The Setup

You start with 100 noodles, each with 2 loose ends—so a total of 200 ends.

You then repeatedly grab two free ends at random and tie them together. You do this blindly and continue until no free ends remain. By the end, you’ll have formed several closed loops of various lengths.

The Question

What is the expected number of loops you’ll have when all 200 ends have been tied?

Key Insight: Random Pairings and Cycles

This problem is equivalent to the problem of forming a random pairing of 200 elements (the ends). Each such pairing corresponds to a decomposition of a set of 100 edges (or 200 nodes) into disjoint cycles.

Each tied pair reduces the number of free ends by 2, and eventually all 200 ends are paired.

This process results in a random pairing of the 200 endpoints into 100 edges, and the resulting structure can be seen as a collection of cycles (loops of connected noodle segments).

Known Result

This is a classic result in combinatorics: the expected number of loops (cycles) formed when randomly pairing 2n endpoints is:

∑k=1n12k−1≈ln⁡(n)+γ+ln⁡(2) \boxed{\sum_{k=1}^{n} \frac{1}{2k - 1}} \approx \ln(n) + \gamma + \ln(2)

For n=100 n = 100 , the expected number of loops is:

∑k=110012k−1=11+13+15+⋯+1199≈ln⁡(100)+ln⁡(2)+γ≈4.605+0.693+0.577≈5.875 \sum_{k=1}^{100} \frac{1}{2k - 1} = \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{199} \approx \ln(100) + \ln(2) + \gamma \approx 4.605 + 0.693 + 0.577 \approx 5.875

But this is only approximate. The exact expression is:

∑k=110012k−1 \boxed{\sum_{k=1}^{100} \frac{1}{2k - 1}}

Final Answer

∑k=110012k−1≈5.877 \boxed{\sum_{k=1}^{100} \frac{1}{2k - 1} \approx 5.877}

So on average, you will end up with about 5.88 loops after all the ends are tied.

Reference