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

[Maze1125]

stuff

The problem here is complete lack of rigour on both sides.

So, let us restart from a completely rigorous perspective.

We start by being given a simple sample space:

BB
BG
GB
GG

Or equivalently:

FB intersect SB
FB intersect ¬SB
¬FB intersect SB
¬FB intersect ¬SB

With each of these events having probability of 1/4.

Then the problem asks us to find the value of P(FB intersect SB | OB) within the given sample space. At this point OB is only defined by the phrase "One is a boy." which you claim is ambiguous. I, on the other hand, claim the only reasonable mathematical definition for that phrase within the sample space is (FB union SB) and such a definition gives an answer of 1/3.

You also claim that if a child is selected and then found to be a boy this gives a probability of 1/2. This is true, but to be rigorous, we need to find a way of including this possibility.
There are two ways of doing this:
1. Defining the possibility from outside the given sample space or
2. Defining it using existing properties of the given sample space.

If we choose 1, then we need a new disjoint sample space with the events "First child is selected" (FCS) and "Second child is selected" (SCS) and then take the product of that space with our original space. Now, if we do that, there is an easy definition of OB that gives 1/2 as the answer, but creating such a sample space is a blatant addition of information.

If we choose 2, then we need to define FCS and SCS using existing elements of the space. The only consistent way I see to do this is to define FCS = FB and SCS = SB. If we do this and look at your table (using my notation) we get:

1) FB SB FB
2) FB SB SB
3) FB ¬SB FB
4) FB ¬SB SB
5) ¬FB SB FB
6) ¬FB SB SB
7) ¬FB ¬SB FB
8) ¬FB ¬SB SB

Now 4,5,7 and 8 are blatant contradictions, and 1,2,3 and 6 have extraneous information.

So, reducing the table to account for these problems we get:
1) FB SB
2) FB SB
3) FB ¬SB
4) ¬FB SB

Which is all well and fine except 1 and 2 are identical and you can't rigorously apply Draco's Law of Counting to a table that has two identical events.

Therefore I see no completely rigorous way to produce and answer of 1/2 without using information the question has not provided.

Also, I would ask that you try to refrain from making comments like this:

I'm sorry, you're too wrapped up in your confirmation bias to actually discuss this problem on a meaningful mathematical level.

Because I feel exactly the same way about you, but neither of us has the needed lack of bias to justify such claims about the other.

 

[DracoSuave]

double post wtf

 

[DracoSuave]

The problem here is complete lack of rigour on both sides.

So, let us restart from a completely rigorous perspective.

We start by being given a simple sample space:

BB
BG
GB
GG

Or equivalently:

FB intersect SB
FB intersect ¬SB
¬FB intersect SB
¬FB intersect ¬SB

With each of these events having probability of 1/4.

Then the problem asks us to find the value of P(FB intersect SB | OB) within the given sample space. At this point OB is only defined by the phrase "One is a boy." which you claim is ambiguous. I, on the other hand, claim the only reasonable mathematical definition for that phrase within the sample space is (FB union SB) and such a definition gives an answer of 1/3.

You also claim that if a child is selected and then found to be a boy this gives a probability of 1/2. This is true, but to be rigorous, we need to find a way of including this possibility.
There are two ways of doing this:
1. Defining the possibility from outside the given sample space or
2. Defining it using existing properties of the given sample space.

If we choose 1, then we need a new disjoint sample space with the events "First child is selected" (FCS) and "Second child is selected" (SCS) and then take the product of that space with our original space. Now, if we do that, there is an easy definition of OB that gives 1/2 as the answer, but creating such a sample space is a blatant addition of information.

If we choose 2, then we need to define FCS and SCS using existing elements of the space. The only consistent way I see to do this is to define FCS = FB and SCS = SB. If we do this and look at your table (using my notation) we get:

1) FB SB FB
2) FB SB SB
3) FB ¬SB FB
4) FB ¬SB SB
5) ¬FB SB FB
6) ¬FB SB SB
7) ¬FB ¬SB FB
8) ¬FB ¬SB SB

Now 4,5,7 and 8 are blatant contradictions, and 1,2,3 and 6 have extraneous information.

