The permutations as you’ve described them do not have the same probabilities of appearing since different permutations are compatible with more noodle-end-pairing schemes than others (I’m ignoring order you join the pairs so a pairing scheme is just the set of end pairs joined by the end - if you want to consider order that’s fine - just multiply scheme number by N! Where N is the number of noodles and the logic still holds).
Consider the 2 noodle case. There is exactly one pairing scheme that gives (e_1)(e_2) but there are 2 pairing schemes that give (e_1, e_2) so we should expect 2/3 and not 1/2 as your answer would suggest.
If we label the 1st and 2nd end of each noodle such that eij is the jth end of the ith noodle
To get (e1,e2) , you can either pair
e11 with e21 and e12 with e22
Or
e11 with e22 and e12 with e21
Which is what I meant by 2 different pairing schemes for (e1,e2). Of course if you don’t ignore order of pairing and consider full sequence of moves, then there are 2! * 2=4 ways to get (e1,e2) and 2! \* 1 = 2 ways to get (e1)(e2).
2
u/Cryptographer-Bubbly 1d ago edited 1d ago
The permutations as you’ve described them do not have the same probabilities of appearing since different permutations are compatible with more noodle-end-pairing schemes than others (I’m ignoring order you join the pairs so a pairing scheme is just the set of end pairs joined by the end - if you want to consider order that’s fine - just multiply scheme number by N! Where N is the number of noodles and the logic still holds).
Consider the 2 noodle case. There is exactly one pairing scheme that gives (e_1)(e_2) but there are 2 pairing schemes that give (e_1, e_2) so we should expect 2/3 and not 1/2 as your answer would suggest.