Wonderfully logical illogical probability problem

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jokester, go read about degeneracy. you are wrong on this one.

edit. a bit late with that maybe.

edit again:
"I have two children, this is my daughter..." <-- if this statement allows for the case of GG to be a possibility, then it is no different to "I have two children, if at least one is female... "
 
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Ah sorry didn't realise you were the OP, no they are both case 1, because you have stated "this is my daughter" in both cases, the age of the child is irrelevant to the probability. By stating "this is my daughter" removes the opposite case of being possible and it's irrelevant what one was born first. If you KNOW (ie probability = 1, not 0.5) which child is female there are only two options possible, BG, or GG. the alternative boy girl combination of GB is NOT possible. Younger or older makes no difference as all it does is change GB to BG and BG to GB, it's restricted the sample space by removing the third option that gives the 2/3.

The correct question you need to ask is "I have two children, if at least one is female (this allows for both combinations of BG and GB) what is the probability that the other is male"

The probabilities are linked to what options (from the four:BB BG GB GG (with order mattering, ie BG and GB are different)) are left once the information given in the statement is taken into account.
For this case let the first child be the youngest ie in the option GB the girl is younger than the boy.

Copied from the OP, the first statement is "I have two children, this is my daughter Lisa".

Here we know at least one is a girl so the BB option is out. I think you're happy here that the prob therefore or the other being a boy is 2/3.

The second statement is "I have two children, this is my youngest child, Lisa". This obviously rules out the options BB and BG (as we know the first child is a girl) This is where this second statement differs from the first, you are given more information. You are told that not only is at least one of his two children a girl, but also that it is the youngest child which is a girl. So with optios BG abd BB rules out the only remaining are GB and GG which gives the probability of 1/2.


You're link has some interesting stuff about what (in a real life situation) the person (tutor in this case) would actually say depending on how many daughters/sons they had. In this example this has to be ignored, i copied it verbatim from the sheet of problems i was set, personally i would phrase it slightly clearer.

Hopefully you can see that being told whether she is the younger or older does matter (but not which one, merely that you are told one, seems bizarre to me even now). Heh, what a great question 'eh!! :D
 
Quoting for proof for the naysayers.

I'm presuming those boy-girl combinations were generated randomly? Good effort though and it shows exactly what need to be shown. That is when the position of the child is fixed (child1 orchild2 or youngest or oldest) as well as the gender, the probability drop from 2/3 to 1/2 (or in that model 1/3 to 1/2).

Nice one Zogger
 
yeh they are all random.. 5 minute jobby last night

PHP:
$c1girl = 0;
$c2boyGivenGirl = 0;
$atLeast1 = 0;
$otherGirl = 0;

echo "<p>This is like some twisted school seating plan</p>";

for($i = 0; $i < 10000; $i++) {
	$child1 = rand(0,1);
	$child2 = rand(0,1);
	echo $child1 ? "girl" : "boy";
	echo " ";
	echo $child2 ? "girl" : "boy";
	echo ", ";
	if ($child1) {
		$c1girl++;
		if (!$child2) {
			$c2boyGivenGirl++;
		}
	}
	if ($child1 || $child2) {
		$atLeast1++;
		if ($child1 && $child2) {
			$otherGirl++;
		}
	}
}

echo "<p>At least 1 child is a girl in $atLeast1 cases. Of these, the other child was a girl $otherGirl times (" . ($otherGirl/$atLeast1) * 100 . "%) .<br>";
echo "Child 1 is a girl in $c1girl cases. Of these, the other sibling was a boy $c2boyGivenGirl times (" . ($c2boyGivenGirl / $c1girl) * 100 . "%) .<br>";
echo "Both children were boys in " . ($i - $atLeast1) . " / 10,000 cases</p>";
?>

:)
 
I though somebody would have mentioned the difference between Bayesian theory and classical statistics. It was the only thing in 'O' level maths which sent me off to sleep!

If the parent of the children was a language pendent the the other child is a boy, if they are not then you cannot tell as the standard of grammar in the population is terrible (and this is from somebody who failed 'O' level English 3 times who now has to correct drafts from somebody at work who got a A GCSE!).
 
