Tangled Threads

Solution & Explanation

The problem involves a probabilistic model where we have a bowl containing $N$ strands of spaghetti. Each spaghetti has two ends, resulting in a total of $2N$ ends. The goal is to understand the expected number of complete loops formed when randomly tying two ends together repeatedly until all ends are paired.

Key Concepts:

1. Random Pairing: Each time you pick an end and pair it with another, two possibilities arise:

   • You form a loop if you tie an end to its corresponding end.
   • You extend a spaghetti if you tie an end to a different spaghetti's end.

2. Probability of Forming a Loop:

   • For the first end picked, there are $2N - 1$ other ends to choose from.
   • The probability that the chosen end forms a loop is $\frac{1}{2N - 1}$.

3. Recursive Nature:

   • After each pair is formed, the problem reduces in size by one spaghetti, i.e., $N-1$ spaghettis and $2(N-1)$ ends remain.
   • The probability of forming a loop in subsequent trials follows a similar pattern: $\frac{1}{2(N-1) - 1}$, $\frac{1}{2(N-2) - 1}$, etc.

Expected Number of Loops:

The expected number of loops, $E[N_{\text{loops}}]$, can be determined by summing the probabilities of forming a loop at each stage:

$E[N_{\text{loops}}] = \frac{1}{2N-1} + \frac{1}{2N-3} + \frac{1}{2N-5} + \ldots + \frac{1}{3} + 1$

This is a telescoping series where each term represents the probability of forming a loop at each stage of the process.

Detailed Calculation:

• First Trial:

  • Probability of forming a loop: $\frac{1}{2N-1}$

• Second Trial:

  • After the first loop or extension, $N-1$ spaghettis remain.
  • Probability of forming a loop: $\frac{1}{2(N-1)-1}$

• Continue this process until all ends are paired.

Example Calculation:

For $N = 3$:

• Total Ends: 6
• Expected Loops: $E[3_{\text{loops}}] = \frac{1}{5} + \frac{1}{3} + 1 \approx 1.733$

Conclusion:

The expected number of loops formed in the bowl is the sum of the probabilities of forming a loop at each stage, which is mathematically represented as a series of reciprocals of odd numbers starting from $\frac{1}{2N-1}$ to 1. This solution leverages the symmetry and probabilistic nature of the problem, providing a clear formula to compute the expected value for any given $N$.