Super Duper Extra Hard Brainteaser - Page 3

jimdmcevoy · #21 03-17-2005, 03:14 PM

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in case anyone is waiting for an answer, i don't know, i'm trying to work it out now, this is similar to something my friend asked me a while ago, and i think i figured out the answer to his question, but haven't yet got one for this one

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you bastard. youre not allowed to ask brain teasers and not mention you dont know the answer already.MFer!! [img]/images/graemlins/cool.gif[/img]
that being said, nodody knows yet if im an idiot or not with my 50% answer!!

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[img]/images/graemlins/grin.gif[/img] in my defense I thought I knew the answer, but this is slightly different to another problem I solved, and then I did feel a little bit bad for people guessing when I realized i didn't know the answer, and then i thought it was a little funny

__Q__ · #22 03-17-2005, 03:15 PM

I really don't see how the second question is any different then the first. It doesn't seem to matter that the girls name is sarah. I think that info is just provided to confuse you.

Correct me if I'm wrong, but if there are two kids and your using a 50/50 breakdown between girls and boys, then the 4 possiblities are BB BG GB and GG. We know she has a girl, so there are three remaining possibilities. BG GB and GG. All three of which are equally possible. Therefore 1/3 of the time its GG.

Patrick del Poker Grande · #23 03-17-2005, 03:16 PM

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I really don't see how the second question is any different then the first. It doesn't seem to matter that the girls name is sarah. I think that info is just provided to confuse you.

Correct me if I'm wrong, but if there are two kids and your using a 50/50 breakdown between girls and boys, then the 4 possiblities are BB BG GB and GG. We know she has a girl, so there are three remaining possibilities. BG GB and GG. All three of which are equally possible. Therefore 1/3 of the time its GG.

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This is exactly what I'm saying, but that could all just be an effect of my post-Chipotle haze.

Huskiez · #24 03-17-2005, 03:18 PM

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.

__Q__ · #25 03-17-2005, 03:19 PM

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This is exactly what I'm saying, but that could all be just an effect of my post-Chipotle haze.

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Well I just finished off some Taco Bell, so what do I know.

partygirluk · #26 03-17-2005, 03:19 PM

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We know she has a girl, so there are three remaining possibilities. BG GB and GG. All three of which are equally possible. Therefore 1/3 of the time its GG.

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Do you think the italicised bit is true given that you know she has a Sarah?

Sporky · #27 03-17-2005, 03:19 PM

i'm re-submitting my answer after deleting it.

there are 3 piles of options.
B = boy
G = girl

BB GG GB

i dont' think that the 1% has anything to do with the problem. once you know that you have a girl, regardless of name...you end up with the two groups: GG (1/3rd chance) and GB (2/3rd chance). the BG group has twice as many people as the GG group, there are the same number of girls in each. which would mean 1/2 chance there is two girls with one named sarah and 1/2 chance there is 1 girl and she is named sarah.

totally wrong?

mostsmooth · #28 03-17-2005, 03:20 PM

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Now I'm getting 199/399, but still not sure. Give me 5 mins.

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yah i'm pretty sure this is right

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whats wrong with 50%?

jimdmcevoy · #29 03-17-2005, 03:20 PM

what your saying seems to make sense, and it'd be weird if it wasn't 1/3, but the way i see it before talking to the woman this is true:

boy/sarah with probability 1/200
non-sarah girl/sarah with probability 199/40000

and various other combinations without sarah in it

so after you find out she has a girl named sarah, you know it's one of these two possibilities, and hence you get the 199/399 number

partygirluk · #30 03-17-2005, 03:20 PM

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there are 3 piles of options.
B = boy
G = girl

BB GG GB

i dont' think that the 1% has anything to do with the problem. once you know that you have a girl, regardless of name...you end up with the two groups: GG (1/3rd chance) and GB (2/3rd chance). since you are twice as likely to find a girl named sarah in the GG group, wouldn't it be twice as likely that if you find sarah, you have two girls? that bumps the GG group to be even with the BG group. which would mean 1/2 chance there is two girls with one named sarah and 1/2 chance there is 1 girl and she is named sarah.

totally wrong?

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Almost right.