From http://mathforum.org/library/drmath/view/52186.html
Supposing that we randomly pick a child from a two-child family. We
see that he is a boy

and want to find out whether his sibling is a
brother or a sister. (For example, from all the children of two-child
families, we select a child at random who happens to be a boy.) In
this case, an unambiguous statement of the question could be:

From the set of all families with two children, a child is
selected at random and is found to be a boy. What is the
probability that the other child of the family is a girl?


Note that here we have a pool of kids (all of whom are from two-child
families) and we're pulling one kid out of the pool. This is like the
problem you're talking about. The child selected could have an older
brother, an older sister, a younger brother or a younger sister.

Let's look at the possible combinations of two children. We'll use B
for Boy and G for girl, and for each combination we'll list the older
child first, so GB means older sister while BG means younger sister.
There are 4 possible combinations:

{BB, BG, GB, GG}

From these possible combinations, we can eliminate the GG combination
since we know that one child is a boy. The three remaining possible
combinations are:

{BB, BG, GB}

In these combinations there are four boys, of whom we have chosen one.
Let's identify them from left to right as B1, B2, B3 and B4. So we
have:

{B1B2, B3G, GB4}

Of these four boys, only B3 and B4 have a sister, so our chance of
randomly picking one of these boys is 2 in 4, and the probability is
1/2 - as you have indicated.
Case 1 in OP, except with girl swapped with boy.

The key part is that you're specifically picking out a child (a girl in the OP). But in the other case you're not. As you see it makes no difference whether the kid is younger or older.

Edit: The 2/3 case totally relies on you not specifically identifying one child as being female as it removes the possibility of that child being a boy!
 
I'm not a probability expert, but to me it is irrelevant whether the girl is older or younger. If you factor in the possibility that the other child could be a twin brother or sister, I think this again highlights that the age order has no bearing on the probability of the sex of the other child?


edit - ignore the twin bit - forgot one is always older even if only by a few minutes.
 
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From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?

1/2

From the set of all families with two children, a family is selected at random and is found to have a boy. What is the probability that the other child of the family is a girl?

2/3


After reading that link, this makes sense to me now. What I still find ambiguous is which of these applies to the OP. :)
 
From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?

1/2

From the set of all families with two children, a family is selected at random and is found to have a boy. What is the probability that the other child of the family is a girl?

2/3


After reading that link, this makes sense to me now. What I still find ambiguous is which of these applies to the OP. :)

I wholeheartedly agree
 
From http://mathforum.org/library/drmath/view/52186.html

Case 1 in OP, except with girl swapped with boy.

The key part is that you're specifically picking out a child (a girl in the OP). But in the other case you're not. As you see it makes no difference whether the kid is younger or older.

Edit: The 2/3 case totally relies on you not specifically identifying one child as being female as it removes the possibility of that child being a boy!

This is actually a different situation and a different question is answered, this siblings problem is not the same as the two daughter problem i posted.

To remove the ambiguity from the OP:

Situation one:
You know - Man has two children
You know - One is a girl


Situation two:
You know - Man has two children
You know - One is a girl
You know - It is the youngest which is the girl


Now situation one answered:
4 combos {BB, BG, GB, GG} (let the first child in the combo be the eldest).
We know one is a girl therefore BB is out, we are left with {BG, GB, GG}.
Out of these three possible situations two contain boys, giving rise to the 2/3 answer.

Now situation two answered:
We know not only that one child is a girl (ruling BB out) but also that it is the youngest which is a girl which therefore also rules out GB.
We are therefore left with {BG, GG}.
Now you can see that the prob of the other child being a boy is 1/2.


This has to be clear enough, i can't see there's any flaw in that and this removes all linguistical ambiguity from the OP (which i admit was confusing but that was only as i copied it verbatim).
 
Mmm so if 2/3 is correct, if I have 1 daughter already and my wife is about to give birth, the odds on it being a boy are 2/3 and being a girl are 1/3 ?

That cant be right it must be 1/2
 
Mmm so if 2/3 is correct, if I have 1 daughter already and my wife is about to give birth, the odds on it being a boy are 2/3 and being a girl are 1/3 ?