However, that extraneous information comes into account for the possibility that she selected a child then told the gender, rather than simply selected a boy. I agree that this is extraneous information, however because this possibility exists within the problem space, It must be taken into account in calculations. Otherwise the calculations are not truly accounting for all outcomes.

So, reducing the table to account for these problems we get:
1) FB SB
2) FB SB
3) FB ¬SB
4) ¬FB SB

At which point you've simply stated 'The problem didn't say that could have happened' and used that as an excuse to eliminate outcomes -that could have happened.-

So, in other words, you have just eliminated outcomes that are possible and distinct in a probability spread... thusly you have tainted the probability spread.

Bad form.

Which is all well and fine except 1 and 2 are identical and you can't rigorously apply Draco's Law of Counting to a table that has two identical events.

Only after chicanery noted above.

Therefore I see no completely rigorous way to produce and answer of 1/2 without using information the question has not provided.

However my answer isn't 1/2. There is also no completely rigorous way, by similiar logic, to produce an answer of 1/3 without using information the question has not provided. 1/3 presumes the woman is making a statement of cardinality of boys, rather than selecting a child and determining the gender.

In otherwords, it presumes a boy-bias in the situation before the given event has been declared. The question does not state this, therefore -this- is not a safe assumption either.

Also, I would ask that you try to refrain from making comments like this:

I'm sorry, you're too wrapped up in your confirmation bias to actually discuss this problem on a meaningful mathematical level.

Because I feel exactly the same way about you, but neither of us has the needed lack of bias to justify such claims about the other.

Here's the skinny.

The premise 'the unknown child is a boy' is B.
IF the known child is selected before gender identified, the probability is 1/2. Let this premise be X, and the probability of being a boy in this instance PX(B).
IF the known child was selected based upon gender, the probability is 1/3. Let this premise be Y, and the probability of being a boy in this instance PY(B).

Now, as previously calculated, the probability of the gender being a boy -must- be equal to P(X)PX(B) + P(Y)PY(B) as X and Y are mutually exclusive situations.

Now, in order for the answer to be PX(B), then P(Y)PY(B) must be equal to 0, and P(X) must be equal to 1. PY(B) is nonzero, therefore P(Y) must be equal to zero. Y however, is a possibility, and therefore has a nonzero probability. Therefore, the probability CANNOT be equal to 1/2.

In order for the answer to be PY(B), or 1/3, then P(X)PX(B) must be equal to 0, and P(Y) must be equal to 1. PX(B) is nonzero, therefore P(Y) must be equal to zero. X, however, is a possibility, and therefore has a nonzero probability. Therefore, the probability CANNOT be equal to 1/3

P(X) and P(Y) are functions across the domain [0,1] and thus it can be calculated that P(X)PX(B) + P(Y)PY(B), or P(X)/2 + P(Y)/3, is a function across the possible domain [1/3,1/2].

However, because we've just proven the answer cannot be 1/3 or 1/2, we eliminate those, leaving a domain of (1/3, 1/2).

P(X) and P(Y) are unknown, however they are functions of each other, in that P(X) = 1 - P(Y). Thusly, P(X)/2 + P(Y)/3 can be expressed as a function of a single variable, P(Y). Thusly, P(B) (the probability of the other child being a boy) is a function f(x) = [1-x]/2 + x/3 where x is a variable with the range (0, 1) and the function has the domain (1/3, 1/2)

The answer simply -cannot be- 1/3.

 

[Maze1125]

There is also no completely rigorous way, by similiar logic, to produce an answer of 1/3 without using information the question has not provided. 1/3 presumes the woman is making a statement of cardinality of boys, rather than selecting a child and determining the gender.

That is utter utter bollocks.
I have shown explicitly how to rigorously find an appropriate probability space and definition of OB that gives exactly P(FB intersect SB|OB) = 1/3

I have also shown that there is no way to produce and answer of 1/2 without a greater amount of information or a lack of rigour.

The only place that your argument is the slightest bit rigorous jumps directly past those points and assumes that both are valid claims using equal information without proving that claim in the slightest.
Yes, the deduction you make is completely correct if that assumption is true, but you have completely failed to give any rigorous reason to back that assumption up.

And, so you don't misunderstand me, I am asking you to do these two things:

Prove this rigorously:

IF the known child is selected before gender identified, the probability is 1/2.

And prove rigorously that it can be done with as little information as my construction that gives 1/3.

