BMAD Posted November 24, 2015 Report Share Posted November 24, 2015 Choose, at random, three points on the circle x^{2}+y^{2}= 1. Interpret them as cuts that divide the circle into three arcs. Compute the expected length of the arc that contains the point (1, 0). Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted November 25, 2015 Report Share Posted November 25, 2015 Choose, at random, three points on the circle x^{2}+y^{2}= 1. Interpret them as cuts that divide the circle into three arcs. Compute the expected length of the arc that contains the point (1, 0).First thoughtsFirst, agree that "random" is satisfied by the following procedure. Find two random real numbers on (0, 1]. Append "0" to them, multiply all three numbers by 2pi and call them a, b and c. Mark the points r=1 and theta={a b c} on the plane. Agree that the points are randomly spaced on the circle x^{2}+y^{2}= 1. Now spin the circle as if it were a wheel, fastened at the origin, and with the points attached. When it stops (from some sort of non-mathematical friction) let's agree that the three points, in their new locations, satisfy a random choice of three points on the unit circle. Now it is clear why the answer to the puzzle is not the intuitive 2pi/3. The odds favor a longer arc. Even though 2pi/3 is the expected length of an arc chosen at random from the three arcs, the point (1, 0) does not make a random choice. It makes a biased choice; the bias is precisely the length of the arc. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted November 25, 2015 Author Report Share Posted November 25, 2015 Hmmm i am not sold that the answer Isnt 2pi/3 Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted November 25, 2015 Report Share Posted November 25, 2015 What ... my description of selecting random points does not overwhelmingly convince you? How about this.Instead of making the cutting points random and (1,0) fixed, assign the cutting points. and let the point in question be random. Let the cutting points subtend arcs of pi/3. 2p/3 and 3pi/3. Then the probability that the arcs contain a randomly chosen point are respectively 1/6, 2/6 and 3/6. The weighted outcomes are then (1/6)(pi/3) + (2/6)(2pi/3) + (3/6)(3pi/3) = pi(14/18), which is bigger than pi(12/18) = 2pi/3. This one example shows that when one arc is larger than the other two, its probability of containing the point (any random point) is larger than (1/3) so it contributes a disproportionate amount to the weighted average (expectation) of the arc lengths.. This happens in every case that one arc length is larger than the others, and those constitute 100% of the cases. (The case with equal arc lengths has probability 0.) There's probably a way to average over all possible sets of arc lengths. And it will take that kind of calculation to gives the answer. Someone else may do that calculation. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted November 25, 2015 Author Report Share Posted November 25, 2015 Where is my flaw...*contains what i believe to be the solution*Let L1, L2, and L3 be the lengths of those three arcs. We have L1 + L2 + L3 = 2π. Onthe other hand the expected value of several random variables is additive:E[L1 + L2 + L3] = E[L1] + E[L2] + E[L3] .By symmetry E[L1] = E[L2] = E[L3], and the sum must be 2π, hence each expectedvalue is 2π/3. So, this is the answer, the expected value of the arc containing the point(1, 0) is 2π/3 Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted November 25, 2015 Report Share Posted November 25, 2015 This is the point I would question.By symmetry E[L1] = E[L2] = E[L3]I would say there is symmetry only if L1 = L2 = L3 which happens with probability zero. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted November 28, 2015 Report Share Posted November 28, 2015 Here's another try at distinguishing our ways of looking at this problem. Spoiler Analysis 1. Pick three points at random. Color the arc from point 1 to point 2 red. From point 2 to point 3 green and from point 3 to point 1 blue. Do this a million times. What is the expected length of the red arc? Clearly it's 2π/3. Of the green arc? Again, 2π/3. And also 2π/3 for the blue arc. The point (1,0) will be colored red, green or blue. Whichever color it receives, the arc that colors it will therefore have an expected length of 2π/3. Analysis 2. Pick three points at random and color their arcs red green and blue, but this time color the longest arc red and the shortest arc blue. Now ask: what is the probability that (1,0) is red? Well, because red is the longest of the three arcs, it has the greatest probability. (1,0) is more likely to be red than green and more likely to be green than blue. The probabilities are simply p(red) = L(red)/2π, p(green) = L(green)/2π and p(blue) = L(blue)/2π. Now ask: if we do this a million times, what will be the expected color of (1,0)? Will it be a neutral gray (equal amounts of RGB) or will it have a reddish tinge? In Analysis 1 color is not related to length. The flaw comes when we say "whichever color it receives" and think that we're saying "whichever arc (long or short) it lands on," as if all cases are equally likely. They're not. After many trials, colors will occur in equal numbers, but lengths will not. And length is what we are concerned with. Analysis 2 puts length into the picture. (1,0) can still get any of the three colors, but not with equal likelihood. Because its arc is the longest, red is the favored color -- just as the largest sector on a Wheel of Fortune gives the favored outcome. For the same reason, (1,0)'s bias toward red means the expected length of its arc will be greater than the average 2π/3. It's given by E[L of arc that covers (1,0)] = p_{1}L_{1} + p_{2}L_{2 }+ p_{3}L_{3} = p(red)xL(red) + p(green)xL(green) + p(blue)xL(blue). Which is easy to simulate, and the result is π. Quote Link to comment Share on other sites More sharing options...