That cant be right it must be 1/2

No, that's the other scenario. You know the girl is oldest so the odds of having a boy are 50% and this is an independent event.

The odds that you have a boy given that you also have a girl are 2/3. Note that the girl can be younger or older.








I think
 
No, that's the other scenario. You know the girl is oldest so the odds of having a boy are 50% and this is an independent event.

The odds that you have a boy given that you also have a girl are 2/3. Note that the girl can be younger or older.

But surely if a woman has 5 boys in a row the odds on the 6th being a boy are still 50% ?
 
But surely if a woman has 5 boys in a row the odds on the 6th being a boy are still 50% ?

Zogger you are correct.

The probability that the 6th child will be a boy is 50%. As this is an independent event.

With your wife about to give birth the prob that its a boy is still 50%. However given that you (one) have (has) two children at least one of which is a girl, the prob that the other is a boy is 2/3.

This is due to the combinations of B and G possible and the fact that BB is obv not poss as you (one) already have (has) a girl.

Out of {BB, BG, GB, GG} BB is not poss which leaves {BG, GB, GG} where two out of those three contain boys.

Tis just the strange counter intuitive workings of probability.
 
But surely if a woman has 5 boys in a row the odds on the 6th being a boy are still 50% ?

True, but if you know a woman has 6 children and that at least 5 are female the chances of the 6th being male are not 50%.

Back on the original question, I'm still of the opinion that it's 2/3 from the question asked.
I can see how the wording can confuse, but from the math forum page Jokester linked to we are looking at the second version, where we choose the family first IMO.

But now let's look at a different way of selecting the "boy" in the
problem. Suppose we randomly choose the two-child _family_ first. Once
the family has been selected, we determine that at least one child is
a boy. (For example, from all the mothers with two children, we select
one and ask her whether she has at least one son.) In this case, an
unambiguous statement of the question could be:

From the set of all families with two children, a family is
selected at random and is found to have a boy. What is the
probability that the other child of the family is a girl?

Note that here we have a pool of families (all of whom are two-child
families) and we're pulling one family out of the pool. Once we've
selected the family, we determine that there is, in fact, at least one
boy.

Since we're told that one child (we don't know which) is a boy, we can
eliminate the GG combination. Thus, our remaining possible
combinations are:

{BB, BG, GB}

Each of these combinations is still equally likely because we picked
one of the four families.

Now we want to count the combinations in which the "other" child is a
girl. There are two such combinations: BG and GB.

Since there are three combinations of possible families, and in two of
them one child is a girl, the probability is 2/3.


You meet your new tutor in town accompanied by a young girl. He says to you "I have two children, this is my daughter Lisa". What is the probability that his other child is a boy. Would it make a difference if he had said "I have two children, this is my youngest child, Lisa"? (You may assume a 50% chance of any one birth being a boy or a girl.


The fact that you meet your tutor determines the family, you then find out that one of that family's children is a girl.
 
True, but if you know a woman has 6 children and that at least 5 are female the chances of the 6th being male are not 50%.

This is true, as long as order does not matter, ie you don't know which of the 6 is the male (not in some weird hermaphroditic way) but in the ordering sense, So you can't say the boy is the oldest, otherwise the probability of him being a boy will again drop to 1/2. Strange i know!
 

You meet your new tutor in town accompanied by a young girl. He says to you "I have two children, this is my daughter Lisa". What is the probability that his other child is a boy. Would it make a difference if he had said "I have two children, this is my youngest child, Lisa"? (You may assume a 50% chance of any one birth being a boy or a girl.


The fact that you meet your tutor determines the family, you then find out that one of that family's children is a girl.
In the OP these are two distinct cases, one where we are told whether she is the youngest or the oldest (which it actually is is not important) where in this case the prob is 2/3. Then the other case where we are not told whether she is the elder or the younger, where the prob is then 1/2.
 
In the OP these are two distinct cases, one where we are told whether she is the youngest or the oldest (which it actually is is not important) where in this case the prob is 2/3. Then the other case where we are not told whether she is the elder or the younger, where the prob is then 1/2.

Well, yeah I was just looking at the first part as that's the one that seems to be causing the issues!
 
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