Super Duper Extra Hard Brainteaser

[jimdmcevoy]

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?

[partygirluk]

I'm getting just under 1/5 but I could well be out.

[Asufiji2004]

For a 140+ IQ you don't seem too confident.

[partygirluk]

I am wrong.

[__Q__]

i'd say, 1/3

[Patrick del Poker Grande]

I say 1/3 for both, assuming an equal distribution of boys and girls (that a woman is equally likely to have a boy as a girl, not 60/40 or 55/45 or whatever it is in reality).

[jimdmcevoy]

[ QUOTE ]
For a 140+ IQ you don't seem too confident.

[/ QUOTE ]

Should a person with 140+ IQ necessarily be confident?

[daryn]

</font><blockquote><font class="small">In risposta di:</font><hr />
</font><blockquote><font class="small">In risposta di:</font><hr />
For a 140+ IQ you don't seem too confident.

[/ QUOTE ]

Should a person with 140+ IQ necessarily be confident?

[/ QUOTE ]

no

[partygirluk]

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

[Asufiji2004]

[ QUOTE ]
[ QUOTE ]
For a 140+ IQ you don't seem too confident.

[/ QUOTE ]

Should a person with 140+ IQ necessarily be confident?

[/ QUOTE ]

in his ability to figure out this xtra hard brain teaser?....yes!!

[partygirluk]

The intuitive answer is 1/3 btw, I just need to make sure my maths coincides with it /images/graemlins/smile.gif

[mostsmooth]

without thinking- 1/3
with some crazy miracle thinking- 31%

[jimdmcevoy]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
For a 140+ IQ you don't seem too confident.

[/ QUOTE ]

Should a person with 140+ IQ necessarily be confident?

[/ QUOTE ]

in his ability to figure out this xtra hard brain teaser?....yes!!

[/ QUOTE ]

this is not an xtra hard brain teaser, this is a

Super Duper Extra Hard Brainteaser

[partygirluk]

Still getting 199/399 but my logic gut is not confident /images/graemlins/smile.gif

[mostsmooth]

[ QUOTE ]
without thinking- 1/3
with some crazy miracle thinking- 31%

[/ QUOTE ]
ok my last crazy miracle thinking guess is 50%
i want credit if any of those are correct /images/graemlins/cool.gif actually if 1/3 isnt correct, i think 50% is
/images/graemlins/cool.gif

[mostsmooth]

screw it, im going with %50

[mostsmooth]

theres 50 girls in the GG pile and 50 girls in the BG/GB pile
shes got at least one of those girls.
50-50! no?

[jimdmcevoy]

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

[mostsmooth]

[ QUOTE ]
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

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

[jimdmcevoy]

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

[/ QUOTE ]

yah i'm pretty sure this is right

[jimdmcevoy]

[ QUOTE ]
[ QUOTE ]
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

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

[/ QUOTE ]

/images/graemlins/grin.gif 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__]

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]

[ QUOTE ]
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.

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

[Huskiez]

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__]

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


[/ QUOTE ]

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

[partygirluk]

[ QUOTE ]
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.

[/ QUOTE ]

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

[Sporky]

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]

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

[/ QUOTE ]

yah i'm pretty sure this is right

[/ QUOTE ]
whats wrong with 50%?

[jimdmcevoy]

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]

[ QUOTE ]
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?

[/ QUOTE ]

Almost right.

[Patrick del Poker Grande]

[ 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.

[__Q__]

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%

[jimdmcevoy]

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

[Slow Play Ray]

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

[partygirluk]

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?

[__Q__]

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

[jimdmcevoy]

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

and 99% original brain teaser is 1/3

[Patrick del Poker Grande]

[ 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.

[Patrick del Poker Grande]

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.

[DMBFan23]

[ 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.

[jimdmcevoy]

i'm guessin 201/40000

[partygirluk]

[ QUOTE ]
i'm guessin 201/40000

[/ QUOTE ]

I wasn't really looking for an answer, and haven't worked it out. My point to Patrick was that is is quickly obvious that it is not 3/4, and thus the information about the girl is relevant, be it her preference for dolls, or her name.

[jimdmcevoy]

[ QUOTE ]
[ 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.

[/ QUOTE ]

that sounded good, but then you got an answer that no one has gotten yet

[DMBFan23]

what can I say, I'm a rebel.

[jimdmcevoy]

[ QUOTE ]
[ QUOTE ]
i'm guessin 201/40000

[/ QUOTE ]

I wasn't really looking for an answer, and haven't worked it out. My point to Patrick was that is is quickly obvious that it is not 3/4, and thus the information about the girl is relevant, be it her preference for dolls, or her name.

[/ QUOTE ]

oh, yes i agree

[Tyler Durden]

How are BG and GB different?

[partygirluk]

[ QUOTE ]
How are BG and GB different?

[/ QUOTE ]

So you think 2 boys, 2 girls and 1 of each are all equally likely?

[mostsmooth]

there are 4 ways a woman can have 2 kids
bb
bg
gb
gg

[tpir90036]

Don't think of them as combinations of items from a group... think of them as a sequence of independent events.

[Patrick del Poker Grande]

[ QUOTE ]
How are BG and GB different?

[/ QUOTE ]
If it makes you feel better, we can just say that there's a 50% chance of having exactly one boy and one girl and 25% of having either two boys or two girls.

[DMBFan23]

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?

[mostsmooth]

[ QUOTE ]
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 ]
1/4
1/2

[NoTalent]

This is my favorite brainteaser:

[ QUOTE ]
Three logicians, A, B, and C, are wearing hats, which they know are either black or white but not all white. A can see the hats of B and C; B can see the hats of A and C; C is blind. Each is asked in turn if they know the color of their own hat.

Their answers are:
A: "No."
B: "No."
C: "Yes."
What color is C's hat and how does he know?

[/ QUOTE ]

[tpir90036]

FWIW, I am really struggling to see how the Sarah part sends us from 1/3 all the way to ~50%. I don't think the answer to the 2nd questions is exactly 1/3... but I don't see how this slightly different way of saying "I have one girl" tilts things so drastically. The math that one poster listed made sense when I read it... but it goes against common sense.

[DMBFan23]

[ QUOTE ]
[ QUOTE ]
[ 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.

[/ QUOTE ]

that sounded good, but then you got an answer that no one has gotten yet

[/ QUOTE ]

sorry,

it should be 99/299. a pubic hair's length from 1/3.

[k_squared]

[ QUOTE ]
[ 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.

[/ QUOTE ]

But once you know she has a daughter named sarah and that she won't name her other daughter sarah that just means that 100% of the times she has a girl they won't be named sarah, not that she will only have a girl 99% of the time...

Whatever name you give your child won't change how often you have a girl. There isn't a smaller chance that she has a girl because she named her first girl Sarah that would require some weird stipulations about the world that don't make sense. And intuitively (and logically), rather than adjusting our entire world view to adjust for this fact it is better to consider the precept that exactly 1% of all children are named Sarah as flawed.

Take it a step further.
Half the girls in the world are named 'Ash-tray-head' and no one has two girls named 'Ash-try-head' (we can all agree that would be confusing).

I have a daughter named Ash-tray-head then does that mean that it is less likely that my other child is a girl because my one daughter is named Ash-tray-head? Is it true that NOW I have a 75% chance of haveing a boy because I have used up half of the set of possible girls by picking that name? Absolutely not! What it does mean is that I can't name my next girl that, but that is not a controlling factor of whether or not I have a girl.

To be perhaps a bit more concise these two things are not causally related! They are instead coincidental truths about the world that are true, but not causally related.

Know, if you really wanted to make this a logical puzzle that made sense then be very clear about what the rules are. Say something like at an orphanage with an equal number of boys and girls, 1% of the girls at the orphanage are named Sarah. Further, if you adopt multiple children you never adopt two with the same name. Given that you have adopted one girl named Sarah what is the chance that you will adopt another girl rather than a boy.

The problem though seems to be that to really figure this out you have to either make more assumptions (that either the orphanage immediately gets another orphan named Sarah and so has the same overall ratio of Sarahs to total girls, or we need to know the total number of Sarahs (or girls, or orphans) to solve because when you take that Sarah out you change the ratio slightly and less than 1% of the orphan girls are now Sarah...

So, all I am saying is that logic puzzles are great... but this one seems to me to be lacking in clear rules and boundaries which are the only way that you can come up with an answer that is anything but arbitrarily figured out based on a number of assumptions you have to make.

-k_squared

[Patrick del Poker Grande]

[ QUOTE ]
[ QUOTE ]
[ 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.

[/ QUOTE ]

But once you know she has a daughter named sarah and that she won't name her other daughter sarah that just means that 100% of the times she has a girl they won't be named sarah, not that she will only have a girl 99% of the time...

Whatever name you give your child won't change how often you have a girl. There isn't a smaller chance that she has a girl because she named her first girl Sarah that would require some weird stipulations about the world that don't make sense. And intuitively (and logically), rather than adjusting our entire world view to adjust for this fact it is better to consider the precept that exactly 1% of all children are named Sarah as flawed.

Take it a step further.
Half the girls in the world are named 'Ash-tray-head' and no one has two girls named 'Ash-try-head' (we can all agree that would be confusing).

I have a daughter named Ash-tray-head then does that mean that it is less likely that my other child is a girl because my one daughter is named Ash-tray-head? Is it true that NOW I have a 75% chance of haveing a boy because I have used up half of the set of possible girls by picking that name? Absolutely not! What it does mean is that I can't name my next girl that, but that is not a controlling factor of whether or not I have a girl.

To be perhaps a bit more concise these two things are not causally related! They are instead coincidental truths about the world that are true, but not causally related.

Know, if you really wanted to make this a logical puzzle that made sense then be very clear about what the rules are. Say something like at an orphanage with an equal number of boys and girls, 1% of the girls at the orphanage are named Sarah. Further, if you adopt multiple children you never adopt two with the same name. Given that you have adopted one girl named Sarah what is the chance that you will adopt another girl rather than a boy.

The problem though seems to be that to really figure this out you have to either make more assumptions (that either the orphanage immediately gets another orphan named Sarah and so has the same overall ratio of Sarahs to total girls, or we need to know the total number of Sarahs (or girls, or orphans) to solve because when you take that Sarah out you change the ratio slightly and less than 1% of the orphan girls are now Sarah...

So, all I am saying is that logic puzzles are great... but this one seems to me to be lacking in clear rules and boundaries which are the only way that you can come up with an answer that is anything but arbitrarily figured out based on a number of assumptions you have to make.

-k_squared

[/ QUOTE ]
This is exactly my line and is the most intelligent thing said in this whole thread.

[Asufiji2004]

reminds me of a song..

When your slidin into third and you feel a sudden turd...DIARRHEA

[incognito]

It's only relevant in your case because the answer was no. In the original question, the answer was yes. So we don't have to worry about the probability that the child is a girl whose name happens not to be Sarah. We know the child is a girl. At that point, we no longer care about her name.

Said another way: in your case, the child is either a boy or a girl that doesn't like barbies. In that case, the preference is relevant because the question of gender is still up in the air.

