Re: Super Duper Extra Hard Brainteaser
 

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[ QUOTE ]

yes, ever so slightly more than 50-50 the other kid is a boy, it doesnt make it anywhere near 1/3.

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awesome we agree.

now consider all the ways she can have two kids.

boy boy - no.

so that leaves

sarah, boy
boy, sarah
sarah, girl.


find out the probability of sarah, girl. I think it's 99/299. what do you think?

[/ QUOTE ]
i think the answer ends at 50-50(approximately)and if im wrong, you can buy me a beer

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

yes, ever so slightly more than 50-50 the other kid is a boy, it doesnt make it anywhere near 1/3.

[/ QUOTE ]

awesome we agree.

now consider all the ways she can have two kids.

boy boy - no.

so that leaves

sarah, boy
boy, sarah
sarah, girl.


find out the probability of sarah, girl. I think it's 99/299. what do you think?

[/ QUOTE ]
i think the answer ends at 50-50(approximately)and if im wrong, you can buy me a beer

[/ QUOTE ]
What if you're both wrong?

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]

yes, ever so slightly more than 50-50 the other kid is a boy, it doesnt make it anywhere near 1/3.

[/ QUOTE ]

awesome we agree.

now consider all the ways she can have two kids.

boy boy - no.

so that leaves

sarah, boy
boy, sarah
sarah, girl.


find out the probability of sarah, girl. I think it's 99/299. what do you think?

[/ QUOTE ]

I might be wrong, but I think you should also include girl, sarah.

Because the chance of having a Sarah is so unlikely, having two girls gives you two shots at having a Sarah.

Note in my math if you asked for the chance of two girls given that the mom has a girl not named Sarah, it's ~1/3.

Re: Super Duper Extra Hard Brainteaser
 

if I'm proven wrong, I'll be incredibly happy that we know the answer.

if I'm wrong I'll buy myself a beer.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

yes, ever so slightly more than 50-50 the other kid is a boy, it doesnt make it anywhere near 1/3.

[/ QUOTE ]

awesome we agree.

now consider all the ways she can have two kids.

boy boy - no.

so that leaves

sarah, boy
boy, sarah
sarah, girl.


find out the probability of sarah, girl. I think it's 99/299. what do you think?

[/ QUOTE ]
i think the answer ends at 50-50(approximately)and if im wrong, you can buy me a beer

[/ QUOTE ]
What if you're both wrong?

[/ QUOTE ]
then you buy us both a beer

Re: Super Duper Extra Hard Brainteaser
 

sarah girl and girl sarah can be considered the same.

BB
BG
GB
GG

are all equally likely. but, we know that one girl is a sarah. for the boy cases, it's obvious who sarah is. for the sarah and other girl case, it doesnt matter which girl is sarah, we just know that one of them is. I should have specified, but it doesnt change the answer IMO.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
sarah girl and girl sarah can be considered the same.

BB
BG
GB
GG

are all equally likely. but, we know that one girl is a sarah. for the boy cases, it's obvious who sarah is. for the sarah and other girl case, it doesnt matter which girl is sarah, we just know that one of them is. I should have specified, but it doesnt change the answer IMO.

[/ QUOTE ]

I think it changes when you factor in that a girl named Sarah is very unlikely.

This is the way I see it.

B B
B G1
G1 B
G1 G2

Note the last combination has two shots at getting a girl named Sarah. Either G1 or G2 could be Sarah.

Re: Super Duper Extra Hard Brainteaser
 

yeah, that's what I'm saying too. word, we agree, now I just have to say it the right way.

have you ever seen conditional probability? the jist of it is that we KNOW that there is a girl named sarah. now we just have to worry about the other one.

the question isn't "what are the odds she has a girl named sarah and another girl", it's "GIVEN that we have a sarah, what are the odds we ALSO have a girl (knowing that we can't have two sarahs)"

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
yeah, that's what I'm saying too. word, we agree, now I just have to say it the right way.

have you ever seen conditional probability? the jist of it is that we KNOW that there is a girl named sarah. now we just have to worry about the other one.

the question isn't "what are the odds she has a girl named sarah and another girl", it's "GIVEN that we have a sarah, what are the odds we ALSO have a girl (knowing that we can't have two sarahs)"

[/ QUOTE ]

Scroll up to my post with the math (the latter one). It has all probabilities in it explicitly shown.

I think this riddle is over. Done with this thread until we get an answer from OP.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
I think this riddle is over. Done with this thread until we get an answer from OP.

