2

Mila has four ropes. She chooses two of the eight loose ends at random (possibly from the same rope) and ties them together, leaving six loose ends. She again chooses two of these six ends at random and joins them, and so on, until there are no loose ends. At this point she has somewhere between one and four loops of rope (inclusive). Find the expected value of the number of loops Mila ends up with.

Should the probability just be $4$ as it doesn't matter in which order we tie the ropes? I really stuck up on this.

WolverineA03
  • 277

1 Answers1

2

As she holds one end of the first rope, there is a $\dfrac{1}{7}$ chance the second end she picks up belongs to the same rope. So after the first knot, she has three lengths of rope remaining that are not loops plus zero or one loop.

Next she takes another end and ties it. There is a $\dfrac{1}{5}$ chance she makes a loop. Regardless of whether or not she made a loop, she has two lengths of unlooped ropes.

The next two she ties has $\dfrac{1}{3}$ chance of making a loop. And finally her last knot will always make a loop.

That's an expectation of $\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{3}+1$ loops.

SlipEternal
  • 10,555
  • 1
  • 17
  • 38