 

[Ildecia]

It's variable probablity, but I'm not sure the question is phrased right,

The oldest kid is a boy?

anyway

Boy Boy is in
Boy Girl is in
Girl Girl is out
Girl Boy is in

So it's 33%

so... that still doesn't make sense... i understand the probability logic... but every birth has a 50/50 probability.

births aren't conditional probability so idk what your looking at

Yeah, we ain't talking about boy boys and boy girls etc, the child who we know to be a boy isn't important. We are just talking is the one child a Boy or a Girl. The existence of the another child doesn't have anything to do with the other being a boy or a girl.

you validated my point.

it isnt variable probability. they are independant events

 

[Maze1125]

It's variable probablity, but I'm not sure the question is phrased right,

The oldest kid is a boy?

anyway

Boy Boy is in
Boy Girl is in
Girl Girl is out
Girl Boy is in

So it's 33%

so... that still doesn't make sense... i understand the probability logic... but every birth has a 50/50 probability.

births aren't conditional probability so idk what your looking at

Yeah, we ain't talking about boy boys and boy girls etc, the child who we know to be a boy isn't important. We are just talking is the one child a Boy or a Girl. The existence of the another child doesn't have anything to do with the other being a boy or a girl.

you validated my point.

it isnt variable probability. they are independant events

No they aren't.
The phrase "One is a boy." rather than "The first is a boy." means that the probabilities have become dependent.

 

[DracoSuave]

You are seriously being absurd in your counter-argument. Absolutely, 100% positively absurd.

There is also no completely rigorous way, by similiar logic, to produce an answer of 1/3 without using information the question has not provided. 1/3 presumes the woman is making a statement of cardinality of boys, rather than selecting a child and determining the gender.

That is utter utter bollocks.
I have shown explicitly how to rigorously find an appropriate probability space and definition of OB that gives exactly P(FB intersect SB|OB) = 1/3

No, what you did was start with the assumption 'The parent selected the child without regard to gender' and then went 'But selecting the child without regard to gender is irrelevant' and eliminated it.

What you DID was take the assumption, and then say 'But since the assumption doesn't apply, we now remove it.'

That is a dishonest argument. You've taken the argument 'The assumption never applies' and disguised it as 'This is what happens if we apply the assumption.'

That's not mathematical rigor, that is a strawman argument disguised as math.

I have also shown that there is no way to produce and answer of 1/2 without a greater amount of information or a lack of rigour.

And I have shown that, if a certain case happens to be true, than that particular case produces a probability of 1/2. All that is required for my argument to be true, is that particular case be -possible-.

You don't understand my argument. That's the problem.

The only place that your argument is the slightest bit rigorous jumps directly past those points and assumes that both are valid claims using equal information without proving that claim in the slightest.

Actually I have. What you did was assume that the case 'She selected her child before hand' was the full argument towards the original statement, not a single possibility within that statement. Therefore, you rejected a possible outcome with a fully calculated possibility because it did not fit -your- offered argument, which is not the same as mathematical rigor.

What -I- did was offer both it, and yours as -outcomes- within the given problem, as both are possibilities. If your situation is true, the probability is 1/3. So I went and proved that your situation is -possible- but that another possible situation also fits the criteria, and under -that- situation, the probability changes.

Yes, the deduction you make is completely correct if that assumption is true, but you have completely failed to give any rigorous reason to back that assumption up.

I don't need to prove that the assumption is true. I need to prove that it is -possible-. And if it is possible, then my argument holds.

To disprove the argument, you need to prove the assumption is impossible.

And, so you don't misunderstand me, I am asking you to do these two things:

Prove this rigorously:

IF the known child is selected before gender identified, the probability is 1/2.

I already did. We know that there is a selection. That is not additional information.
And -if- that selection exists, then we can account for it in probability.

And prove rigorously that it can be done with as little information as my construction that gives 1/3.[/quote]

Your construction also neglects information that we have in the problem, i.e., that there is a selection process, and that the manner performed can have an effect on the final probability. This is information that we have managed to deduce from the question.

-Neglecting- information that we -do- have is only mathematically valid if you prove it is irrelevant

But, we know there is an unknown selection process, and that all selection processes are divided into two general isomorphisms:

Where both children are considered when declaring gender, which is the 1/3 situation, and where one child is considered when declaring gender.