[jimdmcevoy]

[ QUOTE ]
[ QUOTE ]
[ 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.

[/ QUOTE ]

But once you know she has a daughter named sarah and that she won't name her other daughter sarah that just means that 100% of the times she has a girl they won't be named sarah, not that she will only have a girl 99% of the time...

Whatever name you give your child won't change how often you have a girl. There isn't a smaller chance that she has a girl because she named her first girl Sarah that would require some weird stipulations about the world that don't make sense. And intuitively (and logically), rather than adjusting our entire world view to adjust for this fact it is better to consider the precept that exactly 1% of all children are named Sarah as flawed.

Take it a step further.
Half the girls in the world are named 'Ash-tray-head' and no one has two girls named 'Ash-try-head' (we can all agree that would be confusing).

I have a daughter named Ash-tray-head then does that mean that it is less likely that my other child is a girl because my one daughter is named Ash-tray-head? Is it true that NOW I have a 75% chance of haveing a boy because I have used up half of the set of possible girls by picking that name? Absolutely not! What it does mean is that I can't name my next girl that, but that is not a controlling factor of whether or not I have a girl.

To be perhaps a bit more concise these two things are not causally related! They are instead coincidental truths about the world that are true, but not causally related.

Know, if you really wanted to make this a logical puzzle that made sense then be very clear about what the rules are. Say something like at an orphanage with an equal number of boys and girls, 1% of the girls at the orphanage are named Sarah. Further, if you adopt multiple children you never adopt two with the same name. Given that you have adopted one girl named Sarah what is the chance that you will adopt another girl rather than a boy.

The problem though seems to be that to really figure this out you have to either make more assumptions (that either the orphanage immediately gets another orphan named Sarah and so has the same overall ratio of Sarahs to total girls, or we need to know the total number of Sarahs (or girls, or orphans) to solve because when you take that Sarah out you change the ratio slightly and less than 1% of the orphan girls are now Sarah...

So, all I am saying is that logic puzzles are great... but this one seems to me to be lacking in clear rules and boundaries which are the only way that you can come up with an answer that is anything but arbitrarily figured out based on a number of assumptions you have to make.

-k_squared

[/ QUOTE ]

hmm, well i reckon that this is balanced by the fact that if you have a girl and give her a non-sarah name, you cannot name the second girl that same non-sarah name, therefore you increase slightly the chance that the second girl's name is Sarah.

But if you want to be nitpicky, go with the orphanage rules, and give me your best answer.

[DMBFan23]

I think you're arguing in a circle. you're taking a fact about the world and applied it to a decision that caused that fact to come about. the mother didn't use the 1% rule when her daughter was born and decide "oh crap .99999% of girls are named sarah, we NEED a sarah". the fact that the second daughter is NOT named sarah, and 1% of all the girls in the world at the time of the questioning are named sarah (assuming equal distributions of girls and boys, + 1 girl to leave an equal distribution AFTER factoring sarah into the equation) makes it less likely from a probabilistic standpoint that the second child will also be a girl.

do it this way. you have 10 marbles in a bag, 5 red (boys)
4 blue (girls not named sarah) and 1 black (sarahs).

what is the probability you get a red marble if you pick?

what about if you remove the black marble first?

[Patrick del Poker Grande]

[ QUOTE ]
I think you're arguing in a circle. you're taking a fact about the world and applied it to a decision that caused that fact to come about. the mother didn't use the 1% rule when her daughter was born and decide "oh crap .99999% of girls are named sarah, we NEED a sarah". the fact that the second daughter is NOT named sarah, and 1% of all the girls in the world at the time of the questioning are named sarah (assuming equal distributions of girls and boys, + 1 girl to leave an equal distribution AFTER factoring sarah into the equation) makes it less likely from a probabilistic standpoint that the second child will also be a girl.

do it this way. you have 10 marbles in a bag, 5 red (boys)
4 blue (girls not named sarah) and 1 black (sarahs).

what is the probability you get a red marble if you pick?

what about if you remove the black marble first?

[/ QUOTE ]
This only applies in the orphanage case. Nature doesn't care about what you name your child and this woman is 50/50 to have a girl each time she's pregnant regardless of whether she has another child, be it boy or girl or what its name is.

[tpir90036]

Common sense tells me that this answer makes more sense... but the math that Huskiez presented makes sense too if you step through it. Can someone reconcile the differences between them? Why does one come up with ~1/3 and one come up with ~1/2. Which is missing what?

[DMBFan23]

[ QUOTE ]

hmm, well i reckon that this is balanced by the fact that if you have a girl and give her a non-sarah name, you cannot name the second girl that same non-sarah name, therefore you increase slightly the chance that the second girl's name is Sarah.

But if you want to be nitpicky, go with the orphanage rules, and give me your best answer.

[/ QUOTE ]

luckily we don't need to worry about which order the daughters were born in, or whether the mom know at the time the percentage of girls named sarah, or whether someone pointed a gun to her head and told her not to name both kids sarah, or whatever. we know that a certain percentage of girls are named sarah, and we can't pick from those girls for our second girl because her second daughter is not named sarah.

[mostsmooth]

[ QUOTE ]
This is my favorite brainteaser:

[ QUOTE ]
Three logicians, A, B, and C, are wearing hats, which they know are either black or white but not all white. A can see the hats of B and C; B can see the hats of A and C; C is blind. Each is asked in turn if they know the color of their own hat.

Their answers are:
A: "No."
B: "No."
C: "Yes."
What color is C's hat and how does he know?

[/ QUOTE ]

[/ QUOTE ]
c-black
if c were white then:
b would know his own hat couldnt be white otherwise A would have known his hat was black
if c were black, b wouldnt be able to figure out his own hat because A couldnt tell his own regardless if b was black or white, and B couldnt tell his own by looking at either black/black or black/white
thus, C's hat is black
did i win?

[Patrick del Poker Grande]

[ QUOTE ]
Common sense tells me that this answer makes more sense... but the math that Huskiez presented makes sense too if you step through it. Can someone reconcile the differences between them? Why does one come up with ~1/3 and one come up with ~1/2. Which is missing what?

[/ QUOTE ]
I agree more with the math/logic that produces the ~1/3 answer than the math/logic that produces the ~1/2 answer, but I believe they're both wrong because the answer is exactly 1/3.

[DMBFan23]

[ QUOTE ]
[ QUOTE ]
I think you're arguing in a circle. you're taking a fact about the world and applied it to a decision that caused that fact to come about. the mother didn't use the 1% rule when her daughter was born and decide "oh crap .99999% of girls are named sarah, we NEED a sarah". the fact that the second daughter is NOT named sarah, and 1% of all the girls in the world at the time of the questioning are named sarah (assuming equal distributions of girls and boys, + 1 girl to leave an equal distribution AFTER factoring sarah into the equation) makes it less likely from a probabilistic standpoint that the second child will also be a girl.

do it this way. you have 10 marbles in a bag, 5 red (boys)
4 blue (girls not named sarah) and 1 black (sarahs).

what is the probability you get a red marble if you pick?

what about if you remove the black marble first?

[/ QUOTE ]
This only applies in the orphanage case. Nature doesn't care about what you name your child and this woman is 50/50 to have a girl each time she's pregnant regardless of whether she has another child, be it boy or girl or what its name is.

[/ QUOTE ]

I agree that at the time of birth you'd have to predict 50/50, but the second daughter (or son) is already born, and we know that it's less likely that she's a girl because assuming otherwise equal distributions, she cannot possibly be a sarah and thus it's just more likely from a random sampling standpoint that she's a boy. there are more boys left in the potential answer pool to choose from.

one aside, I think that the sample sizes here are large enough that removing this sarah from the pool does not significantly alter the % of girls named sarah after removing her from consideration. so even when she is not considered, 1% of all remaining girls (roughly) are named sarah.

[tpir90036]

[ QUOTE ]
what about if you remove the black marble first?

[/ QUOTE ]
Doesn't this ignore the fact that we might remove the black marble second?

The two brainteasers are not different enough to make one have an answer of ~1/3 and one have an answer of ~1/2 unless we are solving one of them "incorrectly" when compared to the other one. Am I totally off?

[Slow Play Ray]

Not to repeat myself, but...

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

[/ QUOTE ]

[DMBFan23]

I think the first one is clearly 1/3, and by logic the second one is less than 1/3

to answer your point in case you use threaded mode, I'll copy this from my last reply to PatrickDelPokerGrande

"I agree that at the time of birth you'd have to predict 50/50, but the second daughter (or son) is already born, and we know that it's less likely that she's a girl because assuming otherwise equal distributions, she cannot possibly be a sarah and thus it's just more likely from a random sampling standpoint that she's a boy. there are more boys left in the potential answer pool to choose from."

[mostsmooth]

[ QUOTE ]
[ QUOTE ]

hmm, well i reckon that this is balanced by the fact that if you have a girl and give her a non-sarah name, you cannot name the second girl that same non-sarah name, therefore you increase slightly the chance that the second girl's name is Sarah.

But if you want to be nitpicky, go with the orphanage rules, and give me your best answer.

[/ QUOTE ]

luckily we don't need to worry about which order the daughters were born in, or whether the mom know at the time the percentage of girls named sarah, or whether someone pointed a gun to her head and told her not to name both kids sarah, or whatever. we know that a certain percentage of girls are named sarah, and we can't pick from those girls for our second girl because her second daughter is not named sarah.

[/ QUOTE ]
look at my solution that gave me %50, the math isnt as in depth as the 49.98% answer, but its close enough, and simple enough
theres 50 girls in the GG pile and 50 girls in the BG/GB pile
shes got at least one of those girls.
its 50-50 she has 2 girls
the combinations to generate at least one girl are GG, BG, GB, thats 4 girls. multiply by 25 to get 100 girls.
25 x GG =50 girls (group A)
25 x BG =25 girls (group B)
25 x GB =25 girls (group B)
or
25 x BG/GB =50 girls (group B)
sarah is from either from group A or group B. if its from group A, she has 2 girls.

[Bluffoon]

[ QUOTE ]
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.

[/ QUOTE ]

This.

[DMBFan23]

[ QUOTE ]

look at my solution that gave me %50, the math isnt as in depth as the 49.98% answer, but its close enough, and simple enough
theres 50 girls in the GG pile and 50 girls in the BG/GB pile
shes got at least one of those girls.
its 50-50 she has 2 girls
the combinations to generate at least one girl are GG, BG, GB, thats 4 girls. multiply by 25 to get 100 girls.
25 x GG =50 girls (group A)
25 x BG =25 girls (group B)
25 x GB =25 girls (group B)
or
25 x BG/GB =50 girls (group B)
sarah is from either from group A or group B. if its from group A, she has 2 girls.

[/ QUOTE ]

mostsmooth,

I made some assumptions that I listed in another post.

