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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 :/).
This is not right. When you see one girl BG, GB and GG are no longer equal 25% odds. I've said this before. There's several ways you can look at this.
Without order, after Either child #1 or child #2 is G
GG
[GB, BG]- Girl you saw cannot act as both placeholders. GB eliminates BG. The only instance when the girl can act as both placeholders is in the question: What are the odds that at least one is a girl? However, this question refers to after you KNOW one is a girl, therefore the probabilities change.
Another way:
GB <---Girl answers door 50% here
BB <---0% chance of answering
BG <---Girl answers door 50% here
GG <---Girl answers door 100% here.
Therefore GG = GB + BG
Another way:
Parents decide that first son will be named Mike
Second son will be named Chris
First daughter will be named Anna
Second daughter will be named Krista
possibilies:
Mike Chris (25%)
Anna Mike (25%)
Mike Anna (25%)
Anna Krista (25%)
After we eliminate Mike Chris as impossible, there remain
Anna Mike
Mike Anna
Anna Krista
If the question was: You see anna at the door. What are the odds that the other child's name is Mike, then indeed, there would be a 2/3 chance.
However the question doesn't state this. It simply states that you see a girl. What are the odds that the other child's name is Mike, it is 50% because Krista is added to the equation
Or you can simply state that
A woman has 2 children, 1 is known
2-1 = 1 unknown, therefore random.
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Catyoul
France2377 Posts
Wow, you actually had strength left to retype all that yet another time or you just copy/pasted ? :p
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IIRC, read my post on bottom of page 8, on page 10 for a more detailed version and on top of page 12 for an east way of seeing how it would have been 2/3.
Ah yes I understand know, the original question was poorly worded. Now i can sleep calmly, thx 
edit: wtf! I suck at qouting :/
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On November 23 2004 15:32 KarlSberg~ wrote:
ok
*************** SPOILER ******************
When a lemming meets an other lemming they both change directions. When lemming A and B meet, after they meet lemming A will go in the direction lemming B went before they met. Lemming B will go in the direction lemming A went before they met.
So as a whole when 2 lemmings meet, it's like nothing happens. The problem would be the same if lemmings could go across each other without ever changing direction.
As a consequence the maximum time is the time a lemming takes to go across the whole platform: 100 seconds.
That makes no sense. Let's suppose there are 6 lemmings
LemmingE---><---Lemming CLemming B ---> <---Lemming ALemmingD---><---LemmingF
<---LemmingELemming C---><---Lemming BLemmyA---><---LemmingDLemmingF--->
<---LemmingCLemmingB---><---LemmingALemmingD--->
<---LemmingBLemmingA--->
When two lemmings meet, they begin walking in opposite directions, that's what happens. Therefore assuming they are all placed equidistantly apart, they will go back and forth, if they are placed facing ><><><><><><><, they will continue bouncing back and forth until there are no more lemmings to their right/left
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wow, you didn't have to type all that Moltke, page reference would have been fine but thx anyhow =]
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Koennen Sie sprechen Deutsch?
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Catyoul
France2377 Posts
Let's say when Lemming E and C meet in your example. Lemming E was going to the right and Lemming C to the left. Look at what happens just after they met and replace the letter 'E' by 'C' and the letter 'C' by 'E'. Tada ! it's just as if the lemmings go straight through each other.
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The easiest way to explain it:
The odds of having a boy/girl is the exact same as the odds of having 2 children of the same gender. Either boy/boy or girl/girl. The odds that both of the woman's kid both being girls/boys is the same regardless of you knowing the gender of one of them because she had the kids long before you knew their gender
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On November 23 2004 16:05 Catyoul wrote:
Let's say when Lemming E and C meet in your example. Lemming E was going to the right and Lemming C to the left. Look at what happens just after they met and replace the letter 'E' by 'C' and the letter 'C' by 'E'. Tada ! it's just as if the lemmings go straight through each other.
Originally:
Lemming E---><---LemmingC
Then:
<---Lemming ELemmingC--->
Switcheroo:
<----LemmingCLemmingE--->
Okay...still different.
The point is that Lemming E in the new case still cannot go through Lemmming A. If we switch Lemming A with lemming E now, all that means is Lemming C dies, then lemming A 10 seconds later (assuming lemming E was originally on edge of plank) and lemming E 15 seconds later.
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On November 23 2004 16:00 LinYu_roy wrote:
Koennen Sie sprechen Deutsch?
Nur ein bisschen :/
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Here's how I'm solving the lemming problem
There are 100 meters separating 20 lemmings.
In the worst possible scenario, the lemmings would be 5m apart from each other with a pair facing toward each other like so:
><><><><><><><><><
Let's say we're on left edge of the plank.
Lemming1---><---Lemming2 meets 2.5 seconds in
<---Lemming1Lemming2---> Lemming 1 dies 5 seconds in, lemming 2 returns to his original position 5 seconds in moving rightwards
Lemming2---><---Lemming3 Lemming 2 and lemming 3 bump 7.5 seconds in (as lemming 3 is returning from his bump in with lemming 4)
<---Lemming2Lemming3---> Lemming 2 will walk off the plan within another 7.5 seconds, making it 15 seconds in. While lemming 3 will meet with lemming 4 again 10 seconds in
Conclusion: Each subsequent lemming will take an additional 5 seconds from the last lemming to die, as the plank gets more concentrated around the center since the lemmings at the edges fall off
There are 10 pairs of lemmings, since 2 fall off at the same time on either side.
The time for the last pair to fall off are 5x10=50 seconds after the 2nd to last.
