I’ve
looked at this a little less rigorously in A Circle, a Triangle and Some Random Lines and The Circle, Triangle and Random Line with an Answer. If
you are unaware of the Bertrand Paradox, please look at the first of these to
see a framing of the problem and the latter for a description of the solutions.
I
don’t intend to look again at the method of solving the problem that uses a
randomly selected radius and results in a probability of p=1/2, since that is
not contentious (unless you follow the thinking of Božur).
Instead I am going to focus on the two solutions which, normally result
in probabilities of p=1/4 and p=1/3, with a particular focus on the latter.
In
The Circle, Triangle and Random Line with an Answer, I showed that the standard solution to the
problem that results in p=1/4 makes an assumption as to how one approximates
points that are selected at random and then used as midpoints of a chord.
A
very rough approximation of points would be like this:
For a circle of radius r, midpoints that lie inside a smaller circle of radius r/2 will define a chord of greater than √3.r (the length of the sides of an equilateral triangle with all three corners lying on the circumference of the larger circle). Thus:
If the midpoint lies in the light green area, the resultant chord will be longer than √3.r. With this very rough level of granularity, the apparent probability of selecting a midpoint which results in a longer chord appears to be zero (although we might quibble about the placement of the point).
There
are two ways to further divide the area inside the circle in order to more
accurately approximate points (one implying polar coordinates, the other cartesian coordinates):
These intermediate approximations give you p=1/2 (8/16) and p=1/4 (4/16), but neither seems perfect. The approximation by radial division has much smaller areas within the r/2 circle and the approximation by Cartesian division has variable sizes and some segments which cross zones. Nevertheless, let us crack on and look at finer approximations (from which the dots are eliminated and I will instead mark with orange those areas which correspond with the midpoint of a chord which is longer than √3.r):
These approximations result, again, in p=1/2 (64/128) and p=1/4 (12/48). (I eliminated any segment under half the regular size and considered the halfhalf segments to be on the borderline and eliminated those too.) These results don’t vary with increasing segmentation. For example, using a finer granularity:
If
you’re really keen, you might want to try counting the segments on the one
below. I personally could not be
bothered to count them but I do know that the radially segmented equivalent has
2048 segments, 1024 of which are in the inner circle.
That
all said, noone seems to agitate for the p=1/4 result to be the correct one. The
same cannot be said for the p=1/3 result, which has a champion in Božur.
My
argument against the p=1/3 result is that is relies on a special case, the
selection of two points on the circumference of the circle to define a chord.
If
this case is generalised such that any two points anywhere may be selected, so long as the resultant line crosses the circle. I illustrated this
thus:
I
provided a piece of graphical (albeit slightly wonky looking) proof:
Say
we have a circle of radius r and select a point at R*r from the locus of the
circle. Further say that the circle’s
locus lies on the origin (0,0) and the distant point is on the xaxis at (R*r,0). Then we take a tangent from the circle that
passes through this distant point. Then
draw another line between the tangent and the nominal xaxis.
It’ll
look a bit like this (click on it to see a larger version):
m = y(when x=0)/x(when y=0)
c = y(when x=0)
So:
m = r/2/√(11/R^{2})/(R*r) = 1/2R/√(11/R^{2})
c = r/2/√(11/R^{2})
But
as R>∞, 1/R>0, so:
m = 0
c = r/2
Which
is a line with zero gradient passing through the circle at half the radius,
like this:
To
calculate a few results from different values of R, we need to use the
following facts (included for completeness  if you are mathematically inclined,
you’ll either know these or remember them from those glorious days in
mathematics classes/lectures):
A line intersects a circle when the
values of x and y satisfy both
y^{2} + x^{2} = r^{2}
and
y = mx + c

Thus
(mx + c)^{2} + x^{2} = r^{2}
and so
(m^{2} + 1).x^{2} +
(2mc)x + (c^{2} – r^{2}) = 0

And a quadratic equation in the form
α.x^{2} + β.x + γ = 0
is given by
x = (β ± √(β^{2}  4α.γ)/2α

And the distance between two points is
given by
chord = √((y_{2}y_{1})^{2}
+ (x_{2} – x_{1})^{2})

From
these, the following table was populated:
r

10


R

2

4

8

16

32

64

128

256

m

0.288675

0.129099

0.062994

0.031311

0.015633

0.007813

0.003906

0.001953

c

5.7735027

5.1639778

5.0395263

5.0097943

5.0024432

5.0006105

5.0001526

5.0000381

α

1.0833333

1.0166667

1.0039683

1.0009804

1.0002444

1.0000611

1.0000153

1.0000038

β

3.333333

1.333333

0.634921

0.313725

0.156403

0.078144

0.039065

0.019532

γ

66.66667

73.33333

74.60317

74.90196

74.97556

74.99389

74.99847

74.99962

x_{1}

6.455619

7.862545

8.309818

8.495063

8.579956

8.620656

8.64059

8.650455

y_{1}

7.6370794

6.179028

5.5629957

5.2757851

5.1365705

5.0679676

5.0339059

5.0169337

x_{2}

9.5325422

9.1740204

8.9422293

8.808481

8.7363204

8.698795

8.679654

8.6699867

y_{2}

3.0216948

3.9796169

4.4762188

4.7339901

4.8658715

4.9326428

4.9662467

4.9831045

chord

16.641006

17.17795

17.286244

17.312024

17.318392

17.319979

17.320376

17.320475

r.√3

17.32050808


delta

0.67950

0.14256

0.03426

0.00848

0.00212

0.00053

0.00013

0.00003

%

4.0833%

0.8299%

0.1982%

0.0490%

0.0122%

0.0031%

0.0008%

0.0002%

In
other words, you don’t have to get that close to infinity before the
probability of a chord being longer than r.√3 begins to approach p=1/2.

Just
out of interest, I ran a simple spreadsheet simulation and over 20 samples of
50,000 random values of R between ±1x10^{15} and random multiplications
of 2θ between 0 and 1, 49.99062% of the resultant chords were longer than r.√3.
It
looked like this, with many more rows (click to see a larger version):
My
results had an experimental error of 0.0938%.
For a 99.9999% certainty that my hypothesis is true with a population of
one million samples, to the best of my knowledge, I am allowed an experimental
error of 0.098%.
Therefore,
I consider my hypothesis that p=1/2 to be true.
No comments:
Post a Comment
Feel free to comment, but play nicely!
Sadly, the unremitting attention of a spambot means you may have to verify your humanity.