1) the sample sizes AFTER REMOVING SARAH are equal.
2) the sample sizes are large enough that the percentage of sarahs to non-sarahs does not chance by removing this particular sarah. (note that this also implies that the sample size in relation to girls and boys does not change significantly by removing this sarah.

if you only work from a sample size of 10, then removing ANY girl gives 9 girls, and it's more likely we have a boy since there are 10 boys left in the answer pool and 9 girls. in that case, the answer to the first question would be 9/29, and not 1/3. that's messy, so IMO we should treat the sample sizes as close enough to infinite that removing this particular sarah form the answer pool does not change any of the percentages

[mostsmooth]

why are you removing sarah from anything?
also, the answer to the first question is absolutely 1/3
you tried to refute that using your coinflip idea and you proved yourself wrong. i have no math training, as can be seen by the way i solve these problems, but my answer is close enough that 1/3 is obviously wrong for the second question
/images/graemlins/cool.gif

[DMBFan23]

no, the coinflip idea was just to prove that BG and GB were different responses, and had to be considered in addition to BB and GG. This was in response to Tyler Durden's question.

I'm taking sarah out of things because we can't have any sarah's as our second girl. "sarah" can be thought of as 1% of all girls. so we have to take her out of the potential girls to have as the second child. this makes it more likely than normal that the second child is a boy. I hope you see why.

I cross posted this in probability, hopefully someone from there will chime in.

[mostsmooth]

[ QUOTE ]

I'm taking sarah out of things because we can't have any sarah's as our second girl. "sarah" can be thought of as 1% of all girls. so we have to take her out of the potential girls to have as the second child. this makes it more likely than normal that the second child is a boy. I hope you see why.

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

[ QUOTE ]
I cross posted this in probability, hopefully someone from there will chime in.

[/ QUOTE ]
let us hope

[Huskiez]

My math earlier I think was wrong.

I assumed also that there are infinite names to choose from and infinite girls and boys, so naming a girl Sarah would not really affect any probability.

Here's another crack at it.

Here are all the possibilities.

B = Boy, Gs = Girl named Sarah, Gns = Girl not named Sarah

P(B,B) = 1/4

P(B,Gs) = 1/2 * 1/200 = 1/400
P(B,Gns) = 1/2 * 99/200 = 99/400

P(Gs,B) = 1/400
P(Gns,B) = 99/400

P(Gs,Gns) = 1/200 * 1/2 = 1/400
P(Gns,Gs) = 99/200 * 1/200 = 99/40000
P(Gns,Gns) = 99/200 * 99/200 = 9801/40000
P(Gs,Gs) = 0 as defined by game

Each combination still adds up to 1/4, as it should.

So now we're given that Gs exists, and want to find out how often the mother has two girls given that info.

Number of occurrences with Gs = 1/400 + 1/400 + 1/400 + 99/40000 = 399/40000.
Number of occurrences of Gns and Gs = 1/400 + 99/40000 = 199/40000.

Probability then is (199/40000) / (399/40000) = 199/399.

[censored].

I'd assume OP would identify a right answer, so I guess this is not right.

[DMBFan23]

[ 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?

[Slow Play Ray]

[ QUOTE ]
I'm taking sarah out of things because we can't have any sarah's as our second girl. "sarah" can be thought of as 1% of all girls. so we have to take her out of the potential girls to have as the second child. this makes it more likely than normal that the second child is a boy. I hope you see why.

[/ QUOTE ]

I don't get why you guys are all stuck on this. If the first question was phrased "...What's the probability that she has two girls with different names?" would that have changed the answer?

The fact that one of her daughters is named Sarah does not make her less likely to have a second girl, it only makes it impossible for her to have named that second girl Sarah.

My guess, especially based on the seemingly sarcastic tone of the subject, is that the OP is just enjoying the hell out of this discussion...

[DMBFan23]

[ QUOTE ]
The fact that one of her daughters is named Sarah does not make her less likely to have a second girl

[/ QUOTE ]

agree to disagree. I think you are wrong though.

[mostsmooth]

[ 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

[Patrick del Poker Grande]

[ 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?

[Huskiez]

[ 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.

[DMBFan23]

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.

[mostsmooth]

[ 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

[DMBFan23]

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.

[Huskiez]

[ 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.

[DMBFan23]

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)"

[Huskiez]

[ 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.

[Patrick del Poker Grande]

[ 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.

[ethan]

[ QUOTE ]
What are the chances she has two girls?

[/ QUOTE ]

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.

[DMBFan23]

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

good luck all,
me

[DcifrThs]

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

[Vince Young]

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

[DMBFan23]

[ QUOTE ]
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 ]

[sammysusar]

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.

[gaming_mouse]

[ 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.

[jason_t]

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;
}

[partygirluk]

[ 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?

[DMBFan23]

[ QUOTE ]
[ 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?

[ethan]

[ 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.

[DMBFan23]

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

[partygirluk]

[ 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?

[Patrick del Poker Grande]

[ 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 (http://forumserver.twoplustwo.com/showflat.php?Cat=&amp;Number=1949103&amp;page=0&amp;view=colla psed&amp;sb=5&amp;o=&amp;fpart=all&amp;vc=1). All of it.

[partygirluk]

[ QUOTE ]
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

[jason_t]

[ 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.

[sam h]

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.

[Patrick del Poker Grande]

[ 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.

[jason_t]

[ 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....

[partygirluk]

[ 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 (http://forumserver.twoplustwo.com/showflat.php?Cat=&amp;Number=1949103&amp;page=0&amp;view=colla psed&amp;sb=5&amp;o=&amp;fpart=all&amp;vc=1). All of it.

[/ QUOTE ]

Please explain which part of what I said was wrong.

[Patrick del Poker Grande]

[ 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.

[Patrick del Poker Grande]

[ 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 (http://forumserver.twoplustwo.com/showflat.php?Cat=&amp;Number=1949103&amp;page=0&amp;view=colla psed&amp;sb=5&amp;o=&amp;fpart=all&amp;vc=1). 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.

[partygirluk]

That is not what your link did.

[DMBFan23]

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?

[gaming_mouse]

[ 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?

[DMBFan23]

[ 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.

[partygirluk]

How much are you willing to bet?

[gaming_mouse]

[ QUOTE ]


I will bet money that I am right.

[/ QUOTE ]

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

[DMBFan23]

[ 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.

[Patrick del Poker Grande]

[ 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.

[partygirluk]

[ QUOTE ]
[ 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.

[/ QUOTE ]

I'm tempted to take it as I am pretty sure the real answer is 199/399

[Patrick del Poker Grande]

[ QUOTE ]
[ QUOTE ]
[ 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.

[/ QUOTE ]

I'm tempted to take it as I am pretty sure the real answer is 199/399

[/ QUOTE ]
I am willing to entertain doubts that it is not 1/3, even that it's 199/399, but I am quite quite sure it is not 1/2.

[sam h]

[ 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 ]

I'm don't buy it yet but could be pursuaded if you explained more.

This might be where we disagree: You say that there are equal probabilities of the four outcomes but I'm not so sure.

B, G(s)
G(s), B
G, G(s)
G(s), G

I just don't see it. If I took a random sample of the population that had two kids, one of which was a girl named Sarah, I could not justify expecting 75% of the firstborn children to be girls.

[DMBFan23]

[ 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 ]

gaming mouse,

I hate to be a dick, but even if your logic is correct that the answer must be closer to 1/2 than 1/3, the answer CANNOT be exaclty 1/2 due to the stipulation that both kids cannot be sarah.

[Patrick del Poker Grande]

[ QUOTE ]
[ 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 ]

I'm don't buy it yet but could be pursuaded if you explained more.

This might be where we disagree: You say that there are equal probabilities of the four outcomes but I'm not so sure.

B, G(s)
G(s), B
G, G(s)
G(s), G

I just don't see it. If I took a random sample of the population that had two kids, one of which was a girl named Sarah, I could not justify expecting 75% of the firstborn children to be girls.

[/ QUOTE ]
That's because you're giving equal weight to G,G(s) and G(s),G to other combinations like B,G(s), i.e. that G,G(s) has a 25% likelihood, as does G(s),G, when really it's both of them put together that have a 25% likelihood.

[partygirluk]

OK right. Lets assume 1,000,000 families.

250,000 will be 2 boys
250,000 willl be 2 girls
500,000 will be one of each.

Of the 500,000 with 1 boy, 1 girl, there will be 5,000 instances where the girl is Sarah.

Of the 250,000 instances with 2 girls, there will be 4975 instances with a girl named Sarah.

Therefore there is a 4975/9975 (199/399) chance that the mother has 2 girls.

[gaming_mouse]

[ QUOTE ]
[ 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.

[/ QUOTE ]

All calcs assume exactly 2 children.

P(2 girls | girl named sara) = P(2 girls &amp; 1 named Sara)/P(girl names sara)

P(girl named sara) = P(1st child sara OR 2nd child sara) = 2*P(1st child sara) = 2*.5*.01

P(2 girls and one named sara) = 2*P(1st child sara, 2nd child girl) = 2*.5*.01*.5

Plugging back in, we get .5. Where is the mistake?

[jason_t]

[ QUOTE ]

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


[/ QUOTE ]

Not sure who'd take that bet; I won't as I believe the answer is 1/2 also. My math confirms it, as does my simulation.

I'll make the same bet.

[sam h]

[ QUOTE ]
That's because you're giving equal weight to G,G(s) and G(s),G to other combinations like B,G(s), i.e. that G,G(s) has a 25% likelihood, as does G(s),G, when really it's both of them put together that have a 25% likelihood.

[/ QUOTE ]

That's basically what I've been thinking. I think it might be that, among the population that has two kids and one girl named Sarah, G, G(s) + G(s), G still equals 33%.

[DMBFan23]

you are allowing both daughters to be named sarah, at the least.

[Patrick del Poker Grande]

[ QUOTE ]
[ QUOTE ]

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


[/ QUOTE ]

Not sure who'd take that bet; I won't as I believe the answer is 1/2 also. My math confirms it, as does my simulation.

I'll make the same bet.

[/ QUOTE ]
I'll take your $100 too. Any other takers... err... givers?

[gaming_mouse]

[ QUOTE ]
you are allowing both daughters to be named sarah, at the least.

[/ QUOTE ]

No I'm not. I'm assuming it's mutually exclusive.

[partygirluk]

I'll bet you it ain't 1/3

[jason_t]

Here is the output from three executions of my simulation:

I simulated this 10000000 times, 3 times.

oak.3&gt; ./a.out
49976 / 99301
0.503278
oak.4&gt; ./a.out
49675 / 99222
0.500645
oak.5&gt; ./a.out
50154 / 99591
0.503600

[partygirluk]

[ QUOTE ]
OK right. Lets assume 1,000,000 families.

250,000 will be 2 boys
250,000 willl be 2 girls
500,000 will be one of each.

Of the 500,000 with 1 boy, 1 girl, there will be 5,000 instances where the girl is Sarah.

Of the 250,000 instances with 2 girls, there will be 4975 instances with a girl named Sarah.

Therefore there is a 4975/9975 (199/399) chance that the mother has 2 girls.

[/ QUOTE ]

Why has no one responded to this? It seems to pretty much end the debate, unless I something very wrong.

