Spoiler for Here are the details of my solution
Consider N strands of spaghetti. Some of these strands may comprise original strands tied end-to-end. Now, it is easy to see that
there are N ways to pick a pair of strand ends which, when tied together would cause a loop to form. Since there are 2NC2 ways
to pick a pair of ends, it follows that the probability of randomly picking ends which would result in a loop is N/2NC2 which
is equal to 1/(2N-1). So, with probability 1/(2N-1), we will form a loop AND decrease the number of strands by 1. With probability 1-(1/(2N-1)),
we will not form a loop but we will decrease the number of strands by 1. Let E(N) be the expected number of loops formed in this way.
Then, E(N) = (1/(2N-1))*(1+E(N-1)) + (1-(1/(2N-1)))*E(N-1) = 1/(2N-1) + E(N-1). Using this recursive formula, we can telescope out to
E(N) = 1/(2N-1) + 1/(2N-3) + ... + E(1). But E(1)=1. So, E(N) = 1/(2N-1) + 1/(2N-3) + ... + 1/1. This says that the expected number of
loops formed is the sum of reciprocals of all odd integers from 1 to 2N-1.
there are N ways to pick a pair of strand ends which, when tied together would cause a loop to form. Since there are 2NC2 ways
to pick a pair of ends, it follows that the probability of randomly picking ends which would result in a loop is N/2NC2 which
is equal to 1/(2N-1). So, with probability 1/(2N-1), we will form a loop AND decrease the number of strands by 1. With probability 1-(1/(2N-1)),
we will not form a loop but we will decrease the number of strands by 1. Let E(N) be the expected number of loops formed in this way.
Then, E(N) = (1/(2N-1))*(1+E(N-1)) + (1-(1/(2N-1)))*E(N-1) = 1/(2N-1) + E(N-1). Using this recursive formula, we can telescope out to
E(N) = 1/(2N-1) + 1/(2N-3) + ... + E(1). But E(1)=1. So, E(N) = 1/(2N-1) + 1/(2N-3) + ... + 1/1. This says that the expected number of
loops formed is the sum of reciprocals of all odd integers from 1 to 2N-1.





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