Therefore the total length can be calculated as follows:
t=5*1+5*2+5*3....5*10
t=5+10+15...+40+45+50
t=(5+50)+(10+45)+(15+40)...(25+30)
t=55*10
t=550 seconds
If anyone gets a figure over 550 and it works, they would have it better than my answer. Otherwise im pretty sure my answer is right.
Edit: I misread. There are 20 lemmings not 50. Edited problem to fit.
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Catyoul
France2377 Posts
On November 23 2004 15:15 KarlSberg~ wrote:France raised me in the love of fine maths too. :p
Yes, one of the things France still does well 
Where did (or do) you study btw ?
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On November 23 2004 15:32 KarlSberg~ wrote:
ok
*************** SPOILER ******************
When a lemming meets an other lemming they both change directions. When lemming A and B meet, after they meet lemming A will go in the direction lemming B went before they met. Lemming B will go in the direction lemming A went before they met.
So as a whole when 2 lemmings meet, it's like nothing happens. The problem would be the same if lemmings could go across each other without ever changing direction.
As a consequence the maximum time is the time a lemming takes to go across the whole platform: 100 seconds.
Bravo! :-)
the fly problem before leads into this observation
let's see if others understand what you mean
Oh and treat each Lemming as a point, with no width.
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On November 23 2004 16:50 twsan wrote:Show nested quote +On November 23 2004 15:32 KarlSberg~ wrote:
ok
*************** SPOILER ******************
When a lemming meets an other lemming they both change directions. When lemming A and B meet, after they meet lemming A will go in the direction lemming B went before they met. Lemming B will go in the direction lemming A went before they met.
So as a whole when 2 lemmings meet, it's like nothing happens. The problem would be the same if lemmings could go across each other without ever changing direction.
As a consequence the maximum time is the time a lemming takes to go across the whole platform: 100 seconds. Bravo! :-) the fly problem before leads into this observation let's see if others understand what you mean Oh and treat each Lemming as a point, with no width.
Address my objection :/
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Edited my original post. Changed the mistake that assumed 50 lemmings. Now its corrected to match 20 lemmings
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On November 23 2004 16:49 Catyoul wrote:Yes, one of the things France still does well Where did (or do) you study btw ?
Just graduated from Enseeiht computer science engineering school in Toulouse.
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Catyoul
France2377 Posts
On November 23 2004 16:47 MoltkeWarding wrote:
Lemming1---><---Lemming2 meets 2.5 seconds in
<---Lemming1Lemming2---> Lemming 1 dies 5 seconds in, lemming 2 returns to his original position 5 seconds in moving rightwards
Lemming2---><---Lemming3 Lemming 2 and lemming 3 bump 7.5 seconds in (as lemming 3 is returning from his bump in with lemming 4)
<---Lemming2Lemming3---> Lemming 2 will walk off the plan within another 7.5 seconds, making it 15 seconds in. While lemming 3 will meet with lemming 4 again 10 seconds in
Conclusion: Each subsequent lemming will take an additional 5 seconds from the last lemming to die, as the plank gets more concentrated around the center since the lemmings at the edges fall off
There are 10 pairs of lemmings, since 2 fall off at the same time on either side.
The time for the last pair to fall off are 5x10=50 seconds after the 2nd to last.
Therefore the total length can be calculated as follows:
t=5*1+5*2+5*3....5*10
t=5+10+15...+40+45+50
t=(5+50)+(10+45)+(15+40)...(25+30)
t=55*10
t=550 seconds
First you're saying 10 seconds in, which means at absolute time, then you're adding the times ?
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Catyoul
France2377 Posts
Ok, here's the correct result with your calculating method (though you don't prove it is the worst case scenario) :
1. Lemming 1 meets 2 at t=2.5
1'. Lemming 1 dies at t=2.5+2.5=5
2. Lemming 2 meets 3 at t=2.5+5=7.5
2'. Lemming 2 dies at t=7.5+7.5=15
3. Lemming 3 meets 4 at t=7.5+5=12.5
3'. Lemming 3 dies at t=12.5+(3*5+2.5)=30
...
10. Lemming 10 meets 11 at t=(10-1)*5+2.5=47.5
10'. Lemming 10 dies at t=47.5+(10*5+2.5)=100
For the lemmings at the right, the calculation is slightly different since the lemming 20 is actually 5 meters from the right
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On November 23 2004 16:51 MoltkeWarding wrote:
Address my objection :/
In the case you describe the 1st pair will fall after 5 seconds then a new pair every 10 seconds.
The total time will be 95 seconds.
Which is not 100 !!!
But 20 lemmings define 19 intervals. So with 5 meters intervals the total leng th is 95 meters.
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On November 23 2004 17:06 Catyoul wrote:Ok, here's the correct result with your calculating method (though you don't prove it is the worst case scenario) : 1. Lemming 1 meets 2 at t=2.5 1'. Lemming 1 dies at t=2.5+2.5=5 2. Lemming 2 meets 3 at t=2.5+5=7.5 2'. Lemming 2 dies at t=7.5+7.5=15 3. Lemming 3 meets 4 at t=7.5+5=12.5 3'. Lemming 3 dies at t=12.5+(3*5+2.5)=30 ... 10. Lemming 10 meets 11 at t=(10-1)*5+2.5=47.5 10'. Lemming 10 dies at t=47.5+(10*5+2.5)=100 For the lemmings at the right, the calculation is slightly different since the lemming 20 is actually 5 meters from the right
wow, I was just trying to analyze his explanation... heh saved me some thinking
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EPT
