Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
I got 199/399 too.

P(G, B) = P(B, G) = 1/4.
Probability of G being named Sarah is 1%.
So given she has two children, 1/100 * 2/4 = 2/400 she has exactly one girl named Sarah.

P(G, G) = 1/4.
First girl named Sarah happens 1/100 times.
If the first girl isn't named Sarah, then the next one will be named Sarah 1/100 times. Probability of this is 99/100 * 1/100.

Probability of her having two girls, one of which is named Sarah is 1/4 * (1/100 + 99/100 * 1/100) = 199/40000.

So probability of her having two girls, given she has one named Sarah, is:

199/40000 / (199/40000 + 2/400) = 199/399.

Either I am missing something or that wasn't a super duper extra hard brainteaser. Just a simple math problem.

[/ QUOTE ]
This was my answer 1b, but I think I'm gonna stick with my 1a, which is that the Sarah crap is all a line of bull and the answer is 1/3. In my view, this is somewhat similar to the Let's Make A Deal puzzle we just talked about but with the random door picking in the second round.

In the end, I think all these other calculations would be necessary if the answer to whether or not she had a child named Sarah was no. Then you have to worry about whether or not she has a girl but her name just isn't Sarah. Once the answer to this question is yes, it's essentially the same as the first puzzle - she just told you she has a girl and that's all you need to know.

Re: Super Duper Extra Hard Brainteaser
 

It doesn't matter that the girls named sarah.

Besides, you know her name is sara. The probabitly of this woman haveing a girl named sara is 100%

Re: Super Duper Extra Hard Brainteaser
 

i think it's got something to do with the fact that had she had a girl but not one named sarah this situation would not take place

maybe, i dunno, just throwin something out there

Re: Super Duper Extra Hard Brainteaser
 

It's 1/3 right? It's the same as the first problem, who cares what the 1st girl's name is?

Re: Super Duper Extra Hard Brainteaser
 

Chew on this.

Easy problem 1:

What is probability that a mother of 2 has a girl? (Answer in white below)

<font color="white"> 3/4 </font>

Super duper hard problem:

99 per cent of girls like Barbie. You ask a mother of two whether she has a daughter who likes barbie and she says no. What is the probability that she has a daughter? Still 3/4?

Re: Super Duper Extra Hard Brainteaser
 

Wait a second, Are we sure that the answer to the first question is 1/3.

We know that one kid is a girl. the other kid is 50/50, Boy or girl. So 1/2 GB and 1/2 GG.

Both answers are 1/2

Re: Super Duper Extra Hard Brainteaser
 

nah, I'm 95% sure it's 199/399

and 99% original brain teaser is 1/3

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
Wait a second, Are we sure that the answer to the first question is 1/3.

We know that one kid is a girl. the other kid is 50/50, Boy or girl. So 1/2 GB and 1/2 GG.

Both answers are 1/2

[/ QUOTE ]
We're sure. There are four equally possible combinations of children:

BB BG GB GG

By her answering that she does have a girl, that eliminates the first possibility and leaves three remaining. We're now down to three equally possible combinations, of which GG is one:

BG GB and GG

There is a 1/3 chance that the answer is GG.

Re: Super Duper Extra Hard Brainteaser
 

Note that if her answer were "yes, my second child is a girl" or "yes, my first child is a girl" then it would be 50/50 that both her children are girls.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
I'll open with a well know semi-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?

Answer in white:

<font color="white"> 1/3 </font>

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?

[/ QUOTE ]

the answer isn't 1/3. we're drawing from 99% of the sample of girls, but 100% of the sample of boys. knowing that the second girl is NOT a sarah means that 1% of girls are unavailable for us to grab.

so say we have 99 girls and 100 boys, and we want to know the probability, given that one is a girl named sarah, that the other is a girl.

overall we can have

G, B
B, G
G, G

given that we have at least one girl. there are 100 ways to choose the boy in case 1, 100 ways to choose the boy in case 2, and 99 ways to choose the girl in case 3. so, I get 99/300.

initially it looks like conditional probability reduces this problem to the first one, but the "sarah stipulation" controls the sampling of the SECOND girl, not the sarah that we know to exist.