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A spinoff to a classic probability question:

You have 100 noodles in your soup bowl. Being blindfolded, you are told to take two ends of some noodles (each end of any noodle has the same probability of being chosen) in your bowl and connect them. You continue until there are no free ends. What's the expected length (in noodles) of the longest noodle loop?

The original question asks for the expected number of loops.

Simulation for my question predicts the answer is around 75, but I have no idea where to take this.

APerson
  • 85
  • You could calculate it recursively. Start with the case of two noodles of length $l_1, l_2$, what is the expected length of the longest final noodle then? Now suppose you know the answer for $k$ noodles of lengths $l_1, \dots l_k$, what is the answer for $k+1$ noodles? Work back to $100$ noodles of length $1$. – Paul Sinclair Jun 27 '23 at 14:45
  • Turns out that idea gets complicated very quickly. Sorting the lengths to be increasing, for $k = 3$, you get different answers when $l_1 + l_2 < l_3$ than when it is not. And as $k$ gets higher, cases diverge any time a noodle is longer than the sum of all shorter noodles. And to do the full calculation, you would have to keep track of all of them. – Paul Sinclair Jun 27 '23 at 15:30

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