EPTPuzzleing Question! (HARD) - Page 55
| Forum Index > General Forum |
|
TheGreenBeret
United Kingdom548 Posts
| ||
|
Beamo
France1279 Posts
On November 24 2004 07:18 Catyoul wrote: Short answer : no you are wrong. Long answer : well the thread is long enough and filled with the explanation, and I and others have already requoted ourselves way too many times. No kidding ? :p Develop your proof/thought and I will discuss it. Oh crap, I can't resist. You are considering the sex of the two children as two dependent things, or as one single probabilistic event if you prefer. Up to you, nothing wrong with that. But you forget that in this situation, the girl/boy combo has its chances of showing a girl at the door of only 50% while the girl/girl one has 100%. Very schematically : GG -> 25% GB -> 25% BG -> 25% BB -> 25% GG -> girl at door = 100% GB -> girl at door = 50% BG -> girl at door = 50% BB -> girl at door = 0% Multiply that and you have : GG and girl at door = 25% GB and girl at door = 12.5% BG and girl at door = 12.5% -> BG or GB and girl at door = 25% BB and girl at door = 0% (the remaining 50% are the cases with boy at door.) Thus the chances for the other child being a boy or a girl are equal. edit : man, gotta love this problem ^^ But we know the girl oppened the door, so there is 100% chance she did open the door... Lets try with 4 child. same question but you have 3 girls that open the door. Do you still think you have 50% chance for the last child to be a boy ??  | ||
|
BigBalls
United States5354 Posts
The child gender problem IS NOT ISOMORPHIC to the door prize problem. In the door prize problem WE KNOW WHAT DOORS ARE NOT WINNERS. In this problem, we dont know whether the girl is the older child or the younger child. and even if we did, it wouldnt matter, it would still be 50% | ||
|
BigBalls
United States5354 Posts
Ok, someone give me some of the better puzzles from this? I missed a lot of math related ones cause I had to drive home for thanksgiving  Karlsberg, what are you working on? Im still in university, but im beginning to study in more depth algebra and difference sets. Ive also done a little bit of advanced work on graph theory/topology. I think when i go for phd i will work in algebra though | ||
|
Catyoul
France2377 Posts
On November 24 2004 09:30 Beamo wrote: But we know the girl oppened the door, so there is 100% chance she did open the door... Lets try with 4 child. same question but you have 3 girls that open the door. Do you still think you have 50% chance for the last child to be a boy ?? If you know the girl opened the door and you discard the probability for that event to have happened, you're absolutely not considering the 2 children as a single event but as 2 independent events. The fact that a girl opened the door should not appear anywhere anymore in your reasoning, don't bother with the GG/GB/BG/BB combinations, you are seeing it as (G/B) followed by another (G/B). With 4 childs, the easy method still works (the 4th child is f* independent from the first 3, how hard is that to grasp :p), let's see with the complicated ones. First the number of cases for each distribution to appear : 4G 0B : 1 3G 1B : 4 2G 2B : 6 1G 3B : 4 0G 4B : 1 Probability to have 3 girls open the door : 4G 0B : 1 3G 1B : 1/4 (it's C_4_3, that is 3! (4-3) ! / 4!, or you could see it as (3/4) * (2/3) * (1/2) = 1/4) Total number of cases : 4G 0B : 1*1 = 1 3G 1B : 4*1/4 = 1 => same chance to happen It works with as many children as you want, since the 4 for number of cases for 3G 1B is 1/C_4_3, and for (N-1)G 1B it's 1/C_N_(N-1) which will get perfectly compensated by the C_N_(N-1) (oh magic at work ! ... no it's just that it logically gives the same answer as the easy way). | ||
|
Beamo
France1279 Posts
Nonono !!! :D (still with 4 children) Knowing there are 3 Girls => there is a bigger probability that the last child is a boy (G -> 20%, B -> 80%. I know you agree with that) When you see 3 girls openning the door it's a fact not a probability, it gives you the info that there are at least 3 girls in the familly but that's it :/ You never need to calculte the probability of 3 Girls opening the door since it happened When I read the riddle I understand it as -> Knowing there is a girl (since she opened the f* door) what is the probability for the second child to be a boy ( 2/3 ). At least I think that's what it is meant to be | ||
|
Clutch3
United States1344 Posts
On November 24 2004 07:18 Catyoul wrote: Short answer : no you are wrong. Long answer : well the thread is long enough and filled with the explanation, and I and others have already requoted ourselves way too many times. No kidding ? :p Develop your proof/thought and I will discuss it. Oh crap, I can't resist. You are considering the sex of the two children as two dependent things, or as one single probabilistic event if you prefer. Up to you, nothing wrong with that. But you forget that in this situation, the girl/boy combo has its chances of showing a girl at the door of only 50% while the girl/girl one has 100%. Very schematically : GG -> 25% GB -> 25% BG -> 25% BB -> 25% GG -> girl at door = 100% GB -> girl at door = 50% BG -> girl at door = 50% BB -> girl at door = 0% Multiply that and you have : GG and girl at door = 25% GB and girl at door = 12.5% BG and girl at door = 12.5% -> BG or GB and girl at door = 25% BB and girl at door = 0% (the remaining 50% are the cases with boy at door.) Thus the chances for the other child being a boy or a girl are equal. edit : man, gotta love this problem ^^ Yep, you hit it on the head. Well done. | ||
|
MoltkeWarding
5195 Posts
On November 24 2004 11:03 Beamo wrote: To Catyoul Nonono !!! :D (still with 4 children) Knowing there are 3 Girls => there is a bigger probability that the last child is a boy (G -> 20%, B -> 80%. I know you agree with that) When you see 3 girls openning the door it's a fact not a probability, it gives you the info that there are at least 3 girls in the familly but that's it :/ You never need to calculte the probability of 3 Girls opening the door since it happened When I read the riddle I understand it as -> Knowing there is a girl (since she opened the f* door) what is the probability for the second child to be a boy ( 2/3 ). At least I think that's what it is meant to be No that is wrong too. That question is the same as: What are the odds that out of five children, at least one is a girl? Here the possibilities for at least 4 male children are: MMMMF MMMFM MMFMM MFMMM FMMMM MMMMM Therefore 4:1 The way you stated is means: Knowing that four are male, what % is the other child female? MMMMM MMMMF MMMFM MMMMM MMFMM MMMMM MFMMM MMMMM FMMMM 50%. You're getting the question mixed up. | ||
|
Catyoul
France2377 Posts
On November 24 2004 11:03 Beamo wrote: Nonono !!! :D Yes yes yes yes !!! :D On November 24 2004 11:03 Beamo wrote: When you see 3 girls openning the door it's a fact not a probability, it gives you the info that there are at least 3 girls in the familly but that's it :/ It doesn't exactly give you the info that there are 3 girls in the family. It gives you the info that the first 3 children you saw were girls. Not. The. Same. Thing. On November 24 2004 11:03 Beamo wrote: You never need to calculte the probability of 3 Girls opening the door since it happened Ok my other friend has 10 girls. I don't need to take the probability for that into account, hey it has already happened. But you will agree with me that you can't have a friend with 10 girls and not have increased probability that there is a boy right ? Of course not ! If you don't need need to take the probability into account because it has already happened, you cannot take it into account to influence the rest. On November 24 2004 11:03 Beamo wrote: When I read the riddle I understand it as -> Knowing there is a girl (since she opened the f* door) what is the probability for the second child to be a boy ( 2/3 ). At least I think that's what it is meant to be It is "knowing the first child you saw is a girl", not "knowing one of the 2 is a girl". It does not cover the same cases, been there and explained that a couple of times in the thread already, write all the cases if you want to convince yourself  | ||
|
lastas
Sweden1219 Posts
| ||
|
Shiv
France447 Posts
On November 24 2004 12:01 lastas wrote: Its obvious that people with an understanding of logic/math are supporting the 50% theory, and people that just wish things to be cool and hard for people to understand support the 66% theory. =_= It's obvious that you have no clue of what you are talking about. Damn. 55 pages, and still. Okay, I'm going to use one of Travis' post, since he's on the 50%' side, and he explained it in quite a simple way . On November 21 2004 22:22 travis wrote: ok the 2 children are 2 seperate things she can have a: boy, girl boy, boy girl, boy girl, girl the first one is a fucking girl the 2nd one can still be a: girl boy boy girl now how many of those possibilities are girls, and how many are boys. ok thanks Well, he has the first path of the reasoning exact, then he just fuck up, don't really know why though. May I quote again? boy, girl boy, boy girl, boy girl, girl the first one is a fucking girl You see a girl. OKAY. Then, we have: boy, girl <- Possible boy, boy <- Impossible girl, boy <- Possible girl, girl <- Possible Oh and, this post was probably the one you missed : On November 21 2004 22:11 Famouzze wrote: let me put it to you this way. do you believe if a woman has 10000 kids, the more kids she has the closer to a 50/50 layout it will inevitably get? remember how the gender layout of the entire planet goes: close to 50/50. Do you know why is gets SO close to 50/50? It's because it's not 50/50 when you're having series. It could get to 80/20 easily if that was 50/50 each time. That's "La loi des grands nombres". | ||
|
Clutch3
United States1344 Posts
On November 24 2004 12:25 Shiv wrote: It's obvious that you have no clue of what you are talking about. Damn. 55 pages, and still. Okay, I'm going to use one of Travis' post, since he's on the 50%' side, and he explained it in quite a simple way . Well, he has the first path of the reasoning exact, then he just fuck up, don't really know why though. May I quote again? You see a girl. OKAY. Then, we have: boy, girl <- Possible boy, boy <- Impossible girl, boy <- Possible girl, girl <- Possible Your mistake is assuming that all possible outcomes are equally likely. You say "possible" and don't bother to do the math. As Catyoul already pointed out, this is not true. As Catyoul also pointed out, there are two ways to fulfill the initial conditions if the actual situtation is G-G and only one for each of the B-G cases. Therefore, the extra weight you give this situation is how you get the 50%. Of course, there are much simpler ways to arrive at this conclusion. Basic laws of probability do not allow the two children's genders to be probabilistically dependent. The tricky part is seeing that those rules apply here. | ||
|
seeyoulater
970 Posts
| ||
|
Catyoul
France2377 Posts
On November 24 2004 12:25 Shiv wrote: Do you know why is gets SO close to 50/50? It's because it's not 50/50 when you're having series. It could get to 80/20 easily if that was 50/50 each time. That's "La loi des grands nombres". You obviously have no idea of the actual "Loi des grands nombres" (law of large numbers in English IIRC) or how it is proved. Yes, it is 50/50 boy/girl every single time, and yes with a high number of children it converges to a Gaussian distribution centered around 50% boys/girls. | ||
|
twsan
64 Posts
| ||
|
NoSteratu
Turkey770 Posts
if first(older) child is girl answer is 1/2 | ||
|
BigBalls
United States5354 Posts
And someone give me some cool problems that i missed!!! I dont want to go through old posts trying to find them cause im likely to look at solutions as well | ||
|
BigBalls
United States5354 Posts
| ||
|
BigBalls
United States5354 Posts
| ||
|
twsan
64 Posts
On November 24 2004 13:15 BigBalls wrote: Catyoul, what is your solution? And someone give me some cool problems that i missed!!! I dont want to go through old posts trying to find them cause im likely to look at solutions as well start from page 50, the fly and train problem, then the lemmings problem | ||
| ||