Seeing as we -cannot assume both children were considered-, to do so and operate under that precept is USING INFORMATION NOT GIVEN IN THE PROBLEM. That is what you are doing.

Therefore, we must consider -both- situations in our calculations. You've rigorously calculated the situation where both children are considered, and I thank you for that.

However, when one child is being considered for gender, in -that instance- then you have the information 'One child is selected' provided for you by the nature of this outcome itself. To say otherwise is like saying 'You can't use 'I rolled a 6' as valid information in calculating probabilities when rolling dice for outcomes where you rolled a 6!' It shows an utter lack of reasoning, and I reject it based on utter and obvious absurdity. It would be the similiar me countering your argument by saying 'You cannot use outcomes that have the other child as a girl! You don't have the information that girls exist in the problem!'

So, stop being absurd. If you are considering outcomes based on the situation of the selection of a single child, then 'A single child was selected' is a valid piece of information for those outcomes.

Which means my 8 outcomes argument stands for those outcomes as a rigorous argument.

Which means that 1/2 stands for those outcomes.

Which means that the mathematical analysis stands for -all- outcomes.

Which means that the function f(x)=x/3+(1-x)/2 is a function of range (0,1) and domain (1/3,1/2) stands.

Which means that it is PROVEN the answer CANNOT BE 1/3. (NOR can it be 1/2.)

------------------------

To simplify the argument. If the gender-identification of one child is fully dependant (either she selected by the criterion boy, or 'One is a boy' is a quantification of boys, not an indentification of one child), then 1/3 is obviously the correct answer. However, if the gender-identification is not dependant, if she just grabbed a child and told you the gender, if her method of selection has rendered the events independant, then as they are independant events, 1/2 would be the correct answer.

Because situations exist where 1/2 is the correct answer, for the total answer to be 1/3, there must exist outcomes that exist where answers -less than 1/3- exist.

What are those outcomes?

If X is 'Selection processes that have less than 1/3 as their probability' and Y is 'The other child is a boy when X' and Z is 'Selection processes that result in the genders of both children being independant with regard to that process' and W is 'The other child is a boy in situations where the probability is proven to be 1/3'

Then all you need to do to prove that the probability over all selection processes is 1/3 is prove that

P(X)P(Y) + P(Z)/2 = 1/3.

Of course, that involves proving that X could exist.

As it is shown right now, X is not proven to exist, tho Z does exist.

 

[Maze1125]

more stuff

I see a whole lot of claims there and not one single piece of mathematical rigour anywhere. I mean you haven't even used to word "define" once, which is practically the word that is used the most in mathematical arguments.

I'm beginning to suspect that, even though you're the one who brought up rigour in the first place, you don't even know how to produce a rigorous probabilistic argument, and maybe not even how probability theory is founded in the first place.

MACM 101, bitches.

What exactly are the details of this course, and what mark did you get?

 

[DracoSuave]

more stuff

I see a whole lot of claims there and not one single piece of mathematical rigour anywhere. I mean you haven't even used to word "define" once, which is practically the word that is used the most in mathematical arguments.

I'm beginning to suspect that, even though you're the one who brought up rigour in the first place, you don't even know how to produce a rigorous probabilistic argument, and maybe not even how probability theory is founded in the first place.

Alright, here you go.

Sigh. Can't believe I'm having to do this. Nevermind the fact that a selection process that causes independance of the outcomes makes the outcomes independant, never mind a gorram tautology...

If the basis of selection of the child is random and the gender determination made after the selection of the child, then (before elimination of non-conforming outcomes) the probability of a particular child being the one selected in a pairing is 50%. Given that each possible pairing has a chance of being selected of 25%, that leaves each child having a 12.5% chance of being selected.

(Bear in mind, this manner of selection -will- satisfy the criteria for the problem. The problem only states that a child is a boy. )

This means that each unique child has an equal chance of being selected. If we arrange the unique pairings as below, we can also give each unique child a unique designation:

b1 b3
b2 g3
g1 b4
g2 g4

And thusly we can identify the eight outcomes for selected children.

b1; b2; g1; g2; b3; g3; b4; g4

Now, we can eliminate the outcomes that do not conform to the criteria of the problem. That leaves:

b1, b2, b3, and b4.

