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Author Topic: Annoying probability question of the day
Aris Katsaris
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The thing that Badvok is arguing (correctly) is that the probability for gender of the second child is independent of the gender of the first child.

What he's arguing wrongly is that he thinks this means the probability of two boys is independent of the probability of one boy.

You can't merge probabilities the way Badvok is doing.

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Badvok
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quote:
Originally posted by Aris Katsaris:
Look at conditional probability please.

Probability(A given B) = (Probability of A AND B) / Probability (B)

Probability(two boys, given atleast one boy) = Probability(two boys) / probability(atleast one boy) = (1/4) / (3/4) = 33%

Not relevant:
quote:
When in a random experiment the event B is known to have occurred, the possible outcomes of the experiment are reduced to B, and hence the probability of the occurrence of A is changed from the unconditional probability into the conditional probability given B.
In this case the events are not linked and therefore conditional probability does not apply.
Unless you are categorically stating that having one boy child affects the likelihood of the other being a boy (though true biologically this is not really the case here).

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DonaldD
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quote:
And if we collapse the problem space by saying Child1 is a boy we get:
Badvok, you went through all that effort, and you just repeated the semantic issue in the end. By limiting the statement to a very specific child, yes you reach the 1/2 value. But your statement above is exactly equivalent to saying "And if we collapse the problem space by saying my son John is a boy we get". Do you see how this is different from the interpretation "at least one of my children is a boy"?

Now take that same interpretation, which in your case is exactly equivalent to 'either child1, child2 or both are boys' and apply it to your 8 combinations and see what you get.

As Aris has so testily observed, been there, done that.

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Badvok
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quote:
Originally posted by Aris Katsaris:
What he's arguing wrongly is that he thinks this means the probability of two boys is independent of the probability of one boy.

The probability of one boy is 1 - this is stated in the question.
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Badvok
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quote:
Originally posted by DonaldD:
Now take that same interpretation, which in your case is exactly equivalent to 'either child1, child2 or both are boys' and apply it to your 8 combinations and see what you get.

But that is not the question! The question is "What is the probability the other child is a boy if one is a boy?".
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DonaldD
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quote:
In this case the events are not linked and therefore conditional probability does not apply.
This conditional probability calculation requires the events to be independent (ie, random, not linked) so it very much could apply.
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DonaldD
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quote:
But that is not the question! The question is "What is the probability the other child is a boy if one is a boy?".
Yes, sort of: if "one of them is a boy" not "child1 is a boy".
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Badvok
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quote:
Originally posted by DonaldD:
quote:
But that is not the question! The question is "What is the probability the other child is a boy if one is a boy?".
Yes, sort of: if "one of them is a boy" not "child1 is a boy".
It is "Child1 is a boy". John has declared the gender of one of his children - this child can be labelled as Child1. The question is then: What is the probability the other child (labelled as Child2 so as to mark it as a separate and distinct entity to Child1) is also a boy?

[ July 27, 2010, 07:38 AM: Message edited by: Badvok ]

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Aris Katsaris
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quote:
Not relevant:
Not relevant?? That's what conditional probability is all about. That's pretty much the ONLY relevant thing.

quote:
What is the probability the other child is a boy if one is a boy?
No, the question is what is the probability that *BOTH* are boys if "at least one" is a boy.

You are misquote the quote to indicate that one particular child was mentioned to be a boy. NO. No particular child was declared a boy. The only thing declared was that ATLEAST ONE OF THEM was a boy.

Not "the other" - "both"
Not "if one" - "if atleast one"

quote:
The probability of one boy is 1 - this is stated in the question.
That's what Probability of (A given B) means. That you assume the B happens and it. To calculate (A given B) with the formula i gave, you calculate the probability that B would have to happen (WITHOUT you knowing it did happen).

You don't understand conditional probability.
You take B by itself. The you take the possibilities of A and B together. Then to calculate (A given B) you divide the probabilities of (A and B) by probability of (B).


If we know one *particular* child is a boy, then the probabilities are:
(Both boys, given one particular child a boy) = (Both boys)/(one particular child a boy)
= (1/4) / (1/2) = 50%.

If we don't know that one particular child is a boy, then:
(Both boys, given atleast one boy)= (Both boys)/(atleast one boy) = (1/4)/(3/4) = 33%

Math proves you wrong.

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Badvok
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quote:
Originally posted by Aris Katsaris:
Math proves you wrong.

