I have n pieces of rope in a bowl. I pick two ends randomly and tie them together, and continue picking two more ends and tying them together until there are no free ends. What is the probability that I end up with one large loop?
It seems to me that this probability is the product of
$\dfrac{(2n-j)}{(2n-j+1)}$
for j$=2,4,6,…$etc… so that, for example, if $n=5$, we have, after picking one end from the 10, it is:
$$\frac{8}{9} \times \frac{6}{7} \times \frac{4}{5} \times \frac{2}{3},$$
but I am having trouble finding a closed-form expression for this sum. Any help would be much appreciated.
Best Answer
You have $$ \prod_{k=1}^n \frac{2k}{2k+1} = \frac{\prod_{k=1}^n (2k)}{\prod_{k=1}^n (2k+1)} = \frac{\prod_{k=1}^n (2k)^2}{\prod_{k=1}^n 2k(2k+1)} = \frac{4^n (n!)^2}{(2n+1)!} $$