That means there are four outcomes for child selection.

At this point, we can prove the resulting probability -cannot be- 1/3, because that would require a situation where we have 4/3 outcomes that are boys. As we are dealing with integers, 4/3 is not a valid number of outcomes.

But I digress.

Each unique boy has their own sibling:

b1 -> b3
b2 -> g3
b3 -> b1
b4 -> g1

Now, to claim that b1 and b3 are the same outcome is ludicrous. It isn't redundant as the criteria for choosing is based on the child, not on the pairing. But even if you continued to claim such, we've also already proven the answer to this cannot be 1/3 as that would require 1.3 outcomes satisfying it.

4 outcomes. 2 of them produce boy results. 2/4 = 50%. Q.E.D.

---------

But -let's say for the sake of argument- it was legal to remove redundancies, going back to your own counter argument.

You claim after removing the nonsatisfying pairings (a legal move, mathematically) and removing the 'redundant pairing' (a questionable move, mathematically) that you'd have 3 outcomes that satisfy the criteria.

You also then show that according to this, the probability yields 1 outcome out of 3 that are boys.

Well, then, let us look at the outcomes that are rejected because they would yield 'There is one girl.'

You have the pairings:

FG SG First Chosen
FG SG Second Chosen
FG SB First Chosen
FB SG Second Chosen

Then removing the 'redundancies due to removing the selection criteria'

FG SG
FG SB
FB SG.

That leaves us with 3 outcomes, each with equal chance of being selected, 1/3.

That means we have a total number of outcomes for this equal to 6, with each outcome with a chance of being selected of 1/6.

That means we have the following six outcomes:

FB SB
FB SG
FB SG
SB FG
SB FG
FG SG.

Now, here we have two choices. We can either remove redundancies (apparently you think this is a legal maneuver), or we can take each instance of differing gender pairings and add their probabilities together.

If we do the former:

FB SB
FB SG
SB FG
FG SG

Four outcomes, each with a previously calculated probability of 1/6. This yields a total probability space for all possible outcomes of 4/6, or 2/3.

2/3 != 1, and therefore cannot be correct.

So we cannot use this technique here. We -must- therefore add the probabilities together in order to arrive at the correct total probability for all possible outcomes (1).

That yields:

FB SB -> 1/6
FB SG -> 2/6
SB FG -> 2/6
FG SG -> 1/6.

However, we have already established that this cannot be either, that all pairings have an equal probability over the total space before elimination.

Therefore, when you 'removed redundancies' it is therefore proven that the probabilities at -that point- for all 3 outcomes you claim are there -cannot be equal.-

However, if you add the probabilities for both occurances of FB SB in your removal of redundancies, you end up with:

FB SB -> 50%
FB SG -> 25%
FG SB -> 25%.

The outcome that has the boy sibling (the first one) is 50%. Q.E.D.

A rigourous proof that you can have a selection method that yeilds 50%. Done. A rigorous proof that method cannot produce 33%. Done.

Also, MACM 101 was Discrete Mathematics, first year. Everything that has been done in this thread is basic level probability, a class I excelled in, if you must know. This isn't even complex level stuff, and watching you make such a fundamental error such as 'reducing redundancies' without taking into account the different probabilities of the outcomes that produces is staggering.

It'd be like if you had a six sided die, with the 6 replaced with a 1. Collapsing redundancies on that wouldn't yield a 1/5 chance of rolling -anything- on that die no matter how many tables you produce. The same principle applied here.

 

[Maze1125]

stuff

Yes, I guess that is the amount of rigour a standard university would expect from a first year student. Which is to say, not much.

Anyway, we'll ignore that and simply focus on the bits you got wrong, rather than the bits you got right without rigour.

And thusly we can identify the eight outcomes for selected children.

b1; b2; g1; g2; b3; g3; b4; g4

Yes, exactly, you cannot get answer of 1/2 without using a probability space that has at least 8 basic elements. Whereas the answer that gives 1/3 can be done using a space of 4 elements. Therefore the answer of 1/2 must necessarily use more information than the question gives as the answer of 1/3 uses at least as much information as the question gives and the 1/2 uses more than that.

As the answer of 1/2 uses more information, it cannot be a valid answer to the given question. Yes it might be a valid answer to a question with more information, but that is not the given question.

