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On November 24 2004 13:13 NoSteratu wrote:
hmm.. IMO if one of 2 children is girl answer is 2/3
if first(older) child is girl answer is 1/2
Ok, im going to calmly explain this even though catyoul did 20 times.
You know that one of the 2 children is a girl. Lets call her abby. Lets call the other child B (bobby or betty). Ill put the older child first and the younger child second.
possible of children outcomes with abby answering the door
abby betty
betty abby
abby bobby
bobby abby
there are 2 scenarios where a girl is the other sibling and 2 scenarios where a boy is the other sibling
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On November 23 2004 13:40 KarlSberg~ wrote:
I have an interesting math question to submit.
I don't think there is much to discuss since you can't find anything if you don't find the good answer and can't give much help with giving to much, still I can answer questions if someone has some.
I liked it so I guess some people could like it too.
Two trains start at the same time frome NY and LA, both going to the other city. (So obviously they will meet each other during their travel)
LA->NY train goes at Speed1
NY->LA train goes at Speed2
The distance between both towns is D.
A fly Starts from LA at the same time too.
The fly is faster than both trains and goes at SpeedF.
When the fly meets train2, it goes backwards until it meets train1, then it turns around again and so on.
When the trains meet, what distance will the fly have flied?
There are 2 ways to find the answer. One involves mathematical series and lots of calculus.
The other only involves thinking.
Well, v*t = d, so likewise v/d = t
the two trains meet at time (speed1+speed2)/D
the fly travels at a constant speed for a certain time. So, his velocity*his time traveled is his total distance flown.
his velocity is speed f, his time is (speed1+speed2)/D, so the total distance of the fly is [SpeedF*(speed1+speed2)]/D
i saw this one right away
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On November 23 2004 10:46 Catyoul wrote:1. I show you the first 9 flips, they happen to be tails. What's the probability for the 10th one ? 2. I tell you there are at least 9 tails. What's the probability for the 10th one (edit: the one that isn't included in the 9 tails I told you) ? If you don't understand how these 2 questions are different, sorry I can't say much more. Btw, the correct answers are : 1. 50% tail 50% head 2. 1/11= ~9% tail 10/11 = ~91% head -- edit: Let's add a 3rd question, more tricky :p 3. I tell you there are at least 9 tails. Now, what's the probability of the flip number 10 (not the 10th one like in question 2) to be head or tail ? I'll let you think on that one before giving the answer  --
#2 is 1/11 tail, 10/11 head because....
solutions for tails:
TTTTTTTTTT
solutions for heads:
HTTTTTTTTT
THTTTTTTTT
TTHTTTTTTT
TTTHTTTTTT
TTTTHTTTTT
...
TTTTTTTTTH
10 solutions for heads, 1 solution for tails. that is simple enough
the probability on the last one is 1/11 tails still 10/11 heads
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Catyoul
France2377 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
For the cool problems you missed, as twsan said, page 50. You probably missed the riddles around page 15 too I guess, but it's not as funny if you can't participate.
For the solution, c/p :
For the BigBalls one : I had the same idea as LTT yesterday but couldn't follow through because of tiredness. Doing it now, I find the same thing, that is C=D=5 and J=6 and I don't see (yet) where there might be another possibility. Continuing the search
(...)
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.
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On November 23 2004 15:00 IceLeY wrote:
Another one:
Frank & Daniel meet after some time.
Frank: Hi Daniel, how old are your three children now? I forgot it.
Daniel: The product of their ages is 36.
Frank: Oh, that doesn't get me far.
Daniel. Hm, the sum of their ages equals my house number.
Frank: Well...
Daniel: Not to mention that the oldest one has blue eyes.
Frank: Ah, ok, no i know how old they are.
How old are the three children?
Hmm, if the product of their ages is 36, we have a small number of possibile factorizations of 36 for their ages
1 1 36
1 2 18
1 3 12
1 4 9
1 6 6
2 2 9
2 3 6
3 3 4
The sums of these are 38, 21, 16, 14, 13, 13, 11, 10 respectively.
Since he doesnt know the answer after the house number, it must be the case that the house number is 13.