[Patrick del Poker Grande]

[ QUOTE ]
I'll bet you it ain't 1/3

[/ QUOTE ]
To be honest, I believe whether it's 1/3 or something more like 199/399 ends up boiling down to something like semantics. I am completely sure it's not 1/2.

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
you are allowing both daughters to be named sarah, at the least.

[/ QUOTE ]

No I'm not. I'm assuming it's mutually exclusive.

[/ QUOTE ]

Indeed. It is this stipulation that separates the problem from the original and makes my solution correct.

[jason_t]

[ QUOTE ]

I'll take your $100 too. Any other takers... err... givers?

[/ QUOTE ]

What is your solution or reasoning that gaming mouse and I are incorrect? Do you see an error in my simulation?

[partygirluk]

[ QUOTE ]
[ QUOTE ]
I'll bet you it ain't 1/3

[/ QUOTE ]
To be honest, I believe whether it's 1/3 or something more like 199/399 ends up boiling down to something like semantics. I am completely sure it's not 1/2.

[/ QUOTE ]

Ok, I'll bet you $100 it is between .49 and .51

[jason_t]

[ QUOTE ]
To be honest, I believe whether it's 1/3 or something more like 199/399 ends up boiling down to something like semantics. I am completely sure it's not 1/2.

[/ QUOTE ]

1/3 and 199/399 aren't even close to each other.

[jason_t]

[ QUOTE ]
Ok, I'll bet you $100 it is between .49 and .51

[/ QUOTE ]

This is clear.

[Patrick del Poker Grande]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I'll bet you it ain't 1/3

[/ QUOTE ]
To be honest, I believe whether it's 1/3 or something more like 199/399 ends up boiling down to something like semantics. I am completely sure it's not 1/2.

[/ QUOTE ]

Ok, I'll bet you $100 it is between .49 and .51

[/ QUOTE ]
That's the same bet in my eyes and is not provable - it's a matter of something on the order of semantics.

[partygirluk]

199/399 is pretty damn close to 1/2 though /images/graemlins/smile.gif

[partygirluk]

This is very provable, it is just a maths problem! I want my $100!

[Patrick del Poker Grande]

[ QUOTE ]
[ QUOTE ]
To be honest, I believe whether it's 1/3 or something more like 199/399 ends up boiling down to something like semantics. I am completely sure it's not 1/2.

[/ QUOTE ]

1/3 and 199/399 aren't even close to each other.

[/ QUOTE ]
Semantics does not require that they be close to each other. It's not a rounding error we're talking about here.

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]

I'll take your $100 too. Any other takers... err... givers?

[/ QUOTE ]

What is your solution or reasoning that gaming mouse and I are incorrect? Do you see an error in my simulation?

[/ QUOTE ]

Let's call in BruceZ and let him decide this. I am confident we're right.

[Patrick del Poker Grande]

[ QUOTE ]
This is very provable, it is just a maths problem! I want my $100!

[/ QUOTE ]
I never bet you it wasn't 199/399 or that it's not between .49 and .51. I only said it's not 1/2. Like I said before, I'm willing to entertain a 199/399 answer.

[jason_t]

[ QUOTE ]
Let's call in BruceZ and let him decide this.

[/ QUOTE ]

Who is that?

[ QUOTE ]
I am confident we're right.

[/ QUOTE ]

I am too.

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
This is very provable, it is just a maths problem! I want my $100!

[/ QUOTE ]
I never bet you it wasn't 199/399 or that it's not between .49 and .51. I only said it's not 1/2. Like I said before, I'm willing to entertain a 199/399 answer.

[/ QUOTE ]

You need to refute me proof using the conditional probability formula. I have yet to hear any arguments against it, except for DMB's erroneous one.

[Patrick del Poker Grande]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
This is very provable, it is just a maths problem! I want my $100!

[/ QUOTE ]
I never bet you it wasn't 199/399 or that it's not between .49 and .51. I only said it's not 1/2. Like I said before, I'm willing to entertain a 199/399 answer.

[/ QUOTE ]

You need to refute me proof using the conditional probability formula. I have yet to hear any arguments against it, except for DMB's erroneous one.

[/ QUOTE ]
When it all comes down to it, I'm pretty sure nobody's going to let go of their answer. I'd say the probability of me paying you $100 is exactly zero because 1/2 is wrong and of me ever seeing the $200 from the two betting it's 1/2 is very small because you're not going to let this one go. If you haven't accepted that it's not 1/2 yet, I doubt I'll ever be able to convince you.

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
Let's call in BruceZ and let him decide this.

[/ QUOTE ]

Who is that?


[/ QUOTE ]

seriously? he is the most heavyweight probability poster, IMO. i'm surprised you don't know his posts.

[gaming_mouse]

[ QUOTE ]

When it all comes down to it, I'm pretty sure nobody's going to let go of their answer. I'd say the probability of me paying you $100 is exactly zero because 1/2 is wrong and of me ever seeing the $200 from the two betting it's 1/2 is very small because you're not going to let this one go. If you haven't accepted that it's not 1/2 yet, I doubt I'll ever be able to convince you.

[/ QUOTE ]

Well, nice way of skirting my proof.

It's very simple, though. We agree on an arbitrator. I'm fine with BruceZ.

[Patrick del Poker Grande]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
This is very provable, it is just a maths problem! I want my $100!

[/ QUOTE ]
I never bet you it wasn't 199/399 or that it's not between .49 and .51. I only said it's not 1/2. Like I said before, I'm willing to entertain a 199/399 answer.

[/ QUOTE ]

You need to refute me proof using the conditional probability formula. I have yet to hear any arguments against it, except for DMB's erroneous one.

[/ QUOTE ]
What's your answer to the first puzzle... the one without any of this crap about the girl's name?

[sam h]

[ QUOTE ]
Why has no one responded to this? It seems to pretty much end the debate, unless I something very wrong.

[/ QUOTE ]

I'm not so sure about my approach anymore and need to stop procrastinating with this stuff. But I will note that you do not have a population in which 1% of all girls are named Sarah.

[Patrick del Poker Grande]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
This is very provable, it is just a maths problem! I want my $100!

[/ QUOTE ]
I never bet you it wasn't 199/399 or that it's not between .49 and .51. I only said it's not 1/2. Like I said before, I'm willing to entertain a 199/399 answer.

[/ QUOTE ]

You need to refute me proof using the conditional probability formula. I have yet to hear any arguments against it, except for DMB's erroneous one.

[/ QUOTE ]
The problem with your solution is that you're completely leaving out the BB combination. Perhaps you could argue this is something on the order of semantics as well, as if the starting point was either before or after the question was asked of the mother and you're somehow starting after the question is asked.

[gaming_mouse]

[ QUOTE ]


What's your answer to the first puzzle... the one without any of this crap about the girl's name?

[/ QUOTE ]

1/3. I will solve it using the same method.

P(both girls | at least one girl) = P(both girls)/P(at least one girl)

Note that the in the numberator "both girls AND at least one girl" is the same as "both girls".

P(at least 1 girl) = P(1st child girl) + P(2nd child girl) - P(both girls)

Note the subtraction term which is not present in the other solution, because there it is not possible to have 2 saras

= 1/2 + 1/2 - 1/4 = 3/4

P(both girls) = 1/4

Pluggin back in, we get 1/3.

[jason_t]

[ QUOTE ]

The problem with your solution is that you're completely leaving out the BB combination. Perhaps you could argue this is semantics as well, as if the starting point was either before or after the question was asked of the mother and you're somehow starting after the question is asked.

[/ QUOTE ]

The BB combination is irrelevant once we know at least one child is a girl. That is the point to the first question and is equally relevant here.

[gaming_mouse]

[ QUOTE ]

The problem with your solution is that you're completely leaving out the BB combination.

[/ QUOTE ]

Huh? We are told that there is at least one girl, namely Sara. So BB is not possible. Anyway, I don't even see how you are getting that out of the conditional probability formula solution.

[partygirluk]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
This is very provable, it is just a maths problem! I want my $100!

[/ QUOTE ]
I never bet you it wasn't 199/399 or that it's not between .49 and .51. I only said it's not 1/2. Like I said before, I'm willing to entertain a 199/399 answer.

[/ QUOTE ]

You need to refute me proof using the conditional probability formula. I have yet to hear any arguments against it, except for DMB's erroneous one.

[/ QUOTE ]
The problem with your solution is that you're completely leaving out the BB combination. Perhaps you could argue this is something on the order of semantics as well, as if the starting point was either before or after the question was asked of the mother and you're somehow starting after the question is asked.

[/ QUOTE ]

wtf? Am I missing something? What does the BB combination have to do with anything?

[mostsmooth]

so have we agreed on an answer? i cant tell
also, have we glossed over my reply to the logic question?
logic good, math bad /images/graemlins/cool.gif

[gaming_mouse]

[ QUOTE ]


wtf? Am I missing something? What does the BB combination have to do with anything?

[/ QUOTE ]

i sense patrick is backing down and grasping at straws. he has yet to accept my offer of an arbitrator.

[jason_t]

1/3

once we know there is a girl, there are three situations.

B, G
G, B
G, G.

Each of these is equally likely and the latter is the one we are concerned with. Therefore the answer is 1/3.

Or you could use Bayes'.

answer = (1/4) / (3/4) = 1/3.

or you could simulate it to "check" it:

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

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

count1 = count2 = 0;

srand(time());

for(i = 0; i &lt;= 10000000; i++) {
c1 = rand() % 2;
c2 = rand() % 2;

if(c1 == 0 || c2 == 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;
}

oak.11&gt; ./a.out
2499973 / 7499976
0.333331

[partygirluk]

What do you think the odds are of Patrick actually backing down?

[ethan]

[ QUOTE ]
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(?)

[/ QUOTE ]
yes.

If 1% of girls are named Sarah, someone with two girls is more likely to have one named Sarah than someone with one girl.

[Patrick del Poker Grande]

I am willing to accept two answers to this problem.

The first is what I'll call the natural answer (not natural because it's more right, but natural because it's nature, if that makes sense - it doesn't matter, it's just its name), meaning the woman's likelihood of having a boy or a girl is in no way related to the name of her first child (or if it's her first child, what she's intending to name her second child). This solution basically says the whole Sarah name stuff is all a red herring and that her telling you she has a child named Sarah is effectively the same as her saying she has a daughter. The answer to this scenario has been concluded to be 1/3 without much debate and was best explained by k_squared here (http://forumserver.twoplustwo.com/showflat.php?Cat=&amp;Board=exchange&amp;Number=1950064&amp;Fo rum=f20&amp;Words=daughter&amp;Searchpage=0&amp;Limit=25&amp;Main= 1949103&amp;Search=true&amp;where=bodysub&amp;Name=20455&amp;dater ange=1&amp;newerval=1&amp;newertype=d&amp;olderval=&amp;oldertype= &amp;bodyprev=#Post1950064), as was the reason this puzzle is so problematic.

The second answer is what I'll call the orphanage answer and is partygirluk's solution. I relate this one to there being a certain population of children you're basically picking out of an orphanage at random as if they were marbles in a bag. This solution has also been worked over several times in this thread and the answer is 199/399.

