Re: Super Duper Extra Hard Brainteaser
 

i'm guessin 201/40000

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

[ 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

Re: Super Duper Extra Hard Brainteaser
 

what can I say, I'm a rebel.

Re: Super Duper Extra Hard Brainteaser
 

[ 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

Re: Super Duper Extra Hard Brainteaser
 

How are BG and GB different?

Re: Super Duper Extra Hard Brainteaser
 

[ QUOTE ]
How are BG and GB different?

[/ QUOTE ]

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

Re: Super Duper Extra Hard Brainteaser
 

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

Re: Super Duper Extra Hard Brainteaser
 

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

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

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?

Re: Super Duper Extra Hard Brainteaser
 

[ 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

Re: Super Duper Extra Hard Brainteaser
 

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 ]

Re: Super Duper Extra Hard Brainteaser
 

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.

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

[ 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

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

reminds me of a song..

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

Re: Super Duper Extra Hard Brainteaser
 

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.

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

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?

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

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?

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

[ 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?

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

[ 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?

Re: Super Duper Extra Hard Brainteaser
 

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 ]

Re: Super Duper Extra Hard Brainteaser
 

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."

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

Re: Super Duper Extra Hard Brainteaser
 

[ 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

Re: Super Duper Extra Hard Brainteaser
 

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
[img]/images/graemlins/cool.gif[/img]

Re: Super Duper Extra Hard Brainteaser
 

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.

Re: Super Duper Extra Hard Brainteaser
 

[ 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.

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

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

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let us hope

Re: Super Duper Extra Hard Brainteaser
 

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.

Re: Super Duper Extra Hard Brainteaser
 

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

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awesome we agree.

now consider all the ways she can have two kids.

boy boy - no.

so that leaves

sarah, boy
boy, sarah
sarah, girl.


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

Re: Super Duper Extra Hard Brainteaser
 

[ 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...

Re: Super Duper Extra Hard Brainteaser
 

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

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agree to disagree. I think you are wrong though.