So the only two possible combinations left are 2 2 9, 1 6 6
We then know that their is an oldest child, so the children are 2 2 9 years old. Cool little problem. Ill get back to the lemmings one, i think i have a solution but i have to come up with a better justification for it.
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I think for the lemming one the answer is ALWAYS 100 for the maximum, regardless of the amount of lemmings present. Let me think a little more about why
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Catyoul
France2377 Posts
Whoops, I think I let a mistake slip by yesterday btw for my question 3
3. I tell you there are at least 9 tails. Now, what's the probability of the flip number 10 (not the 10th one like in question 2) to be head or tail ? I'll let you think on that one before giving the answer
The cases are :
TTTTTTTTTT
HTTTTTTTTT
THTTTTTTTT
TTHTTTTTTT
TTTHTTTTTT
TTTTHTTTTT
...
TTTTTTTTTH
The flip number 10 has 10/11 to be tail 1/11 to be heads. Sorry I was working with 0's and 1's on paper not with T/H, I got confused
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oh right....i was backwards, thats what i meant. let me look at your solution to my #5 now 
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Catyoul
France2377 Posts
Can't blame you, I did the same yesterday Anyway, it's just an easy illustration of how the question asked can dramatically influence the answer :p
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On November 24 2004 13:16 BigBalls wrote:
HOW MANY TIMES DO PEOPLE HAVE TO EXPLAIN WHY IT IS 1/2 AND NOT 2/3, COME ON, LETS BE SERIOUS HERE, 55 FUCKING PAGES AND YOU THICK HEADED IDIOTS STILL DONT GET IT
1st possibility - Our world is a world where the following can happen, without excluding the fact (A) that we saw a girl standing before the door:
BG <- 1/4 - Can happen - 1/4 stays - becomes 1/3 considering (A)
GB <- 1/4 - Can happen - 1/4 stays - becomes 1/3 considering (A)
GG <- 1/4 - Can happen - 1/4 stays - becomes 1/3 considering (A)
BB <- 1/4 - Can't happen - 1/4 out
So we have 2/3 chances to meet a boy inside. Therefore you're fucking wrong.
2nd possibility - Our world is a world where the following can happen, without excluding the fact that (A) we saw a girl standing before the door:
. . . _
BG = GB <- 1/3 - Can happen - 1/3 stays - becomes 1/2 considering (A)
GG <- 1/3 - Can happen - 1/3 stays - becomes 1/2 considering (A)
BB <- 1/3 - Can't happen - 1/3 out
So we have 1/2 chances to meet a boy inside. Therefore you're fucking right.
I'm sorry. Has many has said, it depends.
>Catyoul: How am I supposed to answer that? I'm studying maths...
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----1. C
D=Ace or J=Ace -> DW => D <= K and J <= K
D=2 and J=2 -> DL => D or J >= 3
Ok, so basically, at least one of the other players dont have a 2 cause if they both did then caroline would say DL. Thus, D or J >=3. Correct.
----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
We know J OR D dont have a 2. Now we know that C OR J dont have a 2. This does not automatically imply that J has a 3 or greater.
Suppose C has a 3, J has a 2 and D has a 3. I think this is where you are messing up, you start your lower bound a little too early.
----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
Now we know that C,D>2. J can still have a 2.
----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.
--------------------------------------------------------------------------
Continuing on in that fashion, C = 5, D = 4 or 5, J = 6.
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On November 24 2004 13:54 Shiv wrote:Show nested quote +On November 24 2004 13:16 BigBalls wrote:
HOW MANY TIMES DO PEOPLE HAVE TO EXPLAIN WHY IT IS 1/2 AND NOT 2/3, COME ON, LETS BE SERIOUS HERE, 55 FUCKING PAGES AND YOU THICK HEADED IDIOTS STILL DONT GET IT
1st possibility - Our world is a world where the following can happen, without excluding the fact (A) that we saw a girl standing before the door: BG <- 1/4 - Can happen - 1/4 stays - becomes 1/3 considering (A) GB <- 1/4 - Can happen - 1/4 stays - becomes 1/3 considering (A) GG <- 1/4 - Can happen - 1/4 stays - becomes 1/3 considering (A) BB <- 1/4 - Can't happen - 1/4 out So we have 2/3 chances to meet a boy inside. Therefore you're fucking wrong. 2nd possibility - Our world is a world where the following can happen, without excluding the fact that (A) we saw a girl standing before the door: . _ BG = GB <- 1/3 - Can happen - 1/3 stays - becomes 1/2 considering (A) GG <- 1/3 - Can happen - 1/3 stays - becomes 1/2 considering (A) BB <- 1/3 - Can't happen - 1/3 out So we have 1/2 chances to meet a boy inside. Therefore you're fucking right. I'm sorry. Has many has said, it depends. >Catyoul: How am I supposed to answer that? I'm studying maths...
