Re: assumptions and consequences
 

[ QUOTE ]
Are you aware how easy this is? It is 50/50 in each example. That we know the first is a girl does not change the odds. I can't believe that I'm posting here, and I didn't read the whole thread, but you guys are way over thinking this.

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That just isn't right. There are 4 scenarios:

1)Boy/Boy
2)Boy/Girl
3)Girl/Boy
4)Girl/Girl

If she says she has at least one girl, that means it is 2, 3, or 4. 3 different ways that she could have at least one girl. Only one of those 3 ways does she have 2 girls. 1/3.

Its kind of like the old gameshow (can't think of name) problem where you are choosing a prize behind 3 different doors. 2 doors have a goat, 1 door has the big prize. After making your choice, the host shows you one of the doors you DIDN'T choose that has a goat behind it. Now you get a chance to stick with your answer or to switch doors. You should definately switch doors, as the door he didn't reveal has a 2/3 chance of being a winner. Not 50/50. Similar concept.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
Question #1
Assuming equal chances of having a boy or a girl born to you, the answer is that 50% of the time she will have girl/boy, and 50% of the time she will have girl/girl. Knowing that she already had one girl changes nothing about the chances concerning the sex of the second child. There must be a girl in the pair, and the other child's sex will be 50/50.

[/ QUOTE ]

Actually, the random chance is 1/4 Boy/Boy, 1/4 Girl/Girl, 1/4 Girl/Boy, 1/4 Boy/Girl. The mother's answer was that she had at least one girl, so that eliminates the boy/boy optioin leaving girl/girl one of three possibilities that already occured, so 1/3rd.

Edit: The key is the probability events already took place, you are just further defining what category the mother fell into originally by specify she had at least one girl. If the probability is for a future event, then if you know the mother had a girl, and it is 50/50 boy/girl, then she will in the future be 50% likely to be boy/girl and 50% likely to be girl/girl, again, the question was posed about probability events that have already occured and you define the branch the person already stepped down, not using a previous branch stepped down to bias a future result from its natural occurance level.

Vern

Re: Super Duper Extra Hard Brainteaser
 

Having not read any posts and doing this all in my head, I think it's 1/2.

The trick is that in the first problem, a girl is always a girl, but in this half she's twice as likely to have a daughter named Sarah with two daughters than she is with just one daughter. So while the second scenario is twice as likely, it's only half as likely to end up with a "Sarah" and thus given that there is a Sarah, it's 50/50 one 1 or 2 daughters.

What do I win?

edit: I had not thought about the issue of how 1% of girls are Sarah, but that no mother would name two different girls Sarah, so we have to start thinking about different scenarios.

Assuming other mothers might name both their daughter's Sarah, but just not this one, I stick with my 50/50.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]
Question #1
Assuming equal chances of having a boy or a girl born to you, the answer is that 50% of the time she will have girl/boy, and 50% of the time she will have girl/girl. Knowing that she already had one girl changes nothing about the chances concerning the sex of the second child. There must be a girl in the pair, and the other child's sex will be 50/50.

[/ QUOTE ]

Actually, the random chance is 1/4 Boy/Boy, 1/4 Girl/Girl, 1/4 Girl/Boy, 1/4 Boy/Girl. The mother's answer was that she had at least one girl, so that eliminates the boy/boy optioin leaving girl/girl one of three possibilities that already occured, so 1/3rd.

Edit: The key is the probability events already took place, you are just further defining what category the mother fell into originally by specify she had at least one girl. If the probability is for a future event, then if you know the mother had a girl, and it is 50/50 boy/girl, then she will in the future be 50% likely to be boy/girl and 50% likely to be girl/girl, again, the question was posed about probability events that have already occured and you define the branch the person already stepped down, not using a previous branch stepped down to bias a future result from its natural occurance level.

Vern

[/ QUOTE ]

I stand by my 50/50 answer to both questions. You're indicating that actual results will be vastly different from the mathematical future probability, which is totally illogical.

If a mother has exactly two children born to her, and we already know that one of those children is a girl, it has no influence on the probability of the sex of the second child (or perhaps the first born, which makes no difference). The answer remains that the second child is a boy 50% of the time and a girl 50% of the time, meaning that 50% of the time the mother will have two girls.

Nothing about the facts in the second question changes this answer.

Re: Super Duper Extra Hard Brainteaser
 

the answer is: 100/99

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
edit: I had not thought about the issue of how 1% of girls are Sarah, but that no mother would name two different girls Sarah, so we have to start thinking about different scenarios.

[/ QUOTE ]

I don't think the 1% Sarah-probability matters in this case, because it's a given condition. The problem is implictly stated as "Given that she has one daughter named Sarah, what are the odds that both her children are girls?" We as the problem-solvers don't care what the odds were for her to get that Sarah in the first place. To put it another way, say a degenerate gambler has a 40% chance of blowing all his money at the racetrack. Tonight he happens to win the lottery at a 1,000,000-to-1 shot. Given that he's won the lottery, what are the odds that he will blow all his money at the racetrack tomorrow?

