Poll: A women has two kids, one is a boy, what are the odds the other is also a boy?

[DracoSuave]

Well, we're dealing with two kids. Yes.

Both kids have an arbitrary ordering. Perhaps you could say 'The older and the youngest.'

You could also say 'The redhead and the blonde'.

The probability table works with any quality we can assign one kid that we do not assign to the other.

For example: 'The child we know, and the child we don't know the gender of.'

This is perfectly legitimate.

This changes the probability table to:

B-B
B-G
G-B
G-G

Oh, wait that doesn't change. But what you exclude does. The first child is the child you know. So G-B and G-G get eliminated.

50%.

And again, you cannot assume this is a randomly sampled woman. All you know is she's saying 'I got a boy.' That's the only information you have to go on. She may be some bint that walked up to you on the street and said 'I got a boy!' in which case, the only random event here is the other child. If you -selected- the woman at random that is a different story.

In this case, however, there is 0% chance (as we can all agree) that we're dealing with a woman without a girl, therefore, we make the assumption that must be part of the question.

Truthfully, you cannot answer the question without assuming important information that is absent. Therefore the question -itself- is bunk.

However, in light of that, it is customary to assume that the only stated random (unknown) element is the only random (unknown) element unless told otherwise, and that is the second child.

It is 1/3 if AND ONLY IF the woman herself is a random or unknown element. As she is not declared as such, the assumption is that she is not, and you only take in the random element declared.

edit: the Law of Large Numbers tells me that Mixed Pair and Same Pair should come out 50/50 over the long haul. So why should revealing information about an already flipped pair change those odds?

Well, it all depends if the woman is the random element involved. If she is, then you're not really dealing with coinflips, you're dealing with the random distribution of a woman who has made two coinflips, which if you remember pascal's triangle, has a distribution of 1 2 1, which is where your 1/3 would arise from. (1+2=3)

 

[DracoSuave]

I just used wikipedia because I know it is correct (at the moment) in this matter.

We're not dealing with an isomorphism of question 1. We're dealing with information regarding any of the two children, but we don't know which one. We're dealing with a random woman, not a random boy. The question isn't: "we have a random boy with a sibling, what is the chance that he has a brother?"
Therefore, we're dealing with this:
Two boys - 1/4
One boy, one girl - 1/2
Two girls - 1/4 - redraw
Here you have the probability-table. You can tell me what the results will be (yes, I know you did).

No, you used wikipedia, and only one part of the article on wikipedia, because it agreed with your answer. You also failed to fully quote that part that 'agreed' with your answer, hense the intellectual dishonesty. And, again, appeal to authority does NOT apply in mathematical rigor.

And you do not know if you are dealing with an isomorphism of question 1. You are simply stating you are and calling it a day. The more I deal with this thread, the more I stand on the 'the question itself is flawed and does not give enough information to give an answer.' But it has forced me to change my answer:

Because now I have to state the answer is equal to n/3+(1-n)/2 where n is the probability that we're talking about a randomly sampled woman, and 1-n is therefore the probability that we are talking about a woman who is not randomly sampled.

n/3+(1-n)/2 is the only answer that has NO assumptions, and applies no bias, and does not presume anything about the question. There is no paradox, this is the correct answer based on -all- of our unknowns.

If you can prove that n = 1 or n = 0 then you can presume to have a numerical answer. Otherwise, you have excluded unknowns and random elements, and must recalculate accordingly.

 

[Maze1125]

Ah--here's the issue. You're not using the information given exactly. The P(OB) isn't 3/4, it's 1: the problem tells us that at least one of the children is a boy. The probability of a pair of children we don't know anything about containing at least one boy is 3/4, but we *do* know something about this pair.

That's not how the equation works.

P(X|Y) = P(X intersect Y)/P(Y) this it true for any events X and Y. The fact that we know Y is given in P(X|Y). But when we look at P(Y) we want to know the probability of Y happening in the first place which means, for that part, we have to forget that we know Y has happened. That's just how the equation works.

