Super Duper Extra Hard Brainteaser

[Hiding]

Everyone has left out hemaphrodites as a possibility.

[CardSharpCook]

[ QUOTE ]
My e-mail to Prof. Blass:
[ QUOTE ]
Hello, Professor Blass. I am a former student at U of M and I have a logic/probability puzzle that a friend and I cannot agree on and I was hoping you wouldn't mind taking a minute to give us your opinion on the answer. I'd thank you in advance for any time you'd care to spend on this - it should be pretty quick for you.

First, an easy one:
You know that a woman has exactly 2 children. You ask her if she has at least one girl, and she says yes. What's the probability that she has two girls?

Now, suppose that 1% of all girls are named Sarah, and that if a mother had two girls that the mother would not name both of them Sarah.

You know a woman has exactly 2 children, you ask her if she has a girl named Sarah, she says yes.

What are the chances she has two girls?


Thanks again for your time,
Patrick

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And his reply:
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Some additional hypotheses are needed to determine these numbers.
For example, you could imagine a society in which each woman keeps having
children until either she has a son or she has 10 daughters. In such a
society, a woman with exactly two children would necessarily have one boy
and one girl.
So let me make some apparently reasonable assumptions: (1) The
probability of any particular child to be a girl is 1/2 and is independent
of the gender of any other children. (This is fairly close to true, but
not exactly true, biologically.) (2) The decision to stop after having
two children is independent of the genders of those children. (This is
needed to exclude examples like the one above; I'm not at all sure it's
true, even in our society. Some families (especially some fathers) really
want a son.)
Under these assumptions, the answer to your first question is 1/3.
(To be really pedantic, let me note that there are other assumptions
implicit in the question --- for example that when "she says yes" she's
telling the truth.) The assumptions imply that the four possible
situations: BB, BG, GB, GG occur equally often. The woman would say yes
in the last 3 of these, and in one of these 3 she has two girls.
For the Sarah question, let me temporarily assume (even though
it's not quite accurate) that each girl is given the name Sarah with
probability 1/100, independently of all others. (This is inaccurate
because no mother names two girls Sarah. I'll come back to this point
later.) Then, in a population of N women with two children each, there
will be (as in the preceding paragraph) N/2 with one boy and one girl, and
among these there will be N/200 whose daughter is named Sarah. In
addition, there will be N/4 with two girls, and among these there will be
approximately N/200 one of whose daughters is named Sarah. (The
approximation is again related to the fact that they won't both be named
Sarah.) So altogether there are N/100 women with a daughter named Sarah.
Of these, half are in the first group (with one boy and one girl) and half
are in the second (with two girls). So, given that a woman has a daughter
named Sarah, the probability that she has two daughters is 1/2.
This makes pretty good sense. Among the women with at least one
daughter, only 1/3 have two daughters. But those who have two daughters
are more likely than the others to have a daughter named Sarah. So when
you look only at women with a daughter named Sarah, you're looking at more
from the two-daughters group than from the one-of-each group.
But I ignored the stipulation that nobody names both of her
daughters Sarah. Taking that stipulation into account will make no
*practical* difference --- the error introduced by ignoring the
stipulation is probably less than the errors introduced by the general
assumptions in the first paragraph. But as a purely *mathematical*
exercise, one can re-do the computation with the stipulation taken into
account. Again, I need an assumption, namely that there is a certain
probability p that a woman will name a daughter Sarah unless she already
has a daughter named Sarah, that this probability is the same for all
women (as opposed, for example, to depending on whether she already has a
son), and that all the naming decisions are probabilistically independent.
(This p will not be 1/100, because of the stipulation, but it will be very
close.) Then among all the women with exactly two children, 1/2 will have
one of each gender, of whom p/2 will have a daughter named Sarah. In
addition, 1/4 will have two daughters, of whom (p+(1-p)p)/4 = (2p-p^2)/4
have a daughter named Sarah. [The calculation of this p+(1-p)p is as
follows: A woman who has two daughters has probability p of naming the
first one Sarah and probability (1-p)p of not naming the first one Sarah
and then naming the second one Sarah.]
If there are altogether N women with exactly two children, then
they will have N daughters (one daughter for each of N/2 women and two
daughters for each of N/4 women), and of these daughters the number named
Sarah is, by the preceding paragraph, [p/2 + (2p-p^2)/4]N = [p - p^2/4]N.
The question says that this is N/100. So you can solve for p; it turns
out to be 2 - squareroot(3.96). Then you can plug this value of p into
the expressions p/2 and (2p-p^2)/4 to find what proportion of women have
one child of each gender with the girl named Sarah and how many have two
girls with one named Sarah. The probability that a woman with a girl
named Sarah has two girls is therefore the second of these proportions
divided by the sum of the two proportions; after some cancellation, this
gives a probability of (2-p)/(4-p). [Since p is very small, close to
1/100, this ratio is close to 1/2, which was the answer when the
stipulation was ignored.]

Andreas Blass


[/ QUOTE ]
Emphasis mine.

There you have it. Congrats to partygirluk and all others who came up with 199/399. I deem GM and jason_t's pwnership of me to be no longer. Further, I deem them to be pwned by me and their locations should suggest as much for the next month plus one day. All monetary matters of the bet were previously called off.

[/ QUOTE ]

Just goes to show that even University Proffesors can fail at logic.

[jimdmcevoy]

awsome, this thread lives on

in case anyone is curious, the people i consider experts (mostly gaming mouse) have concluded that without a doubt the answer to the first simple one is 1/3, and the answer to the second question is either 1/2, or very close to 1/2, and it depends on exactly how you interpret the question

[omahahahaha]

I've got it!!!! I never paid attention during algebra class so give me a little leeway here.


If the total number of boys on earth is equal to B
and the total number of girls on earth is equal to G

I'm pretty sure the answer would look something like this:

G x .99 / B + (G x .99)

Let me know if I'm right and if I am , someone owes me a beer at this year's wsop.

redsoxr1

[chesspain]

I think the answer to #2 is still 1/3. I don't see how guessing the name changes the probability of her already having two girls.

[omahahahaha]

Can any experts out there tell me if I'm correct or not?