Probability theory works by answering questions using the exact information given. If we had to accept every possibility of how the question might have been formulated, we would have to give every answer to any probability question as a function of all the different possibilities and we could never find an exact number without using, at least, functional integration.
If you've ever given an exact answer to a probability question then you know that probability just doesn't work like that.

Also, MACM 101 was Discrete Mathematics, first year. Everything that has been done in this thread is basic level probability, a class I excelled in, if you must know.

Yes, this is a basic question.
That is, until someone who did well in one basic probability course comes along and starts abusing that basic probability course to justify his preconceptions about the problem.

At that point, to justify the answer, a far more rigorous approach, using the analytical basis for probability theory, must be taken. Which might well end up going as far as measure theory if the guy refuses to recognise that he doesn't know anywhere near enough to justify his claims.

This isn't even complex level stuff, and watching you make such a fundamental error such as 'reducing redundancies' without taking into account the different probabilities of the outcomes that produces is staggering.

Except, if you actually check back, I never said that at all.
What I said was that if you have two identical options, you can't rigorously use the Law of Counting to justify a claim about them.

Which is absolutely not, in any way, the same thing as saying you can just collapse the two into one and count them as one.

 

[DracoSuave]

stuff

Yes, I guess that is the amount of rigour a standard university would expect from a first year student. Which is to say, not much.

Anyway, we'll ignore that and simply focus on the bits you got wrong, rather than the bits you got right without rigour.

You should spend a little less time being arrogant and a little bit more time adressing your math.

And thusly we can identify the eight outcomes for selected children.

b1; b2; g1; g2; b3; g3; b4; g4

Yes, exactly, you cannot get answer of 1/2 without using a probability space that has at least 8 basic elements. Whereas the answer that gives 1/3 can be done using a space of 4 elements. Therefore the answer of 1/2 must necessarily use more information than the question gives as the answer of 1/3 uses at least as much information as the question gives and the 1/2 uses more than that.

The 1/3 answer uses the exact same information as the 1/2 answer. 4 pairings. No more, no less.

It results in 8 outcomes, mind you, which is a different matter.

As the answer of 1/2 uses more information, it cannot be a valid answer to the given question. Yes it might be a valid answer to a question with more information, but that is not the given question.

The exact same could be said of your own proof, that it presumes that 'One is a boy' is a quantification of the number of boys, and not simply a statement of identification of a singular boy.

Regardless, by applying more information, the situation exists that the probability is higher than 1/3. That means for the probability to -actually be- 1/3, the Mean Law states that there must also be a situation where the probability is less than 1/3.

Seeing as that situation does not exist, the probability cannot be 1/3.

Probability theory works by answering questions using the exact information given.

On that I agree.

If we had to accept every possibility of how the question might have been formulated, we would have to give every answer to any probability question as a function of all the different possibilities and we could never find an exact number without using, at least, functional integration.

Which is why questions need to be concise so that methods of information gathering that could reasonably affect the calculation of probability do not change the question from one with a real-number answer, to one that is an undefined number within a given range.

The question -is- ambiguous, and has been MATHEMATICALLY proven so. Why can you not accept that?

The problem isn't the math, the problem is the ambiguity of the question which creates a situation where one must apply a slant to the problem in order to adequately solve it.

If you've ever given an exact answer to a probability question then you know that probability just doesn't work like that.

Absolutely not. When you have a question with the parameters adequately defined, a real number answer can be attained in probability.

However not every question in probability can be given a discrete real number answer, sometimes an unknown presented in the question.

For example:

'There is a boy in this room. What are the chances that everyone in this room is a boy?' cannot be answered with a discrete real number answer unless I provide you the number of boys in the room.

In this particular case, the question itself has the ambiguity that 'One is a boy' does not, implicitly, or explicitly, tell you if it's a statement resulting from quantification or from selection. In the case of quantification, then the probability is 1/3, because the statement means 'At least one is a boy.' In the case of selection, then the probability is 1/2, because the statement means 'This one is a boy.'

But 'One is a boy' doesn't tell you -which- of those two senses it means, and the further question 'What are the odds the other is also a boy' does not clear it up.

