24
$\begingroup$

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?

$\endgroup$
6
  • 2
    $\begingroup$ Wow, nice question. Again this shows probability, which is a model and not the real world, is not intuitive, haha $\endgroup$
    – justhalf
    Nov 3 '21 at 4:44
  • 1
    $\begingroup$ I don't follow the relevance of Bob's comment. Since he chose his cuts randomly, why does it matter whether he saw the red mark or not? His statement that "my two random cuts didn't have anything to do with her red mark" would be true regardless of whether or not he saw the red mark. $\endgroup$
    – JBentley
    Nov 3 '21 at 10:35
  • 3
    $\begingroup$ @JBentley Well, one reason is that I tried to make their statements of similar flavor and of similar length ... You may call this a "decorative part" of the puzzle. $\endgroup$
    – WhatsUp
    Nov 3 '21 at 13:03
  • $\begingroup$ Filtering out what's relevant is an important aspect of unravelling the puzzle. $\endgroup$ Nov 4 '21 at 15:35
  • 1
    $\begingroup$ @FirstNameLastName I'm totally fine with your new puzzle as it looks different enough. I definitely don't consider it plagiarism. $\endgroup$
    – WhatsUp
    Nov 29 '21 at 10:01
34
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ A clarification: So you say that (rot13) Obo'f fragrapr ortvaavat jvgu "Gurersber" vf gehr, ohg gur bar ortvaavat jvgu "Juvpurire" vf snyfr. Right? $\endgroup$ Nov 3 '21 at 8:45
  • $\begingroup$ While I agree with the logic here, I'd say that while [the correct person]'s logic looks fine, [the correct person]'s explanation is quite obtuse in that it has nothing to do with [much of story] (yes, this was a distractor, I know). Perhaps a more interesting question may have been "which of the two was closer to the correct answer?" $\endgroup$
    – tmpearce
    Nov 4 '21 at 2:48
  • 1
    $\begingroup$ @GeorgeMenoutis Right. $\endgroup$
    – Gareth McCaughan
    Nov 5 '21 at 1:26
12
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
-6
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Alice's mark is specified to be placed uniformly at random in the question - we absolutely can assume that the three cuts are independently drawn from a uniform distribution. (Not that this, in fact, matters in the end - it's possible to conclude the same result knowing only that Alice's mark is independent from Bob's cuts) $\endgroup$ Nov 2 '21 at 18:39
  • 1
    $\begingroup$ The location of Alice's cut is irrelevant. Bob's two cuts are uniformly distributed around the circle, which are still uniformly distributed with respect to any particular point defined as the first cut, no matter where it is or how it's chosen. Suppose Alice secretly marks her cut, and then Bob chooses where to put his two - nothing about the situation changes if Alice then reveals that she had actually marked a different point. By saying the cuts aren't uniformly distributed, you're suggesting that knowing where Alice's cut is allows you to guess where Bob's are, or vice versa. $\endgroup$ Nov 2 '21 at 19:27
  • $\begingroup$ I agree with this answer. You have two independent uniformly distributed sets. Neither affects the other. Therefore, since Bob's cuts are all that are are apparent in the end, we take only those cuts and the distribution associated with them. The expected length of each is 50cm, and the red mark (along with Alice's cut) may be on either equally. $\endgroup$ Nov 2 '21 at 23:39
  • 4
    $\begingroup$ @gregsdennis The expected length of either piece is indeed 50cm, but one piece of rope will almost surely be longer than the other (Bob has 0 probability to cut the rope exactly in half). The red mark is equally likely to be at any point on the rope, so it's more likely to be on the longer piece of rope. Since the red mark is usually on the longer piece of rope, it cannot have the same expected length as the piece without the mark. $\endgroup$ Nov 3 '21 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.