LOL, I give up, I'll let you all head off to kindergarten now while I get on with some real work.
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Aris Katsaris
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quote:
Originally posted by Badvok:
quote:
Originally posted by DonaldD:
[qb]
quote:
But that is not the question! The question is "What is the probability the other child is a boy if one is a boy?".
Yes, sort of: if "one of them is a boy" not "child1 is a boy".

It is "Child1 is a boy". John has declared the gender of one of his children - this child can be labelled as Child1.
If he did pinpoint to a particular child, you could label it.

But he didn't, so you can't.

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Aris Katsaris
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These are simple highschool math, Badvok. I don't know what country you're from, but it needs an improvement of its educational system.
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Aris Katsaris
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And not just in math, since you don't even seem to understand what the word "given" means.
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TomDavidson
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Aris, could you try to refrain from just straight-up insulting people?
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Aris Katsaris
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For the sake of anyone who's less of an ass than Badvok:

(Both boys, given atleast one boy born on Sunday) =
(Both Boys AND a boy born on Tuesday) / (one boy born on Tuesday) = (13/196) / (27/196) = 13/27

--

Since nobody bothers reading the explanations any more, I'll stick with the so-called kindergarten math.

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Aris Katsaris
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Tom, I told him that math proves him wrong, and in response he told all of us to go to kindergarten.

So, **** him. He's a moron and an *******.

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TomDavidson
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I think what Badvok's saying is that the mere fact that one of the boys can now be distinguished as "the boy you were told about" changes the matrix, even if you don't know which boy you were told about, as follows:

Boy I told you about / Girl
Girl / Boy I told you about
Boy I told you about / Boy
Boy / Boy I told you about

The reason this isn't true is that in the latter case, there is no way to distinguish between the two boys. Ergo, the real space is:

Boy I told you about / Girl
Girl / Boy I told you about
Boys, one of which I told you about

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DonaldD
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I think Badvok just got a little frustrated, Aris. Patience.

Basically, it is not a math issue with Badvok, but a semantic one. It comes down to why the 'child I told you about who I will name Boy 1' is not the same as 'one of the two children' ("I have two kids. One of them is a boy"). Badvok, what you need to get your head around is that the father might also be talking about child2 instead of child1. There is no way for us to know (see Tom's statement above). And this ambiguity does not increase the probability of 2 boys being born out of two.

Again, Badvok, look at the possible combinations from a first born/second born perspective:
1st. 2nd
Girl Girl
Girl Boy
Boy Girl
Boy Boy

That is it. By naming the boys in the last case (labelling them if your prefer) you don't make that case more probable.

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Badvok
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quote:
Originally posted by DonaldD:
I think Badvok just got a little frustrated, Aris. Patience.

Yep, sorry.

quote:
Again, Badvok, look at the possible combinations from a first born/second born perspective:
1st. 2nd
Girl Girl
Girl Boy
Boy Girl
Boy Boy

That is it. By naming the boys in the last case (labelling them if your prefer) you don't make that case more probable.

Nope it doesn't but you have now labelled them 1st and 2nd instead [Smile] I thought labelling would help understanding but it obviously didn't.

If 1st is a boy then 2nd can be either a boy or a girl.
If 2nd is a boy then 1st can be either a boy or a girl.

The probability of both being boys is 1/2.

The error people are making is in using the known to collapse the space [GG,GB,BG,BB] to [GB,BG,BB] which is incorrect. Because one half of the pair has become known the space becomes [kG,kB,kG,kB] or [Gk,Gk,Bk,Bk] where k is the known. In this space the probability that the unknown is B is 2/4 or 1/2.

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Ciasiab
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First let me say that I am not a mathematician / statistician, so I am not conversant in statistical spaces. That being said, as an engineer, I like to check my answer to make sure it is in the right ballpark.


Here is the problem statement:
quote:
John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday." What are the odds that he has two boys?

Since John Doe basically volunteered the information without being probed, to me this would be the same as the following hypothetical situation (ignoring day of the week for now):

I have a room full of 1,000,000 dads. Each is the dad of two children. I ask each dad to tell me the gender of one of their children. Based on my simple statistical training, I would predict that ~500,000 dads would tell me they have a boy and ~500,000 dads would tell me they have a girl. I would also expect ~500,000 dads to have mixed gender families.

Based on the logic by Donald, Aris, and others, if the Dad said that he had at least one boy, then there is a 66% chance he has a boy and a girl. So that gives me 333,333 mixed gender families. Using the same logic on the dad's who said they had at least one girl gives another 333,333 mixed gender families. Totaling the mixed gender families gives 666,666 (impossible!).

The only way the problem works out to 2/3 is if we restate the problem as follows:

John Doe: I have two children.
Me: Is at least one a boy?
John Doe: Yes.

