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Gypsy Chanting Cake
80b883
*Ahem*
"We must first assert ground rules and statements, as to prevent confusion. Each noodle, for all intents and purposes, is either unbroken or treated as unbroken. Second, the creation of a loop is represented in Euclidian space as a singular element with one side and non-concave (between two points in the noodle, there exists a point that is not covered by any matter from said noodle). Third, once two pieces of noodles are tied they will be considered conjoined as part of a potential loop and not an actual loop in the tie. Fourth, at no point may any three noodle ends and/or beginnings tie in the same knot. Finally, chain links between different sets of noodle loops have no effect on this question whatsoever.
There are N noodle beginnings, and N noodle ends. Let us begin by propositioning the fact that this noodle-tying process will always result in at least one loop: If we conjoin any X noodle(s) together, then there will remain exactly N - X noodle beginnings and N - X noodle endings remaining. As we cannot stop while any beginnings or endings remain, we must create at least a single loop, which in the worst-case scenario would consist of every noodle tied from front-to-end and then conjoined with the last pair of beginning and end to form a straight loop.
Let us assume there are 2X noodle ends remaining (it has to be an even number, as each noodle has two ends). You select 1 of these 2X noodle ends. Of any possible noodle end you choose, there are exactly 2X-1 potential noodle ends to tie it to. There are two outcomes: (A) you complete a loop, or (B) you tie two unlooped noodles together. The probability of situation A is exactly 1/(2X-1), as that would require you to select at random the noodle end that is directly connected to the end you originally selected. As A and B are mutually exclusive and the only possible situations, the probability of situation B is therefore 1 - P(A) = (2X-2)/(2X-1). As a direct result, there are now (2X-2) noodle ends remaining. We can now safely assume that any tied noodles can be treated as a single noodle, and any loops will have no impact on the remaining noodle ties. Situation A results in 1 extra noodle loop while situation B results in 0 extra noodle loops.
The answer uses probability and induction: If there is one noodle, then the number of loops that can be created by this process is 1, and therefore the average is 1. If there are N noodles, then the number of loops that are created on average is (Number of noodles created from combining a noodle string with its end * Probability of combining a noodle string with its end + Number of noodles created from combining a noodle string with anything other than its end * Probability of combining a noodle string with anything other than its end)+Y = (1*1/(2N-1)+0*(2N-2)/(2N-1))+Y = (1/(2N-1))+Y, where Y is the number of loops that are created on average for N-1 noodles."
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