0 bubbled Posted November 29, 2015 Report Share Posted November 29, 2015 Here's another try at distinguishing our ways of looking at this problem.Hidden Content I was close to answering, but didn't understand my result yet:I picked three random numbers between 0, 1. I then sorted them from smallest to largest and calculated (1 - largest + smallest). On average, I was getting .5. Translating that to picking three random points between 0, 2pi would thus arrive at pi as the average length of the arc you are looking for. I guess the easiest way to look at it is if you were to pick three random numbers between 0, 2pi on a number line (not a circle), those three numbers would create four sections, and each section would average pi / 2. By picking a point, any point, you are essentially getting two of those sections, on average.Still don't feel like I totally "get" it. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted November 29, 2015 Report Share Posted November 29, 2015 If you pick three numbers on (0,1), sort them smallest to largest, and call them respectively n_{1, }n_{2} and n_{3}, you get <n_{1}> = .25, <n_{2}> = .5 and <n_{3}> = .75. That makes <1 - n_{3} + n_{1}> = .5, which you can see is just the sum of first and fourth average segment lengths.|==========|----------|----------|==========|0 <n1> <n2> <n3> 1If you join the endpoints, you join those segments and get 1/2 for its average length. I agree that this doesn't apply to the conditions of the OP.I took the view of a spinning wheel, divided randomly into three segments, that stops with one of the segments on the top point. OP asks for the expected length of that top segment. It's convenient to think of the wheel having unit circumference. Then the segment lengths L_{i} (along the circumference) are exactly the probabilities p_{i} for that segment to end on top. The expected length E[L_{top}] of the top segment is the sum of the lengths L_{i} weighted by the probability p_{i} of that segment being on top. Thus, E[L_{top}] = p_{1}L_{1} + p_{2}L_{2} + p_{3}L_{3} = L_{1}^{2} + L_{2}^{2} + L_{3}^{2}.Note that if all the segment probabilities are 1/3, this is just the average (L_{1} + L_{2} + L_{3})/3.This is what BMAD is taking to be the answer.Without loss of generality we can pick 0 as a segment boundary. Pick two more numbers n_{1} and n_{2} on (0,1) for the others. Sort them, and get L_{1} = n_{1}, L_{2} = n_{2}-n_{1} and L_{3} = 1-n_{2}. Do this a large number of times, plug those values into the equation above, and you get E[L_{top}] = 1/2. Change back to a unit radius circle by multiplying by 2π. That gives the final answer as π.The difference is to weight the segment lengths before taking the average. Quote Link to comment Share on other sites More sharing options...

1 markweitzman Posted November 29, 2015 Report Share Posted November 29, 2015 (edited) First the problem should be clarified to explicitly make clear what three points at random on a circle mean - otherwise we start to get into trouble with Bertrand's paradox - https://en.wikipedia.org/wiki/Bertrand_paradox_(probability). But assuming the simple definition that the points are distributed uniformly according to equal arc length (or equal angles) in a range from zero to 2*pi, I will give an analytic solution, that is probably way harder than necessary for this problem but very general and useful if the probability distribution is different from uniform. So starting from the point (1,0) let x,y,z be the value of the cut points arc length measured counterclockwise and for simplicity for now use a range of 0-1 instead of 0-2*Pi (afterwards we can scale up the answer by multiplying by 2pi). So essentially we have chosen three points in the 0-1 interval uniformly, this can be visualized as choosing three points from a cube of side 1 uniformly. There are six possible orderings for the values of the three points, so the cube is divided into six regions, but by simple symmetry arguments one can show that the value of integrating the arc length containing the point (1,0) on the circle is the same for all regions of the cube. So assume x<y<z this corresponds to the region of the cube 0<x<1, x<y<1, y<z<1. The arc length of the point containing (1,0) on the circle is given by 1+x-z (easy to see this as total length of circle 1 minus middle interval z-x. So integrating 1+ x - z over the region of the cube given by the prior intervals gives the average arc length corresponding to the x<y<z ordering. The evaluation of the multiple integral is routine, and the result is 1/12. Multiplying by 6 to account for all the orderings, and by 2*pi to rescale as indicated above leads to an answer of pi. First the problem should be clarified to explicitly make clear what three points at random on a circle mean - otherwise we start to get into trouble with Bertrand's paradox - https://en.wikipedia.org/wiki/Bertrand_paradox_(probability). But assuming the simple definition that the points are distributed uniformly according to equal arc length (or equal angles) in a range from zero to 2*pi, I will give an analytic solution, that is probably way harder than necessary for this problem but very general and useful if the probability distribution is different from uniform. So starting from the point (1,0) let x,y,z be the value of the cut points arc length measured counterclockwise and for simplicity for now use a range of 0-1 instead of 0-2*Pi (afterwards we can scale up the answer by multiplying by 2pi). So essentially we have chosen three points in the 0-1 interval uniformly, this can be visualized as choosing three points from a cube of side 1 uniformly. There are six possible orderings for the values of the three points, so the cube is divided into six regions, but by simple symmetry arguments one can show that the value of integrating the arc length containing the point (1,0) on the circle is the same for all regions of the cube. So assume x<y<z this corresponds to the region of the cube 0<x<1, x<y<1, y<z<1. The arc length of the point containing (1,0) on the circle is given by 1+x-z (easy to see this as total length of circle 1 minus middle interval z-x see image below). So integrating 1+ x - z over the region of the cube given by the prior intervals gives the average arc length corresponding to the x<y<z ordering. The evaluation of the multiple integral is routine, and the result is 1/12. Multiplying by 6 to account for all the orderings, and by 2*pi to rescale as indicated above leads to an answer of pi. Edited November 29, 2015 by markweitzman Quote Link to comment Share on other sites More sharing options...