[/ QUOTE ]
The OP has already said he doesn't know the answer, so I'll just fill in and tell you all it's 1/3. Good day.

Re: Super Duper Extra Hard Brainteaser
 

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What are the chances she has two girls?

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A woman with 1 will answer "yes" 1% of the time. A woman with two will answer "yes" (1 - 0.99^2 + 0.01^2) = 2% of the time. (0.99^2 is the probability of both girls _not_ being named Sarah. The 0.01^2 is added back in because "Sarah" can't be given to both.)

There are twice as many women with only one girl as there are with two, but they answer yes half as often.

So, the answer's 1/2.

THE ANSWER IS < 1/3, N/M
 

I am done until someone comes and handles this is a way I cannot.

good luck all,
me

Addendum...ordering
 

now when i first looked at this problem, my gut says that order doesn't matter...but after thinking about it, it wouldn't be a brain teaser b/c then the answer would be 50%...

so order DOES matter and the order of the children (age in this case) affects the answer to the problem...

now what if the woman has fraternal twins and its equiprobable its BG or GB for the fraternal twins...same question...does that make it 50%?

-Barron

Re: Super Duper Extra Hard Brainteaser
 

I still don't understand how #1 is 1/3 instead of 1/2. Are BG and GB not the same thing?

Re: Super Duper Extra Hard Brainteaser
 

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they're the same from the sense of a set of people (when order doesn't matter) but you have to count them as far as equally likely outcomes are concerned.

think about it this way. I flip a coin twice. what are the odds I get two heads? what are odds I get one heads and one tails, in any order?


[/ QUOTE ]

why the first answer
 

i guess im dumb and have taken too many roids but why is the first answer 1/3. seems to this sucker that there are two possibilities 1 boy and 1 girl or 2 girls both equally likely.

Re: THE ANSWER IS < 1/3, N/M
 

[ QUOTE ]
I am done until someone comes and handles this is a way I cannot.

good luck all,
me

[/ QUOTE ]



Given the information, here are the possibilities:

B,Sara
Sara,B
G,Sara
Sara, G

Each of these events is is equally likely -- it's chance is .5*.5*.01. The answer is therefore 1/2.

Re: Super Duper Extra Hard Brainteaser
 

The answer is 1/2. Here's why. Once we have received the information there is a girl and her name is Sarah, we have four possibilities:

Boy, Girl named Sarah
Girl named Sarah, Boy
Girl, Girl named Sarah
Girl named Sarah, Girl.

All of these possibilities are equally likely and there are two possibilities with the situation we are concerned with. So the answer is 2/4 = 1/2.

For those interested in a Monte Carlo simulation to "check" the results, here is C code to do that.

#include &lt;stdio.h&gt;
#include &lt;stdlib.h&gt;

/* c1, c2 are the children; 0 means girl, 1 means boy
n1, n2 are the names; 0 means Sarah, not 0 means not Sarah
count1 is the number of times both children are girls
count2 is the number of times at least one child is Sarah
we are interested in count1 / count2
*/

#define ITERATES 1000000

int main(int argc, char **argv) {
int i;
int count1, count2;
int c1, c2;
int n1, n2;

count1 = count2 = 0;

srand(time());

for(i = 0; i &lt;= ITERATES; i++) {
n1 = -1;
n2 = -1;
c1 = rand() % 2;
c2 = rand() % 2;
if(c1 == 0) n1 = rand() % 100;

if(c2 == 0 &amp;&amp; n1 != 0) n2 = rand() % 100;

if(n1 == 0 || n2 == 0) {
if(c1 == 0 &amp;&amp; c2 == 0) count1++;
count2++;
}

}

printf("%d / %d\n", count1, count2);
printf("%f\n", (float) count1/ (float) count2);

return 0;
}

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]

Girl, Girl named Sarah
Girl named Sarah, Girl.

All of these possibilities are equally likely

[/ QUOTE ]

Are they?

What is the probability of girl not named sarah, girl named sarah?

99/200 * 1/200 which is 99/40,000

What is the probability of girl named sarah, girl?

1/200 * 1/2 which is 100/40,000

Where am I going wrong?

Re: THE ANSWER IS < 1/3, N/M
 

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[ QUOTE ]
I am done until someone comes and handles this is a way I cannot.

good luck all,
me

[/ QUOTE ]



Given the information, here are the possibilities:

B,Sara
Sara,B
G,Sara
Sara, G

Each of these events is is equally likely -- it's chance is .5*.5*.01. The answer is therefore 1/2.

