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On November 24 2004 19:19 wAt-74 wrote:hey guys i hate to bring back an old topic but i hated the fact that the answer was kind of unresolved, well in my book it was. Initially i thought it was 2/3rs then i jump to 1/2 now i think its 2/3rds again. I showed this to one of my stats professor and told him the problem and he told me that he has heard this problem before. He told me that the key was the wording in the problem and that it drastically changed the answer to the problem. But all in all the answer is 2/3rds He even gave me a link to a website that discusses this same problem: http://mathforum.org/dr.math/faq/faq.boygirl.choose.htmlSo now we finally have an answer to this thing and put it to rest
Any math teacher could give any possible answer, there is no way it would change the actual result. Which, all in all, is 1/2.
If I need to be a math teacher to convince people then ok, I'll become one 
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On November 24 2004 19:19 wAt-74 wrote:hey guys i hate to bring back an old topic but i hated the fact that the answer was kind of unresolved, well in my book it was. Initially i thought it was 2/3rs then i jump to 1/2 now i think its 2/3rds again. I showed this to one of my stats professor and told him the problem and he told me that he has heard this problem before. He told me that the key was the wording in the problem and that it drastically changed the answer to the problem. But all in all the answer is 2/3rds He even gave me a link to a website that discusses this same problem: http://mathforum.org/dr.math/faq/faq.boygirl.choose.htmlSo now we finally have an answer to this thing and put it to rest
that web site has the same flawed logic that has been discussed in this forum. gg
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On November 24 2004 19:19 wAt-74 wrote:Show nested quote +On November 22 2004 11:59 wAt-74 wrote:EDIT2: i changed my mine a couple of times on this problem, so i'm not too sure on the answer Hey guys about the boy and girl puzzle the answer IS 2/3. I'm double majoring and one of my majors is probability & statistics and this is a problem that you get in a beginning class. Lemme try to break it down a bit. you are given the fact that you know one of the childs genders, which is a girl. Now you want to find the probability that the other child is either a boy. You can break down this problem into a simple equation: P(the other child is a boy | given that at least one child is a girl) = P(having a girl and boy)/ P(having at least one girl) < -- there is a long proof for this equation but i dont think you would want to hear it. Trust me its true, all statisticians now it, its a basic equation and you can find the proof in any basic statistics book. so what is the probability of having a girl and boy? well your choices are... GG BB GB BG so P(having a girl and boy) = 2/4 // BG GB whats the probability of having at least one girl? P(having at LEAST one girl) = 3/4 // GG BG GB you plug it into the equation and you get (2/4)/(3/4) = 2/3 (if you cant do the division, use a calculator :p) Some people here think BG GB are the same when they are not. Both come in different order and that is very important to accept it. Here is a visual: EDIT: forget the visual i cant get it to line up properly... new visual:1st you have G-->G or G-->B or B-->G or B-->B As you can see they are different, i hope tat helps Oh btw haha GG is still the same as GG(same goes for BB) If you still dont believe me ask a statistician or stats professor about this, its pretty simple he/she should confirm this. AND if you still dont believe me then go to college and spend thousands of dollars so you can solve this. AND BTW: this is a great thread and keep them coming!  hey guys i hate to bring back an old topic but i hated the fact that the answer was kind of unresolved, well in my book it was. Initially i thought it was 2/3rs then i jump to 1/2 now i think its 2/3rds again. I showed this to one of my stats professor and told him the problem and he told me that he has heard this problem before. He told me that the key was the wording in the problem and that it drastically changed the answer to the problem. But all in all the answer is 2/3rds He even gave me a link to a website that discusses this same problem: http://mathforum.org/dr.math/faq/faq.boygirl.choose.htmlSo now we finally have an answer to this thing and put it to rest
Your link is perfectly correct. What's amazing is that you still misinterpreted the type of question that was being asked in this thread. Here is the direct quote.
"Let's say you're friends with a woman, and you know she has two children, but you don't know the gender of either child. Then one day you go over to her house and a girl answers the door and says my mom will be right down. So you now know one of her children is a girl, what are the chances that the other child is a boy?"