You are missing something. If you consider BG different than GB, then you are taking age into account. Therefore Gg is different from gG
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Catyoul
France2377 Posts
On November 24 2004 13:55 BigBalls wrote:
----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
We know J OR D dont have a 2. Now we know that C OR J dont have a 2. This does not automatically imply that J has a 3 or greater.
Suppose C has a 3, J has a 2 and D has a 3. I think this is where you are messing up, you start your lower bound a little too early.
From what C said, either D or J >= 3. If D sees J has a 2, he knows he has a 3, therefore he can't qsdkjfqskldjflqskdjfklqsdjf damn of course he can still lose :p
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Catyoul
France2377 Posts
On November 24 2004 13:54 Shiv wrote:
1st possibility - Our world is a world where the following can happen, without excluding the fact (A) that we saw a girl standing before the door:
BG <- 1/4 - Can happen - 1/4 stays - becomes 1/3 considering (A)
GB <- 1/4 - Can happen - 1/4 stays - becomes 1/3 considering (A)
GG <- 1/4 - Can happen - 1/4 stays - becomes 1/3 considering (A)
BB <- 1/4 - Can't happen - 1/4 out
So we have 2/3 chances to meet a boy inside. Therefore you're fucking wrong.
2nd possibility - Our world is a world where the following can happen, without excluding the fact that (A) we saw a girl standing before the door:
. . . _
BG = GB <- 1/3 - Can happen - 1/3 stays - becomes 1/2 considering (A)
GG <- 1/3 - Can happen - 1/3 stays - becomes 1/2 considering (A)
BB <- 1/3 - Can't happen - 1/3 out
So we have 1/2 chances to meet a boy inside. Therefore you're fucking right.
I'm sorry. Has many has said, it depends.
>Catyoul: How am I supposed to answer that? I'm studying maths...
I'm tired of writing the same thing again. Refer to my posts at : bottom page 8, page 10, page 12, and basically on the last 5-6 pages where I restated them with counter-arguments to specific people.
Ok, you're studying maths. No problem, everyone can make mistakes, especially when wording is so critical. You're not the first one in this thread to have said this though, the last one who used this argument corrected himself 2 pages later and accepted the 50/50. Oh, and btw I have studied maths too and quite a lot of them.
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Catyoul
France2377 Posts
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On November 24 2004 13:56 BigBalls wrote:
You are missing something. If you consider BG different than GB, then you are taking age into account. Therefore Gg is different from gG
Well, I'll try to explain why I think I'm not missing anything. We're talking here about the first world's possibility, more likely ours.
Hum, imagine a thousand houses.
Well, you have 1/4 of them who has 2 girls, 1/4 of them who has 2 boys, and 1/2 of them who has 1 girl 1 boy in no specific order (because we have 1/4 who have 1 boy 1 girl and 1/4 who have 1 girl 1 boy in that specific order) right? We HAVE taken in account their ages right?
So, when we see that girl standing out on the floor of her house, we know we can't take in account the 1/4 of the houses who're hosting 2 boys right?
What's left? 1/3 (1/4 becomes 1/3 when you take out that 1/4 houses hosting boys only) of the houses who has 2 girls, and 2/3 (1/2 becomes 2/3, see above) who have 1 boy and 1 girl.
Thus we do have 2/3 of having a boy in the house.
