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ThudnBlunder
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Random Chord
« on: Aug 22nd, 2003, 9:23pm » |
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A chord is randomly drawn on a unit circle.
What is the probablity that the chord is longer than the radius of the circle?
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TenaliRaman
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Re: Random Chord
« Reply #1 on: Aug 23rd, 2003, 12:39am » |
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Choose one point A on the circumference.Starting with this point let us divide the circumference into six equal parts.
Now if were to find a second point B on the circumference then such that chord length>radius then the two parts on on either side of A are out of question.So the point B must be chosen from the remaining 4 parts.
So the required probability equals 4/6 or 2/3.
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ThudnBlunder
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Re: Random Chord
« Reply #2 on: Aug 23rd, 2003, 4:43am » |
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2/3? Hmm...consider the midpoints of all possible chords.
If the midpoints fall within a concentric circle of radius sqrt(3)/2, the chords will be longer than the radius.
The required probabiity is 'obviously' the ratio of the areas of the two circles:
pi[sqrt(3)/2]^2 = (3/4)pi, and pi(1^2) = pi.
Therefore, the probability is 3/4.
Or is it?
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| « Last Edit: Aug 23rd, 2003, 9:14pm by ThudnBlunder » |
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BNC
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Re: Random Chord
« Reply #3 on: Aug 23rd, 2003, 5:15am » |
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Is it a variant of Bertrand's paradox?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Sir Col
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Re: Random Chord
« Reply #4 on: Aug 23rd, 2003, 5:39am » |
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It took me a bit to figure out the flaw, but I think I've got it...
T&B, it is not possible for the midpoints to occupy the entire concentric circle; the maximum distance from a point on the circumference is 1 unit. I'm sure it would be possible to consider the locus of the midpoints, but taking into account the sweeping arc at the extremity against the concentric overlap would make it very difficult to pursue.
I believe that TenaliRaman's method is both elegant and correct.
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| « Last Edit: Aug 23rd, 2003, 5:42am by Sir Col » |
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Icarus
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Re: Random Chord
« Reply #5 on: Aug 23rd, 2003, 6:19pm » |
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Sir Col - It sounds to me like you are considering one end of the chord to be fixed. But this is not what the problem says. Every point inside a circle other than the center is the mid point of exactly one chord. That chord will be greater than the radius iff its midpoint lies within the circle T&B describes.
So the entire inner circle is the locus of midpoints. The reason that the analysis that T&B offers disagrees with TenaliRaman's solution is something similar, but not that.
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Sir Col
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Re: Random Chord
« Reply #6 on: Aug 24th, 2003, 4:17am » |
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But surely it is quite reasonable to say, "Without loss of generaility, consider a fixed point on the circumference of a circle. By joining this point to the locus of points around the circumference we generate all possible chords in the circle. It follows that 1/3 of those chords will have length less than one radius, hence 2/3 of chords will exceed the radius in length."
Surely I can state W.L.O.G, because of perfect symmetry; we can repeat the method, with the same outcome, from any fixed point on the circumference.
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Icarus
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Re: Random Chord
« Reply #7 on: Aug 24th, 2003, 11:57am » |
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My point was not in objection to TenaliRamen's analysis, but to your claim of a flaw in T&B's.
You said that the midpoints do not occupy all of the inner circle. This is true if you are looking only at chords all sharing a common endpoint that you have fixed. For arbitrary chords, their length exceeds the radius if and on if the midpoint is in the circle, and all points in the circle are midpoints of chords.
While TenaliRamen is justified to choosing an arbitrary "fixed" starting point for his chords, T&B is equally justified in not doing so. And without a fixed endpoint, your claim that the locus of midpoints is not the entire inner circle is false.
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Sir Col
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Re: Random Chord
« Reply #8 on: Aug 24th, 2003, 2:45pm » |
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I think I've found a different answer that seems to work...
By rotational symmetry we can consider a chord which remains perpendicular to the diameter of the circle. If the length of the diameter is 2 units, the chord will exceed the radius (1 unit) for a [sqrt]3 units of the diamter; giving, p=[sqrt]3/2.
The same argument, with the same result, could be applied to a chord which remains perpendicular to the radius of the circle.
What's going on? How can we have three apparently reasonable answers? 