[/ QUOTE ]

I disagree. I think [girl named sarah, girl] and [girl, girl named sarah] are subsets of [girl, girl].

also, how can the probability of this be HIGHER than the first example, which is less restrictive?

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
What is the probability of girl named sarah, girl?

1/200 * 1/2 which is 100/40,000

Where am I going wrong?

[/ QUOTE ]

1/200 * (99/100) * 1/2. They can't both be sarah.

Re: Super Duper Extra Hard Brainteaser
 

jason, by that logic the answer to the first, easier, problem is 1/2, which is wrong.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]
What is the probability of girl named sarah, girl?

1/200 * 1/2 which is 100/40,000

Where am I going wrong?

[/ QUOTE ]

1/200 * (99/100) * 1/2. They can't both be sarah.

[/ QUOTE ]

Hmm. The probability of the first child being a girl called Sarah is just 1/200 and then the probability of the next one being a girl is 1/2 n'est pas?

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
What is the probability of girl named sarah, girl?

1/200 * 1/2 which is 100/40,000

Where am I going wrong?

[/ QUOTE ]

1/200 * (99/100) * 1/2. They can't both be sarah.

[/ QUOTE ]

Hmm. The probability of the first child being a girl called Sarah is just 1/200 and then the probability of the next one being a girl is 1/2 n'est pas?

[/ QUOTE ]
Non. Read this. All of it.

Re: Super Duper Extra Hard Brainteaser
 

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jason, by that logic the answer to the first, easier, problem is 1/2, which is wrong.

[/ QUOTE ]

There are three possibilities once we know there is at least one girl:

girl, girl
boy, girl
girl, boy

Therefore the answer to the first question is 1/3

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
jason, by that logic the answer to the first, easier, problem is 1/2, which is wrong.

[/ QUOTE ]

Wrong, the possibilities in the first question, once we know there is at least one girl are

G, B
B, G
G, G

and all of these are equally likely. So the answer is 1/3.

Re: Super Duper Extra Hard Brainteaser
 

Seems to me that this is only a brainteaser because it induces people to waste a bunch of time with the Sarah red herring when the answer is still 1/3.

Think about it intuitively: If you gathered a population of mothers with two children, at least one of which is a girl, and divided them into subpopulations based on whether they had a child named Sarah, would you expect the subpopulations to have different estimated means on the #ofGirls variable. Basically, the Sarah thing just identifies a subpopulation but shouldn't change any other characteristics.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
Seems to me that this is only a brainteaser because it induces people to waste a bunch of time with the Sarah red herring when the answer is still 1/3.

Think about intuitively: If you gathered a population of mothers with two children, at least one of which is a girl, and divided them into subpopulations based on whether one they had named a child Sarah, would you expect the subpopulations to have different estimated means on the #ofGirls variable. Basically, the Sarah thing just identifies a subpopulation but shouldn't change any other characteristics.

[/ QUOTE ]
Exactly.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
Seems to me that this is only a brainteaser because it induces people to waste a bunch of time with the Sarah red herring when the answer is still 1/3.

Think about intuitively: If you gathered a population of mothers with two children, at least one of which is a girl, and divided them into subpopulations based on whether one they had named a child Sarah, would you expect the subpopulations to have different estimated means on the #ofGirls variable. Basically, the Sarah thing just identifies a subpopulation but shouldn't change any other characteristics.

[/ QUOTE ]

Think about it like this, it might change your mind: knowing that a women has two girls increases the likelihood that she has a child named Sarah. Conversely, knowing she has a child named Sarah....

Re: Super Duper Extra Hard Brainteaser
 

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[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
What is the probability of girl named sarah, girl?

1/200 * 1/2 which is 100/40,000

Where am I going wrong?

[/ QUOTE ]

1/200 * (99/100) * 1/2. They can't both be sarah.

[/ QUOTE ]

Hmm. The probability of the first child being a girl called Sarah is just 1/200 and then the probability of the next one being a girl is 1/2 n'est pas?

[/ QUOTE ]
Non. Read this. All of it.

[/ QUOTE ]

Please explain which part of what I said was wrong.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]
Seems to me that this is only a brainteaser because it induces people to waste a bunch of time with the Sarah red herring when the answer is still 1/3.

