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标题: 绳圈 [打印本页]

作者: constant    时间: 2005-10-28 23:50
标题: 绳圈

口袋里有 n 根绳子。随机选出两个绳头系在一起,这样重复 n 次。求最后连成一个大绳圈的概率。
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作者: 乱弹    时间: 2005-10-29 02:13
标题: 回复:绳圈

What is the probability of not getting a loop of length 1 after one step (which leaves n-1 ropes)?  I think it is (2n-2)/(2n-1) if we assume that all ropes are labeled 1, 2, ..., n and the rope ends of rope k  are labeled 2k-1, 2k (so we think the two ends are different).
 
Then we use the principle of multiplication: p(1)=1, p(n+1)=2n/(2n+1)*p(n)=(2n)!!/(2n+1)!!.
 
 
 Different setting may lead to different answer. For example, under some setting, p(n) could be just 1/n. However,  the one I use might be the right one.
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作者: heavenarch    时间: 2005-10-29 03:35
标题: 回复:绳圈

2^n *n!/(2*n)!www.ddhw.com

 

作者: heavenarch    时间: 2005-10-29 04:21
标题: 回复:回复:绳圈

,wrong, but almost. 2^n *(n!)^2 /(2*n)!
: all possible knots: (2*n)!/(n! *2^n)
: big circles among them: n!www.ddhw.com

hope not wrong againwww.ddhw.com

I am glad to find so many interesting quizs here and so many great men too, respect.www.ddhw.com


 

作者: constant    时间: 2005-10-29 05:07
标题: Induction again? [:X]

  Induction again?





作者: husonghu    时间: 2005-10-29 11:12
标题: 你也是高手。希望你也能多献好题,多显身手[@};-][@};-][>:D<]

  你也是高手。希望你也能多献好题,多显身手





作者: husonghu    时间: 2005-10-29 12:17
标题: 不过你这答案好象不对。同意乱弹P(n)=(2n-2)!!/(2n-1)!!. 我是这样推的....

n根绳时,总的接法数是C(2n,2),不成圈的接法数(每根头尾相接)
是n,能成圈(接成n-1根绳)的概率是(2n-2)/(2n-1);www.ddhw.com

依次,n-1根绳时,能成圈(接成n-2根绳)的概率是(2n-4)/(2n-3);www.ddhw.com

.............www.ddhw.com

所以,P(n)=(2n-2)!!/(2n-1)!!www.ddhw.com

不知做得对不对?www.ddhw.com

 

作者: vicky3712    时间: 2005-10-29 13:07
标题: 回复:不过你这答案好象不对。同意乱弹P(n)=(2n-2)!!/(2n-1)!!. 我是这样推的..

if n = 1,  p(1) = 0!!/1!! = 1; right
   n = 2,  p(2) = 2!!/3!! = 2/6! this is wrong.
the correct approach is
  p(1) = 1;
for n > 1,
  p(2) = (4/C(4, 2)) p(1)  = 2/3;
  p(3) = (C(6,2) - 3)/C(6,2)) P(2) = 8/15;
  p(n) = (C(2n, 2) -n)/C(2n, 2)) p(n-1)
which is a ugly expression.        
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作者: husonghu    时间: 2005-10-29 19:43
标题: Seems nothing wrong with P(n)=(2n-2)!!/(2n-1)!! ..

You were wrong by saying 2!!/3!! = 2/6!www.ddhw.com
Instead, 2!!/3!! = 2/3  
 
In fact, if you simplify your result, you will get exactly
the same thing:
P(n)=(2n-2)!!/(2n-1)!!
 
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作者: 乱弹    时间: 2005-10-30 04:56
标题: [:P]

  





作者: constant    时间: 2005-10-30 06:07
标题: 回复:[:P]

I was thinking about using permutation groups. But somehow the answer was different. Probably "randomness" is defined differently.
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作者: QL    时间: 2005-10-30 06:59
标题: 回复:回复:绳圈

"Different setting may lead to different answer. For example, under some setting, p(n) could be just 1/n. "www.ddhw.com
 
Can you elaborate this?
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作者: 乱弹    时间: 2005-10-30 12:52
标题: 回复:回复:回复:绳圈

My first thought: each permutation is corresponding to a way of tieing. There are n! permutations, with (n-1)! of them being cycles of length n.www.ddhw.com
 
However, later I thought this was wrong. This mapping is not one-one or in any sense constant-constant.
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