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| « Last Edit: Aug 24th, 2003, 4:33pm by Sir Col » |
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ThudnBlunder
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Re: Random Chord
« Reply #9 on: Aug 24th, 2003, 3:30pm » |
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Quote:| What's going on? How can we have three apparently reasonable answers? |
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As BNC rightly points out, this is (a variation of) Bertrand's paradox.
See http://user.sezampro.yu/~seik/indexeng.htm
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Icarus
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Re: Random Chord
« Reply #10 on: Aug 24th, 2003, 5:57pm » |
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on Aug 24th, 2003, 3:30pm, THUDandBLUNDER wrote:
As BNC rightly points out, this is (a variation of) Bertrand's paradox. |
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Yes - though you should understand that the "paradox" is only apparent. That is, there is no real contradiction - it only defies our "common sense".
In this case, the problem lies here: "A chord is randomly drawn on a unit circle."
How one does this "random" drawing decides which probability you get.
If your random choice was made by choosing two points on the circumference (using a uniform distribution with respect to length) and drawing the chord connecting them, you get TenaliRaman's 2/3 probability.
If your random choice was to choose a point out of the interior of the circle (with uniform distribution with respect to area) and draw the unique (if it isn't the center) chord having the point as midpoint, then you get T&B's 3/4 probability.
If your random choice was to choose a diameter (by any means) and then choose a point on the diameter (by uniform distribution with respect to length), and finally to draw the chord having that point as midpoint, then you get Sir Col's ([sqrt]3)/2 probability.
The reason you get three different answers is because each procedure provides different likelyhoods that certain chords will be picked.
The "common sense" idea that Bertrand's Paradox defies is that there is always a natural meaning to the word "random". When you hear someone say "pick it at random", you expect that this fully defines the probabilities. Usually it does (with the assumed uniform distribution), but there are situation where no uniform means of picking is possible. Several puzzles on this site make use of the fact that there is no uniform way of picking a random integer. Other situations, like this one, have too many uniform distributions: there are several distinct ways to "uniformly" make the selection. And the probabilities defined by each do not agree.
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Sir Col
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Re: Random Chord
« Reply #11 on: Aug 25th, 2003, 4:29am » |
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This is fascinating, and entirely revelatory to me.
Is it to do with the lack of uniformity with respect to the density of points within/along our chosen domain: inner region/circumferemce or diameter?
I know it's equivalent, but what we're saying is that,
(i) by selecting a point inside the circle, and making this the midpoint of a chord, 3/4 of those chords you draw will have a length greater than the radius.
(ii) by joining two points on the circumference, 2/3 of those chords you draw will have a length greater than the radius.
(iii) by picking a point on the diameter and drawing a line perpendicular to form a chord, [sqrt]3/2 of those chords you draw will have a length greater than the radius.
When I talk about, lack of uniformity with respect to density, consider (i), for example; is it that there are more points per square unit closer to the edge of the circle than nearer the centre?
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| « Last Edit: Aug 25th, 2003, 4:29am by Sir Col » |
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Sameer
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Re: Random Chord
« Reply #12 on: Aug 25th, 2003, 6:33am » |
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I actually got the sqrt(3)/2 answer.
Here is how I got it.
Consider a chord of lenght 'r' where r is the radius.
Obviously with the chord and the center of circle, we have an equilateral triangle whose area is sqrt(3)*r^2/2 (Considering the mirror equilateral triangle too)
Now the area of the remaining two sectors is 2*pi*r/3
Now probability can be defined as ratio of this area to the entire circle's area with r -> infinity
giving lim r-> inf (sqrt(3)*r^2/2 + 2*pi*r/3)/r^2
giving sqrt(3)/2
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TenaliRaman
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Re: Random Chord
« Reply #13 on: Aug 25th, 2003, 8:38am » |
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Icarus,
So is it ok if i say that required probability is
P = [(2/3)+(3/4)+([sqrt]3/2)]/3
or have we missed any cases which would prevent us from taking the average at this stage ??