Think about intuitively: If you gathered a population of mothers with two children, at least one of which is a girl, and divided them into subpopulations based on whether one they had named a child Sarah, would you expect the subpopulations to have different estimated means on the #ofGirls variable. Basically, the Sarah thing just identifies a subpopulation but shouldn't change any other characteristics.

[/ QUOTE ]

Think about it like this, it might change your mind: knowing that a women has two girls increases the likelihood that she has a child named Sarah. Conversely, knowning she has a child named Sarah....

[/ QUOTE ]
...means she has a 1/3 chance that her other child is also a girl, provided she has exactly 2 children.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
What is the probability of girl named sarah, girl?

1/200 * 1/2 which is 100/40,000

Where am I going wrong?

[/ QUOTE ]

1/200 * (99/100) * 1/2. They can't both be sarah.

[/ QUOTE ]

Hmm. The probability of the first child being a girl called Sarah is just 1/200 and then the probability of the next one being a girl is 1/2 n'est pas?

[/ QUOTE ]
Non. Read this. All of it.

[/ QUOTE ]

Please explain which part of what I said was wrong.

[/ QUOTE ]
I gave you that link implying that this has been explained several times in this thread and that you should read the thread.

Re: Super Duper Extra Hard Brainteaser
 

That is not what your link did.

Re: Super Duper Extra Hard Brainteaser
 

jason,

I don't really think your code checks what we want it to check. you only update the check if both are girls, which leaves out all possibilities of boy, sarah and sarah, boy.

you are checking the ratio of all pairs to pairs with one sarah. we know that this particular pair does NOT have AT LEAST one sarah, it has EXACTLY one.

the correct pseudo code is

c1 = girl (this is sarah)
c2 = rand (1 - 199)

if c2 = 1-100, c2 = boy
if c2 = 101-199, c2 = girl

then take the fraction of c1, c2 pairs that are both girl in relation to the whole. it will be approximately .31

I hate this cliche, but,

do you see why?

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
jason, by that logic the answer to the first, easier, problem is 1/2, which is wrong.

[/ QUOTE ]

No it's not. Let's say she tells us the name of her girl, Sara.

Sara, G
G, Sara
B, Sara
Sara, B

The chance of "Sara, G" is not 1/4. It's 1/8. Do you see why?

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]
Seems to me that this is only a brainteaser because it induces people to waste a bunch of time with the Sarah red herring when the answer is still 1/3.

Think about intuitively: If you gathered a population of mothers with two children, at least one of which is a girl, and divided them into subpopulations based on whether one they had named a child Sarah, would you expect the subpopulations to have different estimated means on the #ofGirls variable. Basically, the Sarah thing just identifies a subpopulation but shouldn't change any other characteristics.

[/ QUOTE ]

Think about it like this, it might change your mind: knowing that a women has two girls increases the likelihood that she has a child named Sarah. Conversely, knowing she has a child named Sarah....

[/ QUOTE ]

decreases the odds of her having two girls given that

a) some % of girls are named sarah
b) boys and girls are equal
c) not both girls can be sarah.

I will bet money that I am right.

Re: Super Duper Extra Hard Brainteaser
 

How much are you willing to bet?

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]


I will bet money that I am right.

[/ QUOTE ]

I will bet up to $100 that my answer of 1/2 is correct.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]
jason, by that logic the answer to the first, easier, problem is 1/2, which is wrong.

[/ QUOTE ]

No it's not. Let's say she tells us the name of her girl, Sara.

Sara, G
G, Sara
B, Sara
Sara, B

The chance of "Sara, G" is not 1/4. It's 1/8. Do you see why?

[/ QUOTE ]

gaming mouse,

there was an original distribution of boys to girls. in that distribution, having a boy and a girl was twice as likely as having two girls. the fact that we now know she has one girl who happens to be named sarah does not change anything - the girl named sarah can be either of the two outcomes of girl, it doesn't add a permutation like property to it, which seems to be what you and jason are doing. if you stipulate that her OLDEST daughter is sarah, then order matters, and the answer is slightly less than 1/2 due to the fact that she can't have two sarahs. but as it is, the answer is slightly less than 1/3.


knowing that this first girl is named sarah means nothing except that it means the other girl cannot be sarah, which makes it slightly less likely that the other person is a girl. so the sarah clause, is, like patrick said, SOMEwhat of a red herring. it still provides useful information though.

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
[ QUOTE ]


I will bet money that I am right.

[/ QUOTE ]

I will bet up to $100 that my answer of 1/2 is correct.

[/ QUOTE ]
I'll take it.