Still 40%, of course. If we said "what are the odds he will win the lottery and then blow the resulting money at the racetrack?" it would be .000001 * .40, but in the former question, we really don't give a rip how he got the money he's taking to the racetrack. By the same token, we don't care how the woman got her daughter Sarah (*wink* *nudge*); we just care what the odds are that she has two girls.

It seems to me that the odds are 1/2 in this case as well, because we have four possibilities (Gs = Girl, Sarah; G = Girl; B = take a guess.)

GsB
BGs
GGs
GsG

Two of these possibilities involve her having two girls, 2/4 = 1/2.

Any thoughts/corrections?

Re: assumptions and consequences
 

[ QUOTE ]
Instead, I feel like a better approach would be to look at all the 2-child families and adjust my first assumption to something like "when a girl is born, there is an X% chance that she will be named Sarah, unless she already has a sister named Sarah" and set X such that the result is 1% of all girls are named Sarah.

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One problem with what you want to do here, is that you can't solve for X unless you assumes soimething about the population, like, only 2-child families are allowed.

(BTW, I suspect if you made this assumption, and solved for X, then when you plugged the X in, the result would be such that the answer to the original question would be 1/2.)

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Question #1
Assuming equal chances of having a boy or a girl born to you, the answer is that 50% of the time she will have girl/boy, and 50% of the time she will have girl/girl. Knowing that she already had one girl changes nothing about the chances concerning the sex of the second child. There must be a girl in the pair, and the other child's sex will be 50/50.

[/ QUOTE ]

Actually, the random chance is 1/4 Boy/Boy, 1/4 Girl/Girl, 1/4 Girl/Boy, 1/4 Boy/Girl. The mother's answer was that she had at least one girl, so that eliminates the boy/boy optioin leaving girl/girl one of three possibilities that already occured, so 1/3rd.

Edit: The key is the probability events already took place, you are just further defining what category the mother fell into originally by specify she had at least one girl. If the probability is for a future event, then if you know the mother had a girl, and it is 50/50 boy/girl, then she will in the future be 50% likely to be boy/girl and 50% likely to be girl/girl, again, the question was posed about probability events that have already occured and you define the branch the person already stepped down, not using a previous branch stepped down to bias a future result from its natural occurance level.

Vern

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I stand by my 50/50 answer to both questions. You're indicating that actual results will be vastly different from the mathematical future probability, which is totally illogical.

If a mother has exactly two children born to her, and we already know that one of those children is a girl, it has no influence on the probability of the sex of the second child (or perhaps the first born, which makes no difference). The answer remains that the second child is a boy 50% of the time and a girl 50% of the time, meaning that 50% of the time the mother will have two girls.

Nothing about the facts in the second question changes this answer.

[/ QUOTE ]
[censored] A, man, this has been beat to death and is over.

Even if you don't accept this particular answer, you have to at least admit that it's a poorly constructed puzzle because it's open to a variety of interpretations. It's more like an inperpretation contest than a logic puzzle. Let it die again and let's forget it.

Re: Super Duper Extra Hard Brainteaser
 

What you are missing is the question is not about the probability of a future event, it is about the probability of events that have already occured. You are assuming that the person that has two children, and answers that "At least one of them is a girl" had the girl first, and therefor since we are assuming the likelihood of having a boy and girl are the same, the person is 50% likely to to have their second child as a boy or girl.

Look at it this what, there are four mothers with two children each, one has boy/boy, one has boy/girl, one has girl/boy and one has girl/girl. Exrapolate this to the full population because it is a normal distribution of a 50/50 event. How many mothers can answer the question "Do you have at least one daughter?" Answer, 3. So now that you have narrowed to the group down to those 3 as a subgroup, you pose a second question, "How many of those three can answer that both children are girls?" The answer is 1, so once you define the sub group as people with two children that can answer yes to "I have at least one daughter" of that group, 1/3 of that group can then also answer yes to the follow up question, "Do you have two daughters?"

This isn't about asking a mother "Did you have a daughter first?" and then asking "Do you have two daughters?" That would be 50/50, but in the question posed, there are 4 equally likely probability branches, b/b,b/g, g/b, g/g.

You may stand by your answer all you want, it doesn't make it right. If you don't understand the way the first question is resolved, the trying to work out the second is wasted effort.

Vern

Re: Super Duper Extra Hard Brainteaser
 

I'm a Math Major; let's see how disgraceful my "logic" is...

[ QUOTE ]
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?

[/ QUOTE ]

Assume Boy/Girl distribution is 50/50.
For 200 families, we get
Boy = 100
Girl = 99
Sarah = 1

Since we already have a Sarah, we can no longer choose it, leaving us with

Boy = 100
Girl = 99

to choose from. So, 99/(99 + 100) => 99/199 chance that mother has two girls.

Re: Super Duper Extra Hard Brainteaser
 

Everyone has left out hemaphrodites as a possibility.

Re: NOT SO FAST
 

[ 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

[/ QUOTE ]

And his reply:
[ QUOTE ]
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.

Re: Super Duper Extra Hard Brainteaser
 

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

Re: Super Duper Extra Hard Brainteaser
 

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

Re: Super Duper Extra Hard Brainteaser
 

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.

Re: Super Duper Extra Hard Brainteaser
 

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