Regardless, if the answer could be 1/3 or 1/2 based on the sense of 'One is a boy' and those are the only two reasonable possibilities, then if the sense of 'One is a boy' is unknown, then it -itself- is a matter of probability, of the sort we do not have the ability to calculate.

Thusly, the answer is an incalculable number between 1/3 and 1/2, but cannot actually be either of them, while the ambiguity exists. Resolve the ambiguity, and then you change the n in my equation to either 0 or 1, and the answer becomes a defined real number solution.

You can claim 'simplicity' all you like, but if simplicity does not provide an adequate answer, then it is not adequate.

Also, MACM 101 was Discrete Mathematics, first year. Everything that has been done in this thread is basic level probability, a class I excelled in, if you must know.

Yes, this is a basic question.
That is, until someone who did well in one basic probability course comes along and starts abusing that basic probability course to justify his preconceptions about the problem.

And that is the problem here. You've used your bias that there must be a discrete numerical answer to this problem to blind you to the ambiguity of the problem. You simply cannot accept that the question could mean 'This one is a boy, what about the other one?' You also simply cannot accept the method of selection is actually relevant in the Boy or Girl Paradox.

However... seeing as you're dismissive of my "First Year Opinion" because it apparently doesn't appeal to your sense of 'expertise', I am forced to resort to an appeal to authority; might I refer you instead to a textbook on the subject, wherein the Boy/Girl paradox is discussed:

Grinstead and Snell's Introduction to Probability; The CHANCE Project; Version dated 4 July 2006 [http://math.dartmouth.edu/~prob/prob/prob.pdf] on Page 183, section 4.3, Paradoxes.

Where it says:

This problem and others like it are discussed in Bar-Hillel and Falk. These
authors stress that the answer to conditional probabilities of this kind can change
depending upon how the information given was actually obtained. For example,
they show that 1/2 is the correct answer for the following scenario.

And later

In the preceding examples, the apparent paradoxes could easily be resolved by
clearly stating the model that is being used and the assumptions that are being
made.

So... while my 'First Year' knowledge doesn't seem to agree with your sensibilities, perhaps you need to go to the textbook itself where it informs you, that yes, the method of information gathering is, in fact, integral to the resolution of paradoxes of this type, expecially when there are ambiguities in how the information was obtained.

Not to mention, the very one who invented the Boy/Girl Paradox also agrees with this, even tho at first he stated it was 1/3.

At that point, to justify the answer, a far more rigorous approach, using the analytical basis for probability theory, must be taken. Which might well end up going as far as measure theory if the guy refuses to recognise that he doesn't know anywhere near enough to justify his claims.

As I said, you are letting your arrogance replace your ability to analyze math. You are using ad hominems in place of add the numbers correctly.

This isn't even complex level stuff, and watching you make such a fundamental error such as 'reducing redundancies' without taking into account the different probabilities of the outcomes that produces is staggering.

Except, if you actually check back, I never said that at all.
What I said was that if you have two identical options, you can't rigorously use the Law of Counting to justify a claim about them.

...without of course using the correct mathematics to show that those options, are, in fact, identical. In fact, while taking two non-identical options (B1->B3 and B3->B1 in the by-the-child based method) and calling them identical options simply because you deleted an inconvenient column.

...while ignoring the further mathematical disproof of the validity of that very tactic.

Which is absolutely not, in any way, the same thing as saying you can just collapse the two into one and count them as one.

And yet, that is exactly what you did. You never said you could, you just went ahead and did it.

Because I am starting to doubt your own mathematical expertise here, as basic algebra (The Mean Law) seems to have gone over your head.

 

[Maze1125]

You should spend a little less time being arrogant and a little bit more time adressing your math.

Yes, maybe you should.

The 1/3 answer uses the exact same information as the 1/2 answer. 4 pairings. No more, no less.

It results in 8 outcomes, mind you, which is a different matter.

To get an answer of 1/2 you need a probability space of size 8, to get an answer of 1/3, you need a space of size 4.

If you understand rigorous probability as well as you act like, I don't understand why you can't seem to comprehend how these two are different.

The exact same could be said of your own proof, that it presumes that 'One is a boy' is a quantification of the number of boys, and not simply a statement of identification of a singular boy.

Whether you say that or not, it doesn't count as an extra piece of information for probability theory.

