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On November 25 2004 08:59 BigBalls wrote:
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
ah yes now i see it but where was the wording a problem in the thread?
I actually thought you were wrong at first, it's harder to take the wording into account in english
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On November 21 2004 20:42 Famouzze wrote:
ok i will just post one more, this is the second hardest riddle i know of and has been the subject of much debate among my friends (some of them still won't accept the truth of the right answer)
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?
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"a girl answers the door"
is very different from
"at least one of the children is a girl" (which is how the problem is supposed to be put)
Because
a girl answers => there is at least a girl
But
there is at least a girl =/=> a girl answers
The difference is exactly the same as in the BigBall's post you quoted
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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
Ya, it's been explained above pretty well. Basically you get confused because we open the door, there's a boy (or girl), and you think "Oh, a two-child family, one child is a boy", must be the same thing. The point of this hypothesis tho is you have to know ahead of time that the family has a boy and that you being the one knocking at the door, insist that the the boy be the one who answers it. That clearly isn't happening here. We don't know who's going to answer the door so we can't insist on it being any particular gender at all.
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Catyoul
France2377 Posts
On November 24 2004 17:38 BigBalls wrote:
Ok, one more problem before I have to go for the night...this one is a good one.
You have a balance (a way of measuring if things are equal) and 12 identically sized silver balls. However, one of the balls is a different weight than the others.
Using 3 or less weighings, find the different weighted ball AND whether it is lighter or heavier.
Have fun.
I had been asked this problem and variants thereof a couple of times. There is a very elegant solution that doesn't require horrible calculs or trees (imagine I ask you for 2004 balls :p). First let's feel the problem. Let's say you have the right to use P weighings. We know that each of these weighings has 3 possible outcomes, so all in all we have 3^P possibilites, that's directly related to the number of balls we can differentiate (with some subtleties depending on the information we have).
Each time, the balance can be either ... balanced (we write it 0), goes to the left (we write it 1) or goes to the right (we write it 2). We do 3 weighings, that we can write as a number in base 3, for example : 121 means 1st went to the left, 2nd went to the right and 3rd went to the left. All there is left now is to code the 12 balls with 2 numbers in base 3, one if it is heavier, one if it is lighter. The 2 numbers are related in the following way : you change all 1 by 2 and all 2 by 1 of course. You must also have, for each weighing, the same number of 1's and 2's (that is if you write all the numbers for all balls one under another, each column will have the same number of 1 and 2).
Here is an example of coding :
ball 1 2 3 4 5 6 7 8 9 10 11 12
heavier 001 010 011 012 100 101 102 220 222 221 210 122
lighter 002 020 022 021 200 202 201 110 111 112 120 211
Now, you determine the weighings to do very easily. For each weighing, you place on the left all the balls with a 1 for this weighing and on the right the balls with a 2 (you can choose if you use the heavier coding or the lighter coding, but then stick to it during the whole process). If it has a 0 for this weighing you don't place it. For this example coding, it goes like this (using the heavier numbers) :
weighing - on the left side - on the right side
1 -- 5+6+7+12 -- 8+9+10+11
2 -- 2+3+4+11 -- 8+9+10+12
3 -- 1+3+6+10 -- 4+7+9+12
Now, you just write down the results of your weighings, for example let's say it gave :
1. balanced
2. left
3. right
That translates to 012, which means that ball number 4 is heavier. Problem owned 
Now, for some maths consideration, you can show that you can differentiate (3^p - 3) / 2 balls. If you only have to identify the wrong ball, not knowing if it's lighter or heavier you can check one more ball, code it with only 0's and you're set. If you have a reference ball that you know for sure has the right weight, you can also check (3^p - 3) / 2 + 1 balls. Code that 1 more ball with only 1's, so it is always placed on the same side. That will make 5 balls on 1 side and 4 on the other, so put the reference ball on the other side.
If you know whether the ball is lighter or heavier, you can have upto 3^p balls, which is the absolute maximum. In that case, just split the balls in 3 groups of the same size, weigh 2 of them and you have determined the group in which the bad ball is. Rinse and repeat.