Take your pick. Neither one is 1/2.

[Patrick del Poker Grande]

[ QUOTE ]
What do you think the odds are of Patrick actually backing down?

[/ QUOTE ]
I don't know what your problem is. I'm now at least halfway agreeing with your answer (and somewhat have been for a while). I actually never ruled out jason_t's latest answer. The only thing I'm saying for certain is that it is not 1/2.

[jason_t]

The answer is definitely not 1/3. My simulations show this.

The answer is 1/2. The use of Bayes's theorem shows this. My simulation supports this.

[partygirluk]

[ QUOTE ]
[ QUOTE ]
What do you think the odds are of Patrick actually backing down?

[/ QUOTE ]
I don't know what your problem is. I'm now at least halfway agreeing with your answer (and somewhat have been for a while). I actually never ruled out jason_t's latest answer. The only thing I'm saying for certain is that it is not 1/2.

[/ QUOTE ]

His latest answer was 1/2

[gaming_mouse]

[ QUOTE ]

meaning the woman's likelihood of having a boy or a girl is in no way related to the name of her first child (or if it's her first child, what she's intending to name her second child).

[/ QUOTE ]

not sure what you mean here, but i assumed (and i think it's implicit in the problem) that:

1. the chance that a child is boy or girl is always 50%
2. given that a child is a girl, there is a .01 chance that she is also named sara.

i have already proved my answer, so if you actually want to own up to this bet, let's talk about an arbitrator that we can both agree on (or 2 arbitrators, or whatever you want).

I already offered up one. So?

[Patrick del Poker Grande]

[ QUOTE ]
The answer is definitely not 1/3. My simulations show this.

The answer is 1/2. The use of Bayes's theorem shows this. My simulation supports this.

[/ QUOTE ]
Your latest post gave an answer of 0.333331. Maybe I was mistaken and you were answering the first puzzle?

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
The answer is definitely not 1/3. My simulations show this.

The answer is 1/2. The use of Bayes's theorem shows this. My simulation supports this.

[/ QUOTE ]
Your latest post gave an answer of 0.333331.

[/ QUOTE ]

that was for the original problem -- the non sara one.

[partygirluk]

[ QUOTE ]
[ QUOTE ]
The answer is definitely not 1/3. My simulations show this.

The answer is 1/2. The use of Bayes's theorem shows this. My simulation supports this.

[/ QUOTE ]
Your latest post gave an answer of 0.333331. Maybe I was mistaken and you were answering the first puzzle?

[/ QUOTE ]

Yes.

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]
[ 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.

[/ QUOTE ]

All calcs assume exactly 2 children.

P(2 girls | girl named sara) = P(2 girls &amp; 1 named Sara)/P(girl names sara)

P(girl named sara) = P(1st child sara OR 2nd child sara) = 2*P(1st child sara) = 2*.5*.01

P(2 girls and one named sara) = 2*P(1st child sara, 2nd child girl) = 2*.5*.01*.5

Plugging back in, we get .5. Where is the mistake?

[/ QUOTE ]

My concern in this - butplease keep in mind I am mainly concentrating on basketball:

If the odds of a girl being named Sarah were 100%, this equation would still result in there being a 50% chance of the second kid being a girl, even though in reality it would be 0.

[Patrick del Poker Grande]

[ QUOTE ]
i have already proved my answer, so if you actually want to own up to this bet, let's talk about an arbitrator that we can both agree on (or 2 arbitrators, or whatever you want).

I already offered up one. So?

[/ QUOTE ]
To be honest, there is nobody here whose answer I would trust any more than I trust yours. I don't read the probability forum, so I don't know BruceZ. I recognize the name, but I'm certainly not familiar enough with him to pick him as an arbitrator. I don't see how I could choose an arbitrator from that.

[Patrick del Poker Grande]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ 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.

[/ QUOTE ]

All calcs assume exactly 2 children.

P(2 girls | girl named sara) = P(2 girls &amp; 1 named Sara)/P(girl names sara)

P(girl named sara) = P(1st child sara OR 2nd child sara) = 2*P(1st child sara) = 2*.5*.01

P(2 girls and one named sara) = 2*P(1st child sara, 2nd child girl) = 2*.5*.01*.5

Plugging back in, we get .5. Where is the mistake?

[/ QUOTE ]

My concern in this - butplease keep in mind I am mainly concentrating on basketball:

If the odds of a girl being named Sarah were 100%, this equation would still result in there being a 50% chance of the second kid being a girl, even though in reality it would be 0.

[/ QUOTE ]
This is true.

[mostsmooth]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ 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.

[/ QUOTE ]

All calcs assume exactly 2 children.

P(2 girls | girl named sara) = P(2 girls &amp; 1 named Sara)/P(girl names sara)

P(girl named sara) = P(1st child sara OR 2nd child sara) = 2*P(1st child sara) = 2*.5*.01

P(2 girls and one named sara) = 2*P(1st child sara, 2nd child girl) = 2*.5*.01*.5

Plugging back in, we get .5. Where is the mistake?

[/ QUOTE ]

My concern in this - butplease keep in mind I am mainly concentrating on basketball:

If the odds of a girl being named Sarah were 100%, this equation would still result in there being a 50% chance of the second kid being a girl, even though in reality it would be 0.

[/ QUOTE ]
the puzzle states that the woman would not name both girls sarah. if every girl was named sarah, the other child would always be a boy, and the equations would have to be altered because there would be no chance of GG. so the equation would be wrong, not because its flawed, but because its the wrong equation, no?

[partygirluk]

That is becuase the equation does not adjust for only one Sarah being allowed. Making this adjustment will cause the 1/2 answer to go to 199/399 AFAIK

[gaming_mouse]

[ QUOTE ]

To be honest, there is nobody here whose answer I would trust any more than I trust yours. I don't read the probability forum, so I don't know BruceZ. I recognize the name, but I'm certainly not familiar enough with him to pick him as an arbitrator. I don't see how I could choose an arbitrator from that.

[/ QUOTE ]

How about we find some math professor and send him an email? Or a math forum where a professor answers the questions?

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ 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.

[/ QUOTE ]

All calcs assume exactly 2 children.

P(2 girls | girl named sara) = P(2 girls &amp; 1 named Sara)/P(girl names sara)

P(girl named sara) = P(1st child sara OR 2nd child sara) = 2*P(1st child sara) = 2*.5*.01

P(2 girls and one named sara) = 2*P(1st child sara, 2nd child girl) = 2*.5*.01*.5

Plugging back in, we get .5. Where is the mistake?

[/ QUOTE ]

My concern in this - butplease keep in mind I am mainly concentrating on basketball:

If the odds of a girl being named Sarah were 100%, this equation would still result in there being a 50% chance of the second kid being a girl, even though in reality it would be 0.

[/ QUOTE ]
the puzzle states that the woman would not name both girls sarah. if every girl was named sarah, the other child would always be a boy, and the equations would have to be altered because there would be no chance of GG. so the equation would be wrong, not because its flawed, but because its the wrong equation, no?

[/ QUOTE ]

I am not sure that is true, but let's say it is. Make it 99% then, the same flaw seems to apply - I can't see how the odds of the second kid being a girl would be 50% if 99% of all girls are named Sarah but the second kid can't be.

But again, I haven't really put a ton of thought into this and could be very, very wrong . . .

[Patrick del Poker Grande]

[ QUOTE ]
That is becuase the equation does not adjust for only one Sarah being allowed. Making this adjustment will cause the 1/2 answer to go to 199/399 AFAIK

[/ QUOTE ]
This I believe is also true.

[jason_t]

[ QUOTE ]

Your latest post gave an answer of 0.333331. Maybe I was mistaken and you were answering the first puzzle?

[/ QUOTE ]

You asked what our answer to the first question was. That post was a reply to that. I intended it to demonstrate that our methods for the second question are correct, given that there is no disagreement on the first one.

[mostsmooth]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ 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.

[/ QUOTE ]

All calcs assume exactly 2 children.

P(2 girls | girl named sara) = P(2 girls &amp; 1 named Sara)/P(girl names sara)

P(girl named sara) = P(1st child sara OR 2nd child sara) = 2*P(1st child sara) = 2*.5*.01

P(2 girls and one named sara) = 2*P(1st child sara, 2nd child girl) = 2*.5*.01*.5

Plugging back in, we get .5. Where is the mistake?

[/ QUOTE ]

My concern in this - butplease keep in mind I am mainly concentrating on basketball:

If the odds of a girl being named Sarah were 100%, this equation would still result in there being a 50% chance of the second kid being a girl, even though in reality it would be 0.

[/ QUOTE ]
the puzzle states that the woman would not name both girls sarah. if every girl was named sarah, the other child would always be a boy, and the equations would have to be altered because there would be no chance of GG. so the equation would be wrong, not because its flawed, but because its the wrong equation, no?

[/ QUOTE ]

I am not sure that is true, but let's say it is. Make it 99% then, the same flaw seems to apply - I can't see how the odds of the second kid being a girl would be 50% if 99% of all girls are named Sarah but the second kid can't be.

But again, I haven't really put a ton of thought into this and could be very, very wrong . . .

[/ QUOTE ]
why are you under the impression that you should get the same results for 1% sarahs and 99% sarahs?
or did i miss something?

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]

Your latest post gave an answer of 0.333331. Maybe I was mistaken and you were answering the first puzzle?

[/ QUOTE ]

You asked what our answer to the first question was. That post was a reply to that. I intended it to demonstrate that our methods for the second question are correct, given that there is no disagreement on the first one.

[/ QUOTE ]

How would you answer if 99% of girls were named Sarah? I feel like I am missing something obvious.

Edit - Mort, plug 99% (or any %) into gaming mouse's equation and you get 50%. Unless I messed up (possible), something is off.

[jason_t]

[ QUOTE ]
How about we find some math professor and send him an email? Or a math forum where a professor answers the questions?

[/ QUOTE ]

FWIW, I

1. Am a graduate student in mathematics.
2. Have asked two fellow graduate students this question, one of which is an expert in probability.
3. Have emailed a professor this question.

1. produced 1/2
2. both produced 1/2
3. has not yet replied.

[gaming_mouse]

[ QUOTE ]
That is becuase the equation does not adjust for only one Sarah being allowed. Making this adjustment will cause the 1/2 answer to go to 199/399 AFAIK

[/ QUOTE ]

The equation EXPLICITLY uses the fact that only one sara is allowed. It is the reason that:

P(1st child sara OR 2nd child sara) = P(1st sara) + P(2nd sara)

It is THE REASON this answer differs from the first. If both girls were ALLOWED TO BE NAMED SARA, then the answer would be:

2*.5*.5*.01/(2*.5*.01 - .5^2*.01^2)

or 0.501253133

But because they aren't, the answer is 50%

[partygirluk]

FWIW I have a Gold award in the international maths olympiad and some of the thinking displayed in this thread makes me want to vomit.