So, in this case, it's true that P(OB|OB) = 1, but the equation doesn't want P(OB|OB), it wants P(OB), the probability that OB might happen at all, which is 3/4.

And your example with having a child and flipping a coin is exactly right.
The fact that the answer is 1/3 may not be intuitive, but that doesn't mean it's wrong. Mathematical results are often unintuitive.

Edit:

edit: the Law of Large Numbers tells me that Mixed Pair and Same Pair should come out 50/50 over the long haul. So why should revealing information about an already flipped pair--information that is necessarily true, Principle of the Excluded Middle and all--change those odds?

Because, by revealing information, you're excluding one of the possibilities, if you don't reveal that information, you're not.

Also it depends how you set up the gambling situation. If you choose which information to reveal after you've flipped, that's different than choosing to say that one has come up heads every time and if both come up tails just discarding the flip.

 

[Maze1125]

I just used wikipedia because I know it is correct (at the moment) in this matter.

We're not dealing with an isomorphism of question 1. We're dealing with information regarding any of the two children, but we don't know which one. We're dealing with a random woman, not a random boy. The question isn't: "we have a random boy with a sibling, what is the chance that he has a brother?"
Therefore, we're dealing with this:
Two boys - 1/4
One boy, one girl - 1/2
Two girls - 1/4 - redraw
Here you have the probability-table. You can tell me what the results will be (yes, I know you did).

No, you used wikipedia, and only one part of the article on wikipedia, because it agreed with your answer. You also failed to fully quote that part that 'agreed' with your answer, hense the intellectual dishonesty. And, again, appeal to authority does NOT apply in mathematical rigor.

And you do not know if you are dealing with an isomorphism of question 1. You are simply stating you are and calling it a day. The more I deal with this thread, the more I stand on the 'the question itself is flawed and does not give enough information to give an answer.' But it has forced me to change my answer:

Because now I have to state the answer is equal to n/3+(1-n)/2 where n is the probability that we're talking about a randomly sampled woman, and 1-n is therefore the probability that we are talking about a woman who is not randomly sampled.

n/3+(1-n)/2 is the only answer that has NO assumptions, and applies no bias, and does not presume anything about the question. There is no paradox, this is the correct answer based on -all- of our unknowns.

If you can prove that n = 1 or n = 0 then you can presume to have a numerical answer. Otherwise, you have excluded unknowns and random elements, and must recalculate accordingly.

Whether or not the woman is randomly sampled is irrelevant. We're not talking about statistics, we're talking about probability.
And, as I have shown, using probability laws, the answer is 1/3.

You can wave your intuitive arguments about where ever you want but, unless you can show your claim using probability laws and mathematics, you're not actually showing anything.

 

[DracoSuave]

Whether or not the woman is randomly sampled is irrelevant. We're not talking about statistics, we're talking about probability.
And, as I have shown, using probability laws, the answer is 1/3.

Interesting, and I have shown, using probability laws, the answer is 1/2.

That, of course assumes you bothered to read them.

You can wave your intuitive arguments about where ever you want but, unless you can show your claim using probability laws and mathematics, you're not actually showing anything.

I suggest you actually read posts that disagree with yours before you claim they do not use probability laws or mathematics.

If the sample is randomized, then the -woman with the boy- is the random element, and therefore, you do your probability analysis based on that. Thusly, you look at the outcomes for a woman-with-a-boy, and use a proper analysis, that is what leads to 1/3.

If the sample is -not- randomized, then the woman is a non-random element. She is simply asking you the probability of her other child. Whether the boy exists or not is irrelevant--she is asking you to determine the probability of the other child. The question becomes the same as 'I have a dog. What is the gender of my child?' In which case, the analysis yields 1/2.

As the information as to whether or not the sample is a random woman or not (or, if you prefer wikipedia, whether the woman has taken both children or only one into consideration for that statement) is absent, -that- is the first unknown. And as the first unknown, it has a probability of n if the woman is taking into consideration one or two children. You are assuming she is taking both into consideration with her statement, however the very part of wikipedia that keeps getting quoted informs you -directly- that is not a valid assumption, that it can be taken both ways.

