That would only be the case if the problem had said 'the ELDER child is a boy, what is the
probability that the YOUNGER is a girl'. Since we don't know which child is being referred to
in the first part, Chrismo is right. If you label the children ELDER CHILD and YOUNGER
CHILD instead of BOY and OTHER ONE, you'll see why there are 3 possibilities (girl/girl being
obviously impossible)
I agree... but we don't know whether the boy is the first child or the
second child. If we were awaiting the birth of the second child, then,
absolutely it would be a 50:50 chance, becuase there are only two
combinations (boy/girl, boy/boy), and one of those two satisfies the
conditions...
But in this case, there are three combinations (boy/girl, boy/boy and
girl/boy), and of those three, only two would satisfy the condition of
one child being a girl and the other child being a boy...
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I disagree. We already know that one is a boy. It doesn't matter which
one. The question is what are the chances of the remaining one being
a girl. It has absolutely nothing to do with the other child or the birth
order. The "other" child is simply either a boy or a girl.
Coyote
Hi Coyote,
I know exactly where you're coming from, and this is what makes this
question so evil!!
But the pure evil facts are that in a situation where one child is a boy,
and considering it doesn't matter whether it is the elder or the
younger, the possibilities are
boy/boy
girl/boy
boy/girl
Two of those three make the other child a girl.... and only one makes
the other child a boy. I agree that it is counter-intuitive!
If I have two children, what are the odds that both are girls...
Certainly, 1 in 4 !!! Therefore, there is a 3 in 4 chance that one of
them is a boy. And, if I tell you that one child is a boy, one of those
combinations (girl/girl) is no longer valid, leaving 3 possibilities, of
which two possibilities allow for a girl as the other child (boy/girl or
girl/boy)...
The catch to the question is that it is not asking what are the
chances of someone being a girl? - it is asking if we know that in a
two child family, one of the children is a boy, what are the chances of
the other one being a girl?. This is a diffrerent question to asking If
my first child is a boy, what are the chances of my second child being
a girl?...
-Chris
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I was in the middle of this reply once before when my screen went
blank and returned to the forum for no apparent reason, so you'll
probably see this 1-1/2 times.
I still think the logic is flawed because in essence you are using a
mathmatical equation to determine a solution to a random set of
possibilities. There is nothing on earth to suggest that because one
child is a boy (first or second) that the other can't equally be either a
boy or a girl. Because "girl" shows up twice on the list of possibilities
doesn't increase the chances that the other child is a girl. It is an
illusion. My reasoning is that if you look at it the other way around,
the same is true for the chances of a boy if one is known to be a girl.
This doesn't match with true statistics that show that in real life
approximately 70% of children are born male, with a much higher
mortality rate than that of young females. The approximate 52% -
48% ratio is among adults.
Respectfully,
Coyote
"There is nothing on earth to suggest that because one child is a boy
(first or second) that the other can't equally be either a boy or a girl."
That statement is absolutely correct!!!
In a way, the issue is a subtlety in the wording of the question; in fact
you get the 'weird' 2/3 ratio BECAUSE having a child of one sex has
absolutely no bearing on what the sex of the other child is. What's
going on is the question is biasing what population of families you are
thinking about.
OK, another way. Take 100 2 children families.
Hopefully, we all expect them in the following ratios:
(For this part, I've listed the children in birth order, however that has
no bearing on the issue. If I ignored birthorder, I would have had the
girl/boy group with 50 families instead of the listed two groups of 25
families. Did that make sense?)
boy/girl - 25 families (so 25 boys and 25 girls)
girl/boy - 25 families (so 25 boys and 25 girls)
boy/boy - 25 families (so 50 boys)
girl/girl - 25 families. (so 50 girls)
Note there are 200 children in consideration: 100 boys and 100 girls.
Now, toss out from consideration all 25 girl/girl families. Remember,
we've told you at least ONE of the children is a boy.
Now you are just considering the 3 sets:
boy/girl - 25 families (so 25 boys and 25 girls)
girl/boy - 25 families (so 25 boys and 25 girls)
boy/boy - 25 families (so 50 boys)
150 children left; but it's NOT 75 of each sex! instead, there are 100
boys and 50 girls! Cause you tossed an above average number of
girls (100 of the original 150) out of the equation by saying it wasn't
an all girl family.
And in this new set of families to consider, 100 out of the 150 of the
kids are boys! So the chance of the other child being a boy, given one
of them is ALREADY KNOWN TO BE A BOY, is 100/150 - or 2/3.
I have no idea if that helped at all.
But here is a new probability question:
Given a long thread in the forums, what is the probability that
maggoteer will add a longwinded response?
🙂
I think I started smoking crack right about here:
"Cause you tossed an above average number of girls (100 of the
original 150)...."
I meant to say:
you've tossed out 50 of the 100 girls from consideration, but kept all
the boys in the equation.
(Not sure were I started getting those new numbers.)
So of the remaing 150 kids, 100 of them are boys. only 50 are girls.
ie 2/3 are boys, so the chance of the other kid being a boy is 2/3.
That's what I meant to say.
Now where's Noelle Bush with my stuff??? See, I just can't think
straight with this awful buzzing in ma'h head.
You could demonstrate the soundness of the maths by doing the
following:
Take 200 bits of paper. On 100 of them write "Boy". On the other
100, write "girl".
Put all 200 bits of paper in a bucket and pick out pairs at random until
you have 100 pairs.
After this, you would expect 25 pairs to be "boy/boy", 25 pairs to
be "girl/girl" and the remaining 50 pairs to be 1 boy and 1 girl.
There are 75 pairs that contain at least 1 girl. Of those 75 pairs, 50
of them contain a boy but only 25 contain another girl.
You could most probably use this knowledge to take people's betting
money... in the same way that most people don't know that there are
more numbers in the world beginning with "1" than with "9"...
-Chris
you can also apply similar reasonings to bridge. When the number of
trumps on the opponent's side is even, the chances on an unequal
distribution (say 2-0 or 3-1) are higher than on an equal one (1-1 or
2-2). The number of combinations with equal (or girl/girl) are smaller
than the number of unequal (girl/boy or boye/girl), basically because
one cannot count the symmetrical combination as a different one. sin.
"in real life approximately 70% of children are born male"
What? Male longevity and infant mortality might be worse than female, but it's not that bad!
Are you seriously saying that the if the AVERAGE NEWBORN GIRL will live to 70 (say), then
the AVERAGE NEWBORN BOY will die in his early thirties?
I think that the male/female ratio is more like 52:48 at birth, but 49.5:50.5 in the overall
population (both very rough estimates, don't complain if they are 1-2% out)
On a side note, show some more respect to the Maths: the SET of possibilities is not
'random', it is clearly defined, so a probabilistic approach is perfectly sound. What is random
is which of the possibilities is actually the case IN A PARTICULAR FAMILY, but that's not what
we're talking about.