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I have $2n$ pieces of string which have each of their ends separated, i.e. left ends on one side and right ends on another, and I can only pair like ends together. I need to work out the probability that I will form a loop of length $2n$ from $2n$ pieces of string. I know the number of different arrangements of knots on one side is $2n\choose n$. I have then said that the probability of selecting strings to pair in such a way so as to form a loop of $2n$ is:

$ \frac1{2n}\cdot \frac1{2n-1}\cdot \frac1{2n-1}…=\frac1{2n!}$

and so the overall probability is just:

$P_n=$$2n\choose n$$ \cdot \frac1{2n!}$

I felt this was correct, but I am having real trouble writing $P_n$ in the form:

$P_n \approx Cn^{-\alpha}$

($C$ and $\alpha$ are constants)

using Stirlings Formula and so I am beginning to doubt myself. Any pointers would be greatly appreciated!

THN
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    The number of possible knot arrangements on one side is $(2n-1)!!$ (that's the double factorial), not $\binom{2n}n$. The first rope can choose between $2n-1$ different ends to be tied with, the next unknotted end can choose between $2n-3$ and so on. – Arthur Oct 01 '18 at 10:47
  • @Arthur I see why the first rope can choose between 2n-1 ends, but I'm not sure why the next unknotted end has 2n-3 ends. Suppose n=2 and the ropes are (A1,A2)...(D1,D2); A1 can choose between any of B2,C2,D2 (not A2). Suppose A1 gets knotted with B2; then, B1 can choose between any of A2,C2,D2, since A1 happened to already eliminate B1's other end. – Cyclopropane May 18 '23 at 04:44
  • @Cyclopropane It's been a while, I don't remember my exact thought process. But I do notice "left ends on one side and right ends on another, and I can only pair like ends together." So $A_1$ can choose between $B_1, C_1$ and $D_1$. And then whichever two are left on the $1$ side must be knotted together. – Arthur May 18 '23 at 07:58
  • @Arthur Thanks for the clarification. I see why the number of arrangements of knots on one side is (2n-1)!!, but I don't see why the probability of having a continuous loop is the reciprocal of this number. Can you clarify? – Cyclopropane May 19 '23 at 18:11

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See Expected Number of Loops from $n$ Ropes or Expected number of loops for computing the expected number of loops. These don't explicitly answer the question of the probability of having one loop, but you can work out that the probability of having one loop is

$$ { 1 \over 2n-1 } \times {1 \over 2n-3} \times {1 \over 2n-5} \times \cdots \times {1 \over 1 } $$

as Arthur has also alluded to in his comment.

This can be written as a double factorial but if you want to set up for Stirling's approximation it's better to write it as

$$ {2n \over 2n(2n-1)} {2n-2 \over (2n-2)(2n-3)} {2n-4 \over (2n-4)(2n-5)} \cdots {2 \over (2)(1)}. $$

Now the denminator is clearly $(2n)!$. The numerator is $n! 2^n$ as you can see by pulling a 2 out of each factor.

Let $$P_n = {n! 2^n \over (2n)!}$$. Let's use Stirling to find an approximation for this. In particular we'll use it in the form $n! \sim \sqrt{2\pi n}(n/e)^n$. This gives

$$ P_n \sim {\sqrt{2\pi n} (n/e)^n \times 2^n \over \sqrt{4\pi n} (2n/e)^{2n}}$$. There won't be an approximation of the form $Cn^{-\alpha}$ because there are $n$ factors of $n$ in the numerator and $2n$ factors of $n$ in the denominator. However with some algebra this can be rewritten as

$$ P_n \sim {1 \over \sqrt{2}} {n^n 2^n \over (2n)^{2n}} e^n = {1 \over \sqrt{2}} \left( {e \over 2n} \right)^n$$

Michael Lugo
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