[partygirluk]

[ QUOTE ]
[ QUOTE ]
That is becuase the equation does not adjust for only one Sarah being allowed. Making this adjustment will cause the 1/2 answer to go to 199/399 AFAIK

[/ QUOTE ]

The equation EXPLICITLY uses the fact that only one sara is allowed. It is the reason that:

P(1st child sara OR 2nd child sara) = P(1st sara) + P(2nd sara)


[/ QUOTE ]

but p(2nd sara) is dependent on whether #1 was sara

[gaming_mouse]

[ QUOTE ]

1. Am a graduate student in mathematics.
2. Have asked two fellow graduate students this question, one of which is an expert in probability.
3. Have emailed a professor this question.

1. produced 1/2
2. both produced 1/2
3. has not yet replied.

[/ QUOTE ]

I also have a MS in statistics. But I'm assuming Patrick wants something more independent than that, which is fair. I'm offering suggestions, but he's not taking...

Welshing on a bet is not cool.

[hoyaboy1]

Mouse, please give me your answer, using your equation, if the chance of a girl being named Sarah was 99% or 100%, so I can stop feeling stupid.

[gaming_mouse]

[ QUOTE ]

The equation EXPLICITLY uses the fact that only one sara is allowed. It is the reason that:

P(1st child sara OR 2nd child sara) = P(1st sara) + P(2nd sara)


[/ QUOTE ]

but p(2nd sara) is dependent on whether #1 was sara

[/ QUOTE ]

The formula does not assume independence. It assumes mutual exclusivitity, which we have.

[jason_t]

[ QUOTE ]
I also have a MS in statistics. But I'm assuming Patrick wants something more independent than that, which is fair. I'm offering suggestions, but he's not taking...

Welshing on a bet is not cool.

[/ QUOTE ]

Yeah, I understand that. That's why I prefaced with "FWIW."

[gaming_mouse]

[ QUOTE ]
Mouse, please give me your answer, using your equation, if the chance of a girl being named Sarah was 99% or 100%, so I can stop feeling stupid.

[/ QUOTE ]

If 99%, the answer is the same.

100% is a contradiction, because then the other child has to be a boy, since you can't have two girls named sara. That is, the answer would be 0.

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]
Mouse, please give me your answer, using your equation, if the chance of a girl being named Sarah was 99% or 100%, so I can stop feeling stupid.

[/ QUOTE ]

If 99%, the answer is the same.

100% is a contradiction, because then the other child has to be a boy, since you can't have two girls named sara. That is, the answer would be 0.

[/ QUOTE ]

Why can't the equation adjust for the 100%?

And I simply can't see how if it is 99% the answer would still be 50%. How could the 99 Sarahs out of 100 have a 50% chance of having a sister who is 1/100?

[jason_t]

The answer won't change because of the conditional probabilities.

[mostsmooth]

[ QUOTE ]
[ QUOTE ]
Mouse, please give me your answer, using your equation, if the chance of a girl being named Sarah was 99% or 100%, so I can stop feeling stupid.

[/ QUOTE ]

If 99%, the answer is the same.

100% is a contradiction, because then the other child has to be a boy, since you can't have two girls named sara. That is, the answer would be 0.

[/ QUOTE ]
now im confused (as if i wasnt already), your saying that if 99% of girls were named sarah, and a woman had a girl named sarah, its ~50/50 the other kid is a girl?
this doesnt seem right, not even close to right

[hoyaboy1]

Seems that way, but I can't quite wrap my head around it. If I take 1000 girls, 990 will be named Sarah - yet 495 will have a sister not named Sarah. How does that work?

[Keats13]

[ QUOTE ]
OK right. Lets assume 1,000,000 families.

250,000 will be 2 boys
250,000 willl be 2 girls
500,000 will be one of each.

Of the 500,000 with 1 boy, 1 girl, there will be 5,000 instances where the girl is Sarah.

Of the 250,000 instances with 2 girls, there will be 4975 instances with a girl named Sarah.

Therefore there is a 4975/9975 (199/399) chance that the mother has 2 girls.

[/ QUOTE ]

In this case, there are 1 million girls. And there are 9975 Sarahs. This is not 1/100. Your solution violates one of the assumptions.

Edit: nevermind, I'm an idiot. The limit as the # of families increases is 1/100. Sorry.

[gaming_mouse]

[ QUOTE ]

And I simply can't see how if it is 99% the answer would still be 50%. How could the 99 Sarahs out of 100 have a 50% chance of having a sister who is 1/100?

[/ QUOTE ]

The sister only has to be a girl, 50%

As for the equation, it's not a matter of adjusting. By the assumption of the puzzle, it is impossible to have two girls named Sara. So if all girls are named Sara, the woman can have only 1 girl. This has nothing to do with probability. It's just logic.

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]

And I simply can't see how if it is 99% the answer would still be 50%. How could the 99 Sarahs out of 100 have a 50% chance of having a sister who is 1/100?

[/ QUOTE ]

The sister only has to be a girl, 50%

As for the equation, it's not a matter of adjusting. By the assumption of the puzzle, it is impossible to have two girls named Sara. So if all girls are named Sara, the woman can have only 1 girl. This has nothing to do with probability. It's just logic.

[/ QUOTE ]

Please read my above post about 1000 girls and clear up whatever I am missing.

[partygirluk]

[ QUOTE ]
[ QUOTE ]
OK right. Lets assume 1,000,000 families.

250,000 will be 2 boys
250,000 willl be 2 girls
500,000 will be one of each.

Of the 500,000 with 1 boy, 1 girl, there will be 5,000 instances where the girl is Sarah.

Of the 250,000 instances with 2 girls, there will be 4975 instances with a girl named Sarah.

Therefore there is a 4975/9975 (199/399) chance that the mother has 2 girls.

[/ QUOTE ]

In this case, there are 1 million girls. And there are 9975 Sarahs. This is not 1/100. Your solution violates one of the assumptions.

[/ QUOTE ]

There cant be 2 sarahs

[mostsmooth]

i vote we wait for bruce

[gaming_mouse]

[ QUOTE ]
Seems that way, but I can't quite wrap my head around it. If I take 1000 girls, 990 will be named Sarah - yet 495 will have a sister not named Sarah. How does that work?

[/ QUOTE ]

Who says they all have to have a sister not named sara?

You are looking at the wrong population. The population you want to look at is all 2 children families with one child named sara, not all the girls named sara.

[gaming_mouse]

[ QUOTE ]
i vote we wait for bruce

[/ QUOTE ]

Patrick has rejected using Bruce as an arbitrator, though I am perfectly willing to do so.

[jason_t]

[ QUOTE ]
Patrick has rejected using Bruce as an arbitrator, though I am perfectly willing to do so.

[/ QUOTE ]

Patrick: Please suggest an arbitrator.

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]
Seems that way, but I can't quite wrap my head around it. If I take 1000 girls, 990 will be named Sarah - yet 495 will have a sister not named Sarah. How does that work?

[/ QUOTE ]

Who says they all have to have a sister not named sara?

You are looking at the wrong population. The population you want to look at is all 2 children families with one child named sara, not all the girls named sara.

[/ QUOTE ]

Ok, that is a good point - my 1000 girls example is wrong. I have a followup in a second, but I figured I acknowledge this first.

Edit - How about this: we take 1000 girls from 2 child families (a small adjustment). 990 of them should be named Sarah, no? That only leaves 10 other girls to be sisters of half of them, which doesn't work.

So are we assuming that somehow Sarahs are really more prevelant in families with more or less than 2 kids so that this works out, or am I still missing something?

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
Patrick has rejected using Bruce as an arbitrator, though I am perfectly willing to do so.

[/ QUOTE ]

Patrick: Please suggest an arbitrator.

[/ QUOTE ]

He's totally backing down. I'm a nice guy, though. I would let Patrick off the hook for $50 and Patrick changing his location for the next month to "Getting pwned by GM &amp; jason_t".

[jason_t]

[ QUOTE ]

He's totally backing down. I'm a nice guy, though. I would let Patrick off the hook for $50 and Patrick changing his location for the next month to "Getting pwned by GM &amp; jason_t".

[/ QUOTE ]

[gaming_mouse]

[ QUOTE ]

Edit - How about this: we take 1000 girls from 2 child families (a small adjustment). 990 of them should be named Sarah, no? That only leaves 10 other girls to be sisters of half of them, which doesn't work.

So are we assuming that somehow Sarahs are really more prevelant in families with more or less than 2 kids so that this works out, or am I still missing something?

[/ QUOTE ]

Given you parameters, you would just not be able to have that many 2 children homes containing Saras.

I think the way your trying to visualize may be misguided -- none of this stuff affects the probability calculation. Don't try to imagine the solution in this sense. Think of infinite population, so you don't have to deal with any of these issues.

If you really want to understand it intuitively, just try to understand the conditional probability theorem intuitivly (this is fairly easy), and then this will make sense.

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]

Edit - How about this: we take 1000 girls from 2 child families (a small adjustment). 990 of them should be named Sarah, no? That only leaves 10 other girls to be sisters of half of them, which doesn't work.

So are we assuming that somehow Sarahs are really more prevelant in families with more or less than 2 kids so that this works out, or am I still missing something?

[/ QUOTE ]

Given you parameters, you would just not be able to have that many 2 children homes containing Saras.

I think the way your trying to visualize may be misguided -- none of this stuff affects the probability calculation. Don't try to imagine the solution in this sense. Think of infinite population, so you don't have to deal with any of these issues.

If you really want to understand it intuitively, just try to understand the conditional probability theorem intuitivly (this is fairly easy), and then this will make sense.

[/ QUOTE ]

I think this is what I am getting at - it only works with an infinite population. If we are talking about earth, for example, 99% would change the answer a TON, as the odds of having a second girl would be waaaaaaaaaaay lower (of course, this whole thing would only work if people were killing off their second kid once they had a Sarah).

But, anyway, assuming we are talking about our planet, the answer is NOT the same for 1% and 99%.

[gaming_mouse]

[ QUOTE ]


I think this is what I am getting at - it only works with an infinite population. If we are talking about earth, for example, 99% would change the answer a TON, as the odds of having a second girl would be waaaaaaaaaaay lower (of course, this whole thing would only work if people were killing off their second kid once they had a Sarah).

[/ QUOTE ]

No, no. It doesn't change the calculation at all. I wasn't suggesting that. It's just that you are making up these numbers, and then trying to imagine realistic situations (ie, population distributins like ours here on earth) where they occur.

But look, I can get around your objection by just having 1 or 2 two person families with a girl named sara. All the rest have 1 kid or 3+ kids.

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]


I think this is what I am getting at - it only works with an infinite population. If we are talking about earth, for example, 99% would change the answer a TON, as the odds of having a second girl would be waaaaaaaaaaay lower (of course, this whole thing would only work if people were killing off their second kid once they had a Sarah).

[/ QUOTE ]

No, no. It doesn't change the calculation at all. I wasn't suggesting that. It's just that you are making up these numbers, and then trying to imagine realistic situations (ie, population distributins like ours here on earth) where they occur.

But look, I can get around your objection by just having 1 or 2 two person families with a girl named sara. All the rest have 1 kid or 3+ kids.