In this case the probability of two boys is 66%.

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threads
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quote:
Originally posted by Badvok:
[GG,GB,BG,BB] to [snip] [kG,kB,kG,kB].

You transformed GG to kG where k is the known BOY.
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Badvok
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quote:
Originally posted by threads:
quote:
Originally posted by Badvok:
[GG,GB,BG,BB] to [snip] [kG,kB,kG,kB].

You transformed GG to kG where k is the known BOY.
It doesn't matter whether the known is a boy or a girl, the known is simply a known and becomes a constant that no longer affects the probability.
If the known is a boy the probability of both being boys is equal to the probability the unknown is a boy.

[ July 27, 2010, 09:28 AM: Message edited by: Badvok ]

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Aris Katsaris
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quote:
I have a room full of 1,000,000 dads. Each is the dad of two children. I ask each dad to tell me the gender of one of their children.
No, no, NO! A million times NO, you don't ask that!

You don't ask each dad to tell the gender of one of their children. YOU ASK THEM WHETHER THEY HAVE atleast one boy.

75% of dads will have atleast one boy, and say "YES!"

This splits into 25% of dads with GB, 25% of dads with BG, and 25% of dads with BB.

So, 1 out of 3 dads that replied YES will have a second boy, and 2 out of 3 dads (out of those that replied YES) will NOT have a second boy.


quote:
Based on the logic by Donald, Aris, and others, if the Dad said that he had at least one boy, then there is a 66% chance he has a boy and a girl.
I'm getting close to tears here. We've all repeated a hundred times that asking dad to name the gender of one of his children is DIFFERENT to asking him about whether he has atleast one boy.

You know why? Because 75% of dads have atleast one boy, but only 50% of dads would select a boy if you ask them to pick randomly a gender of one of their kids.

Do you understand now, please?

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Pete at Home
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Badvok, think of this:

Situation 1: Man A tells you he has 2 kids. What's the possibility that his oldest kid is a boy?

Situation 2: Man B tells you that he has two kids, and that at least one of them is a boy. What's the possibility that the oldest kid is a boy?

But I still don't get boy Tuesday.

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Aris Katsaris
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quote:
The only way the problem works out to 2/3 is if we restate the problem as follows:

John Doe: I have two children.
Me: Is at least one a boy?
John Doe: Yes.

In this case the probability of two boys is 66%.

The probability of two boys here is unambiguously 33%.

Thanks for the rephrasing, btw, it's the exact thing that I suggested needed be done, for the statistics to be clear.

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DonaldD
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Well, this is entertaining.
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Badvok
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quote:
Originally posted by Pete at Home:
Badvok, think of this:

Situation 1: Man A tells you he has 2 kids. What's the possibility that his oldest kid is a boy?

1/2 = the same as whether any individual kid is a boy.
quote:

Situation 2: Man B tells you that he has two kids, and that at least one of them is a boy. What's the possibility that the oldest kid is a boy?

Now that is a totally different question and is more like the Monty Hall problem because you have made a selection (the oldest) that directly affects the probability. You have now narrowed the possibilities to 3 (eByB,eByG,eGyB) and hence the probability that the oldest is a boy is 2/3.
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Aris Katsaris
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quote:
But I still don't get boy Tuesday.
I don't blame people that don't get boy Tuesday, he's counter to intuition. The 7x7x2x2=196 possibilities are btw too much to keep in mind, so it gets even more confusing.

So instead of days of the week, let's say Democrat/Republican. Let's consider even distributions. The 4 possibilities:
BB, BG, GB, GG
now become 16 possibilities:
BD/BD, BD/BR, BR/BD, BR/BR
BD/GD, BD/GR, BR/GD, BR/GR
GD/BD, GD/BR, GR/BD, GR/BR
GD/GD, GD/GR, GR/GD, GR/GR

John Doe: I have two children
Me: Is at least one a republican boy?
John Doe: Yes.

Out of those 16 possibilities the following 7 remain:
BD/BR, BR/BD, BR/BR,
BR/GD, BR/GR,
GD/BR, GR/BR

We know for sure dad is one of these possibilities.
Out of these 7 possibilities however only 3 of them contain a second boy.

So there's a 3/7 chance for a second boy.

--

Again math reaches the same conclusion:
(Probability of two boys, given probability of a republican boy) = (Probability of two boys AND a republican boy)/(Probability of a republican boy) = (3/16) / (7/16) = 3/7

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DonaldD
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quote:
If the known is a boy the probability of both being boys is equal to the probability the unknown is a boy.
Badvok, you still don't know which child is 'known'. The father is a black box. And since you don't 'know' that, the child is not 'known' in fact.