Therefore, you have to accept that in the absence of assumption, that the only correct answer is n/3+(1-n)/2.

You cannot make the statement 'Both children were under consideration when the statement is made' and therefore you cannot assume it to be true.

But, to quote you: Also it depends how you set up the gambling situation.

We don't know how this gambling situation was set up. So we're stuck with an unknown variable, and that unknown variable does not factor out of the real probability. Therefore a discrete integer value is not possible. Your math does not account for the unknown situation. Therefore it is flawed.

 

[Maze1125]

Interesting, and I have shown, using probability laws, the answer is 1/2.

No you haven't.

You've made a lot of hand waving arguments that sounds like probability, but not once have you used a probabilistic law, such as P(X|Y) = P(X intersect Y)/P(Y) or P(X intersect Y) = P(X)P(Y) given X and Y are independent.

Proving it using mathematics, or you haven't proven anything.

 

[Maze1125]

I still don't get why, if I was a Coin Flipping Casino where I allowed patrons to bet on a pair of flipped coins after I let them know if there's at least one head or tail, I should give 66% odds (forgetting about a house edge and the .6 repeating for the sake of simplicity) on the Mixed Pair bet and 33% on the Same Pair bet.

edit: the Law of Large Numbers tells me that Mixed Pair and Same Pair should come out 50/50 over the long haul. So why should revealing information about an already flipped pair--information that is necessarily true, Principle of the Excluded Middle and all--change those odds?

To elaborate on this gambling problem.

If you were a a casino and it had two machines, one called "One is Heads." the other called "One is Known." which use the following methods:

One is Heads

This machine flips two coins. If it gets two tails, it discards and flips again until it gets at least one head. At that point it tells you that one is a head and asks you to guess on the other.
With this machine the probability of the other being a head is 1/3.

One is Known

This machine flips two coins, picks one at random, and tells you what it's value is, it then asks you to guess the other coin.
With this machine, the probability of the other coin being a head is 1/2, regardless of what it tells you the known coin is.

The question posed in this topic is the first machine. The question tells us that one is a boy and asks for the probability of the other. It doesn't simply tell us that the sex for one of them is known. It has chosen beforehand what is going to be known.

 

[DracoSuave]

Interesting, and I have shown, using probability laws, the answer is 1/2.

No you haven't.

You've made a lot of hand waving arguments that sounds like probability, but not once have you used a probabilistic law, such as P(X|Y) = P(X intersect Y)/P(Y) or P(X intersect Y) = P(X)P(Y) given X and Y are independent.

Proving it using mathematics, or you haven't proven anything.

Sorry my math was too plain-language enough for you. Don't assume math isn't happening just cause it's written in a form that non-mathematicians can possibly understand. That's just elitism, and not relevant to the argument.

But--

Probability can be applied to any event that may have multiple outcomes.

P(X) is the probability that X will occur. In this case, X is 'The question above takes both boys into consideration.' P(~X) is the probability that X does not occur. As you are aware, P(~X) = 1-P(X).

P(Y) is the probability of the child being a boy. P'(Y) is the probability of the child being a boy if the question takes both boys into consideration, P''(Y) is the probability of the child being a boy if the question does not.

P'(Y), as you've established, is 1/3. This is not disputed. P''(Y), as I've established, is 1/2. This is also not disputed.

P(Y) is dependant, however, on P(X) and cannot divorce itself from that unknown. P(Y AND X) is mutually exclusive of P(Y AND ~X), and therefore P(Y) = P(Y AND X) + P(Y AND ~X)

P(Y AND X) = P'(Y)P(X) = P(X)/3
P(Y AND ~X) = P''(Y)P(~X) = [1-P(X)]/2

P(Y) = P(X)/3 + [1-P(X)]/2

Let n = P(X)

P(Y) = n/3 + (1-n/2)

Of course, that's just symbolic reiteration of stuff you should know if you know your math at all, or reiteration in symbolic form of stuff I already said.

Or is that not 'mathy' enough for you?