My formulation uses a space of size 4 and a single mathematical definition for OB.
Your formulation uses a space of size 8 and a single mathematical definition for OB (Which you never bothered to state. Go rigour!)

You can wave about statements like "You presumed..." all you want, but they don't change the fact that, by the simple numerical values, mine uses less information than yours.

The question -is- ambiguous, and has been MATHEMATICALLY proven so. Why can you not accept that?

You haven't given one single piece of mathematical rigour, so how exactly could it have been mathematically proven?

The problem isn't the math, the problem is the ambiguity of the question which creates a situation where one must apply a slant to the problem in order to adequately solve it.

There isn't an ambiguity, the question is completely concise, the only reason you think there is ambiguity is due to your instance on thinking of the problem in a real context rather than as a simple mathematical formulation.

However not every question in probability can be given a discrete real number answer, sometimes an unknown presented in the question.

For example:

'There is a boy in this room. What are the chances that everyone in this room is a boy?' cannot be answered with a discrete real number answer unless I provide you the number of boys in the room.

I can easily see a way to give that a single numerical value.

Look at all the possible distributions of people in the room, use functional analysis to produce a combined distribution and then find the probability using that combined distribution.

Of course, such an action would be near impossible to do rigorously, and very hard even non-rigorously, but a single numerical value is still quite possible.

Grinstead and Snell's Introduction to Probability; The CHANCE Project; Version dated 4 July 2006 [http://math.dartmouth.edu/~prob/prob/prob.pdf] on Page 183, section 4.3, Paradoxes.

Did you even read that, or did you Google it for the sake of searching for quotes you like?

It clearly says that if you have more information, then the answer can be 1/2.

In fact it is purely a discussion on how this problem could come up in the real world, and the point is that in the real world, you will always have information on the selection process, which means the problem itself can never occur because you will always have more information and so will be solving a different problem, one with more information.

It even points out that the purely mathematical answer that doesn't attempt to work out how this problem could occur in reality, and just solves the problem itself, is 1/3.

Not to mention, the very one who invented the Boy/Girl Paradox also agrees with this, even tho at first he stated it was 1/3.

Undoubted he does agree that in a real world situation the problem would be different, I agree with that, but that is not the same thing as agreeing with that you're saying. That you can't find a simple numerical answer to the question as posed.

Which is absolutely not, in any way, the same thing as saying you can just collapse the two into one and count them as one.

And yet, that is exactly what you did. You never said you could, you just went ahead and did it.

Yes, I did, way way back when we were still talking naively.
Since I have started using rigorous proofs I have not once done that.

Regardless, if the answer could be 1/3 or 1/2 based on the sense of 'One is a boy' and those are the only two reasonable possibilities, then if the sense of 'One is a boy' is unknown, then it -itself- is a matter of probability, of the sort we do not have the ability to calculate.

Thusly, the answer is an incalculable number between 1/3 and 1/2, but cannot actually be either of them, while the ambiguity exists. Resolve the ambiguity, and then you change the n in my equation to either 0 or 1, and the answer becomes a defined real number solution.

Lets suppose that your idea was an appropriate thing to do in probability.

Let's look at the question "A woman has two children, the first is a boy, what is the probability the other is a boy?"

The answer is clearly 1/2.

Except we don't know how these children were selected.
For all we know they were selected from families that all had 2 boys. Which would give a probability of 1.
Or maybe the wording means what a layman would think it would mean, after all, no-one would say "The first is a boy, and so is the second.", they would say "They are both boys." or "The first was a boy and the second a girl." This is clearly ambiguous, and so we have to consider the possibility that we are only looking at boy-girl families, and so the probability would be 0.
It's even possible that they took all families with boys and orientated them so that a boy was first every time. This would give a probability of 1/3.

Therefore the answer to this question cannot be the naive 1/2, but in fact 0p + q/3 + r/2 +s where p,q,r,s > 0 (as every one is possible) and p+q+r+s = 1.

Which is not a pointlessly absurd answer at all...

In fact, this leads back to your constant point that the answer can't be 1/3.
If you idea is valid, then, as you said before, one of the possibilities is that "One is a boy." means "Only one is a boy." which gives a probability of 0.
Therefore the answer is not n/3 + (1-n)/2 but 0p + q/3 + r/2. Which could well be a 1/3, even by your own claims.