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hmmm actually i take back what I said, i'm talking to a friend who i trust and he says that you are right baal
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beginning to look like I drasticly overrated my own skill on this subject
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Catyoul
France2377 Posts
Well, you're still right about the boy/girl problem and about baal's attitude :p
You even accepted your initial mistake (and everybody makes some at a point or another), which places you higher than a lot of people here
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On November 25 2004 11:16 GroT wrote:
hmmm actually i take back what I said, i'm talking to a friend who i trust and he says that you are right baal
Ask your friend the 2 different ways of asking the question; if he has the same answer for both questions, he is ill equipped to answer them.
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Sweden1225 Posts
I know this brainteaser-story in a much better way! Seems rather useless to start it off again at page 60 tho =( BLEH (cause it's cool)
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Catyoul
France2377 Posts
While discussing this by PM, I think I came up with the ultimate counter-argument against the 2/3 people. Ok, so let's imagine you guys are right. So that means that I go to the door, a girl opens me, it means that the other child has 2/3 chance of being a boy. Using the same line of thought, if a boy opens me, it means that the other child has 2/3 chance of being a girl. The argument for that being the following :
4 different cases :
GG
GB
BG
BB
All have the same 1/4 probability. When a girl comes at the door, it means you have eliminated BB, thus 3 cases left, yadda yadda.
At the same time, you will agree that there is on average a 50% chance that a boy or girl answers the door in all 2 children houses right ? (until now, it's all pretty obvious, no magic yet).
Now, you have this :
- chance that a girl opens : 1/2
- chance that a boy opens : 1/2
- if a girl opens : chance of a boy being 2nd : 2/3
thus : if a girl opens, chance of a girl being 2nd : 1/3
- if a boy opens : chance of a girl being 2nd : 2/3
thus : if a boy opens, chance of a boy being 2nd : 1/3
Total chances :
GG : girl + girl = 1/2 * 1/3 = 1/6
GB : girl + boy = 1/2 * 2/3 = 1/3
BG boy + girl = 1/2 * 2/3 = 1/3
BB : boy + boy = 1/2 * 1/3 = 1/6
That is a direct consequence of the 2/3 and in total contradiction with the fact these 4 cases have an equal probability of 1/4.
Damn, I should have kept count of the number of counter-arguments I came up with on that problem :p
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hey catyoul, did you see my 2 latest problems?
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Catyoul
France2377 Posts
er.. no, going to look 
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Catyoul
France2377 Posts
For the split, I'd say divide as said, coming up from 1710 which is dividible by 18, that gives 855, 570 and 190 as shares, so the right proportions, and the lawyer takes the remaining 102 as retribution, these guys are never free of charge :p
For the first one, I have no clue yet except word tricks
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I don't quite get it.
Since the sum of the shares is 17/18 + the shares have to be even, obviously they will not get all the money.
If there something more to be seen?
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you guys are each on the right track with the 2nd one.....combine your ideas though and i think youve got it!
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as for the first one.....well, its pretty hard to see, i made an observation and it fell out of that pretty easily.........although the solution is really fucking dumb and doesnt seem to make much sense in the real world lol
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Catyoul
France2377 Posts
Well, if the lawyer doesn't take any wages, divide 1717 by 17 and split 9/17 to 1st, 3/17 to 2nd and 1/17 to last
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Catyoul
France2377 Posts
Let me see.
1. They are triplets so there is no younger and youngest :p
2. "to sell his apples at the same price" -> You're not mentioning each apple at the same price, so they just sell their whole pack at the same price
3. They just share at the end lol
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On November 25 2004 20:02 Catyoul wrote:Well, if the lawyer doesn't take any wages, divide 1717 by 17 and split 9/17 to 1st, 3/17 to 2nd and 1/17 to last
9/17 + 3/17+ 1/17 = 13/17?
Where does rest of the money go? To your pocket?
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EPT