Now from your link, the 2/3 situation involves the hypothesis that we know a priori (before hand) that one of the children is a girl. i.e. We have definitively chosen a family with a girl and insist that she be the one to answer the door so that we can ask the gender of the other child. Is that happening here? NO. It says specifically that we don't know the gender of either child. The key isn't that a girl appears at the door, it is that a girl did not HAVE to answer the door. There is no stipulation on who MUST answer the door, we just observe that it so happens to be a girl.
Now let us see what your link has to say about the 1/2 situation. Quote (with boy changed to girl for our purposes):
"Supposing, on the other hand, that we randomly pick a child from a two-child family. We see that she is a girl, and want to find out whether her sibling is a brother or a sister."
That is exactly what is happening in our case. We did not force a girl to answer the door, we simply observed that a girl answered the door. In fact our question specifically states that we don't know the gender of either child. So from our point of view (the observer), the girl who answers the door is randomly picked. The confusion probably arises since the mother very well could have forced the girl to answer the door, but that is irrelevant since we are not considering her perspective (she already knows the gender of the other child). This is key, "We see that she is a girl". The gender of the child must be observed, it is not preknown.
So yes your link is good (and hopefully good enough for everyone else who is still arguing about the correct answer). Just make sure you make the correct interpretation.
Edit: Here's where it's easy to get confused. The 2/3 case states "In a two-child family, one child is a girl." as a hypothesis. And we look at our example and see that when we open the door, there's a girl so it's the same thing. The mistake is that the quote is trying to imply that we know that there is a girl in the family before the door is opened and that we insist that this girl be the one to answer the door.
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Yes LTT, the logical tree took me a while to write out also.
I did the problem when i was about 12, my mom gave it to me when i was younger and it took me some time to figure it all out
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Flipping a coin will always be 50/50 odds. It doesn\'t amtter if you do it 100 times and 99 come up heads, it will be 50% heads and 50% tails
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On November 24 2004 23:39 AMDme wrote:
Flipping a coin will always be 50/50 odds. It doesn\'t amtter if you do it 100 times and 99 come up heads, it will be 50% heads and 50% tails
omg after a huge ass explanation this morons comes here and says this... plz die!
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yea gg2w actually i forgot to give this link too
http://mathforum.org/dr.math/faq/faq.boy.girl.html
"In a two-child family, one child is a boy. What is the probability that the other child is a girl?"
They iterate that the probability is 2/3. I dont know maybe i'm interpreting the material incorrectly or not, maybe you can share your thoughts? But anyways i think given "certain" information the probability can be 2/3 or 1/2
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2/3! how hard is that to see oh god -_-
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I first numbered the balls from 1-12. The AA-path is trivial. Follow the BB-path. The later CC-,DD-,EE-paths are also trivial.
Ok:
1-4 vs 5-8 AA if 1-4 = 5-8 => 9-12 contain an odd ball
BB if 1-4 > 5-8 => 1-4 contains a heavy ball or 5-8 contain a light ball
BB
1-2,5,9 vs 6,10-12 CC if 1-2,5-6 > 7,10-12 => 1-2 contain a heavy ball or 7 is a light ball
DD if 1-2,5-6 = 7,10-12 => 3-4 contain a heavy ball or 8 is a light ball
EE if 1-2,5-6 < 7,10-12 => 5-6 contain a light ball
CC
1,9 vs 2,10
DD
3,10 vs 4,9
EE
5 vs 6
Ps. Its much easier to follow in a graph of some sort but if any of you cant follow just ask.
edit: Now I would be happy for a forum which works like an editor,ie doesnt try to be smart.
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you're missing some cases, but your answer is correct and the other cases are pretty easy to see being that they are basically done WLOG.
dont spoil it again you slyna!!!!!! quit figuring out my puzzles!!!
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On November 25 2004 07:41 BigBalls wrote:
you're missing some cases, but your answer is correct and the other cases are pretty easy to see being that they are basically done WLOG.
dont spoil it again you slyna!!!!!! quit figuring out my puzzles!!!
they are funny and takes my mind of my failure in all current physics courses.