"I have two kids. One of them is a boy". Do you agree that this statement is effectively the same as "I have two kids. They are not both girls"?

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Ciasiab
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quote:
You don't ask each dad to tell the gender of one of their children. YOU ASK THEM WHETHER THEY HAVE atleast one boy.
I guess I can't see how you can possibly get this from the original problem statement of John Doe walking up and declaring he has a boy.

The only way I see your logic working is if the problem is restated with probing questions being asked, instead of information declared. If I declare (unasked) information about one of my children, it does not change the probability of my other child being a particular gender. That just doesn't work as shown above.

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Aris Katsaris
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quote:
"The only way I see your logic working is if the problem is restated with probing questions being asked, instead of information declared."
Which is pretty much what I argued at the whole first page of this thread, so we're in agreement here.

Probing questions is the best way to create unambiguous situations, because declared information has the problem of selection bias on behalf of the father.

IF the father is randomly selecting the gender of one his kids to reveal, there's 50% chance that he has two boys. But IF the father only wants to reveal the existence of a boy, there's 33% chance that he has two boys.

In short, everything I said in lots and lots of detail in the 1st page of this thread.

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Badvok
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I think it is a lost cause trying to explain this yet again but here goes:

There are two distinct children.
Their genders are not interdependent.
Their genders are not determined by birth order.
Their gender is not dependent on the day of the week they were born on.

We know one is a boy.

Therefore there is only one single sole distinct individual solitary unknown gender value and that can have only one of two possible values.

All this collapsing of probability sets/spaces incorrectly just confuses the issue and you are bound to get weird and wacky numbers.

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Ciasiab
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quote:
IF the father is randomly selecting the gender of one his kids to reveal, there's 50% chance that he has two boys. But IF the father only wants to reveal the existence of a boy, there's 33% chance that he has two boys.
It sounds like we don't disagree. From the problem statement, there is no way I would assume that the father only wants to reveal the existence of a boy. All we can tell from the problem statement is that the father has chosen one of his kids, and revealed that this particular kid was born on a Tuesday and happens to be a boy.
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Aris Katsaris
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quote:
"We know one is a boy."
We know the existence of a boy. And the existence of a boy IS interdependent with the probability that a specific child is a boy.

quote:
"Therefore there is only one single sole distinct individual solitary unknown gender value and that can have only one of two possible values."
The combination "one girl and one boy" is twice more likely to occur in a population than the combination "two boys".

Again, man: out of 100% dads with two boys: 75% will have at least one boy. But only 25% of the dads will have two boys.

Please, can you respond whether you agree or not with the above fact?

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Aris Katsaris
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quote:
It sounds like we don't disagree. From the problem statement, there is no way I would assume that the father only wants to reveal the existence of a boy. All we can tell from the problem statement is that the father has chosen one of his kids, and revealed that this particular kid was born on a Tuesday and happens to be a boy.
Yes, that's what I argued in the first page of this thread.
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Ciasiab
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Badvok

If the problem were stated as follows:
John Doe: I have two children.
Me: Is at least one a boy?
John Doe: Yes.

Would you agree the probability of two boys changes from 1/2 to 1/3?

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Badvok
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quote:
Originally posted by Ciasiab:
Badvok

If the problem were stated as follows:
John Doe: I have two children.
Me: Is at least one a boy?
John Doe: Yes.

Would you agree the probability of two boys changes from 1/2 to 1/3?

Nope, sorry. It is still 1/2. No matter how you re-phrase it. There is still only one unknown that can have one of only two values.
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DonaldD
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Badvok:

"I have two kids. One of them is a boy".

Do you agree that this statement is effectively the same as:

"I have two kids. They are not both girls"?

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Badvok
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quote:
Originally posted by Aris Katsaris:
The combination "one girl and one boy" is twice more likely to occur in a population than the combination "two boys".

Yep. But that is the wrong question.
The answer to the original question is 'The combination "one girl and one boy" is just as like to occur in a population (of fathers with at least one son) as the combination "two boys"'.

quote:
Again, man: out of 100% dads with two boys: 75% will have at least one boy. But only 25% of the dads will have two boys.
No I think out of 100% dads with two boys 100% will have at least one [Smile]
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Badvok
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quote:
Originally posted by DonaldD:
Badvok:

"I have two kids. One of them is a boy".

Do you agree that this statement is effectively the same as:

"I have two kids. They are not both girls"?

Well yes, but turning the question around doesn't change the number of unknowns.
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