 

[DracoSuave]

The question posed in this topic is the first machine. The question tells us that one is a boy and asks for the probability of the other. It doesn't simply tell us that the sex for one of them is known. It has chosen beforehand what is going to be known.

But what you do not know is if the machine chose to reveal the gender of one child at random or not. You've assumed that it's started with the 'reveal a boy' order, and not simply revealed a child at random. If you reveal a child at random, regardless of the birth order of that child, then the probability is 1/2.

That's the thing, did the woman decide to reveal a boy, or did she just reveal one of her children based on arbitrary criteria, up to, and including, flip a coin? (For example, the question does not state whether or not the woman could have revealed to you a girl, only that she did reveal a boy. This statement effects EVERYTHING and cannot be simply ignored.) The question does not state that if the woman had two girls, that she is to reveal nothing and/or leave for another mother to speak up.

In other words, all you know is that one of her children is a boy. You do not know if the woman is a random woman, or if she is simply walking up to you and saying "I have a boy. What are the chances of my other/next child being a boy?".

Both lead to completely different trains of thought. Whether or not the woman is selected is absolutely important, as the 1/3 presumes the woman -is- selected, by dint of the algorithm itself involves selecting the woman. The 1/2 argument presumes that the woman is not selected/random, and that the only random event involved is the other child.

But -neither- situation is described or alluded to by the question, therefore assumptions are being made on both sides. Hense why I've changed my stance to n/3+(1-n/2), because it depends on what the chances are of this question refering to the former or the later situation.

And we don't -know- which the question refers to, so we cannot state that it is a known. Therefore, it is an unknown (random) event.

That decision -actually affects the answer- and therefore math that ignores this fundamental question is incomplete at best.

 

[Maze1125]

P(Y AND X) = P'(Y)P(X)
P(Y AND ~X) = P''(Y)P(~X)

You have at no point shown that those are true, you're just claiming them with no backing.

Also, the question does depend on both boys. There's no probability to that at all.

The question posed in this topic is the first machine. The question tells us that one is a boy and asks for the probability of the other. It doesn't simply tell us that the sex for one of them is known. It has chosen beforehand what is going to be known.

But what you do not know is if the machine chose to reveal the gender of one child at random or not. You've assumed that it's started with the 'reveal a boy' order, and not simply revealed a child at random. If you reveal a child at random, regardless of the birth order of that child, then the probability is 1/2.

That's the thing, did the woman decide to reveal a boy, or did she just reveal one of her children based on arbitrary criteria, up to, and including, flip a coin? (For example, the question does not state whether or not the woman could have revealed to you a girl, only that she did reveal a boy. This statement effects EVERYTHING and cannot be simply ignored.) The question does not state that if the woman had two girls, that she is to reveal nothing and/or leave for another mother to speak up.

That decision -actually affects the answer- and therefore math that ignores this fundamental question is incomplete at best.

The woman is utterly irrelevant to the question.

But, to avoid your complete random tangent, I'll rephrase it.

There are two people, one is male. What is the probability they are both male?

That is fundamentally the same question, it's just been made without an attempt at avoiding arbitrariness.

Hell, let's go one further:

There are two identical variables, they can each take state A with probability 1/2 and state B with probability 1/2. At least one of them is in state A, what is the probability they are both in state A?

 

[DracoSuave]

The woman is utterly irrelevant to the question.

But, to avoid your complete random tangent, I'll rephrase it.

There are two people, one is male. What is the probability they are both male?

That is fundamentally the same question, it's just been made without an attempt at avoiding arbitrariness.

Is not the same question as below:

Hell, let's go one further:

There are two identical variables, they can each take state A with probability 1/2 and state B with probability 1/2. At least one of them is in state A, what is the probability they are both in state A?

Because it has to do with the phrase 'one is male.' Now, does 'one is male' mean that you've taken one aside, gender checked him, and said 'This one is male' (50%) or does it mean 'at minimum, one of these two is male, but which one I am not telling you' (33%) or does it mean 'The cardinality of males is equal to one'

Thusly, there is a measure of ambiguity in the statement that has not been addressed that continues to exist in the original question of the woman, leading to the situation where one must consider an unknown variable that is the probability spread of that ambiguity's possible interpretations.