0 bubbled Posted November 30, 2015 Report Share Posted November 30, 2015 Hidden ContentThe more I've thought about my solution, the more I think it does apply to OP. I really like your approach too, but I don't think it's a coincidence my solution also arrived at half the total arc. Since the three cuts are randomly chosen, it doesn't matter whether we pick a fixed point and then pick three cuts, or if we pick the random cuts first, and then pick a random point. Either way, once we have the fixed point and the three cuts, we can unroll the arc and do the calculation I suggest. IMO, the OP cleverly disguises that, indeed, there ARE 4 random arcs and on average they are 25% each. And the fixed point to one random point and the fixed point to another of the random points will capture 50%.I like your idea of spinning the circle with three cuts and realizing that the largest cut, on average, figures to contain the point in question. But seeing that the fixed point carries the same weight as the three cuts, has value as well. Sometimes, there are many ways to skin a cat. I'm sure your solution generalizes better and might be "safer" here, but I think I got it as well. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted December 1, 2015 Report Share Posted December 1, 2015 Hidden ContentHidden Content Hidden ContentHidden Content Expanding on my post of Sunday at 8:09 AM:If you pick three numbers on [0,1), sort them smallest to largest, and call them respectively n_{1, }n_{2} and n_{3}, you get <n_{1}> = .25, <n_{2}> = .5 and <n_{3}> = .75. That makes <1 - n_{3} + n_{1}> = .5, which you can see is just the sum of first and fourth average segment lengths.|==========|----------|----------|==========|0 <n1> <n2> <n3> 1If you join the endpoints, you join those segments and get 1/2 for its average length.After some thought IMO this does give the correct answer to the OP. It works because in picking n_{1, }n_{2} and n_{3} the interval [0,1) was used. That provided a frame of reference that identifies 0 as a special point and automatically put the point (x=1,y=0) onto the circle. Which is OK as it turns out, since (x=1,y=0) fortuitously is in the question. Now, that may all have been in mind, but from the comment that things were not initially clear, fortune may simply have shined.Blathering on,Selecting random points from [0,1) does not produce equal lengths when the endpoints are joined. The endpoints (one of them) are involved in the selection. Joining the ends removes that point. This is why I picked only two points from [0,1): that interval automatically uses 0 as one of the points. If that's not immediately clear, note that <n1> is equidistant from 0 and <n2>, just as <n2> is equidistant from <n1> and <n3>. So ... before a question has even been asked, (x=1, y=0) has been placed on the circle. The "wheel has to be spun" as the final step of random point selection.That is,If the OP had asked what is the expected length of the arc that covers the point (x=cos(100^{o}), y=sin(100^{o})) the answer remains 1/2. But if points are still chosen from [0,1), the interval of interest becomes [<n1>, <n2>) and the above answer naively changes to 1/4. |----------|==========|----------|----------|0 <n1> <n2> <n3> 1That can be remedied (instead of spinning the wheel) by choosing points from [z, 1+z) where z is the new point in question.But IMO this approach adds difficulty to other investigations, like BMAD's second puzzle: finding the probability of a point lying on the longest arc. Quote Link to comment Share on other sites More sharing options...

0 bubbled Posted December 3, 2015 Report Share Posted December 3, 2015 Hidden ContentHidden Content Hidden ContentHidden Content Expanding on my post of Sunday at 8:09 AM:Hidden ContentI can't disagree with any of this. I was completely wrong and the fact that my answer matched with the correct answer blinded me to the fact that I was introducing an extra point into the problem. I now realize that you have to be very careful with how exactly you apply random conditions to problems like these. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted December 3, 2015 Report Share Posted December 3, 2015 I spent a long time on this puzzle. I liked it a lot, (thanks, @BMAD), and I like probability puzzles in general. Precisely for the reason @bubbled told us. Quote Link to comment Share on other sites More sharing options...

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