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James Fingas
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Re: Random Chord
« Reply #14 on: Aug 25th, 2003, 9:23am » |
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on Aug 25th, 2003, 8:38am, TenaliRaman wrote:Icarus,
So is it ok if i say that required probability is
P = [(2/3)+(3/4)+([sqrt]3/2)]/3
or have we missed any cases which would prevent us from taking the average at this stage ?? |
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That is the probability of the chord being longer than the radius if we choose equally one of the three ways so far discussed, and then create the chord. This is no more valid than any of the other methods presented so far.
In fact, there are an infinite number of ways to randomly choose a chord through the circle. Here's one that gives an expectation of zero:
Draw a diameter of the circle. On one end of the diameter, draw a line tangent to the circle. Now pick a point randomly on the tangent line, and connect it to the other end of the diameter. Part of this line passes through the circle, forming a chord.
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Sir Col
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Re: Random Chord
« Reply #15 on: Aug 25th, 2003, 10:00am » |
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I said it before, but I'll say it again, although not in the same words: "This is freaky!"
Where has this crazy paradox been all my life?
I suppose an interesting (and I imagine, difficult) question would be, "What is the maximum probability that a random chord can exceed the length of the radius?"
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Sameer
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Re: Random Chord
« Reply #16 on: Aug 25th, 2003, 10:19am » |
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It is freaky!!
I have a thought!! This problem we got three distinct probabilities. Is it possible to have different values or only this three i.e. are there any other methods that we can use to reach another answer.
Another way of writing this would also be that does this problem have three spikes in probability distribution or more or can it be continuous?
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TenaliRaman
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Re: Random Chord
« Reply #17 on: Aug 25th, 2003, 10:32am » |
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James,
i agree i did assume them to be equally likely .... but i thought this measure of averages agrees more with the intuition (and less with mathematics(maybe!)).
Sameer,
it might have a continuous pf.As james suggests, one can devise many ways of forming chords each giving different probabilities of that method giving chords of lengths>radius.
[edit]
As a variation of the question ... is this anything relevent??
say if i had rulers all of lengths > radius but <=diameter.And i at random make chords on the circle.What is the probability that chords are greater than radius of the circle??
[/edit]
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| « Last Edit: Aug 25th, 2003, 10:36am by TenaliRaman » |
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Sameer
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Re: Random Chord
« Reply #18 on: Aug 25th, 2003, 11:44am » |
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I guess then the correct answer would be to figure out the continuous p.f. which would span all the possible probabilities, and this is beyond my scope as i was always weak at probability
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Sir Col
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Re: Random Chord
« Reply #19 on: Aug 25th, 2003, 12:03pm » |
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That is why I asked the question,
on Aug 25th, 2003, 10:00am, Sir Col wrote:| "What is the maximum probability that a random chord can exceed the length of the radius?" |
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So far, the maximum probability is p=[sqrt]3/2 [approx] 0.866. Another interesting question would be, "Is the set of possible probabilities continuous of discrete?"
I suggest that the solution to that last question would answer the first question too.
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Sameer
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Re: Random Chord
« Reply #20 on: Aug 25th, 2003, 12:24pm » |
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My intuition says it is continuous I can't prove it though, have to use my grey cells
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Icarus
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Re: Random Chord
« Reply #21 on: Aug 25th, 2003, 5:12pm » |
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So many comments...
on Aug 25th, 2003, 4:29am, Sir Col wrote:| is it that there are more points per square unit closer to the edge of the circle than nearer the centre? |
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No. Points are equally distributed across the plane. This should be obvious since any portion of the plane can be translated to any other portion with out distorting shape or area.
The problem here is not that the word Random is not well-defined in and of itself. To make its meaning explicit in any given situation requires the definition of a probability function. In many situations, there is a single "natural" choice, so we don't bother to be explicit. Instead this natural pdf is implicitly used. But other situations do not have any natural choice (for example, unbounded sets of numbers). And in situations such as this there are multiple "natural" choices.
There is no reason to assume that because two different processes are both natural and involve uniform distributions, that the probabilities generated by the two must be the same. And this is one example in which they are not. The reason they produce different results is because they choose chords by significantly different means. Some chords are more likely to be picked by one means than by the others. Which one is "correct"? Well, you can give reasonable arguments for all three. Why should one be considered more correct than the other?
So we are left with multiple answers, based on differing assumptions about missing information. It turns out that the question of this puzzle is similar to asking
"The sum of two real numbers is 10. What are the numbers?"