[/ QUOTE ]

Yea, I mentioned the possibility of that earlier.

But you do acknowledge that with a limited population with a normal distribution of family sizes (let's say our planet), the answer if 1% or 99% of girls is Sarah is not the same, right?

The real problem here, which I fully acknowledge, is that as the % of Sarahs gets higher people will have to artificially adjust to keep naming 99% of girls Sarah and never have 2 girls named Sarah (like killing babies) if they have 2 kids.

Edit - anything above 2/3 and the chance of a second girl starts dropping due to some sort of articial measure needing to be introduced.

[gaming_mouse]

[ QUOTE ]

But you do acknowledge that with a limited population with a normal distribution of family sizes (let's say our planet), the answer if 1% or 99% of girls is Sarah is not the same, right?

[/ QUOTE ]

No, I don't. What I acknowledge is that the following three things are inconsistent logically:

1. normal distribution of family sizes like on our planet.
2. the assumptions of the puzzle.
3. 99% of girls names sara

This does not mean that probability is breaking down. It means that your question is flawed. It's like saying, "There are three marbles. Two of them are black. Two are white. You draw one at random. What is the chance it's white?"

[Keats13]

My first reply was not thought out well enough.

HOWEVER:
Because there can't be more than 1 Sarah per family, the probability of naming your daugher Sarah must increase if you already have a daughter not named Sarah. This counteracts the fact that the probability decreases (to 0) if you already have a daughter who IS named Sarah.

If you have 2 girls,
P(2nd Sarah) = P(2nd Sarah|1st Sarah)*P(1st Sarah) + P(2nd Sarah|1st not Sarah) * P(1st not Sarah)
= 0 + P(2nd Sarah|1st Not Sarah) * P(1st not Sarah)
= P(2nd Sarah|1st Not Sarah) * 99/100
(because there can't be 2 Sarahs)

If 1/100 first daughters are named Sarah, then 1/100 second daughters must also be named Sarah for the stated assumption (1/100 of all girls are named Sarah) to be true.

Therefore,
P(2nd Sarah) = 1/100

Plugging in above,
1/100 = P(2nd Sarah|1st not Sarah) * 99/100
P(2nd Sarah|1st not Sarah) = 1/99

So if you have 2 girls, the probability that one of them is named Sarah is

P(1st Sarah) + P(2nd Sarah|1st Not Sarah) * P(1st Not Sarah)
1/100 + 1/99 * 99/100 = 2/100

Plugging this back in to the original equations gives the answer of 1/2.

Unless I screwed something up again, which is entirely possible.

[hoyaboy1]

Actually, it isn't like that at all. That example is impossible. 99% of girls being named Sarah and never having 2 girls named Sarah could never happen naturally (as I've said a bunch of times) but could happen if, say, the government set up a program to do so.

So, again, assuming the question was talking about earth, and I have to assume it was, we have a a choice: put a cap on the number of Sarahs at 2/3, or accept that above that the odds of a second girl falls below 50%, even if the whole situation is being maintained artificially.

So, once more, if we are talking about earth, changing the number of Sarahs to 99% changes the answer. Please accept this so I can go to sleep, it is 3:40 here.

This isn't an important point or anything, I just felt like pointing this out.

[gaming_mouse]

[ QUOTE ]

So, once more, if we are talking about earth, changing the number of Sarahs to 99% changes the answer. Please accept this so I can go to sleep, it is 3:40 here.

[/ QUOTE ]

This is incorrect. The answer is the answer, period. I'm just saying that on earth, where most families are 2 children, it won't be the case that 99% of girls are named Sara and you never have 2 sisters names sara. That is the way in which the assumptions are like the marble problem.

[mostsmooth]

[ QUOTE ]
[ QUOTE ]

So, once more, if we are talking about earth, changing the number of Sarahs to 99% changes the answer. Please accept this so I can go to sleep, it is 3:40 here.

[/ QUOTE ]

This is incorrect. The answer is the answer, period. I'm just saying that on earth, where most families are 2 children, it won't be the case that 99% of girls are named Sara and you never have 2 sisters names sara. That is the way in which the assumptions are like the marble problem.

[/ QUOTE ]
its just a hypothetical
if the problem was "bill can run 80mph, how far can he run in 2 hours"?, we know nobody can run that fast, but we solve it anyway. no debating how fast people can actually run

[jason_t]

[ QUOTE ]
So, once more, if we are talking about earth, changing the number of Sarahs to 99% changes the answer.

[/ QUOTE ]

If this were the case, math would be useless. I assure you that it is not. The answer is independent of whether or not we are talking about Earth.

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]

So, once more, if we are talking about earth, changing the number of Sarahs to 99% changes the answer. Please accept this so I can go to sleep, it is 3:40 here.

[/ QUOTE ]

This is incorrect. The answer is the answer, period. I'm just saying that on earth, where most families are 2 children, it won't be the case that 99% of girls are named Sara and you never have 2 sisters names sara. That is the way in which the assumptions are like the marble problem.

[/ QUOTE ]

Fine, never mind, even though I said many times the 99% would only work if some sort of program was set up to maintain it, you still won't acknowledge the obvious for some odd reason. This has nothing to do with the validity of your argument with Patrick, so I don't quite get your problem. Anyway, I'm going to bed now.

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]
So, once more, if we are talking about earth, changing the number of Sarahs to 99% changes the answer.

[/ QUOTE ]

If this were the case, math would be useless. I assure you that it is not. The answer is independent of whether or not we are talking about Earth.

[/ QUOTE ]

Wait, before I do go to bed - you think that if 99% of girls on earth were named Sarah, those in 2 children families would still have a 50% chance of having a sister not named Sarah? Because that is wrong. Even mouse admitted this - the answer would either be some much lower %(I haven't bothered calculating it), or you'd have to say it would be impossible (the cheap and pointless answer).

[gaming_mouse]

[ QUOTE ]

Wait, before I do go to bed - you think that if 99% of girls on earth were named Sarah, those in 2 children families would still have a 50% chance of having a sister not named Sarah? Because that is wrong. Even mouse admitted this - the answer would either be some much lower %(I haven't bothered calculating it), or you'd have to say it would be impossible (the cheap and pointless answer).

[/ QUOTE ]

mouse didn't admit this. also, i don't know why you're getting mad -- I know this has nothing to do with the original problem, but your understanding is flawed and I'm trying to help.

The point is that if 99% of girls were named sara, and no girl in a two person family could have a sister named sara, there could only be a small percentage of two person familites. but the probability in the puzzle would not change.

[gaming_mouse]

jason,

btw, can we talk about patrick's flagrant and cowardly slinking off here? what's going on? i thought he was a stand up guy.

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]

Wait, before I do go to bed - you think that if 99% of girls on earth were named Sarah, those in 2 children families would still have a 50% chance of having a sister not named Sarah? Because that is wrong. Even mouse admitted this - the answer would either be some much lower %(I haven't bothered calculating it), or you'd have to say it would be impossible (the cheap and pointless answer).

[/ QUOTE ]

mouse didn't admit this. also, i don't know why you're getting mad -- I know this has nothing to do with the original problem, but your understanding is flawed and I'm trying to help.

The point is that if 99% of girls were named sara, and no girl in a two person family could have a sister named sara, there could only be a small percentage of two person familites. but the probability in the puzzle would not change.

[/ QUOTE ]

I fully understand the situation. As I mentioned earlier, we could assume that the % ended up being balanced out by girls in other size families. However, all I have said is:

1 - We assume normal family size distribution
2 - 99% of girls named Sarah, no 2 Sarahs
3 - This could only happen artificially, probably by killing 2nd born girls
4 - This is a small, meaningless point that doesn't effect the real answer

And all I ask is that you admit that under the pointless, artificial scenario I outlined, that the percentage of 2nd kids being girls would have to be WAY under 50%. Almost everyone who had a second girl would be forced to get rid of it somehow, so there would be very, very few 2 girl families (WAY less than the normal 25%).

[jason_t]

[ QUOTE ]
jason,

btw, can we talk about patrick's flagrant and cowardly slinking off here? what's going on? i thought he was a stand up guy.

[/ QUOTE ]

I was wondering about that too. He just disappeared. I haven't had much interaction with before today, but I always assumed he was a good guy.

[gaming_mouse]

[ QUOTE ]

1 - We assume normal family size distribution
2 - 99% of girls named Sarah, no 2 Sarahs
3 - This could only happen artificially, probably by killing 2nd born girls
4 - This is a small, meaningless point that doesn't effect the real answer

[/ QUOTE ]

If you did something like killing 2nd born girls, then it would no longer be true that in a given family there is a 50% chance that each child is boy or girl. This is one of the assumptions of the puzzle. Maybe this is what's confusing you?

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
jason,

btw, can we talk about patrick's flagrant and cowardly slinking off here? what's going on? i thought he was a stand up guy.

[/ QUOTE ]

I was wondering about that too. He just disappeared. I haven't had much interaction with before today, but I always assumed he was a good guy.

[/ QUOTE ]

the sad thing is... if i had been proved wrong, i would have paid him. i guess there is no honor among del poker grandes, as the saying goes.

[jason_t]

[ QUOTE ]
the sad thing is... if i had been proved wrong, i would have paid him.

[/ QUOTE ]

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]

1 - We assume normal family size distribution
2 - 99% of girls named Sarah, no 2 Sarahs
3 - This could only happen artificially, probably by killing 2nd born girls
4 - This is a small, meaningless point that doesn't effect the real answer

[/ QUOTE ]

If you did something like killing 2nd born girls, then it would no longer be true that in a given family there is a 50% chance that each child is boy or girl. This is one of the assumptions of the puzzle. Maybe this is what's confusing you?

[/ QUOTE ]

This is absurd. Nothing is confusing me, except the fact that you for some reason are being really defensive about the scenario I set up which, is, as I admitted, a little dumb. I really thought that once I said my scenario was artificial you'd admit that the answer wouldn't be 50%, but apparently I overestimated you.

[Patrick del Poker Grande]

[ QUOTE ]
jason,

btw, can we talk about patrick's flagrant and cowardly slinking off here? what's going on? i thought he was a stand up guy.

[/ QUOTE ]
For the record, I am not slinking off. I'm working on a project report that I should've had done yesterday. I also ate dinner. I'm lucky my project manager is too caught up in other things to have kicked me in the nuts yet. To be honest, I've spent too much time dicking around on here and not getting the report done. I'm sorry I had other things to do and failed to continue reporting back every minute on the minute. I [censored] hate writing these things, but I have to get it done and it's late.

In light of this, I'll make this proposal to you. It's painfully obvious neither of us is going to budge from our stance. I don't have an arbiter in mind and I really don't know how I'd find one in short order. As I've got a heavy workload coming up and my wife's already pissed at how much time I've been working plus the fact I'm going out of town starting Sunday, I'm not going to spend a whole lot of time harrassing college profs for an answer. Since neither of us has a solution that we're happy with and I'm the one who appears to have the problem finding a suitable arbiter, I propose calling off the money portion of the bet but I'll change my location to your little pwned message for a month. This way we can all be unhappy about at least two things each and we can all move on.

