There should be infinitely many "solutions" to this if you are allowed to define the sampling method arbitrarily.<p>Intuitively, one would think that random points are uniformly distributed, but that is not maintained in two of these solutions. If you randomly generate two points in [0,2pi) with a uniform distribution then you will quickly get the 1/3 probability empirically or provably by the limit.<p>However, if you are allowed to derive the second point from the first (such as the cross section example) or create some other dependency, then you are given great liberty as to what the resulting "probability" will be. Sure you can even guarantee that every two points on the circle will be present "eventually" but infinity is very large and how you approach it will determine your eventual outcome. Here we could subsample preferred points first and actually derive any specific probability (0,1) and maybe even make it sound reasonable. This behaviour will even be maintained into the limit, but it is certainly not intuitive, at least to me.
I got 1/3, but with a "different" distribution. (I took symmetrized pairs rather than pairs. But I realized after I wrote this comment that folding a square in half preserves relative area so symmetrizing makes no difference.)<p>I started off with a flat torus (two ordered points on the circle) and then quotiented out by (x, y) ~ (y, x) (since the chord corresponding to both is the same) to get a triangle as in the diagram [0].<p>By geometric intuition/trigonometry/etc it's clear that a chord has length < sqrt(3) iff the difference between the angles x and y is less than 2pi/3. The region corresponding to pairs with distance distance less than 2pi/3 is shaded in [0]. One adds it up and sees that the unshaded region is 1/3 of the whole.<p>[0]: <a href="http://i.imgur.com/ow8Dkgd.png" rel="nofollow">http://i.imgur.com/ow8Dkgd.png</a>
This will most likely be helpful as well:<p><a href="https://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29" rel="nofollow">https://en.wikipedia.org/wiki/Bertrand_paradox_%28probabilit...</a>
It appears each approach depends on the way random chords are selected. Each approach yields different probabilities of selecting a chord with a certain property.<p>But my intuition suggests that constructing a proof for <i>all</i> chords should be relatively simple, and indeed 1/3 of them hold the property that they are longer than √3.
One could just as easily make an argument for a ratio of sqrt(3)/2. The chord is selected by randomly picking a length out of the range [0, 2], placing it arbitrarily on the circle, and then rotating it randomly between zero and one revolution[1].<p>Before we can select a random chord, we must first map the set of chords onto the set of real numbers (or, equivalently, tuples of real numbers). This is required in order to meaningfully apply a distribution to them. The problem is that there are multiple ways to do this. Which mapping we select has as much effect on the outcome as the probability distribution.<p>Likewise, if we want to "count" all random chords, we need to do the same thing (map them onto the set of real numbers). We will get different "counts" depending on which mapping we select.<p>[1]- Edit: This is not a very satisfying approach, but it does demonstrate the problem well. Being satisfying isn't a requirement for a mathematical construct.
Years ago I posted a question on CraigsList (<a href="https://forums.craigslist.org/?ID=20211237" rel="nofollow">https://forums.craigslist.org/?ID=20211237</a>) Science and Math discussion board, that I think may be nearly equivalent, and equivalently ambiguous in it's notion of random, to this post. It went on for quite some time, and was even reprised later (<a href="https://forums.craigslist.org/?areaID=4&ID=20432604" rel="nofollow">https://forums.craigslist.org/?areaID=4&ID=20432604</a>) by someone else. Helpful people went back and forth with beautiful 3d renders to illustrate their solutions, though those links are sadly broken now.<p>It is very gratifying to read this post, as I think it explains why we were never able to come to a satisfying answer.
I don't agree with any of these solutions. In my opinion the circle should be positioned in an infinite plane, then random vectors should be drawn on the plane and ignored unless they intersect the circle. Then after making some large amount of intersecting vectors you calculate. I'm guessing that this essentially reduces to the random endpoints method.<p>The other two methods don't make sense at all to me. Random chords is just obviously not a random sampling of possible vectors-- it's essentially the same as sampling a gaussian(x) distribution using an equal sampling in x and ignoring the fact that a gaussian is a PDF and you can't sample that way, you need to invert it into a CDF and take an equal sampling in _y_. Very basic I think...<p>And random midpoint doesn't strike me as random either as it forces the orientations of the vectors to be non-random and systematically smaller than random because of the pi*r^2 area-dependence of concentric circles...<p>But, what do I know.
I came up with a third -- it seemed fairly obvious after some thought. I do not quite get the 1/2 and 1/4 explanations. (Unfortunately these gif analogies do not do such a great job of clarifying.) So I decided to ask good old "Reality".<p><pre><code> N = 1000
SQRT3 = Math.sqrt(3).to_r
def point_on_circle
x = (rand().to_r * 2) - 1
y = Math.sqrt(1 - x**2)
y = -y if rand(2) == 0
return [x,y]
end
tally = 0
N.times do
x1, y1 = point_on_circle
x2, y2 = point_on_circle
d = Math.sqrt( (x2 - x1)**2 + (y2 - y1)**2 )
tally += 1 if d > SQRT3
end
puts "#{tally}/#{N}"
</code></pre>
Oddly, it gives an answer close to a third (yeah me!). But... it is always a bit high at approx 36%.
The 1/4th answer has a typo.<p>"The probability of being hit is the probability of being inside the area of fire divided by the total area"<p>No.<p>The probability of being hit is the probability of being inside the area of fire.<p>The probability of being inside the area of fire is the area of fire divided by the entire area.
You and another incredibly smart human have to chose an answer (of the three) in secret.<p>You differ you die.<p>Which one do you chose?<p>To me this is the correct answer. I don't think the other two would even be in consideration.<p>I also think aliens would agree, although I'm not sure that would be fair since the original question (on this website) is being asked to me, a human.<p>[edit] Interesting this is also not the exact Bertrand paradox since they perhaps have added units to it to get around Edwin Jaynes solution *See Wiki
IMO the most natural interpretation of "random" with no qualifiers is the one that gives you 1/3. Pick a point, pick another point, a third of the time the second point in within the far arc of the triangle.<p>I'm not sure the ninja method gives you what I'd call a representative sample of possible chords. Looks like he only cuts vertically, so he can never give you a skew-slice.<p>Similar with the dragon.
It seems every methodology relies on upon the way arbitrary harmonies are chosen. Every methodology yields diverse probabilities of selecting a harmony with a certain property.
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The Bertrand paradox is a problem within the classical interpretation of probability theory.
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