The "Loop of rope" dilemma

Question

It's a "dilemma" that I encountered in real life problems. I believe that it's not new and is quite easy to explain, but may look puzzling at first glance.

Alice and Bob are playing with a loop of rope (i.e. a piece of rope in the shape of a circle) of length $1$ meter. The process goes like this:

Firstly, Alice paints the entire loop blue, and then uses a red pen to draw a mark on one uniformly random point on the loop.

After that, Bob takes a pair of scissors and cuts the loop at two independently uniformly random points.

In the end, the loop is cut into two pieces of ropes, with a red mark on one of the two pieces.

Now they start to argue about the expected length of the piece of rope containing the red mark.

Bob:"What I didn't tell Alice is that I'm actually color-blind, hence I could not see the red mark that she drew. Therefore my two random cuts didn't have anything to do with her red mark, and after the two cuts, each piece should have expected length $\frac 1 2$ meter. Whichever piece her red mark belongs to, the expected length is $\frac 1 2$ meter."

Alice:"What I didn't tell Bob is that, before drawing the red mark, I secretly cut the rope at that point and then glued the cut back. I then drew the red mark on exactly the same point. Bob didn't notice that during the whole process. Therefore if I reveal my cut in the end, then the final status should be equivalent to cutting the loop at three independently uniformly random points, and hence the expected distance between any two of the three cutting points is $\frac 1 3$ meter. This means that, when I don't reveal my cut, the expected length of the piece containing the red mark is $\frac 2 3$ meter."

Who is correct, and where did the other make a mistake?

Answer 1

Alice

's reasoning looks fine to me.

Bob

is correct that the expected length of each piece is 50cm. However, this doesn't imply that the expectation of (length of piece with the red mark) is 50cm, because the longer piece is more likely to have the red mark on it.

Answer 2

Expanding on Gareth's answer:

Map every point on the rope to a coordinate $x = \tfrac{1}{2\pi}\theta$, where $\theta$ is the clockwise angle from Bob's first cut to the specified point. Hence all points have coordinates $\in [0, 1)$, with Bob's first cut appearing at $x_1 = 0$.

It is then readily seen that if Bob's second cut appears at coordinate $x_2$, the red mark coordinate $x_m$ has probability $p_1 = x_2$ of appearing on the segment traveling clockwise from $x_1$ to $x_2$, which has (normalized) length $\ell_1 = x_2$, and probability $p_2 = 1 - x_2$ of appearing on the segment traveling clockwise from $x_2$ to $x_1$, which has length $\ell_2 = 1 - x_2$.

Since $x_2$ is also uniformly distributed on $[0, 1]$, it follows that $$\begin{aligned}E[\ell] & = \int\limits_0^1 p_1\ell_1 + p_2\ell_2\; dx_2 \\ & =\int\limits_0^1 x_2^2 + (1 - x_2)^2\; dx_2 \\ & = \left.\tfrac{1}{3} x_2^3 - \tfrac{1}{3}(1 - x_2)^3\vphantom{\sum\limits_x^y}\right|_0^1 \\ & = \tfrac{2}{3}\end{aligned}$$ Hence Alice's intuition based on symmetry is confirmed by a more standard probabilistic analysis.

Answer 3

Bob is wrong.

Suppose we reverse the order of operations such that Bob cuts the rope first, and then Alice makes her red mark. In this situation, Alice chooses a point uniformly over two pieces of rope, so the likelihood of the mark being on one piece of rope vs the other is directly proportional to its relative length. It should be clear that the bigger the piece of rope, the more likely it is Alice will mark it. Alice's mark is usually on the larger piece of rope, so it cannot have the same expected length as the piece without the mark. The expected length of the marked and unmarked pieces cannot both be 1/2, since the marked piece is usually the larger one.

Note that no matter how Bob chooses to cut the rope (uniformly, into 2:1 or 10:1 or 100:1 length pieces, etc.), so long as it's independent of Alice's mark, Alice is always more likely to mark the longer piece. All Alice sees is two pieces of rope, from which she uniformly chooses one point. She's more likely to mark the longer piece, so the marked piece is likely to be longer.

Answer 4

The main incorrect logic is Alice's. Her cut has no relation to Bob's cuts, since she made her cut first and Bob didn't know anything about her cut. So we cannot assume that the three cuts are uniformly distributed. Bob's two cuts are uniformly distributed. So we will end up with the first 50cm piece of rope, the second 50cm piece of rope with a red mark, splitting that rope into two random lengths adding up to 50cm.