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. |
2^n *n!/(2*n)! |
,wrong, but almost. 2^n *(n!)^2 /(2*n)! : all possible knots: (2*n)!/(n! *2^n) : big circles among them: n! hope not wrong again I am glad to find so many interesting quizs here and so many great men too, respect. |
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); 依次,n-1根绳时,能成圈(接成n-2根绳)的概率是(2n-4)/(2n-3); ............. 所以,P(n)=(2n-2)!!/(2n-1)!! 不知做得对不对? |
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. |
You were wrong by saying 2!!/3!! = 2/6! 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)!! |
I was thinking about using permutation groups. But somehow the answer was different. Probably "randomness" is defined differently. |
"Different setting may lead to different answer. For example, under some setting, p(n) could be just 1/n. " Can you elaborate this? |
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. However, later I thought this was wrong. This mapping is not one-one or in any sense constant-constant. |
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