How many loops? A girl is holding 6 blades of grass in her hand with the ends protruding above and below. The top ends are tied together in pairs, and then the lower ends are tied together in pairs. What is the probability the all the blades are joined in one loop? What is the probability of obtaining two loops?
Can this be generalised for 2n blades of grass?
(Please correct my if I'm wrong here)
If I labelled the top of each blade with a letter and saw that there were 15 different ways to tie them into pairs, the same applies to the bottom ends, giving a total of 225 different outcomes.
I wrote out all of the 15 combinations (for the top ends, using the letters) and wrote next to it how many loops this gives us. From this I got that there were 8 ways to get 1 loop, 6 ways to get 2 loops and 1 way to get 3 loops.
Is it correct to multiply each of these by 15. So I get, for example, there are 15 different ways to get 3 loops, so the probability of that would be 15/225 (=1/15)
Or am I completely wrong?
I hope this all makes sense
Any help is greatly appreciated
 A: What you’ve done is correct, but you can avoid the multiplication by $15$ by a simple trick. Let the top ends be $A,B,C,D,E$, and $F$, and let the corresponding bottom ends be $a,b,c,d,e$, and $f$. There is no harm in assuming that the tied top pairs are $\{A,B\}$, $\{C,D\}$, and $\{E,F\}$: we can always relabel the blades to make this the case. We now have three strands:
$$\begin{align*}
&a\longleftrightarrow A\longleftrightarrow B\longleftrightarrow b\\
&c\longleftrightarrow C\longleftrightarrow D\longleftrightarrow d\\
&e\longleftrightarrow E\longleftrightarrow F\longleftrightarrow f
\end{align*}$$
Now we can concentrate exclusively on what happens at the bottom. For instance, what has to happen at the bottom in order for us to get a single loop? The end $b$ can be tied to $c,d,e$, or $f$, a total of $4$ choices. The new free end is then $d,c,f$, or $e$, respectively, and it has to be tied to one of the bottom ends of the third strand; that can be done in only $2$ ways. Once that’s done, the new free end must be tied to $a$. Altogether, then, there are $4\cdot2=8$ ways to form the loop.
On the other hand, if we simply pair up bottom ends, there are $5$ ways to choose a mate for $a$. Once that’s been done, there are $3$ ways to find a mate for the alphabetically first of the four remaining bottom ends, and after that there are no choices: there are only two bottom ends left. Thus, there are $5\cdot3=15$ possible ways to pair up the bottom ends, and the probability of getting a single loop is $\frac8{15}$.
This way of looking at it may make the generalization to $2n$ blades a bit easier to think about.
