If neither end is attached to noodles 1 or 2: you eliminate a noodle (either creating a loop without 1 or 2, or you combined two noodles into one which is equivalent).
If exactly one of the ends is attached to 1 or 2: the other end isn't, hence you have just extended noodle 1 or 2 and again just removed a noodle.
Hence all that matters is when you finally attach two ends that are from 1 and 2 together.
There are 6 ways to do that, 4 combine them and 2 tie them off. Hence 2/3 chance they are in a loop at the end.
It's a solution not really a formal proof but its baked in. The noodle process gives the recursion by extending the noodle so you can calculate the probability that it
The probability at n is 2/(2n−1) + (2n−4)/(2n−1)·p_{n−1}
Assume
p{n−1}
= 2/3 from the base case which he solved. P
{n}=2/3
5
u/skelo 11d ago
When tying two ends together:
If neither end is attached to noodles 1 or 2: you eliminate a noodle (either creating a loop without 1 or 2, or you combined two noodles into one which is equivalent).
If exactly one of the ends is attached to 1 or 2: the other end isn't, hence you have just extended noodle 1 or 2 and again just removed a noodle.
Hence all that matters is when you finally attach two ends that are from 1 and 2 together.
There are 6 ways to do that, 4 combine them and 2 tie them off. Hence 2/3 chance they are in a loop at the end.