The second statement, on the other hand, is non-ambiguous and I would agree is 1/3.

However, seeing as we are dealing with an ambiguous statement, either the ambiguity must be cleared up, or accounted for in your probability calculations. It cannot be ignored simply due to convenience.

 

[Maze1125]

The woman is utterly irrelevant to the question.

But, to avoid your complete random tangent, I'll rephrase it.

There are two people, one is male. What is the probability they are both male?

That is fundamentally the same question, it's just been made without an attempt at avoiding arbitrariness.

Is not the same question as below:

Hell, let's go one further:

There are two identical variables, they can each take state A with probability 1/2 and state B with probability 1/2. At least one of them is in state A, what is the probability they are both in state A?

Because it has to do with the phrase 'one is male.' Now, does 'one is male' mean that you've taken one aside, gender checked him, and said 'This one is male' (50%) or does it mean 'at minimum, one of these two is male, but which one I am not telling you' (33%) or does it mean 'The cardinality of males is equal to one'

It means "One is male" that is it.
Which means it can't be "This one is male" because that requires extra information.
Nor can it be "The cardinality of males is equal to one" because that requires extra information too.

We have not been told which one, nor have we been told that there can only be one. All we have been told is "There is one." Therefore, the only equivalent statement from your choices is "At minimum, one of these two is male, but which one I am not telling you."

There is no ambiguity. The statement is merely concise.

The second statement, on the other hand, is non-ambiguous and I would agree is 1/3.

Why?
If there existed the possibility before that we had taken one aside and checked, why can't that possibility, and therefore ambiguity, exist here?

The answer is simple, the ambiguity never existed at all. But due to the attempt at making the situation seem realistic rather than arbitrary, the conciseness of the statement became confused with ambiguity, because in realistic situations language is expected to be used ambiguously rather than concisely.
But when the statement was made completely arbitrary, the language was then expected to be concise, and so the ambiguity apparently disappeared, when in fact it never existed at all.

 

[Maze1125]

One is Heads

This machine flips two coins. If it gets two tails, it discards and flips again until it gets at least one head.

...

The question posed in this topic is the first machine.

Nope: the question says this cannot be a TT/FF pair.

No, the question says that we know it isn't. Not that it can never be.
The family has been picked precisely because they have one boy. That is equivalent to a machine only picking flips with one head. That doesn't mean that two girls can never happen, just that they have been removed from the sample.

But when we look at P(Y) we want to know the probability of Y happening in the first place which means, for that part, we have to forget that we know Y has happened. That's just how the equation works.

Why? Why do we 'forget' that Y has happened? Y *has* happened in the premise of the problem: you can't just 'forget' about it.

Because that's how conditional probability [http://en.wikipedia.org/wiki/Conditional_probability] works. P(X|Y) = P(X intersect Y)/P(Y) is the rule. I think about the probability of Y happening at all because that's what the rule wants. If I was thinking about P(Y|Y) instead I would get the wrong answer because that isn't the rule.

 

[Maze1125]

One is Heads

This machine flips two coins. If it gets two tails, it discards and flips again until it gets at least one head.

...

The question posed in this topic is the first machine.

Nope: the question says this cannot be a TT/FF pair.

No, the question says that we know it isn't. Not that it can never be.

If we know an event came to pass in a specific way, that means it could never be any other way, regardless of the 'probability' of that event.

The family has been picked precisely because they have one boy.

We don't know that. You've just assumed that.

That is equivalent to a machine only picking flips with one head.

No, this is equivalent to a machine *incapable* of flipping two tales. Big difference.

That doesn't mean that two girls can never happen, just that they have been removed from the sample.

What sample? The question never stated that there was a sample: you've assumed that the pair was picked from a sample large enough so that the Law of Large Numbers was operative. That was not stated in the problem.