[hoyaboy1]

[ QUOTE ]
[In light of this, I'll make this proposal to you. It's painfully obvious neither of us is going to budge from our stance. I don't have an arbiter in mind and I really don't know how I'd find one in short order. As I've got a heavy workload coming up and my wife's already pissed at how much time I've been working plus the fact I'm going out of town starting Sunday, I'm not going to spend a whole lot of time harrassing college profs for an answer. Since neither of us has a solution that we're happy with and I'm the one who appears to have the problem finding a suitable arbiter, I propose calling off the money portion of the bet but I'll change my location to your little pwned message for a month. This way we can all be unhappy about at least two things each and we can all move on.

[/ QUOTE ]

I think this is weak. If you agree to the bet you need to figure out a way to settle it and determine a winner.

Edit - and mouse, let's just agree to disagree or something, I think we are getting caught up in the terms of the problem, and it's dumb anyway. But more importantly I have trouble sleeping with something like this still on going. So, if possible, don't even bother responding to my last post.

[Patrick del Poker Grande]

That's fine, you can call me a weak lame pussy or whatever all you want, I'm sorry I can't come to a way to concretely solve this, but I just don't have the time to dick with it. This is 100% honesty and I'm not slinking. It sucks and I understand how much it looks like a cop-out, but it's just the circumstances.

[gaming_mouse]

[ QUOTE ]
I'm sorry I had other things to do and failed to continue reporting back every minute on the minute.

[/ QUOTE ]

forgiven.


[ QUOTE ]
Since neither of us has a solution that we're happy with and I'm the one who appears to have the problem finding a suitable arbiter, I propose calling off the money portion of the bet but I'll change my location to your little pwned message for a month. This way we can all be unhappy about at least two things each and we can all move on.

[/ QUOTE ]

well, if you can live with yourself, I guess it's fine. I will offer you another suggestion if you are feeling more honorable. Write an email to a math professor at a school of your choice with this exact problem on it. I'll leave you on the honor system. If he writes back saying 1/2, pay us what you choose.

gm

[gaming_mouse]

[ QUOTE ]


Edit - and mouse, let's just agree to disagree or something, I think we are getting caught up in the terms of the problem, and it's dumb anyway. But more importantly I have trouble sleeping with something like this still on going. So, if possible, don't even bother responding to my last post.

[/ QUOTE ]

that's fine. for the record, though, I'm really not trying to be defensive. I am quick and very willing to admit when I'm wrong, and I'm sure you can find many posts that prove that.

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]


Edit - and mouse, let's just agree to disagree or something, I think we are getting caught up in the terms of the problem, and it's dumb anyway. But more importantly I have trouble sleeping with something like this still on going. So, if possible, don't even bother responding to my last post.

[/ QUOTE ]

that's fine. for the record, though, I'm really not trying to be defensive. I am quick and very willing to admit when I'm wrong, and I'm sure you can find many posts that prove that.

[/ QUOTE ]

Yea, I realize you are a standup guy, which is why I am perfectly content to just end this.

[Patrick del Poker Grande]

[ QUOTE ]
well, if you can live with yourself, I guess it's fine. I will offer you another suggestion if you are feeling more honorable. Write an email to a math professor at a school of your choice with this exact problem on it. I'll leave you on the honor system. If he writes back saying 1/2, pay us what you choose.

gm

[/ QUOTE ]
For the record, I do feel quite shitty about how this is ending. I'm not convinced I'm wrong, though, so I feel this is a fair solution. I'll give the e-mail a shot when I get a chance. I'm sure you'll get an opportunity to pwn me again some time when we can actually come to some concrete solution and then it will get paid off.

[BOTW]

Assume there are 400 two child families, we can have BB, BG, BS(arah), GB SB, SG or GS:

100 BB
99 BG
1 BS
99 GB
1 SB
98 GG
1 SG
1 GS

She says that she has a child named Sarah, so we are choosing among: 1 BS, 1 SB, 1 SG, and 1 GS. So it is 50% right?

[ QUOTE ]
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.

[/ QUOTE ]
The next one will be named Sarah more often to compensate for the fact that the first was not, no? Isn't it 1/99 times? Doesn't the math show it is 50% instead of 199/399, then?

This is interesting: www.wiskit.com/marilyn/boys.html (http://www.wiskit.com/marilyn/boys.html)

You meet a woman, and ask how many children she has, and she replies "two." You ask if she has any girls, and she replies "yes." After this brief conversation, you know that the woman has exactly two children, at least one of whom is a girl. When the question is interpreted this way, the probability that both of her children are girls is 1/3.

You meet a woman and her daughter. You ask the woman how many children she has, and she replies "two." So now you know that this woman has exactly two children, at least one of whom is a girl. When the question is interpreted this way, the probability that both of her children are girls is 1/2.

[DMBFan23]

[ QUOTE ]
[ QUOTE ]
I'm sorry I had other things to do and failed to continue reporting back every minute on the minute.

[/ QUOTE ]

forgiven.


[ QUOTE ]
Since neither of us has a solution that we're happy with and I'm the one who appears to have the problem finding a suitable arbiter, I propose calling off the money portion of the bet but I'll change my location to your little pwned message for a month. This way we can all be unhappy about at least two things each and we can all move on.

[/ QUOTE ]

well, if you can live with yourself, I guess it's fine. I will offer you another suggestion if you are feeling more honorable. Write an email to a math professor at a school of your choice with this exact problem on it. I'll leave you on the honor system. If he writes back saying 1/2, pay us what you choose.

gm

[/ QUOTE ]

MY DRUNK EPIPHANY:

party girl is right. you were more right than me. we were both wrong.

it must be less than 1/2. it cannot be 1/2. partygirluk is right.

[jimdmcevoy]

Okay so there may be a slight problem in the original question, it seems gaming mouse is not happy with it, although I don't agree with gaming mouse that there is neccesarily a problem with it.

But I'll make this addition:

there are exactly 100 different names for girls, and a random girl is equally likely to be any of these names (1% each just like Sarah)

are you happy with this gaming mouse?

So now I am going to abandon my original answer of 199/399 and jump into the 1/2 camp. my new answer:

with only knowing the woman has two children the combinations and corresponding probabilities are:

boy boy - 1/4
boy sarah - 2/400
boy non-sarah - 198/400
non-sarah non-sarah - 98/400
sarah non-sarah - 2/400

so now when you find out she has a girl named sarah it is clear it's a 50/50 the other child is a boy/non-sarah

my mistake before was figuring the odds of the last two terms, I didn't take into account the fact that if you didn't name the first girl sarah, there will now be a 1/99 chance you name the second girl sarah, not 1/100

[DMBFan23]

partygirl is right. anything else is posturing.

[DMBFan23]

[ QUOTE ]
[ QUOTE ]

He's totally backing down. I'm a nice guy, though. I would let Patrick off the hook for $50 and Patrick changing his location for the next month to "Getting pwned by GM &amp; jason_t".

[/ QUOTE ]

[/ QUOTE ]

you guys were owned by party girl, and so was I. I have admitted I was wrong, and so should you both. and change your locations to reflect your arrogance.

[jimdmcevoy]

It's funny, at the same time you switched to agree with partygirl, I switched away from party girl

One thing I have going for me is i'm on gaming mouse's team now

[DMBFan23]

I need to post an owned picture of myself. that way I get monopoly on owning DMBFan. but DMBFan was surely owned.

[jason_t]

[ QUOTE ]
you guys were owned by party girl, and so was I.

[/ QUOTE ]

We were?

[ QUOTE ]
I have admitted I was wrong, and so should you both.

[/ QUOTE ]

I will admit I am wrong if convinced that I am. Frankly, neither gaming mouse nor I were d[/i]icks at all during this thread. Neither was anyone else with the exception of...

[ QUOTE ]
and change your locations to reflect your arrogance.

[/ QUOTE ]

Dave Matthews Band sucks.

[DMBFan23]

dude, uncalled for. you appealed to authority which was total BS and you know it. partygirl's math in irrefutable. a monte carlo simulation will only show the answer is closer to 1/2 than 1/3, and I was wrong about that. but it will never show exactly 1/2, and thats ok because the answer is not 1/2. again partygirl is right. I was a math major too...you and me and your grad school dudes are wrong.

no one was a dick? gaming mouse insists that patrick should change his avatar to say owned by you guys, which is ironic becaus gaming mouse bet that he was right, and was not right. sadly I was wronger than both of you, so I can only critique so much.

"dave matthews band sucks"

haha childish dude. awesome. you have graduated college? I'm sorry you cant do better.

[DMBFan23]

Quote:
OK right. Lets assume 1,000,000 families.

250,000 will be 2 boys
250,000 willl be 2 girls
500,000 will be one of each.

Of the 500,000 with 1 boy, 1 girl, there will be 5,000 instances where the girl is Sarah.

Of the 250,000 instances with 2 girls, there will be 4975 instances with a girl named Sarah.

Therefore there is a 4975/9975 (199/399) chance that the mother has 2 girls.



Why has no one responded to this? It seems to pretty much end the debate, unless I something very wrong.

[jason_t]

[ QUOTE ]
dude, uncalled for. you appealed to authority which was total BS and you know it.

[/ QUOTE ]

First, I said "FWIW."

[ QUOTE ]
partygirl's math in irrefutable.

[/ QUOTE ]

I stand by the Bayes' Theorem math.

[ QUOTE ]
a monte carlo simulation will only show the answer is closer to 1/2 than 1/3, and I was wrong about that. but it will never show exactly 1/2, and thats ok because the answer is not 1/2.

[/ QUOTE ]

That's why I repeatedly said "check" and "support." I never said "confirm." I am well aware of the uses of a Monte Carlo simulation.

[DMBFan23]

[ QUOTE ]
First, I said "FWIW."

[/ QUOTE ]

ah. that makes it valid.

[ QUOTE ]
I stand by the Bayes' Theorem math.

[/ QUOTE ]
so you cant refute an answer different from yours, ie party girl's math which I included in my recent response to you?


[ QUOTE ]
a monte carlo simulation will only show the answer is closer to 1/2 than 1/3, and I was wrong about that. but it will never show exactly 1/2, and thats ok because the answer is not 1/2.

[/ QUOTE ]
if you want to show the answer is not 1/3 or close to it, then a correct sim will do that. but that sim doesnt refute partygirl's answer. I still want to see that answer refuted.

btw, now 2 people have been a dick in this thread.

[jimdmcevoy]

[ QUOTE ]
Of the 250,000 instances with 2 girls, there will be 4975 instances with a girl named Sarah.

[/ QUOTE ]

okay, I will respond, there will be 5000 instances with a girl named Sarah

[DMBFan23]

yeah but some of those instances will be two sarahs, which is not allowed.

[jimdmcevoy]

not the way i see it, i do realize there is no sarah-sarah combo

[BOTW]

[ QUOTE ]
yeah but some of those instances will be two sarahs, which is not allowed.

[/ QUOTE ]
You need that many to have 1% named Sarah. There are no two girls named Sarah.