Bertrand's Paradox

I vote for door #3. However, I see that "random" admits of several definitions.

I've seen at least part of this paradox before, but I've forgotten whether or how it was resolved. I haven't looked at part 2 yet, so that will be next.

Hot dog! Now I'm ready for "Let's Make a Deal".

Draw the circle radius R with it's equilateral triangle where the ends of all the sides touch the circumference. Add a radius from the circle center to a junction of two of the sides. Add a perpendicular from the circle center to one of those sides. You have constructed a right angle triangle with 60^o angle at the center of the circle and a 30^o angle where it meets the junction of the two sides. The length of the perpendicular is R Cos60^o which is 0.5 R. Extend the perpendicular to the circumference gives you a second radius with length 0.5 R from the center to the chord and 0.5 R from the chord to the circumference. Any new chord which passes closer to the circle center than the center of any side of the equilateral triangle will be longer than it's sides. All other new random chords will be shorter. So any new chord drawn at random stands an equal chance of being longer or shorter than a side of the equilateral triangle and a small but mathematically indeterminate chance of being the same length.

Here's what I think...

The key is "what is meant by a random chord".

1. You can assume that one point can be fixed and the other oriented at different angles from the first.

2. You can assume that the orientation of the chord can be fixed at a particular and the distance from the center is varied. All other angles should give you the same result.

3. You can assume that the center point of the chords defines the chords and the chords longer than the triangle side fall in a circle 1/4 the area of the unit circle.

It's easier to analyze if you replace the random variable, chord angle in #1, linear distance in #2 and #3, with a range of equally spaced values. This should give the same answer as putting in random values, and is much more controllable.

For example, in #1, the angle can be varied from -90 degrees to 0 degrees to +90 degrees in 1000 steps.

In #2 and #3, the chord is varied in distance from the center of the unit circle from 0 to 1.0 in 1000 steps.

Now you can see that the variable that is varied (angle or distance) determines how the chords are sampled. In #1, 1/3 of the chords sampled are longer than the triangle side. In #2, 1/2 of the chords sampled are longer. In #3, 1/2 of the chords are longer, and associating the center points to a circle area makes it 1/4.

On the question of which is right...

In #1, the longest chords, inside the triangle, are furthest apart. The longest chords are under represented, driving the average length to a lower level. We compute 1/3, the angle subtended by the triangle, but the true percentage of chords should be higher. Number 1 is not correct.

#1

#2, I think, is correct. If the chords were straws with width 0.01, 100 would fill half the circle. Since the base of the triangle is at r=0.5, 50 would be longer, or 50% chance for a random chord to be longer or shorter than the triangle side length.

#2

#3 doesn't make sense. The center points of the chords have zero area. So the fact that the longer chord center points fall in a circle of 1/4 the area doesn't imply that the probability of the chords being longer is 1/4. The longer chord center points may just be packed tighter in the center.

#3

In #1 all the chords only cut one side of the triangle. If you include chords that cut two sides of the triangle you would get the correct distribution. As it is drawn the chords are not random so #1 is not valid.

I think that #2 is a sub set of the correct answer that I postulated in post 3. If you draw a new green radius to the lowest junction of the triangle sides you create a right angle triangle with the green radius as the hypotenuse, half the grey side of equilateral as the base and part of your original black radius as the height. The length of that part is calculated from L = R Cos 60^o = R/2 So 1/2*¹ of a set of random chords cutting cut your radius inside the equilateral side and 1/2*¹ outside (*¹ with a small adjustment for chords which are on the equilateral side so are neither inside nor outside).

I think that #3 is the full answer. If you plot the point where your black radius cuts the side of the triangle for an infinite number of radii each displaced by a small angle then the locus of those points creates the inner circle in #3. Now any chord drawn must have it's center point on one of that infinite number of radii, and that chord will be inside or outside of the halfway point on that radius. So as shown in #2 the theoretical mathematical probability is 50%^*1

However if you draw an odd number of chords then there must be an extra one on inside or outside so at least half the time, and owing to the random distribution likely much more than half the time, the theoretical mathematical probability will only be an approximation which may (this is after all random) become more accurate as the number of chords drawn increases.

I haven't watched it, but clearly the problem lies in defining what a "random chord" is.

Clearly any chord has to pass through two points on the circumference, and, if we choose one of those points to be at one of the corners of the triangle (you can do that because wherever the "real chord" was originally you can always rotate the triangle until the points coincide; the triangle is really only defining a length): then if we choose the other end of the chord to be anywhere on the circumference the probability that it is longer than the triangle side is clearly 1/3.

You're right, it's the assumption of what random variable has a uniform distribution. I'm partial to #2, which works if you have a circle sitting on top of a random set of lines, and changing the scale or position maintains the same result. But it seems there is no consensus since the puzzle was originally posed.

Here is the follow up video:

Imagine a 2-dimensional space with an infinite number of lines going across it at an infinite number of angles. Now limit this to a still infinite number of lines that are all parallel. Now draw in this space a circle with an inscribed equilateral triangle, and have one side of this triangle be parallel to the orientation of this infinite number of lines. Any line passing through the circle between the side of the triangle and its center will be longer than the side. The same applies to any line that is equally far from the center beyond the side we have mentioned. Since the distance from the midpoint of the inscribed side to the center is 1/4 of the diameter of the circle, the total portion of the circle's diameter for which a line is longer than the triangle's side is 1/2 the diameter. By extension, then, if the lines can be at any angle, the same probability exists--0.5 or 50%.

The proposed solution in which one draws lines between two random points along the circumference of the circle forces the lines to be closer together when the points are nearby and allows them to be further apart when the points are further apart. This violates the requirement that the lines be random, and this method and its probability of 1/3 or 33% is false. You can visually see this by drawing lines between points around the circle that are 2, 4, 6,...174, 176, 178, 180 degrees apart--you will see that the lines are not spaced equally apart, which is a requirement if the lines were random.

I have not looked at the approach that gives a probability of 1/4 or 25%, but I believe it can be also shown to be non-random.

JMM

I tried a Monte Carlo approach and used a circle of radius 1 and then picked up two angles randomly and calculated pairs (X1, Y1), (X2, Y2) on the circumference of the circle. I then formed a chord between the two points and calculated the length and compared it to the length of the chord formed by the side of the equilateral triangle.

I ran the simulation for various numbers of chords; I was surprised at how close the ratio came, even for small numbers of chords.

For 10 the ratio was 0.4; 100 was 0.27; 1,000 was 0.325 and a 2nd run at a 1,000 was 0.327 and lastly for 10,000 the ratio was 0.3293. So, I'm thinking a ratio of 1/3 is a good approximation.

I liked the analysis of the various approaches everyone has used. I'm on the lazy side when I look at infinities of possibilities in problems and find Monte Carlo method of finding a reasonable value more satisfying and less time consuming.

As I see it, the problem is in setting-up the equations and points. Therein comes the hidden or un-recognized skewing of the results. Did you first randomly pick an angle starting from a random point along the circumference of the circle? If so, you will get the 1/3 result. I prefer a different approach which is the third one discussed in the OP.

This is an interesting study in how to set-up a problem for statistics. JMM

I had the program pick two angles between 0 and 360 degrees. the X1, Y1, points were the cos and sin of the first angle and then X2 and Y2 were the cos and the sin of the second angle (since I set the radius at unity, I didn't need to do r*cos or r*sin.) I then calculated the length of the cord between the two points. This way I was always on the circumference of the circle. I could have picked random angles on 0.1 deg increments but I thought one degree was enough to check things out.