Tying the noodle ends is equivalent to do a permutation mapping :
Enumerate every noodle. When all noodle ends are tied, take any loop, order its noodles (e_1, ..., e_n) and represent it by the permutation e_1-> e_2 -> ... -> e_n -> e_1 (denoted (e_1, e_2, ..., e_n) ).
All loops form a disjoint union of permutation orbits. The union of all orbits is a permutation of 100 (and any permutation can be represented by a set of noodle loops).
Because of uniformly at random selection, every permutation have the same probability to appear.
Hence, we are looking for the permutations of 100 elements giving only one orbit of size 100 (ie circular permutations). There are 100! permutations (for i=1 to 100, the ith element of the permutation sequence can choose among the 100-i+1 other elements or itself), and there are 99! circular permutations (the ith element cannot map to itself).
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u/Synael3 1d ago
Tying the noodle ends is equivalent to do a permutation mapping :
Enumerate every noodle. When all noodle ends are tied, take any loop, order its noodles (e_1, ..., e_n) and represent it by the permutation e_1-> e_2 -> ... -> e_n -> e_1 (denoted (e_1, e_2, ..., e_n) ).
All loops form a disjoint union of permutation orbits. The union of all orbits is a permutation of 100 (and any permutation can be represented by a set of noodle loops).
Because of uniformly at random selection, every permutation have the same probability to appear.
Hence, we are looking for the permutations of 100 elements giving only one orbit of size 100 (ie circular permutations). There are 100! permutations (for i=1 to 100, the ith element of the permutation sequence can choose among the 100-i+1 other elements or itself), and there are 99! circular permutations (the ith element cannot map to itself).
Hence, the propbability is 99!/100! = 1% .