Originally posted by wolfgang59Somehow I doubt it. If you can prove that more than 1 answer is correct, and none of the answers are equivalent to each other, then there is a flaw in the reasoning. Most likely, some operation used to get to the answer is undefined for a particular operation. The most famous example of this is proving that 1 = 0 by sneakily dividing by 0. Of course, this type of thing happens a lot in problems like this one that involve infinity.
Before everyone gets all het-up over this beware of mtthw's warning!
I vaguelly remember a similar problem at school (a long time ago!) when the Maths teacher proved three different answers!
No tricks. All were valid!
Originally posted by wolfgang59Well, let's hear 'em then. 😕
ALL WERE VALID.
Its because the question doesn't define precisely what RANDOM means.
mtthw had the idea first!
"Random" means there is no discernible bias or pattern to the list of chords chosen during the experiment, such that it is impossible to predict with certainty which chord will be chosen next based on the previous choices. However, this does not apply to the method of construction of the sample space unless you're using a Monte Carlo method (which is statistical and not analytical in nature).
Speaking of which, I revised my Excel spreadsheet. Originally, I had generated an "x" between -1 and 1, and then generated the corresponding "+/-y" so that the point was guaranteed to lie on the circle. I did the same thing for the other point in the pair, and then took the distance between the two. If the answer was greater than SQRT(3), then the chord fulfilled the criteria and got a "1" score. If not, it got a "0". Then I added up the 1's and divided by the number of trials to get the probability. Unfortunately, this method tended to leave holes near x=1 and x=-1, so it wasn't a very good representation. However, the probability after several thousand trials ended up being around 36%, +/- about 2%.
I just revised it so that now I generate random angles between 0 and 2*pi, plot the points where they land on the circle, and find the distance between the pairs as before. This time, the answer comes out to 33% +/- about 1%, which is a much tighter result. Also, upon graphing the pairs I didn't notice any holes so I think the distribution is pretty good.
Originally posted by PBE6Why does every other post have a finite weighting? There are an inifinite number of chords passing through any point in the interior of the circle.
I agree you can normally leave out any single point without affecting the final answer in these types of questions, but you can't leave out a singularity with infinite weighting while every other point has finite weighting!
Originally posted by PBE6I think this supports my point - you've chosen random points, but you've not used the same probability distribution, so you end up with a different answer. I don't fancy trying to work out the analytical solution to your first method though!
Speaking of which, I revised my Excel spreadsheet. Originally, I had generated an "x" between -1 and 1, and then generated the corresponding "+/-y" so that the point was guaranteed to lie on the circle. I did the same thing for the other point in the pair, and then took the distance between the two. If the answer was greater than SQRT(3), then the chord fulf ...[text shortened]... phing the pairs I didn't notice any holes so I think the distribution is pretty good.
Having thought some more about it, I do think that the most intuitive answer is definitely 1/3.
Originally posted by zzywChoose a random radius. Rotate the triangle so that one of it's sides is perpendicular to the radius. Now pick a point on the radius to be the mid-point of the chord. This leads to conclusion p=1/2.
One could also make a case for p=1/2, but I'm afraid my English isn't good enough to explain it well🙁
But I'm not saying too much as I think I've run into this puzzle before so it's like cheating.
Originally posted by mtthwTrue, but each "point" in the interior of the circle, apart from the exact center, corresponds to the mid point of one and only one chord. Pick any point, roate the circle so that if the centre of the circle is the origin then the point lies on the y-axis. then if the point you chose has position (0, n) the line x=n is the only line that has mid point (0, n). (Note that we are assuming the line stops and starts at the ends of the circle.)
Why does every other post have a finite weighting? There are an inifinite number of chords passing through any point in the interior of the circle.
Originally posted by luskinI'm slightly confused by what you mean here. Name the triangle ABC. without loss of generality, take the radius that is perpendicular to BC and so goes through the point A. But now what? Taking the line with midpoint A we meerly have a point of length 0.
Choose a random radius. Rotate the triangle so that one of it's sides is perpendicular to the radius. Now pick a point on the radius to be the mid-point of the chord. This leads to conclusion p=1/2.
Originally posted by geniusProbably the problem should say "randomly chosen chord that isn't a diameter", but I'm not sure it makes nuch difference.
ah-I was cut off. You have calculated the probability of chords of length greater than or equal to the length of the side of the triangle and strictly less than the diameter of the circle.
Originally posted by mtthwYes, the new probability distribution I used doesn't give me any holes! I just checked the average values of all my points, and they are all between -0.01 and +0.01 (i.e. clustered around and very close to 0) with a Gaussian distribution between -1 and 1, so I'm quite confident that these numbers are sufficiently random and unbiased.
I think this supports my point - you've chosen random points, but you've not used the same probability distribution, so you end up with a different answer. I don't fancy trying to work out the analytical solution to your first method though!
Having thought some more about it, I do think that the most intuitive answer is definitely 1/3.
This is good, because the numerical analysis generates random chords in a different way than I generated the sample space of all possible chords in my analytical solution. My next task will be to try generating random chords by placing a point within the circle and then determining the chord length, and seeing if the distribution is Gaussian as well.
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Originally posted by mtthwTrue, but except for a mid-point at the centre of the circle, a chord is completely defined by its mid-point.
Why does every other post have a finite weighting? There are an inifinite number of chords passing through any point in the interior of the circle.
EDIT: Oops, genius already pointed that out. 😕
Originally posted by mtthwArgh!! I think you're right. I just tried using a random midpoint selected 2 ways (one with polar variables and one with Cartesian variables), and got different answers for each. The polar one gave me 0.5 and the Cartesian one gave me 0.25. As you say, "uniform randomness" needs some explanation.
I think this supports my point - you've chosen random points, but you've not used the same probability distribution, so you end up with a different answer. I don't fancy trying to work out the analytical solution to your first method though!
Having thought some more about it, I do think that the most intuitive answer is definitely 1/3.
Apparently this is a well-known puzzle (unknown by me until now 😕) called Bertrand's Paradox. I found a link with some neat Java applets that demonstrate how the probability distribution changes depending on the method used to generate random points:
http://www.cut-the-knot.org/bertrand.shtml
I still feel that the sweep method generates the proper sample space (the analytical solution for the sample space, not randomly generated) because it generates every possible chord drawn from a fixed point on the circumference without double counting, and every rotation of this construction is symmetrical so they do not impact the weighting. The mid-point construction rests on the "fact" that a mid-point defines a chord, which is true except when the mid-point coincides with the centre of the circle (as discussed before).
Of course, I don't have any justification for my feeling outlined above, so I can't even pretend it's conclusive. Is there anyone with an advanced mathematics background who can help us sort out whether some of these random chord generators are right and some are wrong, or whether I'll just have to accept that there is no reason to disregard any particular type of random distribution in this solution?