You're just making arbitrary complaints now.
The question is simple. We have two variables that could be B or G with equal probability, we know at least one has the value B, what is the probability the other takes the value B too.
Everything else is just an analogy.

But when we look at P(Y) we want to know the probability of Y happening in the first place which means, for that part, we have to forget that we know Y has happened. That's just how the equation works.

Why? Why do we 'forget' that Y has happened? Y *has* happened in the premise of the problem: you can't just 'forget' about it.

Because that's how conditional probability [http://en.wikipedia.org/wiki/Conditional_probability] works.

The fact that one sibling is male is an independent event in regards to the other sibling being male. Therefore, it's 50/50.

If the pair of siblings were picked from a large enough pool that the Law of Large Numbers came into effect, I would agree. But that's not what the problem asks. You're just adding information to the problem that is not there.

No I'm not.

We want to know P(FB intersect SB | OB)
We can use probability laws to deduce this value.
P((FB intersect SB)|OB) = P(FB|OB)P(SB|(FB intersect OB)) = (P(FB intersect OB)/P(OB))(P(SB intersect FB intersect OB)/P(FB intersect OB)) = (P(FB)/P(OB))(P(FB intersect SB)/P(FB)) = ((1/2)/(3/4))((1/4)/(1/2)) = 1/3

In fact we can use the definition of conditional probability rather more directly.
P(FB intersect SB | OB) = P(FB intersect SB intersect OB)/P(OB) = P(FB intersect SB)/P(OB) = (1/4)/(3/4) = 1/3

 

[DracoSuave]

The woman is utterly irrelevant to the question.

But, to avoid your complete random tangent, I'll rephrase it.

There are two people, one is male. What is the probability they are both male?

That is fundamentally the same question, it's just been made without an attempt at avoiding arbitrariness.

Is not the same question as below:

Hell, let's go one further:

There are two identical variables, they can each take state A with probability 1/2 and state B with probability 1/2. At least one of them is in state A, what is the probability they are both in state A?

Because it has to do with the phrase 'one is male.' Now, does 'one is male' mean that you've taken one aside, gender checked him, and said 'This one is male' (50%) or does it mean 'at minimum, one of these two is male, but which one I am not telling you' (33%) or does it mean 'The cardinality of males is equal to one'

It means "One is male" that is it.
Which means it can't be "This one is male" because that requires extra information.
Nor can it be "The cardinality of males is equal to one" because that requires extra information too.

We have not been told which one, nor have we been told that there can only be one. All we have been told is "There is one." Therefore, the only equivalent statement from your choices is "At minimum, one of these two is male, but which one I am not telling you."

There is no ambiguity. The statement is merely concise.

The second statement, on the other hand, is non-ambiguous and I would agree is 1/3.

Why?
If there existed the possibility before that we had taken one aside and checked, why can't that possibility, and therefore ambiguity, exist here?

The answer is simple, the ambiguity never existed at all. But due to the attempt at making the situation seem realistic rather than arbitrary, the conciseness of the statement became confused with ambiguity, because in realistic situations language is expected to be used ambiguously rather than concisely.
But when the statement was made completely arbitrary, the language was then expected to be concise, and so the ambiguity apparently disappeared, when in fact it never existed at all.

Saying that there isn't any ambiguity, when said ambiguity has been pointed out to you directly, does not prove there is no ambiguity.

'I have two ice cream cones, one is chocolate.' That statement implies that there is a different flavor of the other cone, i.e. that it is non-chocolate, i.e. there is a cardinality of one.

'I have two unopened presents, and one was socks. Do you think the other could be socks too?' This statement implies that one was tested, and that the other was not.

If you feel there is no ambiguity, perhaps you are not as familiar with the English language as you believe. Without context to clarify that ambiguity, you cannot clearly make such statements as 'At least one of the two is a boy' or anything else. And in this case, that context -does not exist- as such.

The only reason that we do not accept the 'cardinality of one' situation is that renders the question mathematically trivial and not worth the discussion... otherwise it would be -correct- to assume this is the case.

 

[DracoSuave]

doublepost, crap