[ The puzzle itself is not such a cheat, since its point (and solution) is that the question is ill-defined. Do not mistake me as saying this is a poor puzzle. It is not! ]
on Aug 25th, 2003, 6:33am, Sameer wrote:Consider a chord of lenght 'r' where r is the radius.
Obviously with the chord and the center of circle, we have an equilateral triangle whose area is sqrt(3)*r^2/2 (Considering the mirror equilateral triangle too)
Now the area of the remaining two sectors is 2*pi*r/3
Now probability can be defined as ratio of this area to the entire circle's area with r -> infinity
giving lim r-> inf (sqrt(3)*r^2/2 + 2*pi*r/3)/r^2
giving sqrt(3)/2
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Sorry, Sameer, but I cannot find any way to justify this as a procedure for calculating the probability. You may have got the same probability as Sir Col, but in this case, it looks more like a "(16/64) = 1/4 by cancelling the 6s" type coincidence.
For starters, though T&B did say "unit circle" in his description, the question itself is stated in a way that is size-invariant. Therefore the fact that you got a probability expression that depends on the radius is a dead giveaway that something is wrong. And there is certainly no reason for letting the radius go to [infty], and then proclaiming the limit is in fact the probability for all radii.
I appreciate your effort, but you need to re-examine this.
on Aug 25th, 2003, 8:38am, TenaliRaman wrote:Icarus,
So is it ok if i say that required probability is
P = [(2/3)+(3/4)+([sqrt]3/2)]/3
or have we missed any cases which would prevent us from taking the average at this stage ?? |
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on Aug 25th, 2003, 9:23am, James Fingas wrote:That is the probability of the chord being longer than the radius if we choose equally one of the three ways so far discussed, and then create the chord. This is no more valid than any of the other methods presented so far.
In fact, there are an infinite number of ways to randomly choose a chord through the circle. Here's one that gives an expectation of zero:
Draw a diameter of the circle. On one end of the diameter, draw a line tangent to the circle. Now pick a point randomly on the tangent line, and connect it to the other end of the diameter. Part of this line passes through the circle, forming a chord. |
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James is correct that these three processes are just three out of infinitely many. Unfortunately, I must take exception to his example. It suffers the opposite problem! "Now pick a point randomly on the tangent line". The tangent line is an unbounded set. There is no natural way to "pick a point randomly"! So I cannot agree with his claim that the probability of getting a chord longer than the radius is zero. It would depend on whatever means one devises for picking that point.
But there are infinitely many ways of picking chords, with probabilities for getting chords longer than the radius ranging from 0 to 1. To how many of these the label "Uniform" could be reasonably applied is another question, and not one readily answerable (or even well-defined). It is possible that these three are the only ones, but it seems unlikely to me.
on Aug 25th, 2003, 10:00am, Sir Col wrote:Where has this crazy paradox been all my life?
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Lurking, waiting for its chance to foul you up!
Quote:| I suppose an interesting (and I imagine, difficult) question would be, "What is the maximum probability that a random chord can exceed the length of the radius?" |
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Not difficult at all, but also not interesting. The answer is 1.
Okay - the remainder of the comments are all on the same subject, so I am going to stop quoting them (with apologies to Sameer for failing to quote his excellent comments in these posts after panning his earlier post).
Towr and I had a long go-around recently over this very subject - pdfs of pdfs, or "Metaprobability" in the Gold or Silver thread. Like here, that puzzle has an ill-defined probability. It is possible there to come up with a "uniform" meta pdf, which towr did. Most of the discussion is over the question of whether or not this was justifiable. (To me it is, but only if you specify that your result is based on the assumption of the meta pdf.) And if it was necessary (I say no).
I do not see any natural way myself of defining a meta probability here. The base processes are too complex. (There are plenty of "unnatural ways", but I frown on such goings on!  )
In the end, my answer to this riddle is: "The question is ill-posed". Until a means of picking the chord is specified, the probability is not defined.
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Re: Random Chord
« Reply #22 on: Aug 26th, 2003, 6:18am » |
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Icarus thanks for pointing out the mistake... I shall dig up my probability book and reproof the problem... Probability has always deluded me 
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