Boy-Girl). There is of course the other option in a two child family (Boy-Boy) but this is not an option in this case because one of the children has to be a girl.

So we have three options that it could be: Girl-Girl, Girl-Boy and Boy-Girl. Giving us the 1/3 chance.

[bonanova]

It may be time for us to simply agree with Enigma.

Enigma states that s/he disagrees.

In all likelihood that is a true assertion.

We can all agree with it.

Remember: You can lead a horse to drink, but you can't make him water.

[Guest]

This all comes down to how you look at the scenario. The riddle states...

Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!

This implies a random couple off the street. We don't care who.

They have two kids, one of them is a girl,

Now to me this to is a random event that applies to them alone. So when I work it out I work it out with a series of random events. However all the 1/3 people out their and say that "one of them is a girl" is the same as "given any random couple with at least one girl" and take a look at all the couples and find only the one with girls and work the math out to 1/3. However in probability how you get your information is just as important as the information you get. This being that all the 1/3 people out their don't think that the information being supplied is a random event. The scenario being you walk up to a guy and ask him if he "has at least one girl" and he responds "yes" or "no". However I believe that the information being supplied is as random as the couple, and their is nothing that implies otherwise. I walk up to some guy and ask him if he as any kids and he says "I have two, one of them is a girl". With this information the odds of the other kid being a girl is 1/2 because it is just as likely that we could have been told that one of them is a boy. How dose

They have two kids, one of them is a girl.

turn into

Given any random couple with at least one girl.

I don't see how given information we have, this can be anything but random event. Perhaps you can tell me?