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Catyoul
France2377 Posts
On November 23 2004 16:59 KarlSberg~ wrote: Just graduated from Enseeiht computer science engineering school in Toulouse. Oh cool ! Were you in Toulouse too for "prepa" years ? | ||
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KarlSberg~
731 Posts
Where do you study? | ||
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KarlSberg~
731 Posts
The first argument I gave to explain why 100 seconds is simple and there is no obvious contradiction. The calculus you did to find a different result is a lot more complicated. Since the results contradict each other, the most complicated one is most likely flawed somewhere. | ||
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Catyoul
France2377 Posts
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MoltkeWarding
5195 Posts
On November 23 2004 17:13 KarlSberg~ wrote: One last tip to Moltke. The first argument I gave to explain why 100 seconds is simple and there is no obvious contradiction. The calculus you did to find a different result is a lot more complicated. Since the results contradict each other, the most complicated one is most likely flawed somewhere. Wait I see it. I should have drawn out the 3rd lemming  The sequence was arithmatical, not multiplicative. Each Lemming follows 10 seconds after the last, with 5m to spare on the other side = 100. | ||
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Hitokiri
Spain306 Posts
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LTT
Shakuras1095 Posts
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seeyoulater
970 Posts
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LTT
Shakuras1095 Posts
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seeyoulater
970 Posts
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LTT
Shakuras1095 Posts
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seeyoulater
970 Posts
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KarlSberg~
731 Posts
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Catyoul
France2377 Posts
Coming back to BigBalls question 5, I still don't see how to get the other solution. Here's my line of reasoning, and all the choices are forced, or so it seems to me for now. The downward process (from Ace to lower cards) is straightforward, the upward process (from 2 to higher cards) is a bit more subtle. DW = don't win, DK = don't know, DL = don't lose ----1. C D=Ace or J=Ace -> DW => D <= K and J <= K D=2 and J=2 -> DL => D or J >= 3 ----1. D C=Ace -> lose => C <= K C=K or J=K -> DW => C, J <= Q J=2 -> D >=3 (because of condition 1.C : D or J >= 3) -> DL => J >= 3 ----1. J C,D <= Jacks C or D = 2 -> DL (because of 1.D : J >= 3) => C and D >= 3 C and D = 3 -> DL => C or D >= 4 ----2. C D,J <= 10 D=3 -> C >= 4 (1.J) -> DL => D >= 4 ----2. D C,J <= 9 C or J = 3 -> DL (2.C) => C and J >= 4 C and J = 4 -> DL => C or J >= 5 ----2. J C,D <= 8 C=4 -> J >= 5 (2.D) -> DL => C >= 5 ----3. C D,J <= 7 D or J = 4 -> DL => D and J >= 5 D and J = 5 -> DL => D or J >= 6 ----3. D C,J <= 6 J=5 -> D=6 -> DL => J >= 6 => J = 6 ----3. J J=6 => C,D <= 5 3.C : D >= 5 => D=5 2.J : C >=5 => C=5 Thus C=5, D=5, J=6 I fail to see where there is another possibility. edit : omg, just noticed this thread went to over 1000 replies... | ||
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LaptopLegacy
Netherlands602 Posts
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Beamo
France1279 Posts
On November 23 2004 15:29 ANewBeginning wrote: Hmm so much bickering about that child problem :D, Im gonna try it to At first when we don't know anything we have four possible solutions Bb Bg Gb and Gg. Here it is 25% for 2 boys 50% for boy girl and 25 % for 2 girls. BUT when we se the little brat girl at the door we know that there is a 100% possibility for atleast 1 girl since we don't know if she is the older or the younger sibling we Bg Gb and Gg as possible solutions which leaves us with 66% possibility of the other sibling being a boy. The reason why the possibility changes is because when the 100%. So I join the 2/3 crowd, someone from the 50-50 camp tell me where my reasoning is faulty, or atleast copy paste if an answer has already been posted (thread is to big to read :/). You are right The thing the 50/50 fans don't see is that knowing the position of the girl would bring us to a 50/50 ratio, not knowing it leaves it at 2/3. Ordering the possibilities changes the probabilities. You don't have the same probability to get 4 Heads and 1 Tail in a specific order (exemple : HHTHH) as 4 Heads and 1 Tail in any order. knowing the position of the girl would bring us to a 50/50 ratio (ordered possibilities), not knowing it leaves it at 2/3. But you can keep it up, it makes me laugh reading replies like "Famouzze you have troubles understanding probability, and understanding logic as well." when he's the one being right ;D | ||
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YabooBs
34 Posts
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MoltkeWarding
5195 Posts
On November 24 2004 06:55 Beamo wrote: You are right The thing the 50/50 fans don't see is that knowing the position of the girl would bring us to a 50/50 ratio, not knowing it leaves it at 2/3. Ordering the possibilities changes the probabilities. You don't have the same probability to get 4 Heads and 1 Tail in a specific order (exemple : HHTHH) as 4 Heads and 1 Tail in any order. knowing the position of the girl would bring us to a 50/50 ratio (ordered possibilities), not knowing it leaves it at 2/3. But you can keep it up, it makes me laugh reading replies like "Famouzze you have troubles understanding probability, and understanding logic as well." when he's the one being right ;D Are we still on this? The question does not state: What are the odds that there is 1 boy and 1 girl as opposed to 2 girls? The question stats: one of the children is a girl. What are the odds the other is a boy? These questions are not the same. In the former case you're taking probabilities for a pair of genders. In the latter you're only taking the probability for the one unknown gender. It's pretty obvious. | ||
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Catyoul
France2377 Posts
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. On November 24 2004 06:55 Beamo wrote: The thing the 50/50 fans don't see is that knowing the position of the girl would bring us to a 50/50 ratio, not knowing it leaves it at 2/3. Ordering the possibilities changes the probabilities. You don't have the same probability to get 4 Heads and 1 Tail in a specific order (exemple : HHTHH) as 4 Heads and 1 Tail in any order. 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 ^^ | ||
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Alpha
France1495 Posts
but who knows if something odd can happen | ||
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