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Catyoul
France2377 Posts
On November 24 2004 14:16 Shiv wrote:Show nested quote +On November 24 2004 13:56 BigBalls wrote:
You are missing something. If you consider BG different than GB, then you are taking age into account. Therefore Gg is different from gG Well, I'll try to explain why I think I'm not missing anything. We're talking here about the first world's possibility, more likely ours. Hum, imagine a thousand houses. Well, you have 1/4 of them who has 2 girls, 1/4 of them who has 2 boys, and 1/2 of them who has 1 girl 1 boy (or 1 boy one girl) right? We HAVE taken in account their ages right? So, when we see that girl standing out on the floor of her house, we know we can't take in account the 1/4 of the houses who're hosting 2 boys right? What's left? 1/3 (1/4 becomes 1/3 when you take out that 1/4 houses hosting boys only) of the houses who has 2 girls, and 2/3 (1/2 becomes 2/3, see above) who has 1 boy and 1 girl. Thus we do have 2/3 of having a boy in the house.
Nice example. You forgot one part. On average the houses with one girl and one boy will only have a girl at the front door 1 time out of 2. So when you visit 1000 houses, what's going to happen. 100% of the 250 with girl girl will matter because a girl always shows up, that makes 250 houses with a girl at the door. 100% of the 250 with boy boy won't matter for us because a boy always shows up, so that still makes 250. On 50% of the 500 with girl boy you will see a girl at the door, that makes 250. See now ?
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On November 24 2004 14:20 Catyoul wrote:
Nice example. You forgot one part. On average the houses with one girl and one boy will only have a girl at the front door 1 time out of 2. So when you visit 1000 houses, what's going to happen. 100% of the 250 with boy boy won't matter for us because a boy always shows up, so that makes 0 for now. 100% of the 250 with girl girl will matter because a girl always shows up, that makes 250 for now. On 50% of the 500 with girl boy you will see a girl at the door, that makes 250. See now ?
Ok, the problem states correctly that we see a girl at the door. If it didn't, you would've been right in both world 1 and 2. That's not the case.
Hum, I'm gonna explain it better. A girl OPENS the door. So we do not take this parameter in consideration.
If we did, again, you would have been right.
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Catyoul
France2377 Posts
On November 24 2004 14:22 Shiv wrote:
Ok, the problem state correctly that we see a girl at the door. If it didn't specified it, you would've been right in both world 1 and 2. That's not the case.
Hum, I'm gonna explain it better. A girl OPENS the door. So we do not take this parameter in consideration.
Elaborate, you're just reformulating : "no no no, you are wrong."
Let's take the statistical approach with big numbers, which will tend to the correct probability. So we have our 1000 houses again. Let's say you are going to knock on all doors, because that's what you do to get the probability for all the cases.
What happens at the 3 different type of houses :
- The girl / girl houses : a girl answer the door. ALWAYS. That makes 250 houses with a girl always answering
- The girl / boy houses : a girl answer the door, 1/2 the time. That makes 500/2 = 250 houses with a girl answering and 250 with a boy answering
- The boy / boy houses : a boy answers the door, always. That makes 250 houses with a boy answering.
Totals : 500 houses with a girl answering, 500 with a boy answering... of course !
Now the problem tells us, as you underline, that a girl opens the door. So we are in the 500 cases where a girl has opened the door. In those 500 cases, we see that 250 come from girl/girl houses and 250 from girl/boy houses. That makes 1/2 for the second child to be a boy and 1/2 for the second child to be a girl.
Let's see if we agree on basic probabilities just to be sure.
1. I show you the first 9 flips, they happen to be tails. What's the probability for the 10th one ?
2. I tell you there are at least 9 tails. What's the probability for the 10th one (the one that isn't included in the 9 tails I told you) ?
If you don't understand how these 2 questions are different, sorry I can't say much more.
3rd question, more tricky
3. I tell you there are at least 9 tails. Now, what's the probability of the flip number 10 (not the 10th one like in question 2) to be head or tail ?
Btw, the correct answers are :
1. 50% tail 50% head
2. 1/11= ~9% tail 10/11 = ~91% head
3. 10/11 tails, 1/11 heads
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Express the number 100 using 7,7,7,7 and 1. Only /,*,-,+ are allowed.
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EPT