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*****SPOILER*************
AA. 1-4v5-8
If 1-4 = 5-8, odd number ball in 9-12, go to BB
If 1-4 > 5-8 (WLOG), potentially heavy ball in 1-4, potentially light ball 5-8, go CC
BB. 8,9 v 10,11
If 8,9 = 10,11, odd ball is 12, easy to figure out in 1 weighing whether H or L.
If 8,9 > 10,11, 9 PH, 10,11 PL, go to DD
If 8,9 < 10,11, 9 PL, 10,11 PH, basically same case as above, assume above WLOG
CC 1,2,5,6 v 7,10,11,12
If 1,2,5,6 = 7,10-12, we know odd ball among 3,4 (PH), 8(PL), go to EE
If 1,2,5,6 > 7,10-12, odd ball among 1,2 (PH), 7 (PL), go to FF
If 1,2,5,6 < 7,10-12, odd ball among 5,6 (PL), go to GG
DD 1,10 v 2, 11
1,10 = 2,11 then 9 is a light ball
1,10 < 2,11 then 11 is a heavy ball
1,10 > 2,11 then 10 is a heavy ball
EE 2,3 v 4,5
2,3 = 4,5 then 8 is a light ball
2,3 > 4,5 then 3 is a heavy ball
2,3 < 4,5 then 4 is a heavy ball
FF 1,3 v 2,4
1,3 = 2,4 then 7 is a light ball
1,3 < 2,4 then 2 is a heavy ball
1,3 > 2,4 then 1 is a heavy ball
GG 5v6
5<6 5 is light ball
5>6 6 is light ball
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Ok, im about ready to leave for thanksgiving, so ive decided i should post a few problems for when im gone. Have fun with these guys, theyre tricky...
Three youngsters each have some apples to sell. The oldest has 10, the next had 30, the youngest had 50. How is it possible for each to sell his apples at the same price yet receive the same amount of money? There are many solutions to this, but each is difficult to come up with.
Didnt want to type this one out....but oh well
Aunt jenny has 3 greedy nephewss, Phil, Sam and George, who were looking forward to her death so they could split her will. Jenny dies, but she decided to play a trick on them and made out her will as follows.
The total estate amounts to 1,717 and is to be shared as follows:
Philip gets 1/2
Sam 1/3
George 1/9
Each is to receive an amount in even dollars only
The boys have no clue how to split the money, and go to the lawyer who then figures out the problem and splits their money. How does he do it?
edit: messed up, george gets 1/9 NOT 1/6
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On November 25 2004 01:22 wAt-74 wrote:yea gg2w actually i forgot to give this link too http://mathforum.org/dr.math/faq/faq.boy.girl.html"In a two-child family, one child is a boy. What is the probability that the other child is a girl?" They iterate that the probability is 2/3. I dont know maybe i'm interpreting the material incorrectly or not, maybe you can share your thoughts? But anyways i think given "certain" information the probability can be 2/3 or 1/2
The difference is in the problem which has been given we have:
"In a two child family, the child that opened the door is a xxx"
That is not the same as
"In a two child family, one child is a xxx"
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Indeed.
IF the question were there is a family with 10 children and we know at least 9 are girls, what is the probability the 10th were a boy? The answer would be ba bam 10/11.
GGGGGGGGGB
GGGGGGGGBG
GGGGGGGBGG
GGGGGGBGGG
...
BGGGGGGGGG
GGGGGGGGGG
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However, if the question were if the first 9 children are girls, what is the probability the 10th were a boy? the answer is 1/2
the wording of these problems is CRITICAL.
the first 9 children are girls, what is the probability the 10th is a boy? 1/2
there are at least 9 girls, what is the probability the 10th child is a boy? 1/11.
there are at least 9 girls, what is the probability there is a boy? 10/11.
its very tricky, but with the wording of OUR PARTICULAR PROBLEM, the answer is 1/2
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Catyoul
France2377 Posts
Not even considering the fact I could be a math teacher as well, read the link better 
"Supposing, on the other hand, that we randomly pick a child from a two-child family. We see that he is a boy, and want to find out whether his sibling is a brother or a sister." (...) "and the probability is 1/2."
Thanks for having put the problem at rest
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On November 21 2004 23:35 baal wrote:OK HERE IS THE ANSWER!!!!!ill put two very clear examples of how the curtain thing works, most people dont get this before the explanation, but if still after this you insist its 50/50 then you are kinda slow. So people said that they were 50/50, others said it was 1/3, 1/2, both are wrong. Imagine the same situation, but instead of 3 curtains, 100 curtains with 99 donkeys and 1 car. You choose your curtain, that obviously has 1/100 of probability of having the car behind it, so the host opens 98 curtains, revealing 98 donkeys, so there are 2 curtains left, your initial pick and another one. So the car is behind one of them, and your curtain have 1/100 of being right, and according to some of you they have the SAME probability, so you have two 1/100 curtains, that doesnt sounds right does it?, if you add the two curtains you have 2/100?, nope, they have to be 100/100 between the two of them So your initial pick was 1/100 and it always be that, so the odds of the other curtain of having the car behind it its 99 in 100 !!!! So in conclusion changing the curtain will always be better, in the case of the 3 curtains, if you keep your initial pick, you have 1/3 of being right but if you change you have 2/3 (not 1/2) of being right. If you still dont get it, dont worry heres another more visual explanation: ill number the 3 curtains, and the donkeys will be D and the car C, ill always pick curtain 3, and first ill keep my initial curtain (number 3) and see what happens if we do it with the 3 possibilities: 1 2 3c d d----------lose d c d----------lose d d c----------WIN But now ill always INITIALLY pick curtain 3, but ill always change my mind and select the other one, here it goes: 1 2 3c d d I pick 3, he reveals 2, i change to 1 WIN d c d I pick 3, he reveals 1, i change to 2 WIN d d c I pick 3, he reveals (1 or 2), i change to (1 or 2) LOSE That is the answer, if youve taking probability you would know that it works against our brains, common sense contradics probability, actually its called chained probability and its an advanced probability thing that you will only study if you take an Math oriented college degree. If after these simple explanations you still think its 50/50, then what can i say, you are stupid, and uterlly stubborn. Same goes with the babies, as i said before i know every neuron in your brain yells that its 50/50, but its not, the probability is chained to prior events and it switches it values, even if you see it as an isolated event, there are more odds that it will be a boy if the prior one was a girl, even if your lil brain resists it, it true, and i just explained how it works. I hope i illustrated your small lifes. PS: it really frightens me that even after this people wil still belive its 50/50, but oh well, ive always said it, people are stupid
baal:
I have something called the "asperger syndrome" which means a whole bunch of things but for what i'm about to tell you only this is important: my skills are very divided compared to other people. For example, i am less intelligent than the average 10 year old when it comes to finding my way (in a city or wherever) This is something i hope you won't make fun of me for.
One of the things where I score very high (and that "make up" for my big flaws bringing my total IQ back to above average) is solving riddles or questions like this. Not that it's impossible for me to be wrong, but I am not stupid and I am definetly not stupid when it comes to these kinds of questions.
Many times you said things like "if you still don't get it" or "lol ppl still think its 50/50 i lose faith in humanity" From my point of view this makes you look very bad, not because you are wrong but more because you are so arrogant while being wrong. (dear god i just read you even called people who think it's 50/50 stupid.. I must be the dumbest person in the world because this is one of the only things i am good at)
At any rate, if you choose to believe that in the original scenario (1 curtain is picked, one of the others is revealed to contain a donkey, 2 remain, option of re-choosing is offered) choosing a new curtain increases your chances of getting a car, this is your choice. I cannot persuade you otherwise, because a combination of people has already listed all the reasons why you are wrong. However, please stop insulting the rest of us who disagree with you.
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Grot<3
I think exactly the same.
Even the "i am less intelligent than the average 10 year old when it comes to finding my way" I could have said about myself
Edit: oops I re read and no, I don't agree about everything since the curtain problem's solution is 2/3.
Contrary to the boy girl problem the solution of which is 1/2.
Actually the main point